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The Path approach to Lecture Notes for IV

Riccardo Rattazzi

May 25, 2009 2 Contents

1Thepathintegralformalism 5 1.1 Introducing the path ...... 5 1.1.1 The double slit experiment ...... 5 1.1.2 An intuitive approach to the path integral formalism . .. 6 1.1.3 The path integral formulation ...... 8 1.1.4 From the Schr¨oedinger approach to the path integral . .. 12 1.2 The properties of the path integrals ...... 14 1.2.1 Path integrals and state evolution ...... 14 1.2.2 The path integral computation for a free . . . . .17 1.3 Path integrals as determinants ...... 19 1.3.1 Gaussian integrals ...... 19 1.3.2 Gaussian Path Integrals ...... 20 1.3.3 O(!)correctionstoGaussianapproximation...... 22 1.3.4 Quadratic lagrangians and the . . .. 23 1.4 elements ...... 27 1.4.1 The -ordered product of operators ...... 27

2FunctionalandEuclideanmethods 31 2.1 method ...... 31 2.2 Euclidean Path Integral ...... 32 2.2.1 ...... 34 2.3 ...... 35 2.3.1 Euclidean n-pointcorrelators ...... 35 2.3.2 Thermal n-pointcorrelators...... 36 2.3.3 Euclidean correlators by functional derivatives ...... 38 0 0 2.3.4 Computing KE[J]andZ [J]...... 39 2.3.5 Free n-pointcorrelators ...... 41 2.3.6 The anharmonic oscillator and Feynman diagrams . . . . 43

3Thesemiclassicalapproximation 49 3.1 The semiclassical ...... 50 3.1.1 VanVleck-Pauli-Morette formula ...... 54 3.1.2 Mathematical Appendix 1 ...... 56 3.1.3 Mathematical Appendix 2 ...... 56 3.2 The ﬁxed propagator ...... 57 3.2.1 General properties of the ﬁxed energy propagator . . . . .57 3.2.2 Semiclassical computation of K(E)...... 61 3.2.3 Two applications: reﬂection and tunneling through a barrier 64

3 4 CONTENTS

3.2.4 On the of the prefactor of K(xf ,tf ; xi,ti).....68 3.2.5 On the phase of the prefactor of K(E; xf ,xi)...... 72

4Instantons 75 4.1 Introduction ...... 75 4.2 in the double well potential ...... 77 4.2.1 The multi- amplitude ...... 81 4.2.2 Summing up the instanton contributions: results . . . . .84 4.2.3 Cleaning up details: the zero-mode and the computation of R ...... 88

5Interactionwithanexternalelectromagneticﬁeld 91 5.1 Gauge freedom ...... 91 5.1.1 Classical ...... 91 5.1.2 Quantum physics ...... 93 5.2 Particle in a constant magnetic ﬁeld ...... 95 5.2.1 An exercise on translations ...... 96 5.2.2 ...... 97 5.2.3 Landau levels ...... 99 5.3 The Aharonov-Bohm eﬀect ...... 101 5.3.1 Energy levels ...... 105 5.4 Dirac’s ...... 106 5.4.1 On coupling constants, and why monopoles are diﬀerent thanordinarychargedparticles ...... 107 Chapter 1

The path integral formalism

1.1 Introducing the path integrals 1.1.1 The double slit experiment One of the important experiments that show the fundamental diﬀerence between Quantum and is the double slit experiment. It is interesting with respect to the path integral formalism because it leads to a conceptual motivation for introducing it. Consider a source S of approximately monoenergetic , for instance, placed at A.Theﬂuxofelectronsismeasuredonascreen C facing the source. Imagine now placing a third screen in between the others, with two slits on it, which can be opened or closed (see ﬁgure 1.1). When the ﬁrst slit is open and the closed we measure a ﬂux F1,whentheﬁrstslit is closed and the second open we measure a ﬂux F2 and when both slits are open we measure a ﬂux F .

1

S 2

A B C

Figure 1.1: The double slit experiment: an source islocatedsomewhere on A,andadetectorislocatedonthescreenC.AscreenB with two slits 1 and 2thatcanbeopenorclosedisplacedinbetween,sothattheelectrons have to pass through 1 or 2 to go from A to C.Wemeasuretheelectronﬂuxonthe screen C.

In , the ﬂuxes at C in the three cases are expected to satisfy the relation F = F1 + F2.Inreality,oneﬁndsingeneralF = F1 + F2 + Fint, and the structure of Fint precisely corresponds to the interference between two

5 6 CHAPTER 1. THE PATH INTEGRAL FORMALISM passing respectively through 1 and 2:

2 2 2 F = Φ +Φ = Φ + Φ +2Re(Φ∗Φ ) (1.1) | 1 2| | 1| | 2| 1 2 F1 F2 Fint

How can we interpret this result?! "# What\$ ! "# is the\$ ! electron?"# \$ More precisely: Does the behaviour imply that the electron is a delo- calized object? That the electron is passing through both slits? Actually not. When detected, the electron is point-like, andmoreremark- ably, if we try to detect where it went through, we ﬁnd that it either goes through 1 or 2 (this is done by setting detectors at 1 and 2 and considering a very weak ﬂux, in order to make the of coinciding detections at 1 and 2 arbitrarily small). According to the interpretation, F should be interpreted as a probability density. In practice this means: compute the amplitude Φas if dealing with waves, and interpret the intensity Φ 2 as a probability density for apoint-likeparticleposition. | | How is the particle/wave not contradictory? The answer is in the indetermination . In the case at hand: when we try todetectwhich alternative route the electron took, we also destroy interference. Thus, another formulation of the indetermination principle is: Any determination of the al- ternative taken by a process capable of following more than one alternative destroys the interference between the alternatives.

Resuming:

What adds up is the amplitude Φand not the probability density itself. • The diﬀerence between classical and quantum composition of • is given by the interference between classically distinct trajectories.

In the standard approach to Quantum Mechanics, the probability ampli- tude is determined by the Schr¨odinger . The “Schr¨odinger” viewpoint somehow emphasizes the wave properties of the particles (electrons, , ...). Ontheotherhand,weknowthatparticles,whiletheyaredescribedby “a probability wave”, are indeed point-like discrete entities. As an example, in the double slit experiment, one can measure where the electron went through (at the price of destroying quantum interference). This course will present an alternative, but fully equivalent, method to com- pute the . In this method, the role of the trajectory of a point-like particle will be formally “resurrected”, but in awaywhichiscom- patible with the indetermination principle. This is the pathintegralapproach to Quantum Mechanics. How can one illustrate the basic idea underlying this approach?

1.1.2 An intuitive approach to the path integral formalism In the double slit experiment, we get two interfering alternatives for the path of electrons from A to C.TheideabehindthepathintegralapproachtoQuantum Mechanics is to take the implications of the double slit experiment to its extreme consequences. One can imagine adding extra screens and drilling more and more 1.1. INTRODUCING THE PATH INTEGRALS 7 holes through them, generalizing the result of the double slit experiment by the . This is the procedure illustratedbyFeynmaninhis book “Quantum Mechanics and Path Integrals”.

Schematically:

With two slits: we know that Φ=Φ +Φ • 1 2 If we open a third slit, the superposition principle still applies: Φ=Φ + • 1 Φ2 +Φ3

Imagine then adding an intermediate screen D with N holes at positions • 1 2 N xD, xD,...,xD (see ﬁgure 1.2). The possible trajectories are now labelled i by xD and α =1, 2, 3, that is by the slit they went through at D and by the slit they went through at B.

1 2 1 2

N 1 3 N −

ABCD

Figure 1.2: The multi-slit experiment: We have a screen with Nslitsanda screen with three slits placed between the source and the detector.

Applying the superposition principle:

N i Φ= Φ xD,α i=1 α=1,2,3 % % & ' Nothing stops us from taking the ideal limit where N and the holes →∞ ﬁll all of D.Thesum i becomes now an integral over xD. ( Φ= dxDΦ(xD,α) α=1,2,3 % ) D is then a purely ﬁctious device! We can go on and further reﬁne our trajectories by adding more and more ﬁctious screens D1,D2,. . . ,DM

Φ= dx dx dx Φ(x ,x ,...,x ; α) D1 D2 ··· DM D1 D2 DM α=1,2,3 % )

In the limit in which Di, Di+1 become inﬁnitesimally close, we have speciﬁed all possible paths x(y)(seeﬁgure1.3foraschematicrepresentation,wherethe screen B has been dropped in order to consider the simpler case of propagation in empty ). 8 CHAPTER 1. THE PATH INTEGRAL FORMALISM

x

x(yi) x(y1)

x(yf ) x(y2) x(y3)

y yi y1 y2 y3 yf

Figure 1.3: The paths from (xi,yi)to(xf ,yf )arelabelledwiththefunctions x(y)thatsatisfyx(yi)=xi and x(yf )=xf .

In fact, to be more precise, also the way the paths are covered through time is expected to (t (x(y),y(t))). We then arrive at a formal representation of the probability amplitude→ as a sum over all possible trajectories:

Φ= Φ( x )(1.2) { } All trajectories% x(t),y(t) { } How is this formula made sense of? How is it normalized?

1.1.3 The path integral formulation Based on the previous section, we will start anew to formulatequantumme- chanics. We have two guidelines:

1. We want to describe the motion from position xi at time ti to position xf at time tf with a quantum probability amplitude K(xf ,tf ; xi,ti)givenby

K(x ,t ; x ,t )= Φ( γ ) f f i i { } All trajectories%

where γ is the of all trajectories satisfying x(t )=x , x(t )=x . { } i i f f 2. We want classical trajectories to describe the motion in the formal limit ! 0. In other words, we want classical physics to be resurrectedinthe ! → 0limit. → Remarks:

! has the dimensionality [Energy] [Time]. That is also the dimen- • sionality of the S which describes× the classical trajectories via the principle of least action. 1.1. INTRODUCING THE PATH INTEGRALS 9

One can associate a value of S[γ]toeachtrajectory.Theclassicaltrajec- • tories are given by the stationary points of S[γ](δS[γ]=0).

It is thus natural to guess: Φ[γ]=f(S[γ]/!), with f such that the classical trajectory is selected in the formal limit ! 0. The speciﬁc choice → S[γ] Φ[γ]=ei ! (1.3) implying S[γ] i ! K(xf ,tf ; xi,ti)= e (1.4) γ %{ } seems promising for two reasons:

1. The guideline (2.) is heuristically seen to hold. In a macroscopic, classical, situation the δS/δγ is for most trajectories much greater than !. Around such trajectories the phase eiS/! oscillates extremely rapidly and the sum over neighbouring trajectories will tend to cancel. 1 (see ﬁgure 1.4).

γ1 xf γ2

xi

Figure 1.4: The contributions from the two neighbouring trajectories γ1 and γ2 will tend to cancel if their action is big.

xi

γcl

xf

Figure 1.5: The contributions from the neighbouring trajectories of the classical trajectory will dominate.

On the other hand, at a classical trajectory γcl the action S[γ]isstationary. Therefore in the neighbourhood of γcl, S varies very little, so that all trajectories in a tube centered around γcl add up coherently (see ﬁgure 1.5) in the sum over trajectories. More precisely: the tube oftrajectories

∞ 1By analogy, think of the integral R + dxeif(x) where f(x) ax plays the role of S/!: −∞ ≡ the integral vanishes whenever the derivative of the exponent f ′ = a is non-zero 10 CHAPTER 1. THE PATH INTEGRAL FORMALISM

in question consists of those for which S Scl ! and deﬁnes the extent to which the classical trajectory is well deﬁned.| − | We≤ cannot expect to deﬁne our classical action to better than !.However,innormalmacroscopic ∼ situations Scl !.Intheexactlimit! 0, this eﬀect becomes dramatic and only the classical≫ trajectory survives.→

Once again, a simple one dimensional analogy is provided by the inte- + 2 gral ∞ dxeix /h,whichisdominatedbytheregionx2 < h around the stationary−∞ point x =0. ∼ *

2. Eq. (1.3) leads to a crucial composition property. Indeed the action for apathγ12 obtained by joining two subsequent paths γ1 and γ2,likein ﬁg. 1.6, satisﬁes the simple additive relation S[γ12]=S[γ1]+S[γ2]. Thanks to eq. (1.3) the additivity of S translates into a factorization property for the amplitude: Φ[γ12]=Φ[γ1]Φ[γ2]whichinturnleadstoacomposition property of K,whichweshallnowprove.

Consider indeed the amplitudes for three consecutive ti

x

γ1 y xi

x γ2 f

t ti tint tf

Figure 1.6: The path from (xi,ti)to(xf ,tf )canbeobtainedbysummingthe paths γ1 from (xi,ti)to(y,tint)withγ2 from (y,tint)to(xf ,tf ). 1.1. INTRODUCING THE PATH INTEGRALS 11

Thanks to eq. (1.3) we can then write:

dyK(xf ,tf ; y,tint)K(y,tint; xi,ti) ) i = dye ! (S[γ1(y)]+S[γ2(y)]) γ ,γ %1 2 ) i = dye ! S[γ(y)]

γ(y)=γ2(y) γ1(y) ) % ◦ = K(xf ,tf ; xi,ti) . (1.5)

Notice that the above composition rule is satisﬁed in QuantumMechanics as easily seen in the usual formalism. It is the quantum analogue of the classical composition of probabilities

P1 2 = P1 αPα 2 . (1.6) → → → α % In quantum mechanics what is composed is not probability P itself but the amplitude K which is related to probability by P = K 2. | | It is instructive to appreciate what would go wrong if we modiﬁed the choice S[γ]/! in eq. (1.3). For instance the alternative choice Φ= e− satisﬁes the compo- sition property but does not in general select the classical trajectories for ! 0. This alternative choice would select the minima of S but the classical trajec-→ tories represent in general only saddle points of S in space. Another 2 alternative Φ= ei(S[γ]/!) ,wouldperhapsworkoutinselectingtheclassical trajectories for ! 0, but it would not even closely reproduce the composition property we know→ works in Quantum Mechanics. (More badly: this particular choice, if S were to represent the action for a of two particles, would imply that the amplitudes for each individual particle do notfactorizeevenin the limit in which they are very far apart and non-interacting!).

One interesting aspect of is that a fundamentalunitofaction (!)isintroduced.Inclassicalphysicstheoverallvalueoftheactioninunphys- ical: if S(q, q˙) S λS(q, q˙)(1.7) → λ ≡ the only eﬀect is to multiply all the of motion by λ,sothatthe solutions remain the same (the stationary points of S and Sλ coincide). Quantum Mechanics sets a natural unit of measure for S.Dependingonthe size of S,thesystemwillbehavediﬀerently: large S classical regime • → small S quantum regime • → In the large S limit, we expect the trajectories close to γcl to dominate K i Scl (semiclassical limit). We expect to have: K(xf ; xi) (smooth function) e ! . In the S ! limit, all trajectories are comparably∼ important: we must· sum them up in a∼consistent way; this is not an easy mathematical task. 12 CHAPTER 1. THE PATH INTEGRAL FORMALISM

Due to the mathematical diﬃculty, rather than going on with Feynman’s construction and show that it leads to the same results of Schr¨odinger equation, we will follow the opposite route which is easier: we will derive the path integral formula from Schr¨odinger’s operator approach.

1.1.4 From the Schr¨oedinger approach to the path integral Consider the transition amplitude:

iH(tf −ti) iHt K(x ,t ; x ,t ) x e− ! x = x e− ! x (1.8) f f i i ≡⟨ f | | i⟩ ⟨ f | | i⟩ where we used time translation invariance to set: ti =0,tf ti = t. To write it in the form of a path integral, we divide t into− N inﬁnitesimal steps and consider the amplitude for each inﬁnitesimal step (see ﬁgure 1.7). We label the intermediate times tk = kϵ by the integer k =0,...,N.Noticethat t0 =0andtN = t. 0 t = Nϵ ϵ 2ϵ (N 1)ϵ − Figure 1.7: The interval between 0 and t is divided into N steps.

We can use the completeness relation x x dx =1ateachstepnϵ: | ⟩⟨ | N 1 * iHt − iHϵ iHϵ ! ! ! xf e− xi = dxk xf e− xN 1 xN 1 e− xN 2 ⟨ | | ⟩ ⟨ | | − ⟩⟨ − | | − ⟩··· ) k=1 + iHϵ iHϵ x e− ! x x e− ! x (1.9) ⟨ 2| | 1⟩⟨ 1| | i⟩ iHϵ/! Consider the quantity: x′ e− x .Using p p dp =1: ⟨ | | ⟩ | ⟩⟨ |

iHϵ * iHϵ x′ e− ! x = dp x′ p p e− ! x (1.10) ⟨ | | ⟩ ⟨ | ⟩⟨ | | ⟩ ) pˆ2 If we stick to the simple case H = 2m + V (ˆx), we can write:

2 2 ϵ pˆ ϵ p i ! 2m +V (ˆx) i ! 2m +V (x) 2 p e− h i x = e− h i p x + ϵ (1.11) ⟨ | | ⟩ ⟨ | ⟩ O where the O(ϵ2)termsarisefromthenon-vanishingcommutatorbetweenˆ& ' p2/(2m) and V (ˆx). We will now assume, which seems fully reasonable, that in the limit ϵ 0thesehigherorertermscanbeneglected.Lateronweshallcome back on→ this issue and better motivate our neglect of these terms. We then get:

′ 2 i p(x x) iHϵ ϵ p e ! ! i ! 2m +V (x) − x′ e− x dp e− h i (1.12) ⟨ | | ⟩≃ · 2π! ) , - x′ x We can deﬁne: − x˙: ϵ ≡ 2 ϵ p iHϵ dp i ! +V (x) px˙ x′ e− ! x e− 2m − (1.13) ⟨ | | ⟩≃ 2π! h i ) 1.1. INTRODUCING THE PATH INTEGRALS 13

By performing the change of variablesp ¯ p mx˙ the integral reduces to simple gaussian integral for the variablep ¯: ≡ −

2 ϵ p¯ mx˙ 2 iHϵ dp¯ i ! +V (x) x′ e− ! x e− 2m − 2 (1.14) ⟨ | | ⟩≃ 2π! h i ) m ϵ 1 2 1 ϵ i ! [ mx˙ V (x)] i (x,x˙) ! = e 2 − = e L (1.15) 2πi!ϵ A . 1 A

At leading order in ϵ we can further! "# \$ identify (x, x˙)ϵ with the action S(x′,x)= ϵ (x, x˙)dt.Byconsideringalltheintervalswethusﬁnallyget:L 0 L * N 1 iHt − 1 i N−1 ! ! S(xl+1,xl) xf e− xi =lim dxk e l=0 ϵ 0 N P ⟨ | | ⟩ → A ) k+=1 N 1 1 − dxk S(xf ,xi) =lim ei ! (1.16) ϵ 0 A A → ) k+=1 Dγ R where Dγ should be taken as! a deﬁnition"# of\$ the functional measure over the space of the trajectories. We thus have got a path integral formula for the transition* amplitude:

Scl(x,x˙ ) i ! K(xf ,t; xi, 0) = D[x(t)]e (1.17) ) We see that we (actually somebody else before us!) had guessedwelltheform of the transition amplitude. The path integral approach to QMwasdeveloped by Richard Feyman in his PhD Thesis in the mid 40’s, following ahintfrom an earlier paper by Dirac. Dirac’s motivation was apparentlytoformulateQM starting from the lagrangian rather than from the hamiltonian formulation of classical mechanics.

Let us now come back to the neglected terms in eq. (1.11). To simplify the p2 notation let us denote the kinetic operator as T = i 2m! and the potential U = iV/!;wecanthenwrite − − ϵ(T +U) ϵT ϵT ϵ(T +U) ϵU ϵU ϵT ϵ2C ϵU p e x = p e e− e e− e x = p e e− e x(1.18) ⟨ | | ⟩ ⟨ | 2 | ⟩ ⟨ | | ⟩ ϵ p 2 i ! 2m +V (x) ϵ C = e− h i p e− x (1.19) ⟨ | | ⟩ where C is given, by using the Campbell-Baker-Haussdorf formula twice, as a of between T and U 1 ϵ C = [T,U]+ [T,[T,U]] + [U, [U, T ]] + ... (1.20) 2 6 { }

By iterating the basic commutation relation [ˆp, V (ˆx)]] = iV ′(ˆx)andexpanding the exponent in a series one can then write − n= r=n s=r ϵ2C ∞ n r s p e− x =1+ϵ ϵ p − P (x)(1.21) ⟨ | | ⟩ n,s,r n=0 r=1 s=0 % % % 14 CHAPTER 1. THE PATH INTEGRAL FORMALISM

where Pn,s,r(x)isahomogenouspolynomialofdegreen +1 r in V and its derivatives, with each term involving exactly r + s derivatives.− For instance P1,0,1(x)=V ′(x). We can now test our result under some simple assumption. For instance, if the derivatives of V are all bounded, the only potential problem to concentrate on in the ϵ 0limitisrepresentedbythepowersofp.This is because the leading contribution→ to the p integral in eq. (1.15) comes from the region p 1/√ϵ,showingthatp diverges in the small ϵ limit. By using p 1/√ϵ the∼ right hand side of eq. (1.21) is 1+O(ϵ3/2)sothat,eventaking into∼ account that there are N 1/ϵ such terms∼ (one for each step), the ﬁnal result is still convergent to 1 ∼

1 3 ϵ lim 1+aϵ 2 =1. (1.22) ϵ 0 → / 0

Before proceeding with technical developments, it is worth assessing the rˆole of the path integral (P.I.) in quantum mechanics. As it was hopefully highlighted in the discussion above, the path integral formulation is conceptually advantageous over the standard operatorial formulation of Quantum Mechanics, in that the “good old” particle trajectories retain some rˆole. The P.I. is however technically more involved. When working on simple quantum like the , no technical proﬁt is really given by path integrals. Nonetheless, after overcoming a few technical diﬃculties, the path integral oﬀers a much more direct viewpoint on the semiclassical limit. Similarly, for issues involving topology like the origin of Bose and Fermi statistics, the Aharonov-Bohm eﬀect, quantization in the presence of a magnetic monopole, etc. . . path integrals oﬀer a much better viewpoint. Finally, for advanced issues like the quantization of gauge theories and for eﬀects like instantons in quantum ﬁeld theory it would be hard to think how to proceed without path integrals!...Butthatisforanother course.

1.2 The properties of the path integrals 1.2.1 Path integrals and state evolution

To get an estimate of the dependence of the amplitude K(xf ,tf ; xi,ti)onits arguments, let us ﬁrst look at the properties of the solutionstotheclassical with boundary conditions xc(ti)=xi, xc(tf )=xf .

Let us compute ∂tf Scl.WhereScl is deﬁned

tf S S[x ]= (x , x˙ )dt (1.23) cl ≡ c L c c )ti with xc asolution,satisfyingtheEuler-Lagrangeequation: ∂ ∂ ∂ L L =0. (1.24) t ∂x˙ − ∂x 1 2x=xc

We can think of xc as a function x f(x ,x ,t ,t ,t) c ≡ i f i f 1.2. THE PROPERTIES OF THE PATH INTEGRALS 15 such that f(x ,x ,t ,t ,t= t ) x = const i f i f i ≡ i f(x ,x ,t ,t ,t= t ) x = const i f i f f ≡ f Diﬀerentiating the last relation we deduce:

∂tf xc + ∂txc =0 t=tf

= 3 ∂tf xc =4 x˙ c (1.25) ⇒ t=tf − t=tf 5 5 Similarly, diﬀerentiating the relation5 at t = ti5 we obtain

∂t xc =0 (1.26) f t=ti 5 And the following properties are straightforward:5

∂xf xc =1 (1.27) t=tf

∂x xc5 =0 (1.28) f 5 t=ti 5 Using these properties, we can now5 compute:

tf ∂ ∂ ∂tf S = (x, x˙) + dt L∂tf x + L∂tf x˙ L t=tf ∂x ∂x˙ )ti 1 2 5 tf tf 5 ∂ ∂ ∂ = (x, x˙) + dt ∂t L∂tf x dt ∂t L L ∂tf x L t=tf ∂x˙ − ∂x˙ − ∂x )ti 1 6 72 )ti 6 7 5 (1.29) 5

All this evaluated at x = xc gives:

t=tf ∂ ∂tf Scl = (xc, x˙ c) + L ∂tf xc L t=tf ∂x˙ 8 5x=xc 9 5t=t 5 5 5 i 5 5 5 ∂ 5 5 = (x , x˙ ) L x˙ = 5 E (1.30) L c c − ∂x˙ c − 8 x=xc 9 5 5 5t=tf 5 5 5 5 We recognize here the deﬁnition of the5 hamiltonian5 of the system, evaluated at x = xc and t = tf . We can also compute:

tf ∂ tf ∂ ∂ ∂ S = dt ∂ L∂ x dt ∂ L L ∂ x (1.31) xf t ∂x˙ xf − t ∂x˙ − ∂x xf )ti 1 6 72 )ti 6 7

Evaluated at x = xc,thisgives:

t=tf ∂ ∂ ∂ S (x )= L ∂ x = L = P (1.32) xf cl c ∂x˙ xf c ∂x˙ 8 x=xc 9 5 x=xc,t=tf 5 5t=ti 5 5 5 5 5 5 5 We recognize here the deﬁnition5 of the5 momentum5 of the system,evaluated at x = xc and t = tf . 16 CHAPTER 1. THE PATH INTEGRAL FORMALISM

At this point, we can compare these results with those of the path integral computation in the limit ! 0. →

S i ! K(xf ,tf ; xi,ti)= D[x(t)]e (1.33) ) In the limit ! we expect the path integral to be dominated by the classical trajectory and thus to take the form (and we shall technicallydemostratethis expectation later on)

Scl i ! K(xf ,tf ; xi,ti)=F (xf ,tf ,xi,ti) e (1.34)

smooth function where F is a smooth function in the ! ! 0limit.Wethenhave:"# \$ → Scl 1 ∂ ei ! large (1.35) t,x ∼ ! ∂ F (1) comparatively negligible (1.36) t,x ∼O so that we obtain

i!∂ K = ∂ S K + (!)=P K + (!)(1.37) − xf xf cl · O cl O i!∂ K = ∂ S K + (!)=E K + (!)(1.38) tf − tf cl · O cl O and energy, when going to the path integral description of particle 1 propagation, become associated to respectively the wave number ( (i∂x ln K)− ) and oscillation ( (i∂ ln K)). ∼ ∼ t The last two equations make contact with usual quantum mechanical rela- tions, if we interpret K has the and i!∂x and i!∂t as respectively the momentum and energy operators −

i!∂ K P K (1.39) − x ∼ · i!∂ K E K (1.40) t ∼ ·

iH(t′ −t) The function K(x′,t′; x, t)= x′ e− ! x is also known as the propaga- tor, as it describes the probability⟨ amplitude| for| ⟩ propagating a particle from x to x′ in a time t′ t. − From the point of view of the Schr¨odinger picture, K(x′,t; x, 0) represents the wave function at time t for a particle that was in the state x at time 0: | ⟩ iHt ψ(t, x) x e− ! ψ x ψ(t) (1.41) ≡⟨ | | 0⟩≡⟨ | ⟩ By the superposition principle, K can then be used to write the most general solution. Indeed:

1. K(x′,t′; x, t)solvestheSchr¨odingerequationfort′, x′ as variables:

i!∂t′ K = dx′′H(x′,x′′)K(x′′,t′; x, t)(1.42) )

where H(x′,x′′)= x′ H x′′ . ⟨ | | ⟩ 1.2. THE PROPERTIES OF THE PATH INTEGRALS 17

2. lim K(x′,t′; x, t)= x′ x = δ(x′ x) t′ t ⟨ | ⟩ − →

Thus, given Ψ0(x)thewavefunctionattimet0,thesolutionatt>t0 is:

iH(t−t0 ) Ψ(x, t)= x e− ! Ψ ⟨ | | 0⟩ iH(t−t0 ) = x e− ! y y Ψ dy ⟨ | | ⟩⟨ | 0⟩ ) = K(x, t; y,t0)Ψ0(y)dy (1.43) ) Having K(x′,t′; x, t), the solution to the Schr¨odinger equation is found by • performing an integral.

