Path Integrals in Quantum Mechanics

Home , Classical limit, Probability amplitude, Propagator, Wave function

Emma Wikberg

Project work, 4p

Department of Physics Stockholm University

23rd March 2006 Abstract

The method of Path Integrals (PI’s) was developed by Richard Feynman in the 1940’s. It offers an alternate way to look at quantum mechanics (QM), which is equivalent to the Schrödinger formulation. As will be seen in this project work, many "elementary" problems are much more difficult to solve using path integrals than ordinary quantum mechanics. The benefits of path integrals tend to appear more clearly while using quantum field theory (QFT) and perturbation theory. However, one big advantage of Feynman’s formulation is a more intuitive way to interpret the basic equations than in ordinary quantum mechanics. Here we give a basic introduction to the path integral formulation, starting from the well known quantum mechanics as formulated by Schrödinger. We show that the two formulations are equivalent and discuss the quantum mechanical interpretations of the theory, as well as the classical limit. We also perform some explicit calculations by solving the free particle and the harmonic oscillator problems using path integrals. The energy eigenvalues of the harmonic oscillator is found by exploiting the connection between path integrals, statistical mechanics and imaginary time. Contents

1 Introduction and Outline 2 1.1 Introduction ...... 2 1.2 Outline ...... 2

2 Path Integrals from ordinary Quantum Mechanics 4 2.1 The Schrödinger equation and time evolution ...... 4 2.2 The propagator ...... 6

3 Equivalence to the Schrödinger Equation 8 3.1 From the Schrödinger equation to PI’s ...... 8 3.2 From PI’s to the Schrödinger equation ...... 12

4 Free Particle 15

5 The Classical Limit 17

6 Harmonic Oscillator 19

7 Imaginary time and statistical mechanics 25 7.1 General considerations ...... 25 7.2 Application to the harmonic oscillator ...... 26

8 Appendix 28 8.1 Free particle the easy way ...... 28 8.2 Details on the computation of the harmonic oscillator . . . . . 29 8.2.1 Why det(P) = det(N), | det(O)| = 1 ...... 29 8.2.2 Calculation of det(N) ...... 29 8.2.3 Calculation of elements of N−1 ...... 31

1 Chapter 1

Introduction and Outline

1.1 Introduction

One of the most often cited experiments of quantum physics is the double slit experiment. Quantum mechanical particles, e.g. electrons, give rise to an interference pattern, just like waves, when they are allowed to pass through a pair of slits. The interference phenomenon occurs even when they are shot through the slits one particle at a time. This strange behavior is due to the wave nature of the particles, and one says that the electrons pass through both slits at the same time, hence interfering with themselves. The fact that quantum mechanical objects, in contrast to classical, do not follow a certain trajectory, but rather a superposition of such, is counterintuitive in itself. However, in the path integral formulation of quantum mechanics, this property is naturally embedded in the equations. Because of this, Feynman’s theory treats quantum mechanical problems in a perhaps more intuitive way than Schrödingers wave equation. Moreover, the PI formalism treats the classical trajectory of a particle in the same way as any other path, leading to an easy way to ﬁnd and understand the classical limit of the theory. These advantageous properties will perhaps make the method appear more useful than the calculations do at this introductory stage.

1.2 Outline

The outline of this project work is as follows. In chapter two we introduce path integrals starting from the well known Schrödinger formalism. We ex- plore the general expression for the time evolution of a state, leading to the deﬁnition of the propagator. We interpret the propagator and discuss the meaning of "integrating over all paths". At the end of the chapter we arrive

2 at the path integral expression of the propagator for a general Hamiltonian. A formal derivation of the equivalence of the Schrödinger equation and the path integral formulation is contained in chapter three. As a ﬁrst example of a PI we calculate the propagator for a free particle in chapter four. In the appendix this problem is solved in an easier way, using ordinary quantum mechanics. Chapter ﬁve covers the classical limit at a qualitative level. More complicated are the calculations done in chapter six, treating the harmonic oscillator. However, almost every step is performed in detail, therefore even this part of the work should be possible to follow for everyone familiar with elementary quantum mechanics and linear algebra. Some details in the derivation have been moved to the appendix to make the text, at least a bit, easier to read. In chapter seven, the interesting connection between path integrals in imaginary time and statistical mechanics is discussed. The similarity between the partition function and the propagator makes it easy to calculate the energy eigenvalues for the harmonic oscillator. This is also done in this chapter.