K(x′,t′; x, t)containsalltherelevantinformationonthedynamicsofthe • system.

1.2.2 The path integral computation for a Let us compute the propagator of a free particle, decribed by the lagrangian (x, x˙)=mx˙ 2/2, using Feyman’s time slicing procedure. Using the result of Lsection 1.1.4 we can write

N 1 1 − dx iS k ! K(xf ,tf ; xi,ti)=lim e (1.44) ϵ 0 A A → ) k+=1 where: t t ϵ = f − i N N 1 − 1 (x x )2 S = m k+1 − k 2 ϵ %k=0 x0 = xi

xN = xf

Let us compute eq. (1.44) by integrating ﬁrst over x1.Thiscoordinateonly appears only in the ﬁrst and the second time slice so we need just to concentrate on the integral

∞ m 2 2 i ! [(x1 x0) +(x2 x1) ] dx1e 2ϵ − − )−∞ ∞ m 1 2 1 2 i 2ϵ! 2(x1 2 (x0+x2)) + 2 (x2 x0) = dx1e h − − i )−∞ ! 2πiϵ 1 i m 1 (x x )2 = e 2ϵ! 2 2− 0 (1.45) . m 2 1 A√ 2 ! "# \$ Notice that x1 has disappeared: we “integrated it out” in the physics jargon. Putting this result back into eq. (1.44) we notice that we haveanexpression 18 CHAPTER 1. THE PATH INTEGRAL FORMALISM similar to the original one but with N 1insteadofN time slices. Moreover the ﬁrst slice has now a width 2ϵ instead− of ϵ:thefactorsof1/2inboththe exponent and prefactor of eq. (1.45) are checked to match thisinterpratation. It is now easy to go on. Let us do the integral on x2 to gain more conﬁdence. The relevant terms are

∞ m 1 2 2 i ! [ (x2 x0) +(x3 x2) ] dx2e 2ϵ 2 − − )−∞ ∞ m 3 1 2 2 1 2 i 2ϵ! 2 (x2 3 x0 3 x3) + 3 (x3 x0) = dx2e h − − − i )−∞ ! 2πiϵ 2 i m 1 (x x )2 = e 2ϵ! 3 3− 0 (1.46) . m 3 2 A√ 3 ! "# \$ It is now clear how to proceed by induction. At the n-th step the integral over xn will be:

m 1 2 2 i ! [ (xn x0) +(xn+1 xn) ] dxne 2ϵ n − − ) m n+1 1 n 2 1 2 i 2ϵ! n (xn ( n+1 x0+ n+1 xn+1)) + n+1 (xn+1 x0) = dxne h − − i ) ! 2πiϵ n i m 1 (x x )2 = e 2ϵ! n+1 n+1− 0 (1.47) . m n +1 n A√ n+1 ! "# \$ Thus, putting all together, we ﬁnd the expression of the propagator for a free particle:

1 1 2 N 1 m 2 1 m 2 i ! (xf xi) i ! (xf xi) K(xf ,tf ; xi,ti)=lim ... − e 2 Nϵ − =lim e 2 Nϵ − ϵ 0 A 2 3 N ϵ 0 √ → . → A N 2 m (xf −xi) m i ! = e 2 tf −ti (1.48) 2πi!(t t ) . f − i where we used the fact that Nϵ = tf ti. We can check this result by computing− directly the propagatorofaone- dimensional free particle using the standard representation of the propagator:

iHt K(x ,t; x , 0) = x p p e− ! x dp f i ⟨ f | ⟩⟨ | | i⟩ ) 2 1 i p (x x ) i p t = e ! f − i e 2m! dp 2π! ) x −x 2 (x −x )2 1 it p m f i +i m f i = e− 2m! − t 2! t dp 2π! “ ” ) m(x −x )2 m i f i = e 2!t 2πi!t . Scl(xf ,t;xi,0) = F (t)ei ! (1.49) 1.3. PATH INTEGRALS AS DETERMINANTS 19

These results trivially generalize to the motion in higher dimensional , as for the free particle the result factorizes.

As we already saw, we can perform the same integral “`ala Feynman” by • dividing t into N inﬁnitesimal steps.

It is instructive interpret the result for the : • dP m = K 2 = (1.50) dx | | 2π!t

Let us interpret this result in terms of a ﬂux of particles thatstartedat xi =0att =0withadistributionofmomenta

dn(p)=f(p)dp (1.51)

and ﬁnd what f(p)is.Aparticlewithmomentump at time t will have travelled to x =(p/m)t.Thustheparticlesintheintervaldp will be at time t in the coordinate interval dx =(t/m)dp.Thereforewehave dn(x)=f(xm/t)(m/t)dx.Comparingto(1.50),weﬁndf =1/(2π!)= 1 1 const.Thus,dn(p)=(1/2π!)dp dn/dp [Length]− [Momentum]− . We recover the result that the wave⇒ function∼ψ(x, t)=K×(x, t;0, 0) satisﬁes ψ(x, 0) = δ(x)whichcorrespondsinmomentumspacetoanexactlyﬂat distribution of momenta, with normalization dn(p)/dp =(1/2π!).

1.3 Path integrals as determinants

We will now introduce a systematic small ! expansion for path integrals and show that the computation of the “leading” term in the expansion corresponds to the computation of the determinant of an operator. Before going to that, let us remind ourselves some basic results about “Gaussian” integrals in several variables.

1.3.1 Gaussian integrals Let us give a look at gaussian integrals:

λij xixj dxie− i,j (1.52) P i ) + With one variable, and with λ real and positive we have: •

λx2 1 y2 π dxe− = dye− = (1.53) λ λ ) . ) . Consider now a Gaussian integral with an arbitrary number of real vari- • ables

λij xixj dxie− i,j (1.54) P i ) + where λij is real. 20 CHAPTER 1. THE PATH INTEGRAL FORMALISM

T We can rotate to the eigenbasis: xi = Oij x˜j,withO λO =diag(λ1,...,λn), OT O = OOT =1.Then, dx = det O dx˜ = dx˜ .Thus: i i | | i i i i

: : 2 : λij xixj λix˜ dxie− i,j = dx˜ie− i i = P P i i ) + ) + π 1 = (1.55) λ 1 ˆ i i ;det( λ) + . π where again the result makes sense only when all eigenvalues λn are posi- tive. Along the same line we can consider integrals with imaginary exponents, • which we indeed already encountered before

iλx2 i sgn(λ) π π I = dxe = e 4 (1.56) λ ) .| | This integral is performed by deforming the contour into the imaginary plane as discussed in homework 1. Depending on the sign of λ we must choose diﬀerent contours to ensure convergence at inﬁnity. Notice that the phase depends on the sign of λ. For the case of an integral with several variables •

i λjkxj xk I = dxie j,k (1.57) P i ) + the result is generalized to

i(n n ) π 1 I = e +− − 4 (1.58) det( 1 λˆ) ;| π | where n+ and n are respectively the number of positive and negative − eigenvalues of the matrix λjk.

1.3.2 Gaussian Path Integrals Let us now make contact with the path integral:

S[x(t)] i ! K(xf ,tf ; xi,ti)= D[x(t)]e (1.59) ) with tf S[x(t)] = (x(t), x˙(t))dt (1.60) L )ti To perform the integral, we can choose the integration variables that suit • us best. The simplest change of variables is to shift x(t)to“center”itaroundthe • classical solution xc(t): x(t)=x (t)+y(t) D[x(t)] = D[y(t)] c ⇒ (the Jacobian for this change of variables is obviously trivial, as we indi- cated). 1.3. PATH INTEGRALS AS DETERMINANTS 21

We can Taylor expand S[xc(t)+y(t)] in y(t)(notethaty(ti)=y(tf )= • 0, since the classical solution already satisﬁes the boundary conditions xc(ti)=xi and xc(tf )=xf ). By the usual deﬁnition of functional deriva- tive, the Taylor expansion of a functional reads in general

δS S[xc(t)+y(t)] = S[xc]+ dt1 y(t1)(1.61) δx(t1) ) 5x=xc 2 5 1 δ S 5 3 + dt1dt2 5 y(t1)y(t2)+ y 2 δx(t1)δx(t2) O ) 5x=xc 5 & ' 5 with obvious generalization to all orders in y.Inourcase5 S = dt (x, x˙) so that each term in the above general expansion can be writtenasasingleL dt integral *

δS ∂ ∂ dt1 y(t1) dt L y + L y˙ (1.62) δx(t1) ≡ ∂x ∂x˙ ) 5x=xc ) < 5x=xc 5x=xc = 5 5 5 5 5 5 5 5 5 1 δ2S dt1dt2 y(t1)y(t2) 2 δx(t1)δx(t2) ≡ ) 5x=xc 5 ∂2 5 ∂2 ∂2 dt L 5 y2 +2 L yy˙ + L y˙2 (1.63) ∂x2 ∂x∂x˙ ∂x˙ 2 ) < 5x=xc 5x=xc 5x=xc = 5 5 5 5 5 5 and so on. 5 5 5 The linear term in the expansion vanishes by the equations of motion:

δS ∂ d ∂ ∂ t=tf y(t)dt = L L y(t)dt + Ly(t) =0 δx ∂x − dt ∂x˙ ∂x˙ ) 5x=xc ) 6 7 5x=xc 5t=ti 5 5 5 (1.64) 5 5 5 5 5 5 Thus, we obtain the result:

i 1 δ2S 2 3 ! S[xc(t)]+ 2 2 y + (y ) K(xf ,tf ; xi,ti)= D[y(t)]e h δx O i (1.65) ) where δ2S/δx2 is just short hand for the second variation of S shown in eq. (1.63). We can rewrite it rescaling our variables y = √! y˜:

i i δ2S 2 ! 3 ! S[xc(t)] y˜ + (√ y˜ ) K(x ,t ; x ,t )= e D[˜y(t)]e 2 δx2 O (1.66) f f i i N· · ) where the overall constant is just the Jacobian of the rescaling. 2 The interpretation of the aboveN expression is that the quantum propagation of a particle can be decomposed into two pieces:

iS[xc]/! 1. the classical trajectory xc(t), which gives rise to the exponent e .

2Given that is a constant its role is only to ﬁx the right overall normalization of the prop- agator, and doesN not matter in assessing the relative importance of each trajectory. Therefore it does not play a crucial role in the following discussion. 22 CHAPTER 1. THE PATH INTEGRAL FORMALISM

2. the ﬂuctuation y(t)overwhichwemustintegrate;becauseofthequadratic term in the exponent, the path integral is dominated by y √! (that ∼O isy ˜ (1)) so that y(t)representsthe quantum ﬂuctuation/ around0 the classical∼O trajectory. In those physical situations where ! can be treated as a small quantity, one can treat O(!)intheexponentofthepathintegralasaperturbation

2 i S[x (t)] i δ S y˜2 K(x ,t ; x ,t )= e ! c D[˜y(t)]e 2 δx2 [1 + (!)] = (1.67) f f i i N· · O ) i = F (x ,t ; x ,t )e ! S[xc(t)] [1 + (!)] (1.68) f f i i O Where the prefactor F is just the gaussian integral around the classical trajec- tory. The semiclassical limit should thus correspond to the possibility to reduce the path integral to a gaussian integral. We will study this inmoredetaillater on. In the small ! limit, the “classical” part of the propagator, exp(iS[xc]/!) oscillates rapidly when the classical action changes, whiletheprefactor(F and the terms in brakets in eq. (1.68)) depend smoothly on ! in this limit.

As an example, we can compute the value of Scl for a simple macroscopic system: a ball weighing one gram moves freely along one meter in one second. We will use the following unity for energy: [erg] = [g] ([cm]/[s])2. × 1 (x x )2 1 S = m f − i = mv∆x =5000erg s cl 2 t t 2 · f − i 27 ! =1.0546 10− erg s · · S /! 1030 ⇒ cl ∼ 1.3.3 O(!) corrections to Gaussian approximation It is a good exercise to write the expression for the O(!)correctiontothe propagator K in eq. (1.68). In order to do so we must expand the action to ordery ˜4 around the classical solution (using the same short hand notation as before for the higher order variation of S)

S[x + √!y˜] S[x ] 1 δ2S √! δ3S ! δ4S c = c + y˜2 + y˜3 + y˜4 + O(!3/2)(1.69) ! ! 2 δx2 3! δx3 4! δx4 and then expand the exponent in the path integral to order !.Theleading!1/2 correction vanishes because the integrand is odd undery ˜ y˜ →− ! 3 2 √ δ S 3 i δ S y˜2 D[˜y(t)] y˜ e 2 δx2 =0 (1.70) 3! δx3 ) and for the O(!)correctiontoK we ﬁnd that two terms contribute

2 4 3 2 i S[x (t)] 1 δ S 4 1 1 δ S 3 i δ S y˜2 ∆K = i!e ! c D[˜y(t)] y˜ + i y˜ e 2 δx2 (1.71) N 4! δx4 2 3! δx3 ) < 1 2 = 1.3. PATH INTEGRALS AS DETERMINANTS 23

1.3.4 Quadratic lagrangians and the harmonic oscillator Consider now a special case: the quadratic lagrangians, deﬁned by the property:

δ(n)S =0 forn>2(1.72) δxn For these lagrangians, the gaussian integral coresponds to the exact result:

2 S[xc] i 1 δ S 2 i ! ! y K(xf ,tf ; xi,ti)=e D[y]e 2 δx2

) 1 2 2 S[xc] δ S − = const ei ! det (1.73) · δx2 6 7 To make sense of the above we must deﬁne this strange “”: the de- terminant of an operator. It is best illustrated with explicit examples, and usually computed indiretly using some “trick”. This is a curious thing about path integrals: one never really ends up by computing them directly. Let us compute the propagator for the harmonic oscillator using the deter- minant method. We have the lagrangian:

1 1 (x, x˙)= mx˙ 2 mω2x2 (1.74) L 2 − 2 It is left as an exercise to show that:

tf mω 2 2 2xf xi Scl = (xc, x˙ c)dt = xf + xi cot(ωT) (1.75) t L 2 − sin(ωT) ) i 1/ 0 2 where T = t t . f − i Then,

tf tf 1 tf 1 (x + y,x˙ +˙y)dt = m x˙ 2 ω2x 2 dt + m y˙2 ω2y2 dt L c c 2 c − c 2 − )ti )ti )ti & ' & (1.76)' And thus, the propagator is:

t Scl i f m 2 2 2 i ! ! (y˙ ω y )dt K(x ,t ; x ,t )=e D[y]e ti 2 − f f i i · R ) Scl(xf ,xi,tf −ti) = ei ! J(t t )(1.77) · f − i

This is because y(t)satisﬁesy(ti)=y(tf )=0andanyreferencetoxf and xi has disappeared from the integral over y.IndeedJ is just the propagator from xi =0toxf =0:

Scl(xf ,xi,tf −ti) K(x ,t ; x ,t )=ei ! K(0,t ;0,t ) (1.78) f f i i · f i J(t t ) f − i ! "# \$ We already have got a non trivial by simple manipulations. The question now is how to compute J(t t ). f − i 24 CHAPTER 1. THE PATH INTEGRAL FORMALISM

1. Trick number 1: use the composition property of the transition amplitude:

K(xf ,tf ; xi,ti)= dx K(xf ,tf ; x, t)K(x, t; xi,ti) ) Applying this to J(t t ), we get: f − i J(t t )=K(0,t ;0,t )= f − i f i i (S (0,x,t t)+S (x,0,t t )) J(t t)J(t t ) e ! cl f − cl − i dx (1.79) f − − i ) We can put the expression of the classical action in the integral, and we get (T = t t , T = t t, T = t t ): f − i 1 f − 2 − i

J(T ) i mω x2(cot(ωT )+cot(ωT )) = dxe 2! [ 1 2 ] = J(T1)J(T2) ) 1 2 2 iπ 2iπ! sin(ωT )sin(ωT ) dxeiλx = = 1 2 (1.80) λ mω sin(ωT) ) . 1 2 The general solution to the above equation is mω J(T )= eaT (1.81) 2πi! sin(ωT) . with a an arbitrary constant. So this trick is not enough to fully ﬁx the propagator, but it already tells us a good deal about it. We will now compute J directly and ﬁnd a =0.Noticethatinthelimitω 0, the result goes back to that for a free particle. → 2. Now, let us compute the same quantity using the determinantmethod. What we want to compute is:

i tf m 2 2 2 ! (y˙ ω y ) J(T )= D[y]e ti 2 − (1.82) R ) with the boundary conditions:

y(ti)=y(tf )=0 (1.83)

We can note the property:

tf m tf m d2 i tf i y˙2 ω2y2 dt = i y + ω2 ydt= yOyˆ dt 2! − − 2! dt2 2 )ti )ti 6 7 )ti & ' (1.84) 2 where Oˆ = m d + ω2 . − ! dt2 Formally, we have/ to perform0 a gaussian integral, and thus theﬁnalresult 1/2 is proportional to det Oˆ − . To make it more explicit,/ 0 let us in Fourier space for y:

y(t)= anyn(t)(1.85) n % 1.3. PATH INTEGRALS AS DETERMINANTS 25

where yn(t)aretheothonormaleigenfunctionsofOˆ:

2 nπ yn = sin( t) ,n N∗ (1.86) .T T ∈

Oyˆ n(t)=λnyn(t)(1.87)

ynymdt = δnm (1.88) ) The eigenvalues are m nπ 2 λ = ω2 (1.89) n ! T − 1/ 0 2 Notice that the number of negative eigenvalues is ﬁnite and given by n = Tω − int( π ), that is the number of half periods of oscillation containedinT . The yn form a complete of the Hilbert space L2([ti,tf ]) mod y(ti)= y(tf )=0.Thus,thean form a discrete set of integration variables, and we can write: da D[y(t)] = n N˜ (1.90) √ · n 2πi + “Jacobian” where the 1/√2πi factors are singled out for!"#\$ later convenience (and also to mimick the 1/A factors in our deﬁnition of the measure in eq. (1.16)). We get: ˆ ˆ 2 yOy dt = amanymOyndt = λnan (1.91) m,n n ) % ) % And thus,

ill deﬁned

1 2 π − i(n+ n−) dan i 2 e − 4 ˜ 2 n λnan ˜ J(T )=N e = # \$!λn " N i(n +n ) π √ P | | + − 4 n 2πi 8 n 9 e ) + + well deﬁned 1 − 2 in π ! "# \$ = λ Ne˜ − − 2 (1.92) | n| 8 n 9 + Notice that in the continuum limit ϵ 0inthedeﬁnitionofthemeasure eq. (1.16), we would have to consider all→ the Fourier modes with arbitrarily large n.InthislimitboththeproductofeigenvaluesandtheJacobian N˜ are ill deﬁned. Their product however, the only physically relevant quantity, is well deﬁned. We should not get into any trouble ifweavoid computing these two quantities separately. The strategy is to work only with ratios that are well deﬁned. That way we shall keep at large from mathematical diﬃculties (and confusion!), as we will now show.

Notice that Oˆ depends on ω,whileN˜ obviously does not: the modes yn do not depend on ω 3.Forω =0,ourcomputationmustgivethefree

3Indeed a stronger result for the independence of the Jacobianonthequadraticaction holds, as we shall discuss in section 3.1. 26 CHAPTER 1. THE PATH INTEGRAL FORMALISM

particle result, which we already know. So we have

1 − 2 in π J (T )=N˜ det Oˆ e− − 2 (1.93) ω | ω| 1 / 0 2 J (T )=N˜ det Oˆ − (1.94) 0 | 0| / 0 We can thus consider 1 1 2 1 2 ˆ 2 Jω(T ) in π det O0 n λn(ω =0) λn(0) e − 2 = = | | = J0(T ) |det Oˆ | λn(ω) | | λn(ω)| 8 ω 9 6 : n | 7 8 n 9 + (1.95) : 2 The eigenvalues of the operator Oˆ = m d + ω2 over the space of ω − ! dt2 functions y(t)havebeengivenbefore: / 0 2 λ (0) nπ 1 1 n = T = = (1.96) nπ 2 ωT 2 a 2 λn(ω) & ' ω2 1 1 T − − nπ − nπ ∞ λn(0) &∞ ' 1 a & ' = = & ' (1.97) λ (ω) a 2 sin a n=1 n n=1 1 + + − nπ By eq. (1.95) we get our ﬁnal& result:'

ωT in π mω in π J (T )=J (T ) e− − 2 = e− − 2 (1.98) ω 0 · sin(ωT) 2πi! sin(ωT) ;| | . | | So that by using eq. (1.75) the propagator is (assume for simplicity ωT<π)

mω 2 2 2xf xi mω i ! (x +x ) cot(ωT) K(x ,T; x , 0) = e 2 f i − sin(ωT ) (1.99) f i 2πi! sin(ωT) h i . At this point, we can make contact with the solution of the eigenvalue prob- lem and recover the well know result for the energy levels of the harmonic oscillator. Consider Ψn acompleteorthonormalbasisofeigenvectorsofthe hamiltonian such that{ H}Ψ = E Ψ .Thepropagatorcanbewrittenas: | n⟩ n | n⟩

iH(tf −ti) K(x ,t ; x ,t )= x Ψ Ψ e− ! Ψ Ψ x = f f i i ⟨ f | n⟩⟨ n| | m⟩⟨ m| i⟩ m,n % iEn(tf −ti) ! Ψn∗ (xi)Ψn(xf )e− (1.100) n % Let us now deﬁne the partition function:

2 iEnT iEnT Z(t)= K(x, T ; x, 0)dx = Ψ (x) dx e− ! = e− ! | n | ) n 6) 7 n % % (1.101) For the harmonic oscillator, using eq. (1.100) we ﬁnd:

1 i ωT 1 ∞ i(2n+1) ωT K(x, T ; x, 0)dx = = e− 2 = e− 2 ωT 2i ωT 2i sin 1 e 2 ) 2 − − n=0 % (1.102) & ' 1.4. OPERATOR MATRIX ELEMENTS 27

1 E = !ω n + (1.103) ⇒ n 2 6 7 1.4 Operator matrix elements

We will here derive the matrix elements of operators in the path integral for- malism.

1.4.1 The time-ordered product of operators Let us ﬁrst recall the basic deﬁnition of quantities in the . Given an operator Oˆ in the Schroedinger picture, the time evolved Heisenberg picture operator is iHt iHt Oˆ(t) e Oeˆ − . (1.104) ≡ In particular for the we have

iHt iHt xˆ(t) e xeˆ − . (1.105) ≡ Given a time the independent position eigenstates x ,thevectors | ⟩ x, t eiHt x (1.106) | ⟩≡ | ⟩ represent the basis vectors in the Heisenberg picture (notice the “+” in the exponent as opposed to the “ ”intheevolutionofthestatevectorinthe Schroedinger picture!), being eigenstates− ofx ˆ(t)

xˆ(t) x, t = x x, t . (1.107) | ⟩ | ⟩ We have the relation:

S[x(t)] x ,t x ,t = D[x(t)]ei ! . (1.108) ⟨ f f | i i⟩ ) By working out precisely the same which we used in section Consider now a function A(x). It deﬁnes an operator on the Hilbert space:

Aˆ(ˆx) x x A(x)dx (1.109) ≡ | ⟩⟨ | ) Using the factorization property of the path integral we can derive the fol- lowing identity:

S[x] i ! D[x]A(x(t1))e = ) S[xa]+S[xb] i ! D[xa]D[xb]dx1A(x1)e =

) iH(t −t ) f 1 iH(t1 −ti) dx x e− ! x x e− ! x A(x )= 1 ⟨ f | | 1⟩⟨ 1| | i⟩ 1 ) iH(t −t ) f 1 iH(t1 −ti) x e− ! Aˆ(ˆx)e− ! x = ⟨ f | | i⟩ iHt f iHti x e− ! Aˆ(ˆx(t ))e ! x = ⟨ f | 1 | i⟩ x ,t Aˆ(ˆx(t )) x ,t (1.110) ⟨ f f | 1 | i i⟩ 28 CHAPTER 1. THE PATH INTEGRAL FORMALISM

Let us now consider two functions O1(x(t)) and O2(x(t)) of x(t)andletus study the meaning of:

S[x] i ! D[x]O1(x(t1))O2(x(t2))e (1.111) )

By the composition property, assuming that t1

S[xa]+S[xb]+S[xc] i ! D[xa]D[xb]D[xc]dx1dx2e O2(x2)O1(x1)(1.112) )

t1

ti a b

tf t2

c

Figure 1.8: The path from (xi,ti)to(xf ,tf )isseparatedintothreepathsa, b and c.Wehavetodistinguisht1

This can be rewritten:

x ,t x ,t O (x ) x ,t x ,t O (x ) x ,t x ,t dx dx (1.113) ⟨ f f | 2 2⟩ 2 2 ⟨ 2 2| 1 1⟩ 1 1 ⟨ 1 1| i i⟩ 2 1 ) Using the equation (1.109), it can be rewritten:

iH(t −t ) f 2 iH(t2 −t1) iH(t1 −ti) x e− ! Oˆ (ˆx)e− ! Oˆ (ˆx)e− ! x = ⟨ f | 2 1 | i⟩ x ,t Oˆ (t )Oˆ (t ) x ,t (1.114) ⟨ f f | 2 2 1 1 | i i⟩

Recall that this result is only valid for t2 >t1.

Exercise: Check that for t1 >t2,onegetsinstead:

x ,t Oˆ (t )Oˆ (t ) x ,t (1.115) ⟨ f f | 1 1 2 2 | i i⟩ Thus, the ﬁnal result is:

S[x] D[x]O (x(t ))O (x(t ))ei ! = x ,t T [Oˆ (t )Oˆ (t )] x ,t (1.116) 1 1 2 2 ⟨ f f | 1 1 2 2 | i i⟩ ) where the time-ordered product T [Oˆ1(t1)Oˆ2(t2)] is deﬁned as:

T [Oˆ (t )Oˆ (t )] = θ(t t )Oˆ (t )Oˆ (t )+θ(t t )Oˆ (t )Oˆ (t )(1.117) 1 1 2 2 2 − 1 2 2 1 1 1 − 2 1 1 2 2 where θ(t)istheHeavisidestepfunction. 1.4. OPERATOR MATRIX ELEMENTS 29

x˙ 2 Exercise: Consider a system with classical Lagrangian = m 2 V (x). Treat- ing V as a perturbation and using the path integral,L derive the− well-known formula for the operator in the

iH t/! i t Vˆ (t′,xˆ)dt′ U(t)=e− 0 Tˆ[e− ! 0 ](1.118) R 2 ! ! pˆ ˆ iH0t/ iH0t/ where H0 = 2m and V (t, xˆ)=e V (ˆx)e− . 30 CHAPTER 1. THE PATH INTEGRAL FORMALISM Chapter 2

Functional and Euclidean methods

In this chapter, we will introduce two methods:

The functional method, which allows a representations of perturbation • theory in terms of Feynman diagrams.

The Euclidean path integral, which is useful for, e.g., statistical mechanics • and semiclassical tunneling.

We will show a common application of both methods by ﬁnding thepertur- bative expansion of the free energy of the anharmonic oscillator.

2.1 Functional method

Experimentally, to ﬁnd out the properties of a physical system (for example an atom), we let it interact with an external perturbation (a source, e.g. an electromagnetic wave), and study its response. This basic empirical fact has its formal counterpart in the theoretical study of (classical and) quantum systems. To give an illustration, let us consider a system with Lagrangian (x, x˙)and let us make it interact with an arbitrary external forcing source: L

(x, x˙) (x, x˙)+J(t)x(t) L →L i ( (x,x˙ )+J(t)x(t))dt K(x ,t ; x ,t J)= D[x]e ! L (2.1) f f i i| R ) K has now become a functional of J(t). This functional contains important physical information about the system (all information indeed) as exempliﬁed by the fact that the functional derivatives of K give the expectation values of xˆ (and thus of all operators that are a function of x):

!δ !δ K(xf ,tf ; xi,ti J) = xf ,tf T [ˆx(t1),...,xˆ(tn)] xi,ti iδJ(t1) ···iδJ(tn) | J=0 ⟨ | | ⟩ 5 5 (2.2) 5 31 32 CHAPTER 2. FUNCTIONAL AND EUCLIDEAN METHODS

As an example, let us see how the eﬀects of a perturbation can betreated with functional methods. Let us look at the harmonic oscillator, with an anharmonic self-interaction which is assumed to be small enough to be treated as a perturbation:

m mω2 λ (x, x˙)= x˙ 2 x2, (x, x˙)= x4 (2.3) L0 2 − 2 Lpert −4!

The propagator K0[J]fortheharmonicoscillatorwithasourceJ(t)is:

i ( +J(t)x(t))dt K [J]= D[x]e ! L0 (2.4) 0 R ) Consider now the of the perturbation. By repeated use of the functional derivatives with respect to J (see eq. (2.2)) the propagator can be written as:

i λ 4 ! ( 0 x +J(t)x(t))dt K[J]= D[x]e L − 4! R ) n 1 iλ 4 4 i ( +J(t)x(t))dt = D[x] − x (t ) x (t )e ! L0 dt dt n! !4! 1 ··· n R 1 ··· n n ) % ) 6 7 1 iλ n !4δ4 !4δ4 = − K [J]dt dt n! !4! δJ4(t ) ···δJ4(t ) 0 1 ··· n n 1 n % ) 6 7 iλ!3 δ4 exp dt K [J](2.5) ≡ − 4! δJ4(t) 0 6 ) 7 Thus, when we take the J 0limit,wegetthepropagatoroftheanhar- monic hamiltonian in terms of functional→ derivatives of the harmonic propagator in presence of an arbitrary source J.

2.2 Euclidean Path Integral

iHt Instead of computing the transition amplitude xf e− ! xi ,wecouldhave computed the evolution: ⟨ | | ⟩

βH K x e− ! x ,βR (2.6) E ≡⟨ f | | i⟩ ∈ + which corresponds to an imaginary time interval t = iβ. − 0 β = Nϵ ϵ 2ϵ (N 1)ϵ − Figure 2.1: The interval between 0 and β is divided into N steps.

It is straightforward to repeat the same time slicing procedure we employed in section 1.1.4 (see ﬁgure 2.1) and write the imaginary time amplitude as a path integral. The result is:

1 ! SE [x] KE(xf ,xi; β)= D[x]E e− (2.7) ) 2.2. EUCLIDEAN PATH INTEGRAL 33

where D[x]E is a path integration measure we shall derive below and where

β m dx 2 S [x]= + V (x) dτ βE (2.8) E 2 dτ ∼ )0 8 6 7 9

Notice that SE is normally minimized on stationary points. From a functional viewpoint KE is the continuation of K to the imaginary time axis K (x ,x ; β)=K(x ,x ; iβ) . (2.9) E f i f i −

Notice that more sloppily we could have also gotten KE by “analytic con- tinuation” to imaginary time directly in the path integral:

i t (x,x˙)dt′ K(t)= D[x]e ! 0 L (2.10) R ) Then, with the following replacements

t′ = iτ − dt′ = idτ − dx x˙ = i dτ t = iβ − we obviously get the same result. Notice that the interval in the 2 2 2 relativistic description ds = dt + dxi ,correspondingtoaMinkovskymetric with signature , +, +, +, gets− mapped by going to imaginary time into a metric − 2 2 2 with Euclidean signature dsE = dτ + dxi .Hencethe‘Euclidean’suﬃx. Let us now compute explicitly by decomposing the imaginary time interval [0,β]intoN steps of equal length ϵ (see ﬁgure 2.1). As in section 1.1.4 we can write:

Hϵ Hϵ ! ! KE(xf ,xi; β)= xN e− xN 1 x1 e− x0 dx1 dxN 1 (2.11) ⟨ | | − ⟩···⟨ | | ⟩ ··· − ) where xf = xN and xi = x0. We can compute, just as before:

Hϵ Hϵ x′ e− ! x = x′ p p e− ! x dp ⟨ | | ⟩ ⟨ | ⟩⟨ | | ⟩ ) 2 1 ip (x′ x) ϵ p +V (x) e ! − − ! 2m dp ≈ 2π! “ ” ) 1 ϵ m ′ 2 m ′ 2 ϵ ! (p i (x x)) ! (x x) ! V (x) = e− 2m − ϵ − − 2ϵ − − dp 2π! ) 2 1 m x′−x m ! 2 ϵ +V (x) ϵ = e− » “ ” – 2πϵ! . ′ m 1 (x,x′)ϵ 1 SE (x ,x) = e− ! LE e− ! (2.12) 2πϵ! ≡ A . E 34 CHAPTER 2. FUNCTIONAL AND EUCLIDEAN METHODS and then proceed to obtain the Euclidean analogue of eq. (1.16)

N 1 − 1 − N−1 ! SE (xl+1,xl) KE(xf ,xi; β)=lim dxk e l=0 ϵ 0 N P → AE ) k+=1 N 1 1 − dxk SE(xf ,xi) =lim e− ! (2.13) ϵ 0 A A → E E ) k+=1 D[x]E R ! "# \$ Synthetizing: to pass from time path integral formulation to the Eu- clidean time formulation, one simply makes the change iϵ ϵ in the factor A deﬁning the measure of integration, and modiﬁes the integra→nd according to

t 1 β 1 iS = i mx˙ 2 V (x) dt S = mx˙ 2 + V (x) dτ 2 − −→ − E − 2 )0 6 7 )0 6 7 iS/! S /! e e− E (2.14) −→

2.2.1 Statistical mechanics

Recalling the result of the previous section we have

βEn/! KE(xf ,xi; β)= Ψn∗ (xf )Ψn(xi)e− (2.15) n % where Ψn is an othonormal basis of eigenvectors of the Hamiltonian. Wecan then deﬁne{ } the partition function:

βEn/! Z[β]= dxKE(x, x; β)= e− (2.16) n ) % We recognize here the statistical mechanical deﬁnition of the thermal parti- tion function for a system at kB T = !/β,wherekB is the Boltz- mann constant. We can observe the equivalence:

SE Z[β]= dxKE(x, x; β)= D[x]e− (2.17) x(0)=x(β) ) ){ } Thus, the thermal partition function is equivalent to a functional integral over a compact Euclidean time: τ [0,β]withboundariesidentiﬁed.The integration variables x(τ)arethereforeperiodicin∈ τ with β. Let us now check that K( iβ)leadstotherightpartitionfunctionforthe harmonic oscillator. Using equation− (1.102), we get:

βω(n+ 1 ) Z[β]= K(x, x; iβ)dx = e− 2 (2.18) − n ) % which is indeed the expected result. 2.3. PERTURBATION THEORY 35 2.3 Perturbation theory

The purpose of this section is to illustrate the use of functional methods in the euclidean path integral. We shall do so by focussing on the anharmonic oscilla- tor and by computing in perturbation theory the following physically relevant quantities

the time ordered correlators on the : 0 T [ˆx(τ ) ...xˆ(τ )] 0 • ⟨ | 1 n | ⟩ the free energy and the ground state energy. • 2.3.1 Euclidean n-point correlators

β Consider the euclidean path integral deﬁned between initialtimeτi = 2 and β − ﬁnal time τf =+2 .Aswedidfortherealtimepathintegral,wecanconsider the euclidean n-point correlator

β x( )=xf 2 SE [x(τ)] ! D[x(τ)]E x(τ1) ...x(τn)e− (2.19) x( β )=x ) − 2 i By working out precisely the same algebra of section 1.4, the euclidean correlator is also shown to be equal to the operatorial expression

βH/2! βH/2! x e− T [ˆx (τ ) ...xˆ (τ )]e− x (2.20) ⟨ f | E 1 E n | i⟩ where Hτ/! Hτ/! xˆE(τ)=e xeˆ − (2.21) represents the continuation to imaginary time of the Heisenberg picture position operator. Indeed, comparing to eq. (1.105) we have,x ˆ (τ)=ˆx( iτ). E − It is interesting to consider the limit of the euclidean path integral and correlators when β .Thisismostdirectlydonebyworkingintheoperator formulation. By using→∞ the complete set of energy eigenstatets the euclidean amplitude can be written as

β β βH K (x , ; x , )= x e− ! x =(2.22) E f 2 i − 2 ⟨ f | | i⟩ βEn/! = ψn(xf )ψn(xi)∗e− n % β βE0/! β(E1 E0)! →∞= ψ0(xf )ψ0(xi)∗e− 1+O(e− − ) > ? where E0 and E1 are the of respectively the ground state and the ﬁrst . From the above we conclude that for β !/(E1 E0)the βH/! ≫ − operator e− acts like a projector on the ground state. We arrive at the same conclusion also for the correlators

βH βH ! ! lim xf e− 2 T [ˆxE(τ1) ...xˆE(τn)]e− 2 xi (2.23) β ⟨ | | ⟩ →∞ βE0 β(E E )/! = 0 T [ˆx (τ ) ...xˆ (τ )] 0 ψ (x )ψ (x )∗e− ! 1+O(e− 1− 0 ) ⟨ | E 1 E n | ⟩ 0 f 0 i > ? 36 CHAPTER 2. FUNCTIONAL AND EUCLIDEAN METHODS where by 0 we indicate the energy ground state (not the x =0positioneigen- state!). By| ⟩ taking the ratio of eq. (2.23) and eq. (2.22) we thus ﬁnd

βH βH xf e− 2! T [ˆxE(τ1) ...xˆE (τn)]e− 2! xi lim ⟨ | βH | ⟩ = 0 T [ˆxE(τ1) ...xˆE(τn)] 0 β ! ⟨ | | ⟩ →∞ xf e− xi ⟨ | | ⟩ (2.24) where the dependence on xi and xf has cancelled out. The time ordered cor- relators on the ground state are also known as Feynman correlators. Using the path integral fomulation we can suggestively write

SE D[x(τ)] x(τ ) ...x(τ )e ! 0 T [ˆx (τ ) ...xˆ (τ )] 0 =lim E 1 n − (2.25) E 1 E n SE ⟨ | | ⟩ β D ! →∞ * [x(τ)]E e− Notice that the structure of above equation is* reminiscent ofthermalaverages in classical statistical mechanics, with the denominator playing the role of a partition function. As we already said, in the β limit the above equation →∞ is independent of xi,xf .Forthesakesimplicityweshallthenfocusonthecase xi = xf =0anddeﬁne,atﬁniteβ,thenormalizedn-pointcorrelatoras

SE D[x(τ)]E x(τ1) ...x(τn)e− ! n xi=xf =0 ! 2 G (τ ,...,τ )= (2.26) D 1 n SE * D[x(τ)]E e− ! xi=xf =0 where the suﬃx D indicates that this* normalized n-point correlator was com- puted by imposing Dirichlet boundary conditions xi = xf =0inthepath n/2 integral. In the deﬁnition of GD, ! is factored out for later convenience. The euclidan Feynman correlators are related to the Feynman correlators in real time 0 T [ˆx(t ) ...xˆ(t )] 0 (2.27) ⟨ | 1 n | ⟩ by performing an analytic continuation τi iti in such a way as to preserve the time ordering. This is most simply done→ via a counter-clockwise rotation, the , in the complex time plane as shown in ﬁg. 2.2. Indeed the time ordered correlators, both for real and imaginary time, are by construction analytic functions of the time coordinates over the connected domains where none of the time coordinates (ex. τ1,τ2,...,τn)coincide.Eachsuchdomain corresponds to a give ordering of the n time coordinates. The non-analiticity is purely due to the step functions that enforce time orderingandislocalized at the points where at least two coordinates coincide. Then bycontinuingthe time coodinates on trajectories that do not cross (like in theWickrotation)one remains within the domain of analyticity. The real time Feynman correlators play an important role in relativistic quantum ﬁeld theory. The n-point correlators are associatedtotheamplitudes describing the of nI initial particles into nF ﬁnal particles with n = nI + nF .

2.3.2 Thermal n-point correlators One can also deﬁne the correlators using the path integral with periodic bound- ary conditions

SE [x(τ)] ! D[x(τ)]E x(τ1) ...x(τn)e− . (2.28) x( β )=x( β ) ) − 2 2 2.3. PERTURBATION THEORY 37

Figure 2.2: Wick rotation from imaginary to real time. The arrows on the coordinate axes indicate the future direction, both for realandimaginarytime

This object should expectedly be related to statistical mechanics. Indeed, in the operator formulation, eq. (2.28) can be written as

βH βH βH x e− 2! T [ˆx (τ ) ...xˆ (τ )]e− 2! x dx =Tr e− ! [ˆx (τ ) ...xˆ (τ )] ⟨ | E 1 E n | ⟩ E 1 E n ) @ (2.29)A βH/! Normalizing by the partition function Z =Tr[e− ]weobtainhethermal average of the time ordered operator product T [ˆxE(τ1) ...xˆE(τn)]

1 βH T [ˆx (τ ) ...xˆ (τ )] = Tr e− ! T [ˆx (τ ) ...xˆ (τ )] (2.30) E 1 E n β Z(β) E 1 E n B C @ A which in the path integral language corresponds to the normalized n-point ther- mal correlator

SE ! β β D x( )=x( ) [x(τ)]E x(τ1) ...x(τn)e− n (2.30) = − 2 2 ! 2 G (τ ,...,τ ) SE P 1 n D ! ≡ * x( β )=x( β ) [x(τ)]E e− − 2 2 * (2.31) where the have deﬁned GP in analogy with the Dirichlet n-point correlator of the previous section. GD and GP only diﬀer by the choice of boundary conditions, respectively Dirichlet for GD and periodic for GP ,appliedtothecorresponding path integrals. Like with G ,byperformingtheWickrotationτ it we obtain D k → k

n 1 βH ! 2 G (it ,...,it )= Tr e− ! T [ˆx(t ) ...xˆ(t )] , (2.32) P 1 n Z(β) 1 n @ A that is the thermal average of the time ordered product in realtime. 38 CHAPTER 2. FUNCTIONAL AND EUCLIDEAN METHODS

Finally, notice that for β we have →∞ βH e ! lim − = 0 0 (2.33) β Z(β) | ⟩⟨ | →∞ This is expected, as β corresponds to a statistical ensemble at zero tem- perature. Then at β →∞both Dirichlet and periodic correlators coincide with the Feynman correlator→∞

n n lim ! 2 GP (τ1,...τn)= lim ! 2 GD(τ1,...τn)= 0 T [ˆxE(τ1) ...xˆE (τn)] 0 . β β ⟨ | | ⟩ →∞ →∞ (2.34) From the path integral viewpoint, the coincidence of GD and GP in the above equation represents the expected fact that for ﬁnite τ1,...,τn the boundary conditions in eqs. (2.26,2.31) become irrelevant for β →∞ 2.3.3 Euclidean correlators by functional derivatives Consider the euclidean path integral for our system coupled to an external source J(τ). For the case of Dirichlet boundary conditions xi = xf =0wedeﬁne

β β K [β,J] K (0, ;0, J)= E ≡ E 2 − 2 | β 2 D 1 [x]E exp ! dτ ( E(x, x˙) J(τ)x(τ)) (2.35) x =x =0 − β L − ) i f , )− 2 - and similarly for periodic boundary conditions

β β Z[β,J] K (0, ;0, J)= ≡ E 2 − 2 | β 1 2 D[x]E exp dτ ( E (x, x˙) J(τ)x(τ)) . (2.36) ! β x =x − L − ) i f , )− 2 - The correlators can then be expressed in terms of functional derivatives. Using the deﬁnitions in eqs. (2.26,2.31) we have

n 1 !δ !δ ! 2 GD(τ1,...,τn)= ... KE[β,J] (2.37) KE[β,0] δJ(τ1) δJ(τn) J=0 5 n 1 !δ !δ 5 ! 2 GP (τ1,...,τn)= ... Z[β,J] 5 (2.38) Z[β,0] δJ(τ1) δJ(τn) J=0 5 5 0 0 5 Like in section 2.1 we deﬁne KE[β,J]andZ [β,J]tobethepathintegrals 0 associated to the simple harmonic oscillator with euclideanlagrangian E = mx˙ 2 mω2x2 L 2 + 2 .Wewanttoconsiderthegeneralanharmonicoscillatorwith lagrangian mx˙ 2 mω2x2 = + +∆V (x)= 0 +∆V (x)(2.39) LE 2 2 LE 0 where ∆V can be treated as a small perturbation of the dynamics of E.Again just following the procedure shown in section 2.1, the path integralsL for the 2.3. PERTURBATION THEORY 39 anharmonic oscillator can be written as 1 !δ K [β,J]=exp dτ∆V K0 [β,J](2.40) E −! δJ(τ) E 1 ) 2 1 & !δ ' Z[β,J]=exp dτ∆V Z0[β,J](2.41) −! δJ(τ) 1 ) 2 & ' Summarizing:tocomputeanyinterestingquantity(correlatororpartition function) we must 0 0 1. compute the unperturbed KE and Z , 2. learn to take their derivatives with respect to J(τ)inaneﬃcientway. The ﬁrst step amounts to a Gaussian integral. The second is basically a problem in combinatorics, whose solution has a convenient diagrammatic representation: the Feynman diagrams.

0 0 2.3.4 Computing KE[J] and Z [J] 0 0 KE[β,J]andZ [β,J]arejustgaussianintegralswithanexternalsource(eu- clidean forced harmonic oscillator). To compute them we can use the method of Green’s functions. First, let us recall the result for a gaussian integral with a linear source (in the discrete ﬁnite case). We want to compute the integral: n 1 i j k x Oij x +Jkx I[J]= dxke− 2 (2.42) ) +k where Oij is a positive deﬁnite n n matrix and we sum over repeated indices. The inverse G of the matrix O×is deﬁned by: ij ji i G Ojk = Okj G = δ k (2.43)

Such an inverse matrix exists, since Oij is assumed positive deﬁnite. We i i ij then perform the change of variablesx ¯ = x G Jj which obviously has trivial Jacobian − dxk = dx¯k (2.44) +k +k and such that 1 1 1 xiO xj + J xk = x¯iO x¯j + J GklJ (2.45) −2 ij k −2 ij 2 k l Thus, by performing the dx¯ integral: 1 I[J]=I[0] exp J Gij J (2.46) 2 i j 6 7 We can easily transpose that discrete example to our continuum case. Con- 0 0 centrating ﬁrst on KE for the sake of clarity (but the discussion on Z will be analogous) we have

1 m d2 2 0 ! x(τ) +ω x(τ) J(τ)x(τ) D − 2 − dτ2 − KE[β,J]= [x]E e R h i ) & ' 1 1 ˆ ! [ x(τ)Ox(τ) J(τ)x(τ)] = D[x] e− 2 − . (2.47) E R ) 40 CHAPTER 2. FUNCTIONAL AND EUCLIDEAN METHODS

Notice that on the space of functions that vanish at the boundaries (that is at τ = β/2) the diﬀerential operator Oˆ = m( d2/dτ 2 + ω2)ishermitianand positive± deﬁnite. As a matter of fact, in terms− of its matrix elements between time coordinate eigenstates ψ (τ) δ(τ τ )andψ (τ) δ(τ τ ) τ1 ≡ − 1 τ2 ≡ − 2

O(τ ,τ ) dτψ (τ)Oˆψ (τ)=m ∂ ∂ + ω2 δ(τ τ ) 1 2 ≡ τ1 τ2 τ1 τ2 1 − 2 ) 2 2 = m ∂ + ω δ(τ1 τ2)(2.48)& ' − τ1 − using integration by parts& and repeatedly' using x( β/2) = 0, we can rewrite the exponent in eq. (2.47) as ±

1 x(τ )x(τ )O(τ ,τ )dτ dτ + J(τ )x(τ )(2.49) − 2 1 2 1 2 1 2 1 1 ) ) in full analogy with 1 xiO xj + J xi in eq. (2.42) We emphasize here that − 2 ij i O(τ1,τ2)isafunction(betteradistribution)inthetwovariablesτ1 and τ2. Like in the discrete case, it is convenient to rewrite the exponent using the inverse of Oˆ,thatistheGreen’sfunctionsatisfyingDirichletboundary conditions GD(τ1.τ2)

2 2 Oˆτ GD(τ1,τ2) m ∂ + ω GD(τ1,τ2)=δ(τ1 τ2)(2.50) 1 ≡ − τ1 − β& ' β G ,τ = G τ, =0. (2.51) D ± 2 D ± 2 6 7 6 7 1 GD(τ1,τ2)isjustthematrixformofOˆ− .Usingthetwoequationsaboveone can also check that O GD O(τ1,τ3)GD(τ3,τ2)dτ3 = δ(τ1 τ2). The solution of eqs. (2.50,2.51) is given· ≡ by − * β β 1 sinh ω 2 + τ< sinh ω 2 τ> G (τ ,τ )= − (2.52) D 1 2 mω / / sinh00 (ωβ)/ / 00 where τ> =max(τ1,τ2)andτ< =min(τ1,τ2). Notice the properties:

1 β τ1τ2 lim GD(τ1,τ2)= τ1 τ2 2 (2.53) ω 0 m 2 −| − |− β → 6 7 1 ω τ1 τ2 lim GD(τ1,τ2)= e− | − | (2.54) β 2mω →∞ Shifting the integration variable

x(τ)= G (τ,τ′)J(τ ′)dτ ′ + y(τ) G J + y (2.55) D ≡ · ) the exponent is rewritten as (we use to represent the in τ) · 1 1 x Oˆ x + J x = (y Oˆ y + J G Oˆ G Jˆ+2y J) (J y + J G J) 2 · · · 2 · · · D · · D · · − · · D · 1 = (y Oˆ y J G J)(2.56) 2 · · − · D · 2.3. PERTURBATION THEORY 41 so that we ﬁnd

0 1 1 S0 [y] K [β,J]=exp dτ dτ J(τ )G (τ ,τ )J(τ ) D[y] e− ! E (2.57) E 2! 1 2 1 D 1 2 2 E 6 ) 7 ) 1 J G J 0 1 J G J mω = e 2! · D· K [β,0] = e 2! · D· (2.58) E 2π sinh(ωβ) . Following precisely the same steps we can compute the partition function Z0[β,J]. The only diﬀerence is that in this case the functional integral is over the space of periodic functions in the interval [ β/2,β/2]. The inverse of Oˆ is − now given by the Green’s function GP satisfying period boundary conditions: G (β/2,τ)=G ( β/2,τ), G (τ,β/2) = G (τ, β/2). Thus we ﬁnd P P − P P −

1 1 0 0 ! J GP J ! S [y] Z [β,J]=e 2 · · D[y]Ee− E (2.59) )periodic 1 J GP J 1 e 2! · · 2! J GP J 0 = e · · Z [β,0] = βω (2.60) 2sinh 2!

As for GP one ﬁnds

β cosh ω( τ τ ′ ) 1 2 −| − | GP (τ,τ′)= (2.61) 2mω > β ? sinh ω 2 > ? Notice that for β with τ and τ ﬁxed 1 we have →∞ 1 2

1 ω τ1 τ2 lim GP =limGD = e− | − | (2.62) β β 2mω →∞ →∞ showing that for β the boundary conditions expectedly do not matter. →∞ 2.3.5 Free n-point correlators Applying the results of the previous section to eqs. (2.38) weobtainauseful expression for the n-point correlators in the free case, that is the harmonic 0 oscillator. The crucial remark is that the J independent prefactors KE[β,0] and Z0[β,0] drop out from the computation and so we simply get

k=n n n δ 1 ! J GD,P J ! 2 GD,P (τ1,...,τn)=! 2 e 2 · · . (2.63) 8 δJ(τk)9 J=0 k=1 5 + 5 5 n Notice that via the rescaling J J/√! we have put in evidence a factor ! 2 . As the formulae are the same for→ Dirichlet and period boundaryconditions,we shall drop the P and D in what follows. Working out eq. (2.63) explicitly we ﬁnd

1 for n odd: G(τ1,...,τn)=0.Thissimplyfollowsbecauseexp(2 J G J) • is an even functional of J and we take the limit J 0aftercomputing· · the derivatives. →

1 More precisely for τ1 τ2 β,aswellas τ1,2 β/2 β for GD. | − |≪ | ± |≪ 42 CHAPTER 2. FUNCTIONAL AND EUCLIDEAN METHODS

for n even •

G(τ1,...,τn)= G(τp(1),τp(2)) G(τp(n 1),τp(n)) . (2.64) ··· − p σ %∈ n

σn is the of all of n elements where the permutations diﬀering by the exchange of two elements in a pair or the exchange of a pair are identiﬁed. This group contains n! (n 1)!! = (n (n/2)!2(n/2) ≡ − − 1)(n 3) 3 1elements. − ··· · Let us see how this works in the simplest cases. Notice ﬁrst of all that the two point correlator (D or P) simplyequalsthe corresponding Green’s function (thus our notation was wisely chosen!)

1 δ δ ! J GD,P J G(τ1,τ2)= e 2 · · (2.65) δJ(τ1) δJ(τ2) J=0 5 δ 1 5 ! J GD,P J = J(τ)G(τ,τ2)dτe52 · · (2.66) δJ(τ1) J=0 ) 5 = G(τ ,τ ) 5 (2.67) 1 2 5 To compute the correlator for higher n it is convenient to introduce the graphical notation

τ1 A) G(τ ,τ )J(τ )dτ 1 n n n ≡ * τ1 τ2 B) G(τ ,τ ) 1 2 ≡

Each functional derivative acts in two possible ways:

on the exponent, i.e. giving a factor like in A) • on a factor of type A) brought down by a previous derivative, thus giving • afactorlikeinB)

Using the above graphical rules to perform the derivatives one easily sees that after setting J 0thetermsthatsurvivegiveeq.(2.64).Forinstance,thefour points correlator→ is given by the graph shown in ﬁgure 2.3

G(τ1,τ2,τ3,τ4)=G(τ1,τ2)G(τ3,τ4)+G(τ1,τ3)G(τ2,τ4)+

G(τ1,τ4)G(τ2,τ3) . (2.68)

The general result we have found is known in quantum ﬁeld theory as Wick’s theorem. It is normally derived by using creation and destruction operators. The path integral allows for a more direct, functional, derivation, which is what we have just shown. 2.3. PERTURBATION THEORY 43

τ1 τ2 τ1 τ2 τ1 τ2

+ +

τ3 τ4 τ3 τ4 τ3 τ4

Figure 2.3: A graphical representation of the four points correlator.

2.3.6 The anharmonic oscillator and Feynman diagrams Using the diagrammatic method outlined in the previous section, and working in a perturbative expansion in ∆V ,wecancomputethen-pointcorrelators,the propagator and partition function as written in section 2.3.3. For illustrative purposes, we shall focus on the simple example of the anharmonic oscillator with 4 λ4 4 an x potential: ∆V = 4! x .WeshallcomputetheleadingO(λ4)correctionto the free energy. Additional examples, involving other perturbations and other , are extensively discussed in the exercise sessions. From statistical mechanics we have that the free energy F (β)ofasystemin equilibrium at ﬁnite temperature is ! β F (β) F (β)= ln Z[β] Z[β]=e− ! (2.69) −β ↔ We also recall that for β the partition function is dominated by the ground state, so that in this limit→∞ the free energy coincides with thegroundstateenergy

lim F (β)=E0 (2.70) β →∞

Expanding eq. (2.41) at ﬁrst order in λ4 we have (notice that we also rescaled J √!J) → ! λ4 δ 4 1 0 ! dτ( ) J GP J Z[β]=Z [β]e− 4! δJ(τ) e 2 · · (2.71) R J=0 45 !λ4 δ 5 1 0 2 2 J GP J = Z [β] 1 dτ 5 + λ4 e · · (2.72) , − 4! δJ(τ) O - J=0 ) 6 7 5 & ' 5 We learned from Wick’s theorem that the free four points corr5 elator factor- izes as

δ δ 1 J G J e 2 · · =(G(τ1,τ2)G(τ3,τ4)+ δJ(τ1) ···δJ(τ4)

G(τ1,τ3)G(τ2,τ4)+G(τ1,τ4)G(τ2,τ3)) (2.73)

And thus, in the case at hand, the linear term in the expansion of Z[β]is

4 β !λ4 δ 1 3!λ4 2 2 J GP J 2 dτ e · · = GP (τ,τ)dτ. (2.74) − 4! δJ(τ) J=0 − 4! β ) 6 7 5 )− 2 5 This result is graphically represented5 by ﬁgure 2.4: it simply corresponds to ﬁgure 2.3 in the limiting case in which the four points are taken to coincide, 44 CHAPTER 2. FUNCTIONAL AND EUCLIDEAN METHODS

τ1 = τ2 = τ3 = τ4 = τ.Thecoordinateτ is called the interaction point, and the four lines associated to this point form what is called an interaction vertex. Notice that the factor 3 precisely counts the number of diagrams in the Wick expansion of G(τ1,τ2,τ3,τ4). This graphical representation of a term in the perturbative expansion of Z[β]iscalledaFeynmandiagram.