3 Chapter 2

Path Integrals from ordinary Quantum Mechanics

2.1 The Schrödinger equation and time evolution

In ordinary quantum mechanics, as formulated by Schrödinger, Heisenberg and Dirac in the 1920’s, we study the Schrödinger equation for a state |ψ(t)i: ∂ Hˆ |ψ(t)i = ih¯ |ψ(t)i ∂t

−iHˆ (t−t0)/¯h ⇒ |ψ(t)i = e |ψ(t0)i (2.1) ˆ pˆ2 where the (time independent) Hamiltonian operator is H = 2m + V (ˆx). We will in this project work only treat the one dimensional case. Here we have also introduced the notation we will use in this report; xˆ and pˆ are the position and momentum operators, respectively. These have corresponding eigenstates and eigenvalues; xˆ|xi = x|xi (2.2) and pˆ|pi = p|pi. (2.3) The two sets of eigenstates form two complete basis and satisfy orthogonality and completeness relations (what is stated for x below is also valid for p): hx0|xi = δ(x0 − x), (2.4)

4 Z dx|xihx| = 1. (2.5)

∂ For later purposes, let us also state that in the x basis, pˆ = −ih¯ ∂x , which together with the normalizing condition (2.4) for the momentum operator implies 1 hx|pi = √ eipx/¯h. (2.6) 2πh¯ Since the set of position (momentum) eigenstates form a complete basis, one can expand any state |ψi in terms of these;

Z |ψ(t)i = ψ(x, t)|xidx, (2.7) where ψ(x, t) = hx|ψ(t)i. In terms of the wave function ψ(x, t) the Schrödinger equation reads ∂ Hψ(x, t) = ih¯ ψ(x, t), (2.8) ∂t where h¯2 ∂2 H = − + V (x). (2.9) 2m ∂x2 In this project work, we will only deal with such time independent Hamil- tonians, for which the potential V is constant in time. It should be familiar from an elementary course in quantum mechanics, that the diﬀerential equation (2.8) is in these cases separable with respect to space and time variables. The separation constants En are the energy eigenvalues and the energy eigenstates associated with them, ψn(x) = hx|ψni satisfy the equation Ã ! h¯2 ∂2 − + V (x) ψ (x) = E ψ (x), (2.10) 2m ∂x2 n n n or in ket notation ˆ H|ψni = En|ψni, ψn(x) = hx|ψni. (2.11) The energy eigenstates form a complete basis and fulﬁll relations similar to (2.4) and (2.5). The solutions to the time part of (2.8) are exponential factors e−iEn(t−t0)/¯h. The general solution thus becomes X −iEn(t−t0)/¯h ψ(x, t) = cne ψn(x) (2.12) n

5 The complex coeﬃcients cn are determined by the initial conditions, i.e. the form of the wave function at t = t0. The orthogonality requirement on the energy eigenstates implies that Z ∗ cn = ψn(x)ψ(x, t0)dx. (2.13)

Thus, once we have found the energy eigenvalues and eigenfunctions, and know the wave function at some time t0, we can ﬁnd the wave function at any later time t using equation (2.12) and (2.13).

2.2 The propagator

In the previous section we used ordinary Schrödinger quantum mechanics to ﬁnd a general expression for a wave function ψ(x, t), presuming that we knew the wave function ψ(x, t0). We will now write down a diﬀerent expression for the time evolution of a wave function, introducing the so called propagator, which is the most important quantity of the path integral formulation. Start from equation (2.1) and multiply by a position eigenket from the left:

−iHˆ (t−t0)/¯h ψ(x, t) = hx|ψ(t)i = hx|e |ψ(t0)i (2.14)

We may also insert an identity operator (eq. (2.5)), yielding Z ˆ ψ(x, t) = hx|e−iH(t−t0)/¯h dx |x i hx |ψ(t )i = 0 0 | 0 {z 0 } ψ(x0,t0) Z Z −iHˆ (t−t0)/¯h = dx0hx|e |x0iψ(x0, t0) ≡ dx0 U(x, t; x0, t0)ψ(x0, t0), (2.15) where ψ(x0, t0) is the (known) wave function at t = t0.