3G(τ,τ)2 τ ≡

Figure 2.4: A graphical representation of the part of the propagator linear in λ4.

Substituting in eq. (2.72) the explicit expression for GP and performing the (trivial) integral we ﬁnd !λ ωβ Z[β]=Z0[β] 1 4 (ωβ)coth2( )+ λ 2 (2.75) − 32m2ω3 2 O 4 1 2 & ' so that the free energy is ! ! ω 1 ωβ λ4 2 ωβ 2 F = 1+ ln(1 e− )+ coth ( )+ λ . (2.76) 2 ωβ − 32m2ω3 2 O 4 6 7 & ' Notice that the second term in the expression in represents the thermal contribution to the harmonic oscillator free energy, while the third term gives the leading anharmonicity correction. By taking the limit β we obtain →∞ the zero point energy of the anharmonic oscillator at lowest order in λ4 !ω !λ E = 1+ 4 + λ 2 . (2.77) 0 2 32m2ω3 O 4 6 7 & ' Some comments on the above results are in order. By eq. (2.77),thedimension- less parameter controlling the perturbative expansion is !λ λ¯ = 4 . (2.78) m2ω3 This result could have been anticipated in various ways. By eq. (2.71) the expansion parameter must be proportional to !λ4,whichhoweverhasdimension mass2/time3.Withtheparametersathandintheharmonicoscillator,λ¯ is the only dimensionless parameter proportional to !λ4.Also,moredirectly:thewave function of the harmonic oscillator ground state is localized within a distance 2 4 2 2 X = x !/mω,andwithinthisdistanceλ4X /(mω X ) λ¯,setsthe relative⟨ size⟩∼ of the perturbation. ∼ D D While λ¯ 1is,aswastobeexpected,necessarytoapplyperturbation theory, the steps≪ that lead to eqs. (2.76,2.77) leave something to be desired. This is because in eq. (2.75) the ﬁrst term in the expansion is λωβ¯ ,and becomes inﬁnite in the limit β ,suggestingthatthehigherordertermsin∼ the expansion also blow up in→∞ this limit. How can we then make sense of our truncation at lowest order? Our procedure to compute E0 makes sense as long as there exists a range of β where 2.3. PERTURBATION THEORY 45

1. λωβ¯ 1 ∼ ≪ 2. the contribution of the excited states to Z(β)isnegligible,thatisfor

β (E E ) β e− ! 1− 0 1 (E E ) 1(2.79) ≪ −→ ! 1 − 0 ≫ which in our case, at lowest order corresponds, to ωβ 1 ≫ The simultaneous satisfaction of 1) and 2) implies again λ¯ 1, and thus our manipulations were technically justiﬁed. ≪ But there is more. As a matter of fact, the Hamiltonian formulation tells us that for β the exact partition function is →∞ β β βω λ¯ ! E0 (1+ +... ) Z[β] →∞ e− = e− 2 32 (2.80) −→ implying that the higher order corrections proportional to λωβ¯ in eq. (2.75) must build up into the exponential function. In the technical jargon one says that they must exponentiate.Thuseventhougheq.(2.75)isnotagoodapproximation for β ,thestructureofthegrowingtermsisﬁxedtogiveanexponential. As a corollary→∞ ln Z[β]/β is well behaved for β and for this quantity we can safely apply− perturbation theory. →∞ In view of the above discussion we are now motivated to developapicture on the structure of the higher order terms in the expansion of Z[β]. The study of the O(λ¯2)contributionisalreadyverysuggestiveofwhathappens.Let us 2 focus on it ﬁrst. Expanding the exponent in eq. (2.71), we ﬁnd at O(λ4)

! 2 4 4 1 λ4 δ δ 1 J G J 4 4 dτ dτ e 2 · P · x (τ )x (τ ) 2! 4! 1 2 δJ(τ ) δJ(τ ) ∼ 1 2 6 7 ) 6 1 7 6 2 7 5J=0 5 B (2.81)C 5 Thus, we have to sum over diagrams that contain two5 points, τ1 and τ2,with four lines ending at each point. We can write three diﬀerent diagrams, as can be seen in ﬁgure 2.5.

τ1 τ2 τ1 τ2

A C B τ1 τ2

Figure 2.5: The three diﬀerent diagrams contributing to the corrections to the ﬁrst of the anharmonic oscillator. Diagrams A and B (at the top- left corner and at the bottom) are connected, while diagram C consists of two disconnected pieces. 46 CHAPTER 2. FUNCTIONAL AND EUCLIDEAN METHODS

The second order term is then shown, by direct computation, tobeequalto:

1 !λ 2 4 dτ dτ 24G4 (τ ,τ )+ 2! − 4! 1 2 P 1 2 6 7 ) 3 2 2 2 72GP (τ1,τ1)GP (τ1,τ2)GP (τ2,τ2)+9GP (τ1,τ1)GP (τ2,τ2) (2.82) where each of these three addenda corresponds to one of the diagrams4 in ﬁgure 2.5. The numerical factors in front of each of the three factors represents the multiplicity of the corresponding . It is a useful exercise to reproduce these numbers. Let us do it for diagram A. To count its multiplicity we have to count all possible ways to connect the 4 lines comingoutoftheτ1 ver- tex with the 4 lines coming out of the τ2 vertex. Let us pick one given line from τ1:itcanconnecttoτ2 in 4 possible ways. After having connected this ﬁrst line, let us pick one of the remaining 3 lines from τ1:itcanconnecttoτ2 in 3 possible ways. After this we pick one of the remaining two lines, which can connect to τ2 in 2 possible ways. The remaining line can connect in just 1 way. Therefore the number of combinations is 4 3 2 1=24.Inthesamewayonecanrepro- duce the factors 72 and 9 associated× × respectively× to diagram BandC.Thetotal number of diagrams is therefore 24+72+9= 105. On the other hand, eq. (2.81) tells us that the number of diagrams should be the same as for the correlator among 8 x’s (more precisely x(τ1)x(τ1)x(τ1)x(τ1)x(τ2)x(τ2)x(τ2)x(τ2) ). In- deed the combinatoric factor⟨ of section 2.3.5 for the 8-pointcorrelatorgives⟩ 8!/(4! 24)=105. In diagrams A and B there exist at least one line (in fact more) connecting τ1 and τ2.Thesediagramsarethereforecalledconnected.Ontheotherhand in diagram C no line exists connecting τ1 and τ2:diagramCconsistsoftwo disconnected copies of the lowest order diagram in Fig.2.4. This implies that the two-dimensional integral corresponding to C factorizes into the square of one- dimensional integral. Diagrams A and B correspond to genuine, non-trivial, two-dimensional integrals. Putting together the diagrams in Figs. 2.4 and 2.5 the partition function expanded at second order can be written as

!λ 1 !λ 2 Z[β]=Z [β] 1 4 G2 (τ,τ)dτ + 4 G2 (τ,τ)dτ + A + B + O(λ3) 0 − 8 P 2 8 P 4 , 6 ) 7 6 ) 7 - (2.83) where we do not have written the explicit expressions of the connected diagrams AandB.Thepointoftheaboveexpressionisthatthedisconnected diagram Chastherightcoeﬃcienttoprovidetheﬁrststepintheexponentiation of the contribution of order λ4.Thispatterncontinuesathigherordersandforall diagrams: the disconnected diagrams just serve the purpose of exponentiating the contribution of the connected ones

( connected diagrams) Z[β]=Z0[β] e . (2.84) P 3 For instance, at O(λ4), in addition to new connected diagrams, there will be disconnected diagrams corresponding to three copies of Fig.2.4andalsodia- grams made of one copy of Fig. 2.4 with one copy of A (or B). Notice, as it should be evident, that the connected diagrams appear in the above exponent 2.3. PERTURBATION THEORY 47 precisely with the same coeﬃcient they have when they appear for the ﬁrst time in the expansion of Z[β]. The exponentiation in Z[β]impliesthatforthefreeenergywesimplyhave

F = F0 + connected diagrams . (2.85) / 0 Important result:inordertocomputethefreeenergy,andthusalsothe% ground state energy, we only need to compute the connected diagrams. One can easily check exponentiation of the O(λ4)correctioninFig.2.4 ! 4 λ4 δ 1 J G J exp dτ e 2 · P · = − 4! δJ(τ) 8 ) 6 7 9 5J=0 5 1 !λ n δ 5 4 δ 4 1 4 dτ dτ 5 exp J G J = n! − 4! 1 ··· n δJ(τ ) ··· δJ(τ ) 2 · P · n 1 n % 6 7 ) 6 7 6 7 6 7 1 3!λ n 4 dτG2 (τ,τ) +(othertopologies) = n! − !4! P n % 16 ) 7 2 3!λ exp 4 dτG2 (τ,τ) [1 + (other topologies)] (2.86) − 4! P 6 ) 7 The general proof of eq. (2.84) is an exercise in combinatorics. We shall not present it, but leave it as a challenge to the student. Insteadinwhatfollows we want to give a simple proof, valid strictly speaking only for β .Itisa reﬁnement of the discussion around eq. (2.80). →∞ Notice ﬁrst of all that connected diagrams like in ﬁgure 2.6 give a contribu- tion that grows linearly in β.

Figure 2.6: Two examples of connected diagrams that contribute to the propa- gator of the anharmonic oscillator. For example, the diagramontheleftcorre- sponds to dτ dτ G4(τ ,τ ) β. 1 2 1 2 ∼ *

To prove that, notice that in the large β limit, G(τ1,τ2)tendstozeroex- ponentially when τ1 τ2.Considerthenthegeneralstructureofadiagram involving n interaction≫ points

λn dτ ...dτ F (τ ,...,τ )(2.87) ∝ 4 1 n 1 n ) where F consists of a product of G(τi,τj). It is graphically obvious that, if a diagram is connected, the associated function F is signiﬁcantly diﬀerent than zero only in the region τ τ τ where all time coordinates τ are 1 ≃ 2 ≃···≃ n j comparable. By integrating ﬁrst on τ2,...τn,theintegralconvergesfastandwe just get a number c so that

dτ dτ ...dτ F (τ ,...,τ )= dτ c = cβ (2.88) 1 2 n 1 n 1 × ) ) 48 CHAPTER 2. FUNCTIONAL AND EUCLIDEAN METHODS therefore connected diagrams give a contribution linear in β.Thisargument easily generalizes to diagrams that consist of 2 or more disconnected pieces. In that case the integrand is non vanishing when the time coordinates are compa- rable within each connected subdiagram. Therefore a diagramconsistingofk disconnected diagrams will produce a contribution growing like βk. Thus for β we have →∞ 2 Z[β]=Z0[β] 1+A1β + A2β + ... (2.89) & ' where Ak is a sum over all diagrams involving k connected subdiagrams. In particular A1 is the sum over the connected diagrams. βE n Now, since Z[β] e− 0 ,thecoeﬃcientsofβ must be such as to exponen- ∼ tiate A1β,i.e. 1 A = (A )n (2.90) n n! 1 and we can write:

A1β lim Z[β]= lim Z0[β]e (2.91) β β →∞ →∞ !ω E = !A ⇒ 0 2 − 1 Once again we found that the ground state energy is given by thesumofall connected diagrams. Chapter 3

The semiclassical approximation

Let us ask an intuitive question. When do we expect the wave of a particle propagation to be well appoximated by the properties of the trajectory of a classical particle? Well, as for the propagation of , when the λ is much smaller than the characteristic physical size of the system, diﬀraction phenomena are unimportant, and we can use geometric . In the quantum case, we expect similar results: if the wavelength λ is short enough with respect to the typical length over which the potential varies L,we can expect to be able to form a of size between λ and L,whose motion should be well appoximated by that of a classical particle.

We have two approximation methods to analytically treat systems that can- not be solved exactly:

Perturbation theory: Let us assume we are dealing with a system whose hamiltonian H can be written as H = H0 + λHp where H0 is exactly solv- able and λHp can be treated as a small perturbation (that is λ n Hp m (0) (0) (0) | ⟨ | | ⟩|≪ En Em ,where n and En are respectively the eigenstates and eigen- | − | | ⟩ values of H0). In this case, we work in perturbation theory in λ and we compute the eigenvalues of the hamiltonian En as a series in λ:

(0) (1) 2 (2) En = En + λEn + λ En + ...

and similarly we expand the eigenvectors H.

Semiclassical methods: We use these methods when a system is close to the ! 0, but more generally also when dealing with tunneling phenomena. These→ methods oﬀer no diagrammatic representation: when small couplings are involved, semiclassical contributions are proportional 1/λ2 to e− and thus they oﬀer non-perturbative corrections to the system.

49 50 CHAPTER 3. THE SEMICLASSICAL APPROXIMATION 3.1 The semiclassical propagator

The goal of this section is to compute the leading contribution to the path integral in the formal limit ! 0. In practice this will correspond to deriving a general expression for the Gaussian,→ or semiclassical, approximation discussed in section 1.3.2. Let us recall it. Starting from the propagator

S[x] i ! K(xf ,tf ; xi,ti)= D[x]e (3.1) ) we expand it around the classical solution xc: x = xc + y and get

2 3 S[xc] i 1 δ S 2 1 δ S 3 i ! ! 2 2 y + 3! 3 y +... K(xf ,tf ; xi,ti)=e D[y]e “ δx δx ” (3.2) ) The rescaling y = √!y˜ makes it evident that the expansion in powers of y truly corresponds to an expansion in powers of !

2 S[xc] i δ S 2 ! 3 i ! y˜ +√ (y ) K(x ,t ; x ,t )= e D[˜y]e 2 δx2 O . (3.3) f f i i N· ) Finally, dropping orders higher than the second in the Taylorexpansionofthe action, we obtain the Gaussian, or semiclassical, approximation to the propa- gator: 2 i S[xc] i δ S y˜2 K (x ,t ; x ,t )= e ! D[˜y]e 2 δx2 (3.4) sc f f i i N· ) Let us now consider the simple case where:

mx˙ 2 S[x]= V (x) dt (3.5) 2 − ) 6 7

We can expand at quadratic order around xc:

mx˙ 2 my˜˙2 2 → 2 1 2 V (x) V ′′(x )˜y → 2 c

2 where we used the notation V ′′(x )=∂ V (x) . c x |x=xc Thus, we must compute:

i ˙2 ′′ 2 dt(my˜ V (xc)˜y ) I[x ]= D[˜y]e 2 − . (3.6) c N R ) This is a Gaussian path integral like the one we found for the harmonic os- 2 cillator in subsection 1.3.4. More precisely if we write V ′′(xc(t)) Ω (t), the computation generalizes that of the ﬂuctuation determinantfortheharmonic≡ oscillator, for which Ω2(t)=ω2 is time independent. We can thus proceed along the same lines of subsection 1.3.4 and ﬁrst perform the changeofvariables

y˜(t)= a˜nyn(t)(3.7) n % 3.1. THE SEMICLASSICAL PROPAGATOR 51

where yn are the orthonormalized eigenmodes of the diﬀerential operator 2 2 d d 2 m V ′′(x (t)) m Ω (t) (3.8) − dt2 − c ≡ − dt2 − 1 2 1 2 which controls the quadratic action. We have again the changeofmeasure1 da˜ D[˜y(t)] = ˜ n (3.9) N· N· √ n 2πi + so that in the end the gaussian integral in eq. (3.6) gives

1 2 2 d − 1 I[x ]= ˜ det m V ′′(x ) = ˜ D− 2 (3.10) c N − dt2 − c N 6 6 77 One ﬁnal, important, remark before going to the computation,isthattheJaco- bian factor ˜ is independent of V ′′(x (t)). Consider indeed choosing a diﬀerent N c yn′ (t) on the space of square integrable functions on the time interval T = t{ t } f − i

y˜(t)= a˜nyn(t)= a˜m′ ym′ (t) . (3.11) n m % % We have a˜n = Unma˜m′ (3.12) where Umn is an (inﬁnite) satisfying U †U = 1 and detU =1as it follows from standard given that the two sets yn(t) , yn′ (t) are orthonormalized. We thus have { } { } da˜ da˜ n = m′ (3.13) √ √ n 2πi m 2πi + + which proves that the Jacobian ˜ is independent of the orthonormal basis and, afortiori,isindependentofwhateverdiN ﬀerentialoperatorhasthatbasisasits eigenbasis. Let us now proceed. The computation of determinants as in eq. (3.10) was ﬁrst carried out by Gelfand and Yaglom. In our derivation we shall follow aclever,butlessformal,derivationduetoColeman.Amoremathematical derivation can be found in [8]. The result we are are going to prove is that D = c ψ0(tf )(3.14) ˜ 2 · N where: ψ (t)satisﬁesthediﬀerentialequation • 0 md2 +Ω2(t) ψ (t)=0 (3.15) − dt2 0 6 7 with the boundary conditions

ψ0(ti)=0 ψ0′ (ti)=1. (3.16) 1Notice that, just to confuse the reader(!), we use a diﬀerent simbol for the Jacobian with respect to eq. (1.90). This is because we are now working with rescaled variablesy ˜ anda ˜n which absorbed a power of !.Thisistoorientthereaderontheoriginofthefactors,but beware that these Jacobians are anyway ill deﬁned when ϵ 0! → 52 CHAPTER 3. THE SEMICLASSICAL APPROXIMATION

c is a constant that does not depend on Ω2(t), and thus we can compute • it in the free particle case, Ω2(t)=0.

2 2 Consider now two diﬀerent time dependent : Ω1(t)andΩ2(t). Consider also the solution of the “eigenvalue” equation, generalizing eq. (3.15):

md2 +Ω2(t) ψ(1)(t)=λψ(1)(t)(3.17) − dt2 1 λ λ 6 7

Oˆ1 with boundary conditions! "# \$

(1) (1) ψλ (ti)=0 ∂tψλ (ti)=1. (3.18) For all λ,thesetwoconditionsﬁxthesolutioneverywhere,andthuswe • (1) can compute ψλ (tf ).

λ is an eigenvalue of Oˆ if, and only if ψ(1)(t )=0.Indeed,remember • 1 λ f that we are looking for the eigenvalues of Oˆ1 over the space of functions that vanish at ti and tf .

2 Consider then the same problem for Oˆ md +Ω2(t) and call ψ(2) 2 ≡− dt2 2 λ ˆ (2) / (2) 0 the solution of the diﬀerential equation O2ψλ = λψλ with the same boundary conditions in 3.18. We have then the following Theorem: ˆ det O1 λ ψ(1)(t ) − = λ f (3.19) / 0 (2) det Oˆ λ ψ (tf ) 2 − λ We shall now prove the above/ by following0 Coleman, and, as it iscustomary in physics, emphasizing ideas above mathematical rigour. Let us deﬁne ﬁrst: (1) ψλ (tf ) f(λ)= (2) (3.20) ψλ (tf ) f satisﬁes the following properties: f(λ)hasobviouslyallitszeroesincorrespondencewiththeeigenvalues of • Oˆ1 and all its poles in correspondence with eigenvalues of Oˆ2.Allthese zeroes and poles are simple as we show in the mathematical appendix of subsection 3.1.2.

f(λ) 1inthelimitλ , λ/R+.Thispropertyfollowsbysimple • solution→ of the diﬀerential→∞ equation∈ 3.17. Indeed, for λ ,wecan 2 →∞ neglect Ω1,2(t), and thus the solution satisfying the boundary conditions 3.18 is written as (see the mathematical appendix 3.1.3 for more detail)

2 (1,2) 1 m λ λ Ω1,2 i√ m (t ti) i√ m (t ti) ψλ (t)= e − e− − + (3.21) 2i λ − O 8 λ 9 . / 0 3.1. THE SEMICLASSICAL PROPAGATOR 53

One of the two exponentials grows for λ/R+.Thus,therearenozeroes 2 ∈ at t = tf ,andwecansafelyneglectΩ1,2 for all t.Ontheotherhand,for 2 λ/m =[nπ/(tf ti)] ,thezerothordersolution(ﬁrsttermineq.(3.21)) − 2 vanishes at t = tf and therefore the Ω1,2 perturbation cannot be treated as small, showing that limλ f(λ) =1forλ R+. →∞ ̸ ∈ Consider now:

ˆ det O1 λ det m∂2 λ Ω2 g(λ)= − = − t − − 1 (3.22) / 0 det ( m∂2 λ Ω2) det Oˆ2 λ & t 2' − − − − / 0 It has exactly the same simple poles and simple zeroes as f(λ). • For λ , λ/R+,wehaveg(λ) 1. Without aiming at mathematical • rigour→∞ (but a more∈ thorough discussion→ is presented in Homework 8) this ˆ 2 result is qualitatively understood as follows. Notice that T m∂t λ has eigenvalues λ =(nπ/(t t ))2 λ.Thisresultforλ ≡−implies that− n f − i − n the λmin of the eigenvalue of smallest norm λmin goes to inﬁnity | | R 2 when λ , λ/ +.InthissituationtheadditiontothisoperatorofΩ1 2 →∞ ∈ or Ω2 is a small perturbation with negligible eﬀects as λ .Therefore we expect g(λ) 1forλ , λ/R .Obviouslythisreasoningdoes→∞ → →∞ ∈ + not work for λ R+,asthefunctiong has zero and poles for arbitrarily large λ. ∈

Putting together we have that the ratio

f(λ) h(λ)= g(λ) is analytic everywhere (the singularity at the simple poles of f is neutralized by simple zeroes of 1/g and conversely the singularity at the simple zeroes of g is neutralized by simple zeroes of f)andtendsto1asλ .ByCauchy theorem the only such function is h(λ) 1, and that proves our→∞ theorem. More precisely, applying Cauchy theorem to ≡h(λ) 1andintegratingoveracircleC with radius R we have − →∞ 1 h(z) 1 h(λ) 1= − dz 0(3.23) − 2πi z λ → EC − where in the last step we used that lim z (h(z) 1) = 0. | |→∞ −

2 2 (Ω) Consider now Ω1(t) Ω (t)andcallψ0 the solution to the problem in eqs. (3.17,3.18) at λ =0.Asacorollaryofeq.(3.19)wehavethat≡

md2 2 det dt2 Ω (t) 1 − − = c (3.24) / ˜ 2 0 ψ(Ω)(t ) N 0 f where c is independent of Ω(t), and where we have judiciously inserted the normalization Jacobian (which, like c,isindependentofΩ2)inordertodealonly with the physically interesting combination ˜ /Det1/2.Providedweﬁndthe N 54 CHAPTER 3. THE SEMICLASSICAL APPROXIMATION constant c,thesolutiontothediﬀerentialequationproblemineqs.(3.17,3.18) at λ =0,givesusthedeterminantwewereafter!Findingtheconstant c is easy: we just need to use eq. (3.24) for an operator whose determinatwealreadyknow. The easiest option to do so is to consider the free particle case, where Ω(t)=0. In this case the determinat prefactor in K was computed in eq. (1.48)

1 md2 2πi(t t )! det = f − i (3.25) ˜ 2 − dt2 m N 6 7 while the solution to eqs. (3.15,3.16) is

ψ(0)(t)=t t (3.26) 0 − i and thus: 2πi! c = (3.27) m The result for the gaussian integral in the general case is then

m I[x ]= (3.28) c ! (Ω) ;2πi ψ0 (tf )

3.1.1 VanVleck-Pauli-Morette formula (Ω) Let us now derive a useful implicit expression for ψ0 (tf ). The classical equa- tion of motion tells us mx¨c + V ′(xc)=0 (3.29) where the dot denotes derivation with respect to time. By diﬀerentiating it once we get mx¨˙ c + V ′′(xc)˙xc =0 (3.30)

Thus, vc =˙xc satisﬁes the diﬀerential equation that we are trying to solve (Ω) in order to ﬁnd ψ0 (t). However, it doesn’t satisfy in general the boundary conditions ψ0(ti)=0,∂tψ0(ti)=1.ConsidernowtheWronskian:

W (t)=v (t)ψ˙ (t) v˙ (t)ψ (t)(3.31) c 0 − c 0 (Ω) where ψ0(t)=ψ0 (t)isthefunctionwearelookingfor. Since ψ and vc satisfy the same diﬀerential equation, we have W˙ (t)=0,and thus W (t)=W is a constant. We can then write:

d ψ ψ˙ ψ v˙ v 2 0 = v 2 0 0 c = W = v (t ) c dt v c v − v 2 c i 6 c 7 8 c c 9 d ψ 1 0 = v (t ) ⇒ dt v c i v 2 6 c 7 c t 1 ψ (t)=v (t)v (t ) dt′ ⇒ 0 c c i v 2(t ) )ti c ′ tf 1 ψ (t )=v (t )v (t ) dt′ (3.32) ⇒ 0 f c f c i v 2(t ) )ti c ′ 3.1. THE SEMICLASSICAL PROPAGATOR 55

Let us now write ψ0(tf )intermsofenergyandpotential.Wehave,from 2 mvc (t) dx the classical solution: = E V (xc(t)), and dt = .Writingvf = vc(tf ) 2 − vc and vi = vc(ti):

3 xf m 2 ψ (t )=v v dx (3.33) 0 f f i 2(E V (x)) )xi 6 − 7 xf dx v v (3.34) ≡ f i v(x)3 )xi The prefactor is then:

1 I[xc]= xf dx (3.35) 2πi!vf vi√m ; xi 2(E V (x)) − Or in terms of velocities: * m I[x ]= (3.36) c ! xf dx 2πi vf vi 3 ; xi v (x) Let us now express the result in terms of* the classical action.Wehave:

tf mx˙ 2 tf S = c V dt = mx˙ 2 E dt = c 2 − c − )ti 6 7 )ti &xf ' 2m(E V (x))dx E(tf ti)(3.37) xi − − − ) D Thus:

∂Sc = Pf = 2m(E V (xf )) (3.38) ∂xf − F ∂2S ∂P 1 2m ∂E 1 ∂E c = f = = (3.39) ∂x ∂x ∂x 2 E V (x ) ∂x v ∂x i f i ; − f i f i We can note that: tf xf dx xf m t t = dt = = dx (3.40) f − i x˙ 2(E V (x)) )ti )xi )xi . − xf ∂(tf ti) 1 1 ∂E m dx 0= − = 3 ∂xi −vi − 2 ∂xi x 2 (E V (x)) 2 ) i . − ∂E m = xf dx (3.41) ⇒ ∂xi −vi 3 xi v (x) Thus, the derivative of the* action can be written as:

2 ∂ Sc m = xf dx (3.42) ∂xi∂xf −vivf 3 xi v (x) And we get in the end the Van Vleck-Pauli-Morette* determinant:

1 1 ∂2S I[x ]=N = c (3.43) c d2 ! det m 2 V (xc) ;−2πi ∂xi∂xf ; − dt − ′′ & ' 56 CHAPTER 3. THE SEMICLASSICAL APPROXIMATION

And thus we can write the ﬁnal form of the semiclassical propagator:

2 Sc 1 ∂ S i ! c Ksc(xf ,tf ; xi,ti)=e ! (3.44) ;−2πi ∂xi∂xf

3.1.2 Mathematical Appendix 1

Let us now prove that ψλ(tf )onlyhassimplezeroes.Considertheequation:

md2 + V ψ (t)=λψ (t)(3.45) − dt2 λ λ 6 7 with the boundary conditions ψλ(ti)=0,∂tψλ(ti)=1. We will use the notation: d ψ˙ (t) ψ (t) λ ≡ dλ λ d ψ′ (t) ψ (t) λ ≡ dt λ When we diﬀerentiate (3.45) with respect to λ,weget:

md2 + V ψ˙ (t)=λψ˙ (t)+ψ (t)(3.46) − dt2 λ λ λ 6 7 Notice that since we impose the same boundary conditions for all λ,wehave: ˙ ˙ ψλ(ti)=ψλ′ (ti)=0 (3.47) Consider then the “Wronskian”:

W = ψ ψ˙ ′ ψ′ ψ˙ (3.48) λ λ − λ λ W ′ = ψ ψ˙ ′′ ψ′′ψ˙ λ λ − λ λ 1 1 = ψ (V λ)ψ˙ ψ (V λ)ψ ψ˙ λ m − λ − λ − m − λ λ ψ 2> ? = λ ! 0 (3.49) − m On the other hand, we have the boundary conditions:

W (t )=0,W(t )=ψ (t )ψ˙ ′ (t ) ψ′ (t )ψ˙ (t )(3.50) i f λ f λ f − λ f λ f

Thus, if both ψλ(tf )andψ˙λ(tf )vanish,theWronskianiszeroeverywhere, and so is its derivative. Thus, ψ (t) 0. λ ≡ 3.1.3 Mathematical Appendix 2 We want to write explicitly the solution to

2 2 ( ∂ λ Ω (t))ψ =0 ψ(0) = 0 ,ψ′(0) = 1 (3.51) − t − − treating Ω2(t)asaperturbation.Inordertocarryonperturbationtheoryin Ω2 we need the retarded Green’s function for the zeroth order equation G(t, t′)= 3.2. THE FIXED ENERGY PROPAGATOR 57

ˆ 1 ˆ 2 T − where T ∂t + λ.Byasimplecomputationanalogoustotheoneused to ﬁnd the≡ euclidean≡− Green’s function one ﬁnds (see section III of ref. [8] for instance)

1 2i√λ(t t′) 2i√λ(t t′) G(t, t′)=θ(t t′) e− − e − . (3.52) − 2i√λ − / 0 Also deﬁning the zeroth order solution

1 i√λ(t ti) i√λ(t ti) ψ0 = e − e− − (3.53) 2i√λ − / 0 and writing ψ = ψ0 + φ,eq.(3.51)becomes

2 Tˆφ=Ω (ψ0 + φ)(3.54)

1 which by use of G = Tˆ− and by employing standard functional operator no- tation (by we indicate the product of functional operators), can be written as ◦ (1 G Ω2)φ = G Ω2ψ (3.55) − ◦ ◦ 0 and solved iteratively in φ

2 1 2 ∞ 2 n φ =(1 G Ω )− G Ω ψ = (G Ω ) ψ (3.56) − ◦ ◦ 0 ◦ 0 n=1 % and therefore ∞ ψ = (G Ω2)nψ . (3.57) ◦ 0 n=0 % Notice that ψ deﬁned above automatically satisﬁes the boundary conditions ψ(0) = 0, ψ′(0) = 1. For the lowest order correction we have

t 2 1 2i√λ(t t′) 2i√λ(t t′) 2 (G Ω )ψ0 = dt′ e− − e − Ω (t′)ψ0(t′)(3.58) ◦ 0 2i√λ − ) / 0 which for λ/R is O(Ω2/√λ)ψ . ∈ + ∼ 0 3.2 The ﬁxed energy propagator 3.2.1 General properties of the ﬁxed energy propagator The ﬁxed-energy retarded propagator, sometimes simply Green’s function, is deﬁned as the of the retarded propagator θ(t)K(xf ,t; xi, 0)

∞ iEt K(E; x ,x ) K(x ,t; x , 0)e ! dt f i ≡ f i )0 ∞ iHtˆ + iEt = x e− ! ! x dt ⟨ f | | i⟩ )0 ∞ it ˆ ! (E H+iϵ) =lim xf e − xi dt ϵ 0 ⟨ | | ⟩ → )0 i! =lim xf xi (3.59) ϵ 0 ⟨ | E Hˆ + iϵ | ⟩ → − 58 CHAPTER 3. THE SEMICLASSICAL APPROXIMATION

ϵt/! + where, as usual when dealing with distributions, the factor e− with ϵ 0 is introduced to insure converge of the integral at t .Noticethatsince→ we integrate over t>0, the exponent eiEt/! ensures→∞ strong convergence of the integral for any E in the upper complex plane, that is for ImE>0. For ImE>0theretardedpropagatorK(E; xf ,xi)isthusananalyticfunctionof the complex variable E.ThisisanimportantpropertyofretardedGreen’sfunc- tions: analyticity in the upper energy plane is the Fourier space manifestation of , that is of retardation. For the advanced propagator we would have instead analyticity in the lower complex E plane. From the last equation above, the retarded propagator K(E; xf ,xi)isjust the position space representation of the inverse of the Schr¨o d i n g e r o p e r a t o r , Kˆ = i! .Itthussatisﬁestherelation: E Hˆ +iϵ − lim(E Hˆ + iϵ)Kˆ = i!1 (3.60) ϵ 0 → − In coordinate representation, equation (3.60) becomes: !2 ∂ 2 + V (x) E K(E; x, y)= i!δ(x y) . (3.61) −2m x − − − 6 7 This equation, and the boundary conditions we discuss below,fullyﬁxK(E; xf ,xi). Let us discuss this in detail. Given two linearly independentsolutionsψ1 and ψ2 of the homogeneous Schr¨odinger’s equation at energy E !2 ∂ 2 + V (x) E ψ =0 (3.62) −2m x − 1,2 6 7 the following function satisﬁes eq. (3.61) 2mi f(E; x, y)= (θ(x y)ψ (x)ψ (y)+θ(y x)ψ (x)ψ (y)) (3.63) !W − 1 2 − 2 1 where W = ψ1′ (x)ψ2(x) ψ2′ (x)ψ1(x)isthewronskian;werecallthat,by eq. (3.62), W is constant over− x.Howeverinorderforeq.(3.63)tocoincidewith the retarded propagator, we must choose ψ1 and ψ2 satisfying the appropriate boundary conditions. Which boundary conditions are required? The answer is obtained by simple physical considerations. In order to illustrate that, let us work on a case by case basis. Let us focus ﬁrst on motion in a potential V (x)satisfyinglimx V (x)=0, which is relevant for barrier penetration problems and in thecaseofpotential| |→∞ wells of ﬁnite spacial width. In this situation any solution ψ of the Schroedinger equation with positive energy behaves at x as a of outgoing and ingoing plane waves | |→∞

i√2mEx i√2mEx lim ψ = a+e− + b+e− (3.64) x + → ∞ i√2mEx i√2mEx lim ψ = a e− + b e− (3.65) x − − →−∞ The solutions ψ1,2 are selected by requiring K(E; xf ,xi)tohavethephysically correct behaviour at xf .Toﬁndoutwhichbehaviouristhephysically correct one, we must| recall|→∞ that the inverse Fourier transform, the propagator,

∞ iEt/! dE K(x ,t; x , 0) = K(E; x ,x )e− (3.66) f i f i 2π! )−∞ 3.2. THE FIXED ENERGY PROPAGATOR 59

represents the wave function Ψ(xf ,t)att>0forastatewhichatt =0was a δ-function at some ﬁnite position xi,thatislimt 0 Ψ(xf ,t)=δ(xf xi). → − Then at any ﬁnite t>0, for xf + xi,thewavefunctionΨ(xf ,t)should correspond to a superposition of→ waves∞≫ travelling out in the positive x direction, ie with positive momentum and positive energy. This implies

+i√2mExf lim K(E; xf ,xi) e (3.67) x + f → ∞ ∝

Similarly, for xf ,Ψ(xf ,t)shouldbeasuperpositionofwavestravelling out in the negative→−∞x direction, ie with negative momentum

i√2mExf lim K(E; xf ,xi) e− (3.68) x f →−∞ ∝

This gives us all the constraints to fully ﬁx ψ1,2 and write a general expression for K(E; xf ,xi)intheform3.63.Beforediscussingthat,noticethat,analiticity in the upper energy plane, tells us to how to continue √E from E>0toE<0 2.ThenforE<0(ormorepreciselyE =limϵ 0+( E + iϵ)theabove asymptotic behaviours become → −| |

√2m E xf lim K(E; xf ,xi) e− | | 0(3.69) x + f → ∞ ∝ → +√2m E xf lim K(E; xf ,xi) e | | 0(3.70) x f →−∞ ∝ → corresponding to K representing an operator on the space of normalizable wave functions. Indeed to ﬁnd the correct boundary conditions we could have worked the other way around, and consider ﬁrst the asymptoticbehaviourof +√2m E xf K(E,xf ,xi)forE<0. By eq. (3.61) there are two possibilities: e | |

√2m E xf and e− | | .TherequestthatK be square integrable implies the choice in eqs. (3.69, 3.70). Then by analytically continuing eqs. (3.69, 3.70) to E>0 through a path in the Im E>0half-planewegettheoutgoingwaveboundary conditions of eqs. (3.67, 3.68). 3 By eq. (3.63) the behaviour at x + and x is controlled f → ∞ f →−∞ respectively by ψ1 and ψ2.Theneqs.(3.67,3.68)aresatisﬁedif

+i√2m E x lim ψ1(x) e | | (3.71) x + → ∞ ∝ i√2m E x lim ψ2(x) e− | | (3.72) x →−∞ ∝ and provided we have such two solutions we can write the retarded propagator as 2mi K(E; x, y)= (θ(x y)ψ (x)ψ (y)+θ(y x)ψ (x)ψ (y)) (3.73) !W − 1 2 − 2 1

Summarizing: retardation corresponds to outgoing waves at x . →±∞ 2This is because the cut of √E should be chosen at Im E<0toensureanalyticityof K(E; xf ,xi)atImE>0. 3On the other hand, if we continue eq. (3.70) to E>0byapathintheImE<0half-plane we would get incoming instead of outgoing boundary conditions. This would correspond to the advanced propagator, which is analytic in the lower energy half-plane. 60 CHAPTER 3. THE SEMICLASSICAL APPROXIMATION

We can generalize these considerations to other potentials.Forinstance consider the case where limx V (x)=0andlimx + V (x)=+ ,sothat the region x is not accessible→−∞ at ﬁnite energy. It→ is∞ easy to ﬁnd∞ the right →∞ boundary conditions for ψ1,2 for this case too. In the case of ψ2 the condition is the same as before: outgoing . So we have again ψ2 ψ .Forψ1 the two possible behaviours at x + by solving the Schroedinger≡ − equation are exponentially growing or exponentially→ ∞ decreasing. We obviously must choose the second one to make physical sense. Then whe have that ψ1 is just the stationary solution ψstat of the Schroedinger equation with energy E:itbelongs to the physical Hilbert space. Notice on the other hand, that ψ ,anoutgoing wave at x ,willalwaysbe,bycurrentconservation,acombinationof− growing and→−∞ decreasing solution at x .Obviouslyψ does not belong to the physical Hilbert space. But this does→∞ not concern the asy− mptotic behaviour of K because of the θ step functions in eq. (3.63). In this case the propagator is then 2mi K(E; x, y)= (θ(x y)ψstat(x)ψ (y)+θ(y x)ψ (x)ψstat(y)) (3.74) !W − − − −

The last case to consider is the one where motion is fully limited: lim x + V (x)= | |→ ∞ + .Inthiscasebothψ1 and ψ2 must correspond, at any real value of E,to the∞ exponentially dcreasing solutions at respectively x + and x . → ∞ →−∞ To illustrate our results let us consider the simplest possibe case of a free particle: V (x) 0. We have: ≡ ikx ψ1(x)=e ikx (3.75) <ψ2(x)=e− where k = √2mE. We then get the ﬁxed energy retarded propagator for the free particle:

m ik(x y) ik(y x) K(E; x, y)= θ(x y)e − + θ(y x)e − !k − − m /ik x y 0 = e | − | (3.76) !k Which is precisely the expression derived in Homework 6, by inserting a com- plete set of ﬁxed momentum states in eq. (3.59) and performingthemomentum integral using Cauchy theorem. Notice that we recover a result similar to the correlator G (τ ,τ )inthelimitβ . β 1 2 →∞ Now, when we switch on the potential, we will have to ﬁnd the twofollowing solutions to the Schr¨odinger equation:

ikx ψ1 ψ+(x) A+e x + ≡ → ikx → ∞ (3.77) <ψ2 ψ (x) A e− x ≡ − → − →−∞ We will then deﬁne the retarded propagator by:

2mi K(E; x, y)= (θ(x y)ψ+(x)ψ (y)+θ(y x)ψ+(y)ψ (x)) (3.78) !W − − − − 3.2. THE FIXED ENERGY PROPAGATOR 61

Example: Barrier penetration

With the potential V (x)switchedon,assumingthatithasabarrierlocated somewhere along the x axis, we need ψ1,2 to correspond to the solutions to the Schr¨odinger equation with the following asymptotic behaviour:

ikx ikx e + B+e− x ψ1 ψ+(x) →−∞ ≡ → A eikx x + < + → ∞ ikx A e− x − ψ2 ψ (x) ikx ikx →−∞ ≡ − →

m ik(x y) A+ ik(x y) K(E; x ,y )= A e − = e − (3.80) →∞ →−∞ !k + v where v = k/m is the velocity. Thus, K directly gives the transmission coeﬃcient: the propagator (as it should!) directly gives the amplitude for propagation through the barrier. Using the above deﬁnitions, K(E; x, y)iswritten,inthelimitx , y ,in 1 →∞ →−∞ an intuitive and concise way: K(E,x,y) ψ (x)ψ∗ (y). → v out in Similarly, one can consider reﬂexion by a barrier by letting y , x ,andx>y.Faraway,whereV (x) 0, we have: →−∞ → −∞ → 1 K(E; x, y) (ψ (x) + ψ (x))ψ∗ (y) ∼ v in out in direct reﬂected

1 ik(x y) ikx iky = !e"# −\$ !+ B"# e\$− − (3.81) v + / 0 We see in the two contributions in the above equation the relics of the two possible classical paths: a direct path from y to x,andapathfromy to x where the particle is reﬂected by the barrier, as is illustrated in ﬁgure 3.1. This will become more transparent in the semiclassical path integral approach.

3.2.2 Semiclassical computation of K(E) Let us now turn to the computation of K(E; x, y)inthesemiclassicalapprox- imation. This is expected to be a good approximation when we can neglect 62 CHAPTER 3. THE SEMICLASSICAL APPROXIMATION

V

yx

Figure 3.1: The two possible classical paths from y to x.Theﬁrstonegoes directly from y to x,whereasthesecondoneisﬁrstreﬂectedbythepotential barrier before going to x.

derivatives higher than cubic in V (x). Using the semiclassical result in eq. (3.44) we must perform a Fourier transform

2 ∞ 1 ∂ Sc Sc iEt K (E; x ,x )= ei ! + ! dt (3.82) sc f i −2πi! ∂x ∂x )0 ; i f where the classical action Sc in a function of the initial and ﬁnal positions xi and xf ,andthetraveltimet. For ! 0, or equivalently for E large enough, we expect a saddle point computation→ of the integral over t to be a reliable approximation. So we pro- ceed with such a an evaluation. Notice that in doing so we are introducing an additional approximation in computing K(E; xf ,xi)withrespecttothegaus- sian computation of K(xf ,t; xi, 0). In Problem 1 of Homework 7 the condition for the validity of these approximations is worked out. Let us brieﬂy recall how the saddle point approximation works.4 We have to compute the following integral:

∞ g(t) K = f(t) ei ! dt (3.83) 0 ) D In the limit where ! 0, this integral will be dominated by the region → of the stationary point t where g′(t)=0.Inthisregion,wecansubstitute ∗ 1 2 f(t) f(t )andg(t) g(t )+ 2 g′′(t )(t t ) .Wewillgetthefollowing integral:≈ ∗ ≈ ∗ ∗ − ∗ ′′ i g(t∗ ) ∞ i 1 g (t∗) (t t )2 K = f(t )e ! e 2 ! − ∗ dt (3.84) ∗ )0 We will then be ableD to change the integration variable by δt t t ,and extend the lower bound of the integral from t to ,thusgettinganeasily≡ − ∗ integrable Gaussian Integral. − ∗ −∞ Let us therefore look for the stationary point for the integral 3.82: ∂S c + E =0 (3.85) ∂t ⇒ 5t=t∗ 5 E5c(t )+E =0 t = t(E)(3.86) − 5 ∗ ⇒ ∗ where Ec(t)istheenergyoftheclassicaltrajectoryforthemotionfrom xi to xf in a time t,andt is the stationary point. Note that the the stationary point ∗ 4Notice that we have already been using this concept throughout the course (in particular in sections 1.3.2 and 3.1). So the explanation given here should truly have been at the beginning! Better late than never. 3.2. THE FIXED ENERGY PROPAGATOR 63

approximation has selected a classical trajectory with energy Ec = E,showing that we are on the right track to derive a semiclassical approximation to the propagator. We have that the exponent at the stationary point t is given by: ∗

t∗ t∗ xf 2 Sc + Ect = ( + E) dt = mx˙ c dt = p(x)dx (3.87) ∗ L )0 )0 )xi where we used Sc = (mxdx˙ Edt)andwherep(x)= 2m(E V (x)) is the momentum of the classical trajectory− with energy E. − * D Let us now ﬁnd the second order expansion of Sc around t diﬀerentiating ∗ the identity ∂Sc = E (t) ∂t − c ∂2S ∂E c = c (3.88) ∂t2 − ∂t 5t=t∗ 5 5 To explicitly write the result we5 need to use some simple mechanical identi- ties. By usingx ˙ = v(x) 2(E V (x))/m,we have c ≡ c − D xf dx xf m t = = dx x˙ 2(E V (x)) )xi c )xi c . − 3 ∂E xf 1 m 2 1= c dx (3.89) ⇒ − ∂t m 2(E V (x)) )xi 6 − 7 ∂E 1 ∂2S = = v v c (3.90) 1 xf dx i f ⇒−∂t 3 − ∂xi∂xf m xi v * where in the second step we diﬀerentiated with respect to t,andthensett = t∗, Ec = E and where, again, vf = v(xf )andvi = v(xi). The stationary phase integral is therefore:

2 ! i ∂ Sc 2 2πi ! 1 dx dδt exp (δt) = 2 = 2πi (3.91) 2! ∂t2 ∂ Sc m v 3 ) 6 6 7 7 ; ∂t2 ; ) Putting this into eq. (3.82), the ﬁnal result is:

1 xf p(x) K(E; x ,x )= exp i dx ψ (x )ψ∗ (x ) f i v(x )v(x ) ! ∼ WKB f WKB i ; f i 6 )xi 7 (3.92) where we used the result eq. (3.73) to deduce the form of the ﬁxed energy wave functions corresponding to our approximation

x 1 p(x′) ψ (x) exp i dx′ . (3.93) WKB ∼ v(x) ! ; 6 )x0 7 This is the wave function which is derived in quantum mechanics by applying the Wentzel-Kramers-Brillouin (WKB) approximation. Problem 1 of Home- work 7 is devoted to the standard derivation of the WKB approximation in the Schroedinger approach. That standard method also allows a direct study of the 64 CHAPTER 3. THE SEMICLASSICAL APPROXIMATION domain of validity of the WKB approximation. The result is that one necessary requirement for WKB to be valid is

! dp dλ(x) 1(3.94) p2(x) dx ≡ dx ≪ corresponding to a physical situation where the De Broglie wavelength λ varies by a relatively small amount over a distance of order λ.Ofcourse,keepingthe classical parameters of the problem ﬁxed, this requirement would be eventually satisﬁed by formally taking ! 0. This limit, where the wavelength goes to zero, is the quantum mechanics→ analogue of geometric optics limit for the propagation of electromagnetic waves. In that limit the propagation of light can be described by the propagation of particles along rays: the wavelength being small diﬀraction phenomena are negligible. The above equation can also be written as a constraint on the F = V ′(x)actingontheparticle − ! dp mF = ! 1 . (3.95) p2(x) dx p3 ≪ 5 5 5 5 We thus see that at the points5 where5 the5 classical5 momentum is very small the WKB approximation breaks down. In particular this is so attheturning points of the classical motion where E = V and p =0.Now,inallinteresting applications the wave function at the turning points is not directly needed, which is reassuring. However in practically all interestingapplications,inorder to compute the propagator or the wave function at points wheretheWKB approximation applies one still has to deal with matching conditions close to the turning points where the approximation breaks down. In the path integral formulation the issue arises in that one has to consider in thesemiclassical computation of the propagator trajectories that go through turning points. The standard method to overcome the breakdown of the WKB approximation at the turning points is to deform the trajectory into the complex plane for the coordinate x (and for time t as well) in such a way as to avoid the point where v(x)=0.

3.2.3 Two applications: reﬂection and tunneling through abarrier In this section we shall consider two basic applications of the WKB propagator formalism. We shall not worry about the subtleties of reﬂection points, and compute whatever we can compute. Sections 3.2.4 and 3.2.5 will delve with more details into those issues. Let us ﬁrst examine the process of reﬂection by a potential barrier. This will be studied in detail in the exercises. Let us assume for that that we have apotentialV (x)thatvanishesasx ,andthathasabarriersomewhere along the x axis. When we put x ,x→−∞ , x >x,theclassicalaction i f →−∞ f i from xi to xf accepts two solutions: one going directly from xi to xf ,andone reﬂected ﬁrst by the barrier at a and then going to xf ,asshowninﬁgure3.2. Then, the Green’s function will be decomposed into the contributions of the two classical paths: K(E)=K1 + K2.Thecontributionofthereﬂectedpath, 3.2. THE FIXED ENERGY PROPAGATOR 65

V

xi xf a

Figure 3.2: The two possible classical paths from xi to xf .Theﬁrstonegoes directly from xi to xf ,whereasthesecondoneisgoingﬁrstfromxi to a,and then is reﬂected from a to xf .

K2,isinterestingbecauseitcontainsinformationaboutthephase shift induced by the reﬂexion. K1 is the same as without barrier. It turns out that:

xf 1 i π i K (E; x ,x )= e− 2 exp p(x)dx (3.96) 2 f i v(x )v(x ) ! (2) ; i f 6 )xi 7

xf a a where p(x)dx(2) = p(x) dx + p(x) dx is the integral over the re- xi xi | | xf | | ﬂected* path. p(x) is deﬁned* as 2m(*E V (x)) > 0. The origin| of the| extra i factor is that− det m∂ 2 V (x) < 0overthe D t ′′ reﬂected trajectory, or more− precisely this operator− has on−enegativeeigenvalue. Now, if we assume x>y, x, y ,andremembereq.(3.74)wecanwrite:& ' →−∞ K(E; x, y)=K1 + K2 = ψout(y)ψstat(x)

i x π i a a 1 ! p(z) dz i ! ( p(z) dz+ p(z) dz) = e y | | + e− 2 e y | | x | | ;v(x)v(y) R R R / 0

i a π i a π i a π 1 ! p(z) dz i p(z) dz+i p(z) dz i = e y | | − 4 e− ! x | | 4 + e ! x | | − 4 ;v(x)v(y) R R R / 0 a 1 i a p(z) dz i π 1 1 π = e ! y | | − 4 cos p(z) dz (3.97) v(y) R · v(x) ! | | − 4 ; ; 6 )x 7

Thus, we have found the physical eigenstate ψstat of the Schr¨odinger prob- lem:

i a π 1 ! p(z) dz i ψout(y)= e y | | − 4 ;v(y) R 1 1 a π ψ (x)= cos p(z) dz (3.98) stat v(x) ! | | − 4 ; 6 )x 7 Imagine now, that the motion is bounded also to the “far” left at b a. Applying the same argument, to a wave that is reﬂected at b we have≪ the following result for ψstat 1 1 x π ψstat(b)(x)= cos p(z) dz (3.99) v(x) ! | | − 4 6 )b 7 Now, the two stationary solutionsD must coincide up to the sign, as they solve the same problem (energy bound eigenstate in one-dimensional Quantum 66 CHAPTER 3. THE SEMICLASSICAL APPROXIMATION

Mechanics). We have:

1 a π ψ (x)=cos p(z) dz =cos(φ (x)) a ! | | − 4 a 6 )x 7 1 x π ψ (x)=cos p(z) dz =cos(φ (x)) (3.100) b ! | | − 4 b 6 )b 7 Taking the derivatives, we ﬁnd:

∂ φ = ∂ φ φ (x)+φ (x)=θ (3.101) x a − x b ⇒ a b where θ is independent of x. We then get:

ψa(x)=cos(φa(x)) ψ (x)=cos(θ φ (x)) = cos θ cos φ sin θ sin φ b − a a − a θ = mπ, m Z ⇒ ∈ ψ (x)= ψ (x)(3.102) ⇒ a ± b This condition gives a constraint for the integral over the closed contour:

2 a p(z)dz = p(z) dz = ! | | E )b 2 x 2 a 1 p(z) dz + p(z) dz =2π m + (3.103) ! | | ! | | 2 )b )x 6 7 This is the famous Bohr-Sommerfeld semi-classical quantization condition.

Barrier penetration: semiclassical transmission coeﬃcient

Imagine now that the potential V (x)hasabarriersothataclassicalparticle with an energy E cannot go from to + (see ﬁgure 3.3). −∞ ∞ V (x) E

xi a b xf

Figure 3.3: There is no classical trajectory going from xi to xf with an energy E.Thepotentialbarrieristoohightoletaparticlegopastit.Quantum physics in the semi classical limit predict a non-vanishing probability for such a particle to tunnel through the barrier from xi to xf .

However, we can deﬁne a semiclassical contribution to this process, which can be seen in the WKB approximation. The equivalence betweenWKBand Gaussian approximation requires therefore to ﬁnd the corresponding contribu- tion in the path integral approach. Thus, we have to look for the stationary 3.2. THE FIXED ENERGY PROPAGATOR 67 point of the combined D[x]dt integral: * ∞ i ! (Sc+Et) K(E; xf ,xi)= D[x]e dt (3.104) )0 ) The result is that we can ﬁnd a stationary point provided t is complex! Indeed, from the classical equation of motion, we ﬁnd that:

dx 2(E V (x)) = − (3.105) dt . m Considering extending this equation at a

dx 2 E V (x) =+i | − | (3.106) dt . m where we chose the plus sign in order to ensure that the Green’sfunctionremain analytic. Integrating over the full trajectory, we get:

xf m t = dt = dx 2(E V (x)) ) )xi . − v(x) a dx ! b dx"# x\$f dx = i + (3.107) v(x) − v(x) v(x) )xi )a | | )b real imaginary real

The evolution of time! in"# the\$ complex! "# plane\$ along! "# the\$ trajectory is depicted in ﬁgure 3.4. Im(t)

Re(t)

b dx a v(x) − | | * Figure 3.4: The evolution of the time along the trajectory in the complex plane. It acquires a negative imarinary part as the particle travelsthroughthepotential barrier.

Now, regardless of the course of the complex value of t,westillhave:

i i xf (S (t)+Et) = 2m(E V (x)dx (3.108) ! c ! − 5stat )xi 5 D 5 The Jacobian factor from the5 gaussian integration around theextremumof 68 CHAPTER 3. THE SEMICLASSICAL APPROXIMATION t still compensates the determinant:

1 i xf K(E; x ,x )= exp 2m(E V (x)dx f i v(x )v(x ) ! − ; f i 6 xi 7 ) D A+

x i a i f 1 b 1 ! p(x) dx 1 p(x) dx p(x) dx = e xi | | e ! b | | e− ! a | | v(x ) R v(x ) R R ; i ; f # \$! " ∗ ψin ψout (3.109) ! "# \$ ! "# \$ where the transmission coeﬃcient is given by A+. We can then derive the probability P of tunneling:

2 b 2 b 2 ! p(x) dx ! √2m(V (x) E)dx P = A = e− a | | = e− a − (3.110) | +| R R

3.2.4 On the phase of the prefactor of K(xf ,tf ; xi,ti) We have so far been rather cavalier concerning the phase of ourGaussianin- tegrals. However, as our previous discussion makes clear, when multiple tra- jectories contribute to the amplitude it is essential to be incontroloftheir relative phases. In this subsection, we will give the basic idea for the case of K(xf ,tf ; xi,ti). The next subsection is devoted to the phase of K(E; xf ,xi), which is very relevant to the discussion several applications like the computation of energy levels in the WKB approximation (see Homework 7).