The quantity

−iHˆ (t−t0)/¯h U(x, t; x0, t0) = hx|e |x0i (2.16) is the so called propagator. What is the meaning of this quantity? First of −iHˆ (t−t0)/¯h all, notice that e |x0i is the state of some object at time t, supposed −iHˆ (t−t0)/¯h it was in the state |x0i at time t0. Hence, hx|e |x0i = U(x, t; x0, t0) is the probability amplitude that we will find the object in state |xi at time t, if it was in |x0i at t0. To make this even more precise; If we make a measurement of a particle’s position at time t0 and find it to be x0, then the propagator tells us the probability amplitude for finding the particle at the

6 point x if we make a new measurement at a later time t. Looking at equation (2.14) we can thus interpret U(x, t; x0, t0) as something that "propagates" the wave function (of a particle) from the time t0 to the time t. We can write the propagator in terms of the energy eigenstates: Ã ! X −iHˆ (t−t0)/¯h U(x, t; x0, t0) = hx|e |ψnihψn| |x0i = n X −iEn(t−t0)/¯h = hx|ψnie hψn|x0i = n X ∗ −iEn(t−t0)/¯h = ψn(x)ψn(x0)e (2.17) n Substituting this into equation (2.15) one can easily verify that equation (2.10) is reproduced. So far we have only dealt with familiar Schrödinger quantum mechanics. The Schrödinger equation led us to the expression (2.17) for the propagator of a certain Hamiltonian, using the energy eigenstates. The question is: Could we do this in another way, without having to compute the energy eigenfunctions? The answer is (not very surprisingly) yes. In the following chapter we will show how a new expression for the propagator can be obtained, leading us into the "Theory of Path Integrals". It will turn out that the propagator can be written as Z U = Dx(t)eiS[x(t)]/¯h, (2.18) where S[x(t)] is the action, familiar from classical analytical mechanics, and R Dx(t) is the so called measure(this quantity will be explained later).

7 Chapter 3

Equivalence to the Schrödinger Equation

3.1 From the Schrödinger equation to PI’s

In this section we show that the Schrödinger equation leads to the expression for the propagator as written in equation (2.18), and interpret this in terms of path integrals. We start from the ordinary propagator in (2.16), and for simplicity we choose t0 = 0, t = T ;

−iHT/¯h U(x, T ; x0, 0) = hx|e |x0i (3.1)

Let us see what happens if we now divide the time interval T into two smaller parts, t1 and T − t1, and insert the identity operator as in equation (2.5).

−iHT/¯h −iH(T −t1)/¯h −iHt1/¯h U(x, T ; x0, 0) = hx|e |x0i = hx|e e |x0i = Z −iH(T −t1)/¯h −iHt1/¯h = hx|e dx1|x1ihx1|e |x0i = Z −iH(T −t1)/¯h −iHt1/¯h = dx1hx|e |x1ihx1|e |x0i = Z = dx1U(x, T ; x1, t1)U(x1, t1; x0, 0) (3.2)

What does this mean? Of course U(x, T ; x1, t1)U(x1, t1; x0, 0) is the probability amplitude for going from x0 at time t = 0 to x1 at time t1, and then from x1 to x, arriving at t = T . Integrating over all positions x1 we must get the amplitude for traveling from x0 to x, passing any arbitrary point, which

8 is exactly what we started with. This is logical! The particle has to be some- where at the time t1, and the quantum mechanical probability amplitudes for all diﬀerent ways to get between x0 and x add, as illustrated by the double slit experiment. What we have done, basically, is to add the amplitudes for every possible way to go from x0 to x in two steps. Let us now return to equation (3.1) and split the time interval T into N parts of equal size ², such that T = N²;

−iHT/¯h −iH²/¯h −iH²/¯h −iH²/¯h U = hx|e |x0i = hx| e| e {z ··· e } |x0i (3.3) Nfactors