1) Getting the basic idea In order to get an idea, let us go back to the determinant prefactor for the harmonic oscillator which was computed in eq. (1.98). In thatderivationwedid not pay any real attention to the phase factors that arose whensomeeigenvalues became negative. Eq. (1.89) tells us that the number n of negative eigenvalues − equals the integer part of ωT/π (T tf ti). Thus for 0

in π mω J (T )=e− − 2 (3.111) ω 2πi! sin(ωT) . | | rather than by eq. (1.98). Thus when T increases by a half-period π/ω the propagator aquires an extra = i.Withanextrafullperiod2π/ω the propagator just ﬂips sign: ( i)2 = 1. This− result can also be stated as − − 1. For trajectories in which the velocity does not ﬂip sign there is no extra phase factor 2. For trajectories where the velocity ﬂips sign n times the phase is either inπ/2 i(n 1)π/2 e or e − (depending on how many multiples of the half period ﬁt into T ) 3.2. THE FIXED ENERGY PROPAGATOR 69

3. For a trajectory corresponding to exactly k periods of motion the phase factor is ( 1)k − This result is valid for any periodic motion not just for the harmonic oscillator. There is a well developed mathematical theory, called Morse theory, underlying the dependence of the number of negative eigenvalues n on the properties of the trajectory. n is referred to as the Morse index.Wearenotgoingto− linger on this mathematical− connection though. A simple paper with some more details is for instance ref. [9].

2) An explicit non trivial example 5

mω2 2 V (x)=θ(x) 2 x

xi xf x =0

Figure 3.5: .

Let us see now a slightly less straightforward case by considering the motion 1 2 2 in the potential V (x)=θ(x) 2 mω x shown in ﬁg.3.5. The novelty with respect to the harmonic oscillator is that the motion at x<0isnotbounded,so that motion is not periodic. Let us compute the ﬂuctuation determinant using eqs. (3.15, 3.16). We also want to do so by focusing on a 1-dimensional family of trajectories characterized by initial position x(0) = xi < 0andvelocity x˙(0) v>0att =0.Theclassicalmotionischaracterizedby3time,t, intervals≡

xi 1. For 0 t v t1 we have x(t)=xi + vt 0: the particle moves in the V (≤x)=0regionwithpositivevelocity.≤− ≡ ≤

π v 2. For t1 0. −

π π 3. For t t1 + ω the trajectory is again “free”, x(t)= v(t t1 + ω ) < 0 but the≥ velocity is now negative. − −

We want to compute the determinant function ψ0(tf )astf is varied in the above 3intervals.Theoperatorcontrollingthequadraticﬂuctuation is

2 2 d d 2 π m V ′′(x(t)) = m mω θ(t t )θ(t + t)(3.112) − dt2 − − dt2 − − 1 1 ω − where the product of step functions accounts for the fact that V =0onlyfor π ̸ t1

5This problem was proposed in one of the exercise sessions. 70 CHAPTER 3. THE SEMICLASSICAL APPROXIMATION operator 3.112 is solved by patching the solution in the 3 timeregionslisted above. In other words, ﬁrst the solution in region 1 is found. Second, the solution in region 2 is found with initial conditions at t = t1 dictated by continuity of ψ0 and ψ˙0.Thirdthisprocedureisrepeatedtogetthesolutioninregion 3. One ﬁnds respectively in region 1, 2 and 3

1. For 0 t xi t the solution is ψ (t)=t ≤ ≤−v ≡ 1 0 2. For t

For tf t the • ∗ ∗ satisﬁes λ1 < 0 <λ2 <λ3 <...:thereisoneandonlyone negative eigenvalue.

For tf >t the gaussian integral prefactor is then (compare to eq. (3.28)) ∗

i π m I[x ]=e− 2 (3.113) c 2πi! ψ (t ) . | 0 f | Another comment concerns the point at which the determinant ﬂips sign. π Notice that this happens at t strictly larger than the time tr = t1 + at ∗ 2ω which reﬂection by the potential barrier happens. That is to say: ψ0 > 0 before reﬂection. This is a general result that can be easily seen by using the integral formula in eq. (3.32). From that equation is it obvious that ψ0 > 0 for trajectories in which the velocity does not ﬂip sign. Let us show that a stronger result holds, namely limt t ψ0(tf ) > 0whentf approaches tr from f → r below. For t tr the velocity goes to zero linearly: v(t) a( t + tr), with a a positive coeﬃ→cient (so that the acceleration isv ˙ = a<≃0).− Then the integral − in eq. (3.32) diverges but its prefator vf vi goes to zero so that the result is ﬁnite

tf 1 limt t v(tf )v(ti) dt =(3.114) f → r v 2(t) G )ti c H 1 1 v(ti) limt t v(ti)a( tf + tr) +ﬁnite = > 0(3.115) f → r − a2 t + t a 1 − f r 2 and therefore the determinant can only ﬂip sign after reﬂection. re

3.2. THE FIXED ENERGY PROPAGATOR 71

t ϵt t+ ϵ R − R R

Figure 3.6: .

3) The integral expression for ψ0 and reﬂecting trajectories One ﬁnal remark concerns the way to make sense of eq. (3.32) andeq.(3.34) in the case of trajectories where v does change sign. These equations are ill deﬁned in such cases: the factor vf vi is non-zero, but the integral diverges at the points where v 0. This is because in deriving them we had to divide by v → somewhere. Notice however that no singularity should exist in ψ0 is association to the vanishing of the velocity. This is because ψ0 is just given by the solution of an ordinary diﬀerential equation with ﬁnite coeﬃcients. This suggest that be should be able to ‘save’ eq. (3.32) and eq. (3.34). The way todosoisvery simple: we just need to slightly deform the trajectory into the complex (x, t) plane in such a way as to avoid the singularities. When workingwiththetime integral eq. (3.32), by eq. (3.115) we see that the integrand has a double pole: we just need to go around it as shown in ﬁg. 3.6. Notice also thatitdoesnot make any diﬀerence to go around the pole from below, as in the ﬁgure, or from above. This is because the singularity is a double pole and thus the integral over afullorientedcirclearounditgiveszerobyCauchy’stheorem. The integral on the oriented half circle contour Cϵ is easily computed for ϵ 0byparametrizing t t = ϵeiθ,with0 θ π on C ,sothatdt = iϵeiθd→θ.Wehave − r − ≤ ≤ ϵ − dt dt lim = =(3.116) ϵ 0 v2(t) a2(t t )2 → Cϵ Cϵ r ) π ) − 1 iθ 1 ie− dθ = [ 1 1] . (3.117) ϵa2 − ϵa2 − − )0

This result precisely cancels the diverge of the integrals at ttr +ϵ so that we have in the end −

t ϵ t 1 r − dt 1 f dt ψ0 = vivf lim + + + (3.118) ϵ 0 −ϵa2 v2(t) −ϵa2 v2(t) → G1 )ti 2 1 )tr +ϵ 2H We can also change variable t x.Thereﬂectionpointx is deﬁned by → r v(xr)= 2(E V (xr))/m =0,whereE is the energy of the trajectory in consideration. The− equation of motion gives V (x )/m = a.Thenwehavethat D ′ r t ϵ both correspond to x a ϵ2 so that the above equation becomes r ± r − 2 x a ϵ2 x a ϵ2 1 r− 2 dx 1 r− 2 dx ψ = v v + + + 0 i f −ϵa2 v3(x) −ϵa2 v3(x) <, )xi | | - , )xf | | -= (3.119) Using this formula, we can study the conditions on V (x)underwhichψ0 ﬂips sign after reﬂection. Since for a reﬂecting trajectory vivf < 0, we have that ψ0 is negative if and only if the expression in curly is positive. Now, if the potential at x ,i.e.intheclassicallyaccessibleregion,isﬂatenough,then →−∞ 72 CHAPTER 3. THE SEMICLASSICAL APPROXIMATION

it is obvious that, for xi or xf large enough and negative, the expression in curly brakets is dominated by the positive integral in the region away from the turning point. Thus we have ψ0 < 0. However if limx V (x)= fast enough, then dx/v3 is dominated by the region close to→−∞ the turning point.−∞ In this case the expression in curly braket does not necessarily become positive for large * enough xi and xf and ψ0 may never ﬂip sign. A simple and obvious example where this is so is given by the linear potential V = cx.Thisisobviousbecause in this case V ′′ =0andwehavethefreeparticleresultψ (t )=t t > 0. 0 f f − i It is a good exercise to recover the result we found for ψ0 in the potential in ﬁg. 3.5 by using instead the integral representation of ψ0.Weshallnotdothat but just give some suggestions on how to try and do it. Using thespaceintegral representation eq. (3.34), one notices that v(x)hasacutinthecomplexx plane along the x region accessible to classical motion. Physically this corresponds to the velocity being double valued: v(x)= 2(E V (x))/m.Thenψ0 is computed on a trajectory that goes around this± cut...The− willing student can D now go ahead and make a simple remark (on Cauchy integrals) that allows for adirectcomputation.

3.2.5 On the phase of the prefactor of K(E; xf ,xi)

Let us now focus on the phase of K(E; xf ,xi), which is really what in most applications. Again the delicate issue is how to deal with the turning points. Two basic remarks are in order here. The ﬁrst is that for xi or xf close to the turning points, the WKB approx- 6 imation to K(E; xf ,xi)breaksdown because p(x) 0(seeeq.(3.95)).The saddle point approximation can however still be reliable→ if the trajectory can be deformed into the complex C2 plane for (t, x)insuchawaythatp(x)never really crosses zero. In order for this procedure to makes sense two conditions must be satisﬁed (see the discussion in the chapter on the WKB approximation in Landau-Lifshitz [10])

The potential V (x)isanalyticinaregionaroundtheturningpoint.This • way, v(x)= 2(E V (x))/m is also analytic in this region apart from having a cut (due to− the square root) in the physically accessible region D on the real x axis.

The region of analyticity for V (x)overlapswiththeregionwheretheWKB • 3 approximation applies, that is with the region where ! mV ′(x)/p(x) 1. | |≪

When the above two conditions are satisﬁed, a suitable deformation of the tra- jectory can be chosen such that x goes around the turning points in such a way that: 1) Physical quantities, like the action integral, are unaﬀected by the deformation of the trajectory thanks to analyticity. 2) p(x)isnevertoosmall

6 We that the situation is diﬀerent with respect to the apparent singularity of ψ0(tf ) at the turning points which we encountered in the previous section. The gaussian approxima- tion for K(xf ,tf ; xi,ti)doesnotreallybreakdowninthepresenceofturningpoints,witness the fact that this approximation is exact for the harmonic oscillator. Instead the WKB com- putation of K(E; xf ,xi), as we already pointed out, requires an extra approximation, which genuinely breaks down at the turning points. 3.2. THE FIXED ENERGY PROPAGATOR 73 and the WKB is manifestly valid.7 We will explain with an example below how the deformation works. Notice, for the moment, that by deforming the time line into the complex plane (like we did for instance with barrier penetration), the trajectory x(t)whichisdeﬁned to solve d2x m = V ′(x) , (3.120) dt2 − is also, in general, deformed into the complex x plane. The second remark concerns the form of the prefactor itself. By continuing the trajectory into the complex plane we can still make sense of the integral dx/v3 even when the velocity ﬂips sign somewhere. Then, as we already know, all is left in the prefactor is the velocity at the initial and ﬁnal points * 2 1 ∂ Sc 2πi! 1 P (xf ,xi)= − 2 = . (3.121) 2πi! ∂x ∂x · ∂ Sc v(x )v(x ) ; f i ; ∂t2 ; f i

Since v(xf )changessignattheturningpoint,theaboveprefactorgainsa1/√ 1 factor precisely at reﬂection. The only question is that 1/√ 1beingdouble− valued, we cannot yet tell, simply by staring at eq. (3.121), if the− phase is +i or i.Also,justbystaringattheaboveequation,itisnotobvioushowtoproceed −for trajectories with multiple reﬂections. It turns out thattherightansweris:

iπ/2 At each reﬂection the prefactor gains a phase factor ( i)=e− . • − This result can be directly and easily checked for the examplewediscussedin section 3.2.4, using the computations we have already made. There are two contributions to the phase. One is the phase of the VanVleck-Pauli-Morette prefactor in eq. (3.113). We computed this phase factor in section 3.2.4 and iπ/2 found that it is e− for xf and xi negative. The second comes from the phase of the gaussian integral over t and is determined by the sign of ∂2S/∂t2.Itis easy to see that ∂2S/∂t2 > 0. Indeed by eqs. (3.88) and (3.90) we have that ∂2S/∂t2 is just the inverse of the expression in curly brackets in eq. (3.119). In the case at hand that expression is clearly positive since, vivf < 0andsince 2 2 ψ0(tf ) < 0byourdirectsolutionofthediﬀerentialequation.Since∂ S/∂t is positive, like for a non-reﬂecting trajectory, the phase of the t integral is trivial and the only non trivial contribution is that from the determinant, which equals iπ/2 e− .Thisisinagreementwiththegeneralresultwestatedabove There is one quick way to deduce in general that for vivf < 0thephase iπ/2 in eq. (3.121) must be e− .Withoutlossofgeneralityandtosimplifythe discussion let us consider a particle of m =1,letusshiftthedeﬁnitionof the energy such that E =0forourtrajectory,letusshiftx so that the turning point is xr =0andletuschooseV (x) x in the vicinity of x =0.The equation of motion close to x =0isjust ≃

x¨ = 1(3.122) − 1 2 where again, without loss of gererality, we can focus on the solution x = 2 t with v x˙ = t.Forrealt the velocity ﬂips sign at t =0.Ifwedeformt−into ≡ − 7Notice that the time contour shown in ﬁg.3.6 are an example of how to go around turning points. To be manifestly in the WKB approximation ϵ should be large enough. However because of analyticity the result of all integrals is independent of ϵ anyway! 74 CHAPTER 3. THE SEMICLASSICAL APPROXIMATION

t =0 iϵ − (a)

(b)

iϵ t =0

Figure 3.7: .

the complex plane we have two options to go around the point t =0wherethe velocity vanishes. Either we go below like in ﬁg. 3.7a or abovelikeinﬁg.3.7b. These two options lead to diﬀerent (and conjugate) phase shifts for 1/ v(tf ). Which of the two deformations should we choose? It turns out deformation (a) D is acceptable while (b) is not. To understand why, we should gobackandthink what it means to deform t into complex values in our original deﬁnition of the path integral. It means that, rather than doing our simple (and real) slicing dt =(t t )/N ,wearemakingsomeelementdt of the sum dt = t t f − i f − i complex. As long as we write U(tf ti)asaproductofmatricesthisisnota problem. A constraint arises however− when we attempt to writethisproductof* matrices in terms of a path integral as done in section 1.1.4. This is because if ( iHdt/!) Re ( idt) > 0intheexponente − ,thenthep integral in eq. (1.14) is not convergent:− it is growing at p (the ϵ interval of that equation corresponds to our dt). This is just an inﬁnitesimal→∞ way of saying that the euclidean path integral exists only for β>0. Since our semiclassical approximation is based on a path integral representation of the evolution amplitude, we should only consider trajectories (real or complex) for which the path integral representation makes sense. That is to say only trajectories such that Im(dt) 0. This the case for trajectory (a) and not the case for trajectory (b). Wearenowjust≤ left with computing the phase shift in 1/ v(t)=1/√ t along trajectory (a). When t goes from to + iϵ the argument in −√ t goes from + to D + iϵ by passing−∞ in the upper∞− half plane, therefore the− phase of √ ∞+ iϵ −∞is eiπ/2.Theresultingphaseoftheprefactoristhen −∞

1 iπ/2 1 = e− (3.123) v( + iϵ) v( ) −∞ | −∞ | More formally we canD compute the phase shiftD ∆ϕ by integrating over the t contour

+ iϵ dv ∞− dt 1 π ∆ϕ Im ∆ ln(1/√v) = Im = Im = π = ≡ − 2v − 2t −2 − 2 ) )−∞ 3 4 (3.124) Chapter 4

Instantons

4.1 Introduction

In Quantum Mechanics, perturbation theory can miss important features of a physical system. To illustrate this, let us give a quick look at a simple example: a one- mω2x2 3 dimensional particle in a potential V (x)= 2 + λx . The point x =0isaclassicalminimumanditseemsreasonabletostudythe dynamics around this point upon quantization. For λ “small” enough, we can apply perturbation theory and in principle compute the ground state energy to any desired order in λ.Indoingso,however,wewouldnotrealize,orgetany direct indication of one basic pathology of this ground state: its instability! Due to the possibility of tunneling through the barrier, the ground state will be able to decay down to the region where λx3 is negative and dominates the harmonic term (see ﬁgure 4.1). V (x)

tunnel E

Figure 4.1: The stability of the ground state localized around the local minimum of the potential is broken by the possibility for the particletotunnelthrough the barrier to the part of the potential unbounded from below.

The probability of tunneling will be proportional to: P exp 2 √Vmdx ∼ − ! ∼ 3 5 exp m ω .Itthusrepresentsaneﬀectwhichvanishes,asλ / 0,* faster that0 − !λ2 → / 0 75 76 CHAPTER 4. INSTANTONS any power of λ,implyingthatitcannotbeseenatanyﬁniteorderinperturba- tion theory. This is what theorists normally call a non-perturbative eﬀect. As a second example, let us look at the case of a particle in a double well potential: V (x)=λ(x2 a2)2 symmetric under reﬂection x x (see ﬁgure 4.2). − →− V (x)

a a −

Figure 4.2: The double well potential has two relexion symmetric vacua, thus leading to inexistant degenerate eigenstates in perturbation theory.

This potential has two classical ground states: x = a,whichareequivalent because of the reﬂection of the system. In± perturbation theory, we can ignore this doubling and expand around one of these minima, say x = a. We introduce thus the new variable y = x a.Inthisvariable,V (y)isa harmonic oscillator with extra cubic and quartic− terms whichwecantreatas perturbations as long as λ is “small”. We have:

V (y)=λ (y +2a)2 y2 =4λa2y2 +4λay3 + λy4 (4.1) 1 mω2y2 + ω√2mλy3 + λy4 (4.2) ≡ 2 where in the last equation we traded a for the harmonic oscillator frequency ω. For small λ,wecannowcomputethecorrectionstothelowestenergylevels to arbitrary orders in λ without having any direct evidence that there is another minimum with its own localized levels! The situation is here symmetric: a guy sitting at the other classical ground state would compute in perturbation theory exactly the same levels1 ,butcor- responding to a diﬀerent set of states where the wave functionislocalizednear x = a instead of x = a.Thus,inperturbationtheory,theenergylevels,and in particular− the ground state, are all doubly degenerate to all orders. However, from the exact quantum mechanical solution of the problem, we know that for a unidimensional system, there is no degeneracyoftheboundstate energy eigenvalues. Moreover, we know that the ground state wave function has no nodes, and, in the case at hand, is therefore even under x x: →− Ψ (x)=Ψ ( x) 0 0 − 1Technically, this other guy expands the coordinate as x = a y′.Becauseofreﬂection symmetry the potential in y′ satisﬁes V ( a y′)=V (a + y′):− exactly− the form in eq. (4.1). Hence the levels, in perturbation theory,− are− the same. 4.2. INSTANTONS IN THE DOUBLE WELL POTENTIAL 77

On the other hand, the ﬁrst excited state has one , and musttherefore be odd under reﬂection: Ψ (x)= Ψ ( x) 1 − 1 − However, because of the doubling of all the levels that we justdeducedwith our perturbative reasoning, the energy eigenvalues of thesetwolowestlying states mus be degenate to all orders in perturbation theory. That is E1 E0 =0 to all orders in perturbation theory. The goal of this chapteristoshowthat− the leading contribution to E1 E0 as λ 0canbecomputedbysemiclassical methods. More precisely, via semiclassical− → trajectories inimaginarytimecalled instantons.

4.2 Instantons in the double well potential

What we want to study are the properties of the ground state(s)ofthedouble λ 2 2 2 well potential: V (x)= 4! x a .Todothis,letusgobacktotheEuclidean path integral. − & βH ' βH We will compute a e− ! a and a e− ! a for β .Ifthebarrieris high and the the potential⟨ | smooth,| ⟩ we⟨− can| perform| a⟩ semiclass→∞ical approximation in Euclidean time. To compute those two transition amplitudes, we must sum over all possible stationary trajectories (i.e. with stationary action) in Euclideantime. Remember that:

β xf , βH 2 SE [x] ! ! xf e− xi = KE(xf ,xi; β)= D[x]e− (4.3) ⟨ | | ⟩ x , β ) i − 2 β 2 2 mx˙ λ 2 2 2 SE[x]= dτ + x a (4.4) β 2 4! − )− 2 6 7 & ' For each tra jectoryx ¯,wecancomputeitscontributiontoKE in Gaussian approximation:

1 2 2 d − SE [¯x] K = ˜ det m + V ′′(¯x) e− ! (1 + (!)) (4.5) E N − dτ 2 O 6 6 77

Notice that SE is positive deﬁnite. Now,

SE[x] ! KE = D[x]e− (4.6) ) is dominated by the local minima of SE[x]withgivenboundaryconditions:

β x = x f 2 6 7 β x = x (4.7) i − 2 6 7

δSE Imposing now that δx =0leadstotheEuclideanequationofmotion. Notice that the solutions to this equation describe motion intheeuclideantime 78 CHAPTER 4. INSTANTONS

VE(x)

a a −

Figure 4.3: The Euclidean equation of motion corresponds to the real time equation of motion with a reversed potential V (x)= V (x). E −

parameter τ of a classical system with reversed potential VE(x)= V (x)(see ﬁgure 4.3). − In the limit ! 0, we therefore expect K to be dominated by the local → E minima of SE (if there exist more than one). K K + K + K + ... (4.8) E ≈ 1 2 3 Remark: we can make an analogy with the ordinary integrals. Suppose that we have a function S(x), as depicted in ﬁgure 4.4. S(x)

x 123

Figure 4.4: An ordinary function of one variable S(x), accepting three local minima called 1, 2 and 3.

Then, in the limit ! 0, the integral → ∞ S(x) I = e− ! dx (4.9) )−∞ can be approximated by I I + I + I + ... ≈ 1 2 3

S(xi) 2π! ! Ii = e− (4.10) ;S′′(xi)

The anharmonic terms around the local minima give corrections to each Ii of relative size !. 4.2. INSTANTONS IN THE DOUBLE WELL POTENTIAL 79

βH βH βH To compute the transition amplitudes a e− ! a , a e− ! a , a e− ! a βH ⟨ | | ⟩ ⟨− | |− ⟩ ⟨− | | ⟩ and a e− ! a ,wemuststudytheclassicaltrajectoriesintheupsidedown ⟨ | |− ⟩ potential VE (x)= V (x), depicted in ﬁgure 4.3. This potential accepts− two types of solutions: Trivial solutions: x(τ) a which will give the leading contribution to the ﬁrst two amplitudes.≡± Less trivial solutions: the instanton, which starts at x = a at τ = and rolls towards x = a at τ = ,andtheanti-instanton,whichrollsfrom− −∞ a to a.Thesesolutionsareexactinthe∞ β limit. − →∞ The action on the instanton can be computed: as the initial velocityx ˙( ) −∞ is zero, then the Euclidean energy EE is zero, and we can write: mx˙ 2 V (x)=E =0 (4.11) 2 − E This allows us to write an expression for the Euclidean action:

mx˙ 2 2V (x) = V (x) x˙ = 2 ⇒ . m a a ∞ 2 2V (x) SE = mx˙ dτ = m dx = 2mV (x)dx (4.12) a . m a )−∞ )− )− D Thus, in the semiclassical limit, the transition amplitude will be proportional to: βH SE a p(x) ! dx a e− ! a e− ! e− −a 1(4.13) ⟨ | |− ⟩∼ ∼ R ≪ We recover in this result the expected semiclassical tunneling coeﬃcient. λ 2 2 2 For the speciﬁc example of a quartic potential V (x)= 4! x a ,the zero energy solution satisﬁes − & ' ω x˙ = x2 a2 (4.14) ±2a −

&2 ' 2 λa where we used the new variable ω = 3m . The solutions of these equations are the instanton and anti-instanton given by ω x(τ)= a tanh (τ τ ) (4.15) ± 2 − 0 The instanton (with the plus sign) is/ depicted in0 ﬁgure 4.5. Remarks

For the instanton solution, the value of the exponent SE is given by: • ! a 3 SE 2mV (x) 2 λm a ! = ! dx = ! a 3 3 )− D . 2 3m 3/2 ω3 m2ω3 2 = √λm =2 = (4.16) 33/2 λ ! !λ λ¯ 6 7 ¯ !λ where λ = m2ω3 is the dimensionless parameter that characterizes the size of the perturbative corrections to the harmonic oscillator value for 80 CHAPTER 4. INSTANTONS

x(τ) a

τ τ0

a − Figure 4.5: The instanton solution in the double well potential, running from a to a. −

the energy levels. We already encountered λ¯ when we studied perturba- tive corrections to the ground state energy of the harmonic oscillator with a λx4 perturbation. Since Feynman diagrams with k-loops contribute a k 1 correction of order λ¯ − ,thecouplingλ¯ is often called a loop-expansion parameter. The part of the Euclidean propagator due to the instantons is therefore S /! 2/λ¯ proportional to e− E = e− .Thus,eﬀectsassociatedwithinstantons are non-perturbative in λ¯.

1 2 2 ω τ τ0 At τ τ0 " ω ,wehavethat x (τ) a e− | − | 1. Instantons • are| thus− well| localized in Euclidean time:− they≈ exist for≪ a short instant 5 5 ∆τ 1 in Euclidean time (hence5 their name).5 ∼ ω Thus, when β ,uptoexponentiallysmallterms,wehavex a →∞ ≈± away from τ = τ0.Thisisacrucialpropertyconsideringthatweare 1 interested in the limit β :forβ ω the instanton solution looks like a step function (see ﬁgure→∞ 4.6). ≫

a

β τ0 β − 2 2

a −

Figure 4.6: In the large β limit, the instanton solution looks like a step function.

After having jumped from a to a,thereisstillplentyoftimetobounce back! −

More technically, given the instanton solution xI (τ)=af(τ τ0), where ω τ τ 1 − f(τ τ ) =1 e− | − 0| for τ τ ,then | − 0 | − | − 0|≫ ω x (τ)=af(τ τ )f(τ τ)(4.17) IA − 0 1 − 4.2. INSTANTONS IN THE DOUBLE WELL POTENTIAL 81

xIA(τ) a

τ τ0 τ1

a − Figure 4.7: The instanton-anti-instanton quasisolution.

for τ1 τ0 is an approximate solution representing an instanton followed by an≫ anti-instanton. For such a conﬁguration, one can easilycheckthat

δSE ω τ τ e− | 0− 1| . (4.18) δx ∼ 5x=xIA 5 5 As long as we only need to consider5 situations where ω τ0 τ1 is so large that the above exponential is smaller than any eﬀect| we− are| trying to compute, then we should by all means consider the conﬁguration xIA as astationarypointoftheaction.Andthusaddthecontribution of these quasi-solutions to our semiclassical approximation to the path integral. SE Indeed, we shall ﬁnd at the end that τ τ e ! .Thus,neglecting | 1 − 0|∼ that xIA is not an exact solution amounts to neglecting terms of order ω exp(S /!) e− E !Thesetermsdonotexactlyvanishbutareindeedvery small.O Thus we must add the contribution of x to our saddle point & ' IA estimate of the path integral. By extending this argument, we are easily convinced that we must consider • all the approximate solutions with a number N of instantons + anti- instantons. Notice that if we take nI instantons and nA anti-instantons, we must have that nI nA =0fora a,andnI nA =1for a a. (what about a a−and a a ?)→ − − → − →− →− An approximate solution with, e.g. three instantons at τ1, τ3 and τ5 and two anti-instantons at τ and τ will be a good one provided τ τ 2 4 1 ≪ 2 ≪ τ3 τ4 τ5.Wewillchecklaterwhetherthisconditionissatisﬁedfor the≪ leading≪ terms.

4.2.1 The multi-instanton amplitude We will now compute our amplitudes by summing over all possible sequences of widely separated instantons (and anti-instantons): this isthediluted instanton gas approximation. Each of these quasi-solutions contributes to the amplitude in two ways:

Exponent: the contribution to SE is localized at each instanton (or anti- instanton) transition time. Indeed, away from these transition times, the quasi-solution is close to a with a neglectible velocity. SE is therefore zero in these regions. ± 82 CHAPTER 4. INSTANTONS

x(τ) a

τ τ1 τ2 τ3 τ4 τ5

a −

Figure 4.8: A quasi-solution with three instantons localized at τ1, τ3 and τ5, and two anti-instantons localizd at τ2 and τ4.

Each instanton (or anti-instanton) will give a factor SI in the action. Due to the decomposition property of the action, the contribution of an approximate solution with N instantons and anti-instantons will be:

SN N SI S = NS e− ! = e− ! (4.19) N I ⇒

Determinant prefactor: notice again that away from the very short instants over which the transitions take place, the system sits at either one of its mω2 classical vacua at x = a,aroundwhichV ′′(x)= . ± 2 mω2 β β If we had V ′′(¯x) throughout the interval , ,thenthepref- ≡ 2 − 2 2 actor would be the usual one for the harmonic oscillator:> ?

1 1 2 2 2 1 d − mω mω 2 ωβ ˜ det m + V ′′(¯x) = e− 2 N − dτ 2 2π! sinh(ωβ) ≈ π! 6 6 77 6 7 / 0 Given the factorization property of the path integral and, inparticular,of the Gaussian integral we are interested in, we expect that each transition will correct the prefactor of the harmonic oscillator with a factor R. We thus expect

1 2 2 1 d − mω 2 ωβ n ˜ det m + V ′′(¯x) = e− 2 R N − dτ 2 π! (4.20) ⎧ 6 6 77 ⎪ / 0 ⎨ n = nI + nA

for a quasi-solution⎩⎪ with n instantons and anti-instantons. Let us see how this argument can be made sharper, by focussing on the simplest non-trivial case of an xIA solution. (The generalization to an arbitrary sequence of I and A is straightforward.) To do so, weexplicitly write the Gaussian integral in factorized form by considering intermediate times around instanton τ1± = τ1 ∆andaroundanti-instantonτ2± = τ2 ∆ as shown in ﬁgure. We deﬁne the± integration coordinates at intermediate± times as x(τi±)=xi±,fori =1, 2. The Euclidean propagator can thus be 4.2. INSTANTONS IN THE DOUBLE WELL POTENTIAL 83

+ x(τ) τ1 τ2−

τ β τ1 τ2 β − 2 2

+ τ1− τ2

Figure 4.9: Intermediate times τi± = τi ∆havebeenaddedtotheinstanton- ± 1 anti-instanton quasisolution, with τi τj ∆ ω ,sothatatτi±,thesolution is almost in the . | − |≫ ≫

factorized into the convolution of the propagator over ﬁve time intervals (see ﬁgure)

β β + + + β + K ( a, a; , )= dx−dx dx−dx K ( a, x ; ,τ ) E − − 2 − 2 1 1 2 2 E − 2 2 2 × ) (1) + + + + KE(x ,x−; τ ,τ−) K! E(x−,x"#; τ −,τ \$) 2 2 2 2 2 1 2 1 × (2)

+ + β K (x ,x−; τ ,τ−)!K (x−,"#a; τ −, \$ ) (4.21) E 1 1 1 1 E 1 − 1 − 2 (3) ! "# \$ 1 Of course we work under the assumption τ2 τ1 ω .Thefactorization | 1− |≫ turns out to be useful if we also choose ∆ ω .Indeeditisintuitivelyclear that, for ω∆ 1, the amplitudes (1), (2)≫ and (3) should be well approx- imated (exponentially≫ well indeed!) by harmonic oscillatoramplitudes. This is because, during the corresponding time intervals, the trajectory sits very close to one or the other minimum, where the potential is well approximated by the harmonic oscillator one. Moreover, we know that the harmonic oscillator amplitude, for large euclidean time separation, is dom- inated by the ground state (see section 2.3). We can thus approximate, e.g. amplitude (2), by:

H0 − + + + ! (τ2 τ1 ) + ω∆ KE(x2−,x1 ; τ2−,τ1 )= x2− e− − x1 1+O(e− ) H0 (τ − τ +) + + + B0 5e− ! 2 − 1 50 C&Ψ ∗ (x−)Ψ (x' ) ≈⟨+|5 |5 +⟩ 0 2 0 1 + + + ω − + ∗ 2 (τ2 τ1 ) =Ψ0 (x2−)Ψ0 (x1 )e− − (4.22)

where H0 represent the harmonic oscillator Hamiltonian around +a.Here we also decomposed position eigenvectors x into eigenstates of the har- monic oscillator Hamiltonian around +a: |x⟩ = Ψ+(x) n ,andwe n+ n + kept only the ground state contribution. | ⟩ | ⟩ ( By applying the same approximation to (1) and (3) as well, and expanding into the similarly deﬁned harmonic oscillators eigenstatesaround a,we − 84 CHAPTER 4. INSTANTONS

can thus write the Euclidean propagator as

β β + + 2 ωβ K ( a, a; , ) dx−dx dx−dx Ψ−( a) e− 2 E − − 2 − 2 ≈ 1 1 2 2 0 − ) + + + + ω + − + − ∗ ∗ 52 ((τ2 τ2 5)+(τ1 τ1 )) Ψ0−(x2 )Ψ0 (x2−)Ψ0 (x1 )Ψ0− (x1−)e5 − 5 − + + + + KE(x2 ,x2−; τ2 ,τ2−)KE(x1 ,x1−; τ1 ,τ1−) 1 mω 2 ωβ 2 = e− 2 R (4.23) π! / 0 where

+ ω + − + + + + 2 (τ1 τ1 ) ∗ R = dx1−dx1 e − Ψ0 (x1 )Ψ0− (x1−)KE(x1 ,x1−; τ1 ,τ1−)(4.24) ) R will be ﬁxed later on by studying more carefully the one-instanton amplitude. This proves factorization as promised.

We are not done yet! We must still sum over over the locations τ1,...,τn of the instantons. Indeed, as long as τ τ τ ,westillgetanequallygoodapproximatesolution 1 ≪ 2 ≪···≪ n by moving the positions τi around. The summation over instanton and anti- instanton location simply (and obviously) corresponds to the integral:

β τ τ 2 n 2 βn dτn dτn 1 dτ1 = (4.25) β β − ··· β n! )− 2 )− 2 )− 2

How do we normalize the dτi integrals? This is not a problem for the mo- ment: the normalization can be absorbed by the factor R (and this is not by chance!). Finally, one thing we have to be careful about is that instantons and anti- instantons cannot be distributed arbitrarily: for example,startingat a we βH − must ﬁrst encounter an instanton. For a e− ! a ,allcontributionswill ⟨− | |− ⟩ have nI = nA,etc...

We can now put together all the pieces and compute the total semiclassical amplitudes.

4.2.2 Summing up the instanton contributions: results

βH Let us compute a e− ! a ﬁrst. The contributing solutions are depicted in ﬁgure 4.10. ⟨− | |− ⟩

+ + + ··· a a a I A a a I A I A a − − − − − − Figure 4.10: The contributing solutions for the tranition from a to a are solutions with any number of instanton-anti-instanton pairs. − − 4.2. INSTANTONS IN THE DOUBLE WELL POTENTIAL 85

Thus, the amplitude will be:

n 1 SI /! βH mω 2 βω Re− β a e− ! a = e− 2 [1 + (!)] (4.26) ⟨− | |− ⟩ π! · n! · O n=even & ' / 0 %

βH Similarly, we can compute a e− ! a : ⟨ | |− ⟩ n 1 SI /! βH mω 2 βω Re− β a e− ! a = e− 2 [1 + (!)] (4.27) ⟨ | |− ⟩ π! · n! · O n=odd & ' / 0 % Performing the summation, we get:

1 ! ! βH mω 2 βω 1 Re−SI / β Re−SI / β a e− ! a = e− 2 e e− [1 + (!)] ⟨± | |− ⟩ π! · 2 ∓ · O / 0 / 0 (4.28) Recalling that

βH βEn x e− ! x = Ψ∗ (x )Ψ (x )e− ! (4.29) ⟨ f | | i⟩ n f n i n % and comparing to (4.28), we can read out (for β )theenergiesofthelowest energy states: →∞

!ω SI E = !Re− ! (4.30) ± 2 ± Calling + and the corresponding eigenstates, we can write | ⟩ |−⟩

βH βE+ βE− ! ! ! a e− a =Ψ+∗ ( a)Ψ+( a)e− +Ψ∗ ( a)Ψ ( a)e− (4.31) ⟨± | |− ⟩ ± − − ± − − Comparing to (4.28), and calling

1 mω 4 mωx2 ψ (x) e− 2! (4.32) 0 ≡ π! / 0 the harmonic oscillator ground state wavefunction we get

1 2 1 mω 2 1 Ψ∗ ( a)Ψ ( a)= = ψ0(0) − ± − − 2 π! √2 6 7 / 0 1 2 1 mω 2 1 Ψ∗ ( a)Ψ+( a)= = ψ0(0) + ± − ∓2 π! ∓ √2 / 0 6 7 Ψ (x)iseven Ψ (a)=Ψ ( a) − − − − (4.33) ⇒ Ψ (x)isodd Ψ (a)= Ψ ( a) < + + − + − as expected from the usual Schr¨odinger approach. Basically, 1 Ψ (x)= (ψ0(x + a) ψ0(x a)) ± √2 ∓ − Let us now add a few comments about our results. 86 CHAPTER 4. INSTANTONS

1. Notice that in our computation, we have neglected the perturbative cor- rections. Technically speaking, equation (4.30) is “incorrect” because in it, we have only retained the instanton correction which is of order

S/! 2/λ¯ !e− !e− (4.34) ∼ and is smaller than the perturbative corrections (!) we previously com- puted by Feynman diagrams:

∆E !ωλ¯n (4.35) n ∼

The point, however, is that these perturbative correction aﬀect E+ and E by exactly the same amount: they cancel in E+ E . − − − Apurist(asColemanputsit)wouldonlyretaintheinstantoncontribution when writing the diﬀerence

SI /! 2/λ¯ E+ E =2!Re− =2!Re− (4.36) − −

βH One could go ahead and compute the perturbative corrections to a e− ! a with the same Feynman diagram techniques we used before; we w⟨ould± | have |− ⟩ to compute the Green’s function around each solution. For most of the time, the instanton solutions sit at either oneortheother minimum. The perturbative corrections are thus the same as those we have computed before, plus corrections coming from the cubicanharmonic term (see ﬁgure 4.11).

+ + + ···

Figure 4.11: The quantum corrections to the classical solutions. The ﬁrst di- 4 agram is a two-loop correction involving the λ4x correction, the second one 3 is a two-loop correction involving the λ3x correction, and the third one is a three-loop correction.

βH For instance, a e− ! a becomes: ⟨ | |− ⟩ n 1 SI /! βH mω 2 βE0(λ) R(λ)e− β a e− ! a = e− ! ⟨ | |− ⟩ π! n! n=odd & ' / 0 % !ω ∞ E (λ)= 1+ c λ¯n 0 2 n 8 n=1 9 % ∞ n R(λ)=R 1+ dnλ¯ (4.37) 8 n=1 9 % 4.2. INSTANTONS IN THE DOUBLE WELL POTENTIAL 87

where cn are the loop coeﬃcient we have computed previously, and R is the leading contribution to the prefactor we introduced before without computing it explicitly. Moral: we see the two classes of eﬀects at work: Non-perturbative eﬀects, determined by semiclassical methods. They • correspond to the non-trivial stationary points of the Euclidean ac- tion. Perturbative eﬀects. They correspond to small quantum ﬂuctuations • around each stationary point of the action: they are the functional integral analogue of the corrections to the saddle point approximation of ordinary integrals. The perturbative corrections are conveniently organized in a series of Feynman diagrams. The leading λ¯0 term corresponds to evaluating the quantum ﬂuc- tuation determinantO 1 . & ' det(Oˆ) q - λ¯ terms correspond to two-loops diagrams. O - λ¯2 terms correspond to three-loops diagrams. O & ' -Andsoon... & ' 2. We can now also check the validity of our dilute instanton approximation. ! (Re−SI / β)n The exponential series n! is dominated by the terms with n ! ! ∼ Re SI / β.Thus,thetypicalinstantondensityis:ρ n = Re SI / .Itis − ( β − exponentially small! Their average separation in time≈ is:

! 1 eSI / 1 ∆τ = = = (4.38) ρ R ∆E which is independent of β. The sum over the instantons is truly necessary only to reproduce the −βH ! 1 exponential behaviour of e at β>∆E . We can now go back and verifyM thatN the sequences of widely separated in- stanton anti-instanton that are relevant to our result are stationary points of the action up to eﬀects of order

! δS ω SI / ω τ1 τ2 [ e ] e− | − | e− R (4.39) δx ∼ ≈ 5x=xIA 5 5 2/λ¯ which is truly small! Much-much5 smaller than the e− eﬀects we have been computing. 3. Another puzzle that one could have pointed out when we started our com- putation concerned the relevance of multi-instanton trajectories. These have an euclidean action = nSI which is bigger that the single instan- ton one SI .Thecontributionofeachsuchtrajectoryisthusweightedby amuchsmallerexponentialfactor,andonemaywonderifitmakes any βH sense to include them at all. For instance in +a e− a aIAIconﬁgu- 3S ⟨ | |− ⟩ ration contributes with weight e− I ,whichisexponentiallysmallerthan S the one instanton contribution e− I .Now,attheendofourcomputation, 88 CHAPTER 4. INSTANTONS

we obviously know how to answer. Each multi-instanton contributes only nS e− I to the amplitude, however n-instantons have also a higher multi- ∝ n n S plicity R β /n!whichforRβ> e− I compensates for the exponential suppression.∝ We can here draw a simple statistical mechanicsanalogy (just an analogy!): the action exponent plays the role of Boltzmann factor nS E/T n n e− I e− ,themultiplicity R β /n plays the role of an en- ∝tropy factor∼ eS,whiletheamplitudewearecomputingistheanalogueof∝ F/T the partition function e− .Boltzmannsuppressedconﬁgurationscon- tribute signiﬁcantly to the free energy at equilibrium if they have large multiplicity.

4.2.3 Cleaning up details: the zero-mode and the compu- tation of R To wrap things up, we are left with explaining the origin of the dτ integral on the instanton position and with computing R (the two things are obviously related). δ2S We have to study det δx2 ,theﬂuctuationdeterminantaroundthe x=xI one-instanton solution. / 05 5 Surprise: 5 2 δ S 2 = m∂ + V ′′(x )(4.40) δx2 − τ E I 5x=xI 5 possesses one normalizable zero5 mode: 5 2 m∂ + V ′′(x ) y (τ)=0 − τ E I 0 ⎧ ∞ 2 (4.41) & y0(τ) dτ =1' ⎨⎪ | | )−∞ Indeed, we can compute⎩⎪ from the equation of motion: 2 m∂ x (τ)+V ′ (x (τ)) = 0 − τ I E I 2 m∂ x˙ (τ)+V ′′(x (τ))x ˙ (τ)=0 (4.42) ⇒− τ I E I I We can then compute the normalization of this zero mode: 1 S x˙ 2(τ)dτ = mx˙ 2(τ)dτ = I I m I m ) ) m y (τ)= x˙ (τ)(4.43) ⇒ 0 S I . I The presence of Ψ0 originates from the invariance of the action under time ∆ translations: xI (τ +∆)=xI is a solution with same boundary conditions as R ∆ δ2S xI (τ), ∆ ,whichimpliesthatS(xI )=S(xI ). Thus, δx2 =0alongthis direction∀ in∈ the . 2 We can now expand the function x into modes around xI :

x(τ)=xI (τ)+√! yn(τ)˜an (4.44) n % y(τ)

2We are using the same normalization of section! 3.1"# \$ 4.2. INSTANTONS IN THE DOUBLE WELL POTENTIAL 89

Thus, 1 δ2S 1 1 1 exp y2 =exp λ a˜ 2 =exp λ a˜ 2 λ a˜ 2 + −2! δx2 −2 n n −2 1 1 − 2 2 2 ··· 6 7 8 n 9 6 7 % (4.45) as λ0 =0. The integration overa ˜0 should be treated separately and with special care (otherwise we would get a 1 nonsense). λ0 Notice that da˜0 is associated to a shift in the time position of the instanton: !m !m x(τ)=x (τ)+ x˙ (τ) a˜ + x τ +˜a + I S I 0 I 0 S . I · ···≈ 8 . I 9 ··· !m x (τ)=f(τ τ ) da˜ = dτ (4.46) I 0 S 0 0 − ⇒ . I − Let us then choose an N such that the measure for the ﬂuctuation y(τ)is: da˜ D[y]= ˜ n (4.47) N √ n 2π + N is ﬁxed as usual by some simple reference system; for example the har- monic oscillator: my˙2 mω2y2 πn S = + λh(ω)=m ( )2 + ω2 (4.48) E 2 2 → n β 1 2

1 − 2 da˜n 1 λh a˜ 2 h ˜ e− 2 n n n = ˜ λ = N √ P N n n 2π 8 n 9 ) + + mω ˜ = N (4.49) 2π! sinh(βω) det ( m∂ 2 + mω2) . − τ Let us now integrate in our case by separatingD out the integralonthezero mode. We have

1 2 1 2 da˜n λna˜ SI da˜n λna˜ ˜ e− 2 n n = dτ0 ˜ e− 2 n n N √ P mπ! · N √ P n 2π . n 1 2π ) + ) +≥ ˜ 2 2 SI det ( m∂τ + mω ) = dτ0 N − ! 2 2 mπ · det ( m∂ + mω ) · det′ ( m∂ 2 + V (x )) ) . − τ D − τ ′′ I 1 2 2 SI Dmω 2 βω det ( Dm∂τ + mω ) = dτ e− 2 − (4.50) 0 ! ! 2 mπ · π · ;det′ ( m∂τ + V ′′(xI )) ) . / 0 − where we multiplied and divided by the harmonic oscillator determinant, used eq. (4.49) and took the βω 1limit.Bydet′ we indicate the determinant with ≫ the zero mode removed: det′ = λ1 λ2 λ3 ....Now,byastraightforwardbut tedious computation which we shall· not· do· explicitly 3 one ﬁnds

2 2 1 S det ( m∂ + mω ) mω 2 I − τ = v˜ R (4.51) ! 2 ! mπ ;det′ ( m∂τ + V ′′(xI )) π ≡ . − / 0 3To be found in Coleman’s lectures and also proposed in one of the Homeworks to be posted on line (next year...). 90 CHAPTER 4. INSTANTONS wherev ˜ is determined by the asymptotic behaviour ofx ˙ for τ τ 1 : I | − 0|≫ ω

ω τ τ0 x˙ I =˜ve− | − | (4.52)

2 In the speciﬁc case of the λ x2 a2 potential, we have: 4! −

ω(τ& τ0) ' 1 1 e− − τ τ0 − ≫ ω ω(τ τ0) x(τ)=a − ω(τ τ ) a 1 2e− − 1+e− − 0 −→ − τ τ 1 / 0 0 ω ω τ τ x˙ − ≫ 2aωe− | − 0| ⇒ −→ v˜ =2aω (4.53) ⇒ Thus, we can compute:

1 1 1 1 2 3 2 2 5 2 2 mω 2 4ma ω 12m ω 12 v˜2 = = = ω (4.54) π! π! π!λ πλ¯ / 0 6 7 6 7 6 7 We thus ﬁnally obtain the energy splitting between the two lowest energy levels: 1 mω 2 SI ! E+ E =2! ve˜ − (4.55) − − π! / 0 Chapter 5

Interaction with an external electromagnetic ﬁeld

We will now discuss the quantum mechanics of a particle interacting with an external electromagnetic ﬁeld. We will see that the path integral approach oﬀers aneatviewpointontheissueofgaugeinvarianceandontherole of topology.

5.1 Gauge freedom 5.1.1 Classical physics The relativistically invariant action for a neutral spinless particle is x˙ 2 S = mc2 dτ = mc2 1 dt (5.1) − − − c2 )γ )γ . where we have chosen the parametrization so that: 1 1 dτ 2 =(dx0)2 (dxi)2 = (∂ x0)2 (∂ xi)2 dσ2 − c2 σ − c2 σ ⎧ 1 2 (5.2) ⎪ x˙ 2 ⎨⎪ 0 σ = x t dτ = 1 2 dt ≡ ⇒ . − c ⎪ The simplest⎩⎪ relativistic invariant interaction with the electromagnetic quadrivec- tor ﬁeld A =(A , Ai)isobtainedbymodifyingeq.(5.1)to µ 0 − x˙ 2 S = mc2 dτ e A dxµ = mc2 1 + eA eAix˙ i dt (5.3) − − µ − − c2 0 − )γ )γ )γ , . - The coeﬃcient e is the of the particle. Notice that the gauge transformation Aµ Aµ + ∂µα does not aﬀect the equations of motion since S only changes by boundary→ terms. S S e (α(x ,t ) α(x ,t )) → − f f − i i Consequently the classical equations of motion only depend on the electro- magnetic ﬁelds Fµν = ∂µAν ∂ν Aµ,whicharelocalfunctionsofAµ that are not aﬀected by gauge transformations.−

91 92CHAPTER 5. INTERACTION WITH AN EXTERNAL ELECTROMAGNETIC

The classically relevant quantity is the electromagnetic ﬁeld Fµν ,butthe lagrangian involves a redundant quantity: Aµ. Let us consider now what happens in the Hamiltonian description of this system. From equation (5.3), we can ﬁnd the conjugate momenta:

i i ∂ mx˙ i p = Li = + eA (x, t)(5.4) ∂x˙ 1 x˙ 2 − c2 F This leads to the hamiltonian:

H = pix˙ i = ( eAi)2 c2 + m2c4 + eA (5.5) −L − 0 F where A0 = V is the electromagnetic potential.

i Remark: both p and H are not invariant under a change of gauge for Aµ:

i i mx˙ i i i p = + eA (x, t) p′ = p e∂iα x˙ 2 → − ⎧ 1 c2 (5.6) ⎪ − ⎪ F ⎨ i i 2 2 2 4 H = (p eA ) c + m c + eA H′ = H + e∂ α − 0 → 0 ⎪ F ⎩⎪i i (H, p )transformslikee(A0,A ). Notice that the kinetic momentum is gauge invariant (being just a function of the velocity):

i i mx˙ i i i i i Π = p eA p e∂iα e A ∂iα =Π (5.7) ≡ 1 x˙ 2 − → − − − − c2 & ' & ' F Now what is the transformation (5.6) in the Hamiltonian language? Deﬁn- ing F (x, t) eα,wecanwrite: ≡ i i i p p′ = p ∂ F → − i i i i x x′ = x (5.8) ⎧ → ⎨⎪ H(p, x) H′(p′,x′)=H(p, x)+∂ F → 0 ⎪ It is a canonical ⎩transformation, in the sense that the following relations are satisﬁed: i j i j i j ∂p′ ∂p′ ∂p′ ∂p′ p′ ,p′ = = δ ( ∂ F ) ( ∂ F )δ =0 ∂pk ∂xk − ∂xk ∂pk ik − kj − − jk ik O P i j i j i j ∂p′ ∂x′ ∂p′ ∂x′ p′ ,x′ = = δ δ = δ (5.9) ∂pk ∂xk − ∂xk ∂pk ik jk ij O P i i and, moreover, p′ and x′ satisfy the Hamilton equations for H′: ∂p i ∂H ′ = ′ ∂t ∂x i ⎧ ′ (5.10) ∂x i ∂H ⎪ ′ ′ ⎨ = i ∂t ∂p′ ⎪ ⎩⎪ 5.1. GAUGE FREEDOM 93

H and pi are not invariant under gauge transformations, but they are covariant: H, pi H, pi + e (∂ α, ∂ α)(5.11) → 0 − i & ' & ' 5.1.2 Quantum physics Now, what happens to gauge invariance upon quantization? We will study the path integral viewpoint ﬁrst, and the Hamiltonian viewpointlater. In the path integral formalism, we can write very simply:

iS[x] ! K(xf ,tf ; xi,ti)= D[x]e

) x ,t iS[x] ie f f ie ie ! ! dα α(x ,t ) α(x ,t ) D[x]e − xi,ti = e− ! f f K(x ,t ; x ,t )e ! i i (5.12) → R f f i i ) We have the relation:

Ψ(xf ,tf )= K(xf ,tf ; xi,ti)Ψ(xi,ti)dxi (5.13) ) The gauge covariance of this equation leads to:

ie α(x,t) Ψ(x, t) e− ! Ψ(x, t)(5.14) → This is what we should have expected! Indeed, classically, the gauge transformation of the electromagnetic quadrivec- tor ﬁeld is related to a canonical transformation of the Hamiltonian variables:

gauge A transformation A + ∂ α µ −→ µ µ ⎧ xi xi ⎪ canonical (5.15) ⎪ transformation ⎪ pi pi e∂ α ⎨ −→ ⎧ i ⎪ − H ⎨ H + e∂tα ⎪ ⎪ Quantum mechanically,⎩⎪ we can make an⎩⎪ analogy with this representation:

gauge transformation Aµ Aµ + ∂µα −→ (5.16) ⎧ canonical transformation ie α ⎨⎪ Ψ e− ! Ψ −→ ⎪ Indeed, Ψ UΨwith⎩ U †U = UU† = 1 is the most general canon- ical transformation→ in the quantum description. Under the above rotation of the state vectors, the matrix elements of any operator Oˆ transform like Ψ1 Oˆ Ψ2 Ψ1 U †OUˆ Ψ2 .Astheonlythingthatmattersarematrixel- ⟨ements,| | very⟩→⟨ much| like we| do⟩ when going from the Schroedinger to the Heisen- berg picture for time evolution, we could alternatively deﬁne the transformation as one where state vectors are unaﬀected while operators transform according to Oˆ U †OUˆ .Canonicalcommutationrelationsareindeedpreservedbythe that transformation:→

[ˆp, xˆ]= i!1 U †pU,ˆ U †xUˆ = U † [ˆp, xˆ] U = i!U †U = i!1 . (5.17) − → − − 3 4 94CHAPTER 5. INTERACTION WITH AN EXTERNAL

These results immediately tell us how to proceed in the Hamiltonian ap- proach. Let us quantize pi,xj pˆi, xˆj,withthecanonicalcommutationrelations: → pˆi, pˆj = xˆi, xˆj =0 pˆi, xˆj = i!δ (5.18) 3 4 3 − 4ij 3 4 x˙ In the non-relativistic limit where c 1, the classical Hamiltonian takes the form: ≪ 1 2 H = mc2 + pi eAi + eA . (5.19) 2m − 0 Quantizing and neglecting the (now irrelevant)& constant' mc2 we have the Hamil- tonian operator

1 2 H(ˆp, xˆ)= pˆi eAi(ˆx) + eA (ˆx)(5.20) 2m − 0 & ' Let us now work in the coordinate representation, where ∂ xˆi = xi pˆi = i! (5.21) − ∂xi The matrix elements ofp ˆi are not invariant under the gauge transformation ie α Ψ e− ! Ψ: → j 3 Ψ pˆ Ψ = d xΨ∗(x)( i!∂ )Ψ (x) ⟨ 1| | 2⟩ 1 − j 2 ) 3 ie α ie α 3 d xΨ∗e ! ( i!∂ )e− ! Ψ = d xΨ∗( i!∂ e∂ α)Ψ (5.22) → 1 − j 2 1 − j − j 2 ) ) However, the matrix elements of the kinetic momentum Πˆ j =ˆpj eAˆj are gauge invariant. Indeed we have −

j j Πˆ (A)Ψ(x) Πˆ (A′)Ψ′(x) → j ie α = i!∂ eA + e∂ α e− ! Ψ(x) − j − j ie ! α j = e&− i!∂j e∂jα ' eA + e∂jα Ψ(x) − − − ie α j = e− ! Πˆ& (A)Ψ(x)(5.23)' and so the kinetic momentum Πˆ j(A)isacovariant derivative (Πˆ j (A)Ψtrans- forms like Ψ). Thus, when taking matrix elements, the phase factor in eq. (5.23) cancels out and we conclude that the matrix elements of Πˆ j(A)aregaugein- variant. Now, what about the gauge covariance of the Hamiltonian and the Schr¨odinger equation? Under a gauge transformation the Hamiltonian transforms like: 1 2 H α H = pˆi eAˆi + e∂ α + e Aˆ + ∂ α (5.24) −→ α 2m − i 0 t / 0 / 0 Assume now that Ψsolves the Schr¨odinger equation with respect to H:

i!∂tΨ=HΨ 5.2. PARTICLE IN A CONSTANT 95

ie α It is a good exercise to check that Ψα = e− ! ΨsolvestheSchr¨odinger equation with respect to Hα:

i!∂tΨα = HαΨα (5.25)