Now between every factor of e−iH²/¯h we insert an identity operator (N-1 in total), in the same way we did above. We get

−iH²/¯h −iH²/¯h −iH²/¯h ×hx|e |xN−1ihxN−1|e |xN−2i · · · hx1|e |x0i ≡ Z

≡ dx1dx2 ··· dxN−1UxN ,xN−1 UxN−1,xN−2 ··· Ux1,x0 , (3.4)

iH²/¯h where we have identiﬁed U(xj+1, ²; xj, 0) = hxj+1|e |xji ≡ Uxj+1,xj as the probability amplitude for going from the point xj to the point xj+1 in the time interval ², and x ≡ xN . What does (3.4) mean? When we did the splitting into two time inter- vals in the beginning of this section, we ended up with the products of two propagators and an integral over the intermediate point x1, and concluded that we were adding the amplitudes for every possible way to go from x0 to x = xN in two steps. What is done here is exactly the same thing—the only diﬀerence is the number of steps taken between the two end points. Equation (3.4) thus means that we are adding the amplitudes for going from x0 at time t = 0 to x1 at time t = ² etc.... to x at t = T , for all possible combinations of x1, x2, ..., xN−1. i.e. we are summing over all "N-legged" paths, as illustrated in Figure 5.1. We can write this in a more compact way: X Z U = Upath → Upath, (3.5) paths

9 x x 3 x N-2 x 2 x N-1

x N x 1 . . . . x 0

ε (N-1)ε T t

Figure 3.1: Illustration of the "N-legged" paths. where the sum is taken to an integral in the limit N → ∞, ² → 0. In this limit it is clear that the summation over all N-legged paths goes to an integration of the probability amplitudes for every possible continuous path from the starting to the ending point (continous because in the limit ² → 0, −iH²/¯h Uxj+1,xj = hxj+1|e |xji → δ(xj+1 − xj)). This is why Feynman’s theory carries the name of path integrals! Still we are not ﬁnished. It remains to calculate the factors

−iH²/¯h 2 Uxj+1,xj = hxj+1|e |xji = hxj+1|(1 − iH²/h¯ + O(² ))|xji =

i² 2 = hxj+1|xji − hxj+1|H|xji +O(² ). (3.6) | {z } | h¯ {z } I II Here we have Taylor expanded the exponent with respect to ². The ﬁrst term above is just the delta function, according to (2.4). A suitable expression for the delta function to use here is: Z dpj I = δ(x − x ) = eipj (xj+1−xj )/¯h. (3.7) j+1 j 2πh¯ By inserting the unity operator in terms of the momentum eigenstates and using the hermiticity of the potential operator we can write the second term

Z Ã 2 ! i² dpj p = − j + V (x ) eipj (xj+1−xj )/¯h. (3.8) h¯ 2πh¯ 2m j+1

Here we have also used equation (2.6). For simplicity, we switch to

V (¯xj) = V ((xj+1 + xj)/2), (3.9) which will not make any diﬀerence as we take the limit N → ∞ 1. Returning to equation (3.5) this yields    Ã !  Z  2  dpj  i² pj  U = eipj (xj+1−xj )/¯h 1 − + V (¯x ) +O(²2) = xj+1,xj 2πh¯  h¯ 2m j  | {z }  ≈e−i²H(pj ,x¯j )/h¯ Z dpj = eipj (xj+1−xj )/¯he−i²H(pj ,x¯j )/¯h(1 + O(²2)) ≈ 2πh¯

Z Z p2 dpj i²(p x˙ −H(p ,x ))/¯h dpj i²(p x˙ − j −V (¯x ))/¯h ≈ e j j j j = e j j 2m j = 2πh¯ 2πh¯

Z p2 −i²V (¯x )/¯h dpj i²(p x˙ − j )/¯h = e j e j j 2m , (3.10) 2πh¯ where we have omitted terms of order ²2 and higher. Also we have approx- imated the derivative (xj+1 − xj)/² ≈ x˙ j. Again, taking the limit N → ∞, ² → 0 will make this expression exact. The integral over dpj can easily be evaluated. In general (for Re(a) ≥ 0), Z r 2 π 2 e−ax +bxdx = eb /4a (3.11) a 1This manipulation is not always allowed. For more complicated Hamiltonians the choice of x¯j might not arbitrary.

11 r −i²V (¯x )/¯h m i²mx˙ 2/2¯h ⇒ U(x , ²; x , 0) = U = e j e j = j+1 j xj+1,xj 2πi²h¯ r m i²(mx˙ 2/2−V (¯x ))/¯h = e j j . (3.12) 2πi²h¯ As we argued earlier, the propagator for one "path" is a product of these factors; ( ) N−1 N−1 r Y Y m 2 i²(mx˙ j /2−V (¯xj ))/¯h Upath = Uxj+1,xj = e = j=0 j=0 2πi²h¯

µ ¶N/2 P mx˙ 2 m i² N−1 j −V (¯x ) = e h¯ j=0 2 j , (3.13) 2πi²h¯ hence Z U = dx1 ··· dxN−1Upath =

µ ¶N/2 Z N−1 P mx˙ 2 m Y i² N−1 j h¯ 2 −V (¯xj ) = dxk e j=0 = 2πi²h¯ k=1

µ ¶N/2 Z N−1 P m Y i² N−1 h¯ L(x ˙ j ,x¯j ) = dxk e j=0 (3.14) 2πi²h¯ k=1 where we have identiﬁed the Lagrangian L = mx˙ 2 − V (x). 2 P In the limit N → ∞, ² → 0 the sum goes over to an integral as ² N−1 L(x ˙ , x¯ ) → R j=0 j j L(x ˙ j, x¯j)dt = S[x(t)]. Hence, we may write Z U = Dx(t)eiS[x(t)]/¯h, (3.15)

³ ´N/2 R m R QN−1 where the so called measure Dx(t) = 2πi²¯h k=1 dxk. We have thus derived equation (2.18). When calculating an explicit path integral, it is implicit that we are to use equation (3.14) and then take the limit N → ∞, ² → 0.

3.2 From PI’s to the Schrödinger equation

Let us now go the other way around. Starting from the time evolution of a general ket state and using the formula for the propagator we found in the last section, we will derive the Schrödinger equation.

12 As found in section 2.2,

ψ(x, t + ²) = hx| ψ(t + ²)i = hx| e−iH²/¯h| ψ(t)i =

Z = dx hx| e−iH²/¯h| x i hx | ψ(t)i = 0 | {z 0} | 0 {z } U(x,t+²;x0,t) ψ(x0,t) Z = dx0 U(x, t + ²; x0, t)ψ(x0, t). (3.16)

If we consider a very small time interval ², we can make use of equation (3.12); ( " #) µ m ¶1/2 i m(x − x )2 µx + x ¶ U(x, t + ²; x , t) = exp 0 − ²V 0 0 2πhi²¯ h¯ 2² 2

" # µ m ¶1/2 Z im(x − x )2 ⇒ ψ(x, t + ²) = dx exp 0 2πhi²¯ 0 2²h¯ · i² µx + x ¶¸ × exp − V 0 ψ(x , t). (3.17) h¯ 2 0 This is the time evolution according to the path integral theory. Let us now make the substitution η = x − x0, yielding

µ m ¶1/2 Z h i ψ(x, t + ²) = dη exp imη2/2¯h² 2πhi²¯ · i µ η ¶¸ × exp − ²V x + ψ(x + η, t) (3.18) h¯ 2 | {z } x0 Since ² is small, so must η be, and we can expand

dψ η2 d2ψ ψ(x + η, t) = ψ(x, t) + η + + ... (3.19) dx 2 dx2 and · i µ η ¶¸ i² µ η ¶ exp − ²V x + = 1 − V x + + ... = h¯ 2 h¯ 2 i² = 1 − V (x) + ..., (3.20) h¯

13 ³ ´ η where we also have performed a Taylor expansion of V x + 2 and excluded terms of order ²η and higher. Inserting the two expansions into equation (3.18) we thus have

µ m ¶1/2 Z h i ψ(x, t + ²) = exp imη2/2¯h² 2πhi²¯ " # i² ∂ψ η2 ∂2ψ × ψ(x, t) − V (x)ψ(x, t) + η + dη. h¯ ∂x 2 ∂x2

The integral over η can easily be evaluated to give

h²¯ ∂2ψ i² ψ(x, t + ²) = ψ(x, t) − − V (x)ψ(x, t) 2im ∂x2 h¯ " # h¯2 ∂2 ⇔ ih¯ [ψ(x, t + ²) − ψ(x, t)] /² = − + V (x) ψ(x, t). (3.21) 2m ∂x2

As ² → 0 the left hand side turns into a time derivative and we ﬁnally have " # ∂ h¯2 ∂2 ih¯ ψ(x, t) = − + V (x) ψ(x, t). (3.22) ∂t 2m ∂x2

We have derived the Schrödinger equation from the path integral formalism!

14 Chapter 4

Free Particle

We will now do an actual computation of a path integral, starting with the simplest example—the free particle. This is one of those cases where the PI calculation is quite cumbersome compared to that using common QM formalism. However, each step of the PI calculation is just as simple as the one using general quantum mechanics. The diﬀerence is that in the PI calculation we do the same step an inﬁnite number of times. If you want to compare the two calculations, you can have a look in appendix where the problem is solved using ordinary quantum mechanics. With V (x) = 0, we get from equation (3.14):

µ ¶N/2 Z P m i²m N−1 x˙ 2 U = dx dx ··· dx e 2¯h j=0 j = 2πi²h¯ 1 2 N−1

µ ¶N/2 Z P m im N−1(x −x )2 = dx dx ··· dx e 2²h¯ j=0 j+1 j . (4.1) 2πi²h¯ 1 2 N−1

³ ´1/2 m Here it is convenient to change variables. Let yj = 2²¯h xj ⇔ ³ ´1/2 2²¯h xj = m yj, yielding Ã ! µ ¶N/2 (N−1)/2 Z P m 2²h¯ i N−1(y −y )2 U = dy dy ··· dy e j=0 j+1 j (4.2) 2πi²h¯ m 1 2 N−1 | {z } ≡A Now, these integrals are coupled and the expression looks a bit messy. But fortunately there is a pattern to discover if one computes one integral at a time. Starting with the integration over dy1 we get: Z Z 2 2 2 2 2 i[(y1−y0) +(y2−y1) ] i[2y −(2y0+2y2)y1+y +y ] dy1 e = dy1 e 1 0 2 (4.3)

15 Completing the square with respect to y1 in the exponent yields

Z Z 2 2 2 2 2 1 y0+y2 2 y0+y2 2 y0+y2 i[2y −(2y0+2y2)y1+y +y ] − [(y1− ) +( ) + ]= dy1 e 1 0 2 = dy1 e 2i 2 2 2 s s y2+y2 (y2+y2+2y y ) πi − 2 [ 0 2 − 0 2 0 2 ] πi i (y −y )2 = e i 2 4 = e 2 2 0 , (4.4) 2 2 q R −ax2 π where we used that e dx = a . Inserting this into the integral over dy2 and performing the integral in the same way as above, we get s s Z Z y2 2y2 πi i (y −y )2+i(y −y )2 πi 3i [y2− 2 (y +2y )y + 0 + 3 ] dy e 2 2 0 3 2 = dy e 2 2 3 0 3 2 3 3 = 2 2 2 2 s Z y2 2y2 πi 3i [(y − 1 (y +2y ))2+ 0 + 3 − 1 (y +2y )2] = dy e 2 2 3 0 3 3 3 9 0 3 = 2 2 s s s y3 2y2 2 πi 2πi 3i [ 0 + 3 − 1 (y2+4y2+4y y )] (πi) i (y −y )2 = e 2 2 3 9 0 3 0 3 = e 3 3 0 .(4.5) 2 3 2 We now realize that if we compute the N − 1 integrals in this way, we will get Z P N−1 2 i (yj+1−yj ) U = A dy1dy2 ··· dyN−1 e j=0 =

Ã ! N−1 1/2 (πi) i (y −y )2 = A · e N N 0 = N Ã ! Ã ! µ ¶N/2 (N−1)/2 N−1 1/2 m 2²h¯ (πi) 2 = ei(yN −y0) /N = 2πi²h¯ m N

µ ¶1/2 m 2 = ei(yN −y0) /N . (4.6) 2πiN²h¯

Changing back from yi to xi we ﬁnally arrive at

µ ¶1/2 r m im (x −x )2 m im(x −x )2/2T ¯h U(x ,T ; x , 0) = e 2N²h¯ N 0 = e N 0 .(4.7) N 0 2πiN²h¯ 2πhiT¯ Note that the phase of the total propagator here is equal to the phase of the classical trajectory only, which in the case of a free particle corresponds to the action for a straight line. Note also that the limit N → ∞, ² → 0 was very easy taking in this example, since N² = T .

16 Chapter 5

The Classical Limit

The path integral expression for the propagator oﬀers a nice way of obtaining the classical limit of quantum mechanics, although at ﬁrst glance there is no way to distinguish between the classical and the quantum mechanical paths. Have a look at equation (3.15) again: Z U = Dq(t)eiS[q(t)]/¯h

Since |eiγ| = 1 for all γ, every path essentially gives the same contribution to the probability amplitude—in this sense the classical path is no more or no less important than any other path of the particle. However, when integrating, the phases turn out to be crucial. The phase S[x(t)]/h¯ comes with the very small number h¯ in the denominator, so even a quite small change in S[x(t)] for diﬀerent paths will cause great variations in the phase. When the amplitudes for such paths are added, they tend to cancel each- other and cause destructive interference. For macroscopic (classical) particles, S is very large compared to h¯. So even a small (but still macroscopic) change in the action for such a particle will cause great changes in the phase. For such particles, many path amplitudes cancel each-other, making them unimportant for the particles’ way of motion. What we actually discover in reality is that only one certain path is followed—the classical trajectory. Why is this the only path which amplitude is not canceled? The answer is found in the classical theory. From classical mechanics, we know the "principle of least action". The R t2 principle states that the action S[x(t)] = t1 L(x, x˙)dt has a minimum (or sometimes a maximum) for the classical path xcl(t), i.e. δS[xcl(t)] = 0. Thus, a small deviation from the classical path does only cause inﬁnitesimal variations in S[x(t)] and therefore in the phase. The probability amplitudes for the paths extremely close to the classical trajectory thus tend to add

17 up and cause constructive interference! For the paths deviating from the classical one, the change in the action (i.e. the phase) will be so large in terms of h¯ that the interference will be destructive instead. It follows that for macroscopic particles only the classical trajectory will contribute to the motion. Paths interfere x x constructively cl

Paths interfere destructively

Figure 5.1: Emergence of the classical limit. There is constructive interference between paths close to the classical trajectory (thick line) whereas the trajectories further away tend to cancel because of the rapidly changing sign of eiS[x(t)]/¯h.

For microscopic (quantum mechanical) particles on the other hand, the action is typically of the order of h¯. Hence it takes a comparably large change in the action to achieve a significant change in the phase S[x(t)]/h¯. In other words, for very light particles even paths that deviate much from the classical will be of importance. In this case one can no longer talk about the trajectory of the particle, but rather of a superposition of different paths. In the limit h¯ → 0, even very small particles will obviously behave like classical particles, since their action will then become large compared to h¯. The limit h¯ → 0 is also the limit where the quantization of energy etc dis- appears, which is another effect of considering the classical limit of quantum mechanics.

18 Chapter 6

Harmonic Oscillator

Here we show another example of an explicit calculation of a propagator, this time for the harmonic oscillator. I have tried to make the calculation as clear as possible, but one still might have to revive some old knowledge on linear algebra. The reader will perhaps feel comforted by the fact that there is another, much quicker, way to do this derivation, based on Fourier analysis. This method is used e.g. in the book by Feynman and Hibbs, see [7]. However, the approach we have here has the advantage that we need not know the classical trajectory. Again, we start from equation (3.14). In order to make the calculations 2 2 2 xj+1+xj+1xj +xj simpler, we deﬁne x¯j such that x¯j = 3 . We then have the potential mw2 mw2 (x2 + x x + x2) V (¯x ) = x