Thus, the Schr¨odinger equation is gauge covariant. 2 In the same way, we have that the probability density Ψ∗Ψ= Ψ is gauge invariant. Then what is the ? | | It is expectedly given through the velocity operator:

Πˆ j 1 vˆj = = i!∂ eAj (5.26) m m − j − & '

j 1 j j J (x)= Ψ∗(x)ˆv Ψ(x) vˆ Ψ(x) ∗ Ψ(x) 2 − /i &j ' 0 j = − Ψ∗ !∂ ieA Ψ !∂ ieA Ψ ∗ Ψ (5.27) 2m j − − j − / 0 & ' && ' ' 2 j Indeed, by the Schr¨odinger equation, we can check that ∂t Ψ + ∂jJ =0. J j is also manifestly gauge invariant. | | In order to prove current conservation let us deﬁne:

−→D = !−→∂ ieAi i i − i ←−D i = !←−∂ i + ieA (5.28)

Using these deﬁnitions, we can write:

j i J = − Ψ∗−→D Ψ Ψ∗←−D Ψ 2m j − j / 1 0 −→∂ i + ←−∂ i = ! −→D i + ←−D i / 0 −→D 2 D2 = = eA H (5.29) 2m 2m 0 − We can then compute:

j i ∂ J = − Ψ∗ −→D + ←−D −→D Ψ Ψ∗←−D −→D + ←−D Ψ j 2m! i i i − i i i i / / 0/ 0 / 0/ 0 0 = − Ψ∗−→D −→D Ψ Ψ∗←−D ←−D Ψ 2m! i i − i i i / 2 2 0 = − Ψ∗ D Ψ D Ψ ∗ Ψ 2m! − i / & ' & ' 0 = Ψ∗ (H eA )Ψ ((H eA )Ψ)∗ Ψ ! − 0 − − 0 = i &Ψ∗i∂ Ψ (i∂ Ψ)∗ Ψ ' t − t 2 = &Ψ∗∂tΨ ∂tΨ∗Ψ= '∂t Ψ (5.30) − − − | | 5.2 Particle in a constant magnetic ﬁeld

Let us now study the system of a particle in a constant magneticﬁeld. 96CHAPTER 5. INTERACTION WITH AN EXTERNAL ELECTROMAGNETIC FIELD

5.2.1 An exercise on translations Let us ﬁrst study what happens with spatial translations in this particular case. We take the magnetic ﬁeld along the z-axis: B⃗ =(0, 0,B). We can choose the electromagnetic quadrivector as: 1 1 1 A⃗ = B⃗ ⃗r A = By, A = Bx (5.31) 2 ∧ ⇒ 1 −2 2 2 We then recover the magnetic ﬁeld by:

B⃗3 = ⃗ A⃗ = ϵ321∂2A1 + ϵ312∂1A2 = ∂2A1 + ∂1A2 = B (5.32) ∇∧ 3 − / 0 This possible gauge choice leads to the following Hamiltonian, in the coor- dinate representation:

2 2 2 ˆ 1 ˆ i 1 eB eB 2 H = Π (A) = pˆ1 + y + pˆ2 x +ˆp3 (5.33) 2m 2m , 2 − 2 - / 0 6 7 6 7 How are spatial translations realized?

i!∂ ,thenaivegenerators,do not commute with the Hamiltonian. •− i Translations along the x and y axis are described by the • U(a, b)suchthatU(a, b)Ψ(x, y)=Ψ(x + a, y + b). This operator can be iapˆ1 ibpˆ2 ˆ written as U(a, b)=exp ! + ! .ItsactiononoperatorsisO(x +

a, y + b)=U(a, b)Oˆ(x, y)U/ †(a, b). 0

Let us now look at the translational invariance of the spectrum of the Hamil- tonian. We have:

Hˆ (x, y)Ψn(x, y)=EnΨn(x, y)

U(a, b)Hˆ (x, y)Ψn(x, y)=EnU(a, b)Ψn(x, y)

U(a, b)Hˆ (x, y)U †(a, b)U(a, b)Ψn(x, y)=EnU(a, b)Ψn(x, y)

Hˆ (x + a, y + b)Ψn(x + a, y + b)=EnΨn(x + a, y + b)(5.34)

However, Hˆ (x + a, y + b)isrelatedtoHˆ (x, y)byagaugetransformation:

1 1 Bbx A¯ = B(y + b)= By + ∂ 1 −2 −2 x − 2 ⎧ 6 7 ⎪ ¯ 1 1 Bay ⎨⎪ A2 = B(x + a)= Bx + ∂y 2 2 2 6 7 ⎪ ¯ ⎩⎪ Aµ = Aµ + ∂µα B (5.35) ⇒ ⎧ α = (ay bx) ⎨ 2 − Thus, we can put⎩ these results together and write:

ie ie ! α ! α e Hˆ (x, y)e− Ψn(x + a, y + b)=EnΨn(x + a, y + b)(5.36) 5.2. PARTICLE IN A CONSTANT MAGNETIC FIELD 97

We have then the following relation:

Hˆ (x, y)Ψ˜ n = EnΨ˜ n ie eB i (5.37) ! α i ! (bx ay) ! (apˆ1+bpˆ2) < Ψ˜ n = e− Ψn(x + a, y + b)=e 2 − e Ψn(x, y) We have thus obtained the inﬁnitesimal generators of translations: eB T =ˆp y 1 1 − 2 eB T =ˆp + x (5.38) 2 2 2 Notice that these generators commute with Π1 and Π2: eB eB T , Π1 = pˆ y,pˆ + y =0 1 1 − 2 1 2 1 2 3 4 eB eB eB T , Π2 = pˆ y,pˆ x = ([ˆp ,x]+[y,pˆ ]) = 0 (5.39) 1 1 − 2 2 − 2 − 2 1 2 1 2 3 4 However, T1 and T2 do not commute among themselves: eB eB eB [T ,T ]= pˆ y,pˆ + x = ( i! i!)= i!eB (5.40) 1 2 1 − 2 2 2 2 − − − 1 2 The group of translations is thus a projective realization oftheclassical • one. The is related to the enclosed ﬂux. • Previously, we chose one particular way to translate from (x, y)to(x + • a, y + b). Any other way would diﬀer by eiγ ,withγ !eB. ∼ 5.2.2 Motion We want to describe ﬁrst the classical motion of a particle in this constant magnetic ﬁeld, oriented along the z-axis: B⃗ =(0, 0,B). The classical equations of motion give the following result: z¨ =0 z = z + v t. • ⇒ 0 z eB (x, y)describecircularorbitswithfrequencyωc = m .IfeB > 0, the • rotation is clockwise. Notice: harmonicity: the frequency is independent of the radius of rota- tion. We get: x = xc + R cos ωct (5.41) y = y R sin ω t < c − c where (xc,yc)isthecoordinateofthecenterofrotation.Itstimeevolution is described by: 1 1 Π x = x + y˙ = x + y c ω ω m ⎧ c c (5.42) ⎪ 1 1 Πx ⎨⎪ yc = y x˙ = y − ωc − ωc m ⎪ ⎩⎪ 98CHAPTER 5. INTERACTION WITH AN EXTERNAL ELECTROMAGNETIC FIELD

xc and yc are two constants of motion. What is the associated symmetry? The answer is translations in (x, y).

Let us see this directly by working with the quantum Hamiltonian in some gauge. Remember that we have seen that performing a gauge transformation ie α on the vector potential amounts to a phase rotation: Ψ e− ! Ψ. Let us pick Landau’s gauge which is also convenient to→ computeenergylevels:

A = By A =0 (5.43) x − y The Hamiltonian is: 1 Hˆ = (ˆp + eByˆ)2 +ˆp 2 +ˆp 2 (5.44) 2m 1 2 3 / 0 The kinetic momenta are given by:

Πˆ 1 =ˆp1 + eByˆ

⎧ Πˆ 2 =ˆp2 (5.45) ⎪ ⎨ Πˆ 3 =ˆp3 ⎪ Notice that ⎩⎪ • Πˆ , Πˆ = i!eB =0 (5.46) 1 2 − ̸ > ? The two velocity components in the plane perpendicular to B⃗ do not commute. Thus, they cannot be measured simultaneously with arbitrary accuracy.

The combinations • T1 =ˆp1 (5.47) < T2 =ˆp2 + eBx

commute with Πˆ 1,2,3.Thus,[Ti,H]=0,whichimpliesthattheyare conserved. They indeed correspond to the coordinates of the center of theclassical motion. Comparing them to (5.42), we get:

T1 =Π1 eBy = eByc − − (5.48) < T2 =Π2 + eBx = eBxc

Quantum mechanically, also T and T cannot be ﬁxed simultaneously as • 1 2 [T ,T ]= i!eB (5.49) 1 2 − Given a solution of the Schr¨odinger equation Ψ(x, y, z), then • i ! (aT1+bT2) Ψa,b(x, y, z)=e Ψ(x, y, z)(5.50)

is also a solution. What does this operation represent? It is a space translation. 5.2. PARTICLE IN A CONSTANT MAGNETIC FIELD 99

aT1 apˆ1 This fact is obvious for ei ! = ei ! = ea∂x . bT2 For ei ! ,wehave:

bT2 eBx eBx ei ! Ψ(x, y, z)=ei ! eb∂y Ψ(x, y, z)=ei ! Ψ(x, y + b, z)

This is an ordinary translation composed with a gauge transformation.

Notice that [T1,T2] =0impliesthatthetranslationgroupisnon-commutative • in presence of a magnetic̸ ﬁeld. To be stressed and to conclude:

The pairs of canonically conjugated variables are (Πˆ 1, Πˆ 2), which are physical velocities, and (T1,T2)whicharethecoordinatesofthecenteroftheorbit. In a magnetic ﬁeld, the canonically conjugated variables are(velocity,velocity) and (position,position), whereas in ordinary motion in a potential those are (ve- locity,position).

5.2.3 Landau levels In order to study the , let us continue working in the Lan- dau gauge: A = By A =0 (5.51) x − y The rotation invariance around (x, y)=0isnotmanifest. • The translational invariance in the x-direction is manifestly preserved, • whereas it is not manifest in the y-direction. We have the Hamiltonian: 1 Hˆ = (ˆp + eByˆ)2 +ˆp 2 +ˆp 2 (5.52) 2m 1 2 3 / 0 Asp ˆ1 andp ˆ3 commute with Hˆ ,wecanchoose(p3,p1,E)asquantumnum- bers:

pˆ3Ψ=p3Ψ

pˆ1Ψ=p1Ψ Hˆ Ψ=EΨ(5.53) Such a state will satisfy:

i Ψ(x, y, z)=e ! (p1x+p3z)F (y)(5.54) For such a state, the Hamiltonian is equivalent to: 1 Hˆ = (p + eByˆ)2 +ˆp 2 + p 2 (5.55) 2m 1 2 3 / 0 It is the Hamiltonian of a harmonic oscillator in (ˆp2, yˆ)withfrequencyωc = eB p1 m ,whichisthecyclotronfrequencyoftheparticle,andcentered at yc = eB = p1 : − − mωc 1 1 eB 2 p 2 p 2 Hˆ = pˆ 2 + m yˆ + 1 + 3 (5.56) 2m 2 2 m eB 2m 6 7 / 0 100CHAPTER 5. INTERACTION WITH AN EXTERNAL ELECTROMAGNETIC FIELD

By quantizing this harmonic oscillator in (ˆp2, yˆ), we get the energy levels:

p 2 1 E(p ,p ,n)= 3 + !ω n + (5.57) 3 1 2m c 2 6 7 with the wave functions:

i (p x+p z) p1 Ψ (x, y, z)=e− ! 1 3 Ψ y + p3,p1,n n mω 6 c 7 1 2 2 4 mωc y mωc 1 mωc ! Ψn(y)= Hn y e− 2 (5.58) π! √2nn! ! / 0 6 . 7 where Ψn(y)istheordinarywavefunctionofaharmonicoscillatoraround y =0 with frequency ωc,andHn(x)aretheHermitepolynomialsdeﬁnedas:

n n x2 d x2 H (x)=( 1) e e− (5.59) n − dxn

The energy is degenerate in p1 yc.Thisisrelatedtotranslational • invariance. ∼ The result (5.57) is in accord with the classical property of the system: the • frequency of rotation does not depend on the radius: this is harmonicity. The shape of Ψ (x, y) is depicted in ﬁgure 5.1. • | p3,p1,n=0 | x

y yc

Figure 5.1: The shape of the norm of the wavefunction Ψp3,p1,n=0(x, y). We see that it is far from rotational invariant.

Where is cylindrical symmetry gone? To make it manifest, we must su- perimpose diﬀerent wavefunctions:

Ψp3,n2,n(x, y, z)= dp1Ψp3,p1,n(x, y, z)f(p1,n2)(5.60) ) In Landau gauge, we have: •

Πˆ 2 =ˆp2 Πˆ 1 =ˆp1 + eByˆ

T1 =ˆp1 T2 =ˆp2 + eBxˆ (5.61)

We can check that:

Πˆ 1,T2 = eB ([ˆp1, xˆ]+[ˆy, pˆ2]) = 0 (5.62) > ? 5.3. THE AHARONOV-BOHM EFFECT 101

We get:

mω yˆ = Πˆ T c 1 − 1 mω xˆ = T Πˆ (5.63) c 2 − 2

We can thus conclude that in Landau gauge, for the states Ψp3,p1,n, yc = T1 T2 is ﬁxed, whereas xc = is not, and is fully spread. − mωc mωc It is intructive to count the levels for the case where the motion is limited to a region of sides Lx,Ly in the plane. If the motion is limited to 0 ! x ! Lx,thenp1 is quantized: 2π! p1 = n2 n2 Z (5.64) Lx ∈

Thus, the position yc is quantized:

p1 2π! yc = = n2 (5.65) eB eBLx Note that it is the case whether or not the motion is limited in the y direction! Now, if the motion is limited in the y direction as 0 ! y ! Ly,thenthe number of states of quantum numbers (n, p3)wecanﬁtinis:

Ly LxLy N(n, p3)= = eB (5.66) ∆yc 2π!

5.3 The Aharonov-Bohm eﬀect

The study of the vector potential in quantum mechanics opens awindowon deeper aspects of the quantum , some of which have a full description only in quantum ﬁeld theory:

1. The relevance of topology

2. The charge quantization

The Aharonov-Bohm eﬀect is one remarkable and yet simple example. Consider an ideal inﬁnitely long and empty cylindrical shellsuchthatparti- cles do not have access to its interior. The propagation of particles in this setup is characterized by a Schr¨odinger equation with boundary condition Ψ(x, y, z)=0 on the surface of the cylinder. In the path integral approach,thiscorresponds to summing only on paths that do not go through the interior of the cylinder. Imagine now a modiﬁed arrangement where the cylindrical shell encloses a uniform magnetic ﬁeld (see ﬁgure 5.2). This could, for instance, be realized by placing an “inﬁnitely” long solenoid inside the shell. Intuitively, we would think that physics (the propagation ofchargedparti- cles) is unaﬀected outside the shell, as B⃗ =0there.However,quantumme- chanics tells that this conjecture is wrong! To show this, we must consider the vector potential. Outside the shell, B⃗ =0,butA⃗ =0asimpliedbyStokes’theorem(seeﬁgure5.3). ̸ 102CHAPTER 5. INTERACTION WITH AN EXTERNAL ELECTROMAGNETIC FIELD

B =0 B =0 ̸

Figure 5.2: An inﬁnitely long cylinder has a constant magnetic ﬁeld inside it, whereas outside it there is no magnetic ﬁeld.

B⃗

γ

2 S = πr0

γ = ∂σ

A⃗ d⃗x = B⃗ d⃗σ = BS · · Eγ )σ

Figure 5.3: An inﬁnitely long cylinder has a constant magnetic ﬁeld inside it, whereas outside it there is no magnetic ﬁeld.

γ2 xf

xi γ1

Figure 5.4: Two paths γ1 and γ2 from xi to xf that cannot be continuously deformed into each other, due to the between them.

One possible gauge choice (the only one that manifestly respects cylindrical BS symmetry) is Aφ = 2πr , Ar,Az =0incylindricalcoordinates,forr>r0. 5.3. THE AHARONOV-BOHM EFFECT 103

It is immediate to see that Aφ has physical consequences by considering quantum amplitudes in the path integral approach. Consider two paths γ1 and γ2 like in ﬁgure 5.4. γ cannot be continuously deformed into γ . • 1 2 They are topologically inequivalent (they belong to diﬀerent homotopy • classes).

Consider the contribution of γ1 and γ2 to K(xf ,xi; t):

(1) iS1 iS ie γ e ! =exp 0 Aidxi 1 → ! − ! 8 )γ1 9 (2) iS2 iS ie γ e ! =exp 0 Aidxi (5.67) 2 → ! − ! 8 )γ2 9 (i) where S0 is the free particle action for γi. The phase diﬀerence induced by the vector potential is:

e i i i i e i i eBS eΦ ∆φ = ! A dx A dx = ! A dx = ! ! (5.68) γ − γ γ γ − ≡− 6) 2 ) 1 7 E 2− 1 1. Apparently the presence of the solenoid is felt at the quantum level by trajectories that are arbitrarily far from it! 2. Notice that the phase shift vanishes between paths that aretopologically equivalent, since they do not encircle any magnetic ﬂux. The suprising relevance of the vector potential at the quantum level was ﬁrst suggested by Aharonov and Bohm (1959) and later tested by Chambers (1960) and more recently by Tonomura (1982). The ideal “obvious” way to test this eﬀect is to perform the double slit ex- periment in the presence of our inﬁnite and thin solenoid (in the solenoid is replaced by a thin magnetized iron ﬁlament called a whisker), as in ﬁgure 5.5. Two disjoint classes of trajectories contribute to the propagation amplitude from the source S to the screen C:

i S(1) e Aidxi i S(2) e Aidxi D ! 0 − D ! 0 − K(xf ,xi; t)= [x]γ1 e “ R ” + [x]γ2 e “ R ” (5.69) ) ) By Stokes’ theorem, Aidxi is the same for all trajectories in the class 1 γ1 (i.e. passing on the top of the solenoid). The same thing is true for the class 2. * Moreover, we have:

Aidxi Aidxi = Aidxi = BS =Φ(5.70) γ − γ γ γ ) 1 ) 2 E 1− 2 where Φis the magnetic ﬂux through the solenoid. Thus, the total contribution will be:

ie i i ie ! A dx (0) Φ (0) − γ1 ! K(xf ,xi; t)=e R φ1 (xf )+e φ2 (xf )

ie iα (0) / ! Φ (0) 0 = e φ1 (xf )+e φ2 (xf ) (5.71) / 0 104CHAPTER 5. INTERACTION WITH AN EXTERNAL ELECTROMAGNETIC FIELD

xf γ1

xi solenoid S

γ2

A B C

Figure 5.5: The experimental setup to test the Aharonov-Bohmeﬀect.Thetwo paths γ1 and γ2 contribute to the propagation amplitude from the source S to the screen C diﬀerently.

(0) where φi is the contribution of the class i in the absence of the solenoid. The ﬁnal amplitude will be:

2 2 2 (0) (0) (0) ie Φ (0) K = φ + φ +2Re φ ∗e ! φ = F + F + F (5.72) | | 1 2 1 2 1 2 int 5 5 5 5 / 0 5 5 5 5 This result5 tells5 us5 what5 the eﬀect of the solenoid is, no matter how compli- (0) cated φi are. This is because its eﬀect is purely topological: the only property of the trajectory that enters into the phase shift is its topology (whether it belongs to the class 1 or 2). In the case of the free propagation, we have (see ﬁgure 5.6):

(0) (0) 2 i ka z 2 i k z φ ∗φ = e ! b e ! ∥ 1 2 ≡ 2k z ∥ eΦ i ! + ! 2k z eΦ F =2Re e „ « =2cos ∥ + (5.73) ⇒ int ! ! 8 9 6 7 Thus, as Φis varied, the peaks of the interference pattern areshiftedalong the z direction. This eﬀect is periodic in the ﬂux Φwith period: ! 2π 7 2 Φ ∆Φ = =4.135 10− gauss cm (5.74) 0 ≡ e · × which represents a fundamental unit of magnetic ﬂux. The eﬀect of the solenoid vanishes for Φ= nΦ0.

Moral:

1. The Aharonov-Bohm eﬀect shows that Fµν underdescribes electromag- netism; i.e. diﬀerent physical situations in a region may have the same Fµν in that region. 5.3. THE AHARONOV-BOHM EFFECT 105

a

z

a −

b

Figure 5.6: The experimental setup to test the Aharonov-Bohmeﬀect.Thetwo slits are located at a and a along the vertical direction, the two screens are separated by a distance b,andthepositionontheﬁnalscreenisdescribedby− z.

e i i 2. The phase ϕ = ! A dx overdescribes ; i.e. diﬀerent phases in a region may describe the same physics. For example Φ′ = Q Φ+nΦ0.

ie A dxµ 3. What provides a complete description is the factor eiϕ = e ! γ µ over H all possible closed paths in space-time, which after all is what enters in the path integral.

Thus, electromagnetism is the gauge invariant manifestation of a non-integrable phase factor, as stated by Yang and Wu (1975).

5.3.1 Energy levels Let us now do a simple example on the energy levels of a conﬁned on a ring of radius R around a solenoid (see ﬁgure 5.7).

solenoid

R

ring

Figure 5.7: A particle is conﬁned on a ring of radius R around a solenoid of area σ went through by a magnetic ﬁeld of intensity B.

The vector potential describing this situation is such that:

Bdσ = Ad⃗ ⃗x (5.75) )σ E∂σ 106CHAPTER 5. INTERACTION WITH AN EXTERNAL ELECTROMAGNETIC FIELD

The magnetic ﬂux is found to be:

Φ=Bσ = Aθ(R)2πR (5.76)

Which ﬁxes the phase factor:

ie Ad⃗ ⃗x i eΦ iϕ e− ! = e− ! e− (5.77) H ≡ We then ﬁnd the Hamiltonian for the particle:

1 1 i!∂ eΦ 2 H = (p eA )2 = θ 2m θ − θ 2m − R − 2πR 6 7 1 !2 eΦ 2 1 !2 ϕ 2 = i∂ = i∂ (5.78) 2m R2 − θ − 2π! 2m R2 − θ − 2π 6 7 / 0 The eigenstates of this Hamiltonian are:

Ψ (θ)=einθ n Z (5.79) n ∈ with energy levels: 1 !2 ϕ 2 1 !2 E = n = (n α)2 (5.80) n 2m R2 − 2π 2m R2 − / 0 We can notice that when ϕ ϕ +2πm,withm an integer, α α + m,and thus the spectrum is the same.→ → The conclusion is that what is physical is not ϕ itself, but the phase factor iϕ e .Moreover,thephysicalangularmomentumRpθ = !(n α)isnotquantized in integer units of !. −

5.4 Dirac’s magnetic monopole

Let us now very brieﬂy and qualitatively discuss Dirac’s magnetic monopole. Let us take our thin solenoid and cut it in two: the ﬂux will “expand” out. The tip of the solenoid will act more or less as a point source ofmagnetic ﬁeld: a magnetic monopole (see ﬁgure 5.8). Of course ⃗ B⃗ =0isstillsatisﬁed:themagneticﬂuxisconservedasthere is the “incoming”∇· ﬂux from the solenoid. Notice that: g g B⃗ d⃗σ B M (5.81) M ≡ · ⇒ ≈ 4πr2 )σ where σ is a sphere centered at the end of the solenoid, with the surface where the solenoid crosses it excluded. Assume now the ideal limit in which the solenoid is inﬁnitely thin. If for all charged particles the charge ej the ﬂux Φsatisﬁes:

ejΦ ! =2πnj (5.82) there will be no way to tell the presence of the solenoid (also called Dirac’s ). Our object will behave by all means as a point-like magnetic monopole! Let us now turn the argument around: 5.4. DIRAC’S MAGNETIC MONOPOLE 107

B⃗

B⃗

Figure 5.8: The solenoid cut in two will act as a magnetic monopole. Maxwell’s second law is still satisﬁed as the magnetic ﬂux is incoming from the solenoid.

Postulate: magnetic monopoles must exist in order to restore the symmetry between magnetic charge and electric charge which is apparently broken by Maxwell’s equations:

⃗ E⃗ = ρ ⃗ B⃗ =0 (5.83) ∇· E ∇·

Since ⃗ B⃗ =0followsnotfromdynamicsbutfromkinematics:Bi = ϵijk∂j Ak, amonopolestatemustinvolveanincomingﬂuxlocalizedinto∇· some string; but in order to preserve spherical symmetry the string shouldbeunde- tectable. Thus, the Aharonov-Bohm phase should be a multipleof2π for all existing electrically charged particles:

ejgM ! =2πnj (5.84)

Thus, the existence of a single monopole in the implies the quanti- zation of the electric charge!

5.4.1 On coupling constants, and why monopoles are dif- ferent than ordinary charged particles The strength of the electromagnetic interactions is set by the ﬁne structure constant: e2 1 α = 1(5.85) !c ≈ 137 ≪ The electromagnetic interactions are thus “weak”. The quantization condition implies that the minimal magnetic charge of a monopole, if it existed, is: 2π! g = (5.86) min e Thus, the magnetic , αM ,isdeﬁnedas:

g 2c (2π)2!c (2π)2 α = min = = 137 (2π)2 1(5.87) M ! e2 α ≈ · ≫ The magnetic interactions among monopoles are therefore very strong! 108CHAPTER 5. INTERACTION WITH AN EXTERNAL ELECTROMAGNETIC FIELD Bibliography

[1] “Quantum Mechanics and Path Integrals” , R.P. Feynman andA.R.Hibbs, McGraw-Hill, 1965. ( Chapters 1, 2 and 3: Great reading!!!)

[2] “Path Integral Methods and Applications”, R. MacKenzie,arXiv:quant- ph/0004090. (readable overview on the subject)

The following books may be of help in complementing the lecture notes.

[3] “Modern Quantum Mechanics”, J.J. Sakurai, The Benjamin/Cummings Publishing Company, 1985. (chapters 2.5 and 2.6: quick introduction to path integral and nice extended discussion of electromagnetic interactions, Aharonov-Bohm, monopoles)

[4] “Aspects of Symmetry”, S. Coleman, Cambridge UniversityPress,1985. (Sections 1-2 and appendices 1-2 of Chapter 7: The uses of the instantons. IfollowedColemanverycloselyinmylecturesoninstantons)

[5] “Path Integrals in Quantum Mechanics, Statistics and Physics”, , World Scientiﬁc, 1995. Chapter 3, in particular sections 3.9, 3.10 and 3.11 give a good account of Feynman diagrams

[6] “Topics in Advanced Quantum mechanics”, Barry R. Holstein, Addison- Wesley 1982. Chapters 1 and 5 (contains a lot of subjects, which is nice, but sometimes he is too sloppy)

[7] “Techniques and applications of Path Integration”, L.S.Schulman,John Wiley & Sons Inc., 1981 Comprehensive but somewhat formal. Chapters 12, 17 and 18 overlap with presentations in the course.

Other references

[8] H. Kleinert and A. Chervyakov, “Simple Explicit FormulasforGaussian Path Integrals with Time-Dependent Frequencies,” Phys. Lett. A 245 (1998) 345 [arXiv:quant-ph/9803016].

[9] P. A. Horvathy, “The Maslov correction in the semiclassical Feynman in- tegral,” arXiv:quant-ph/0702236.

109 110 BIBLIOGRAPHY

[10] “Quantum Mechanics”, Landau-Lifshitz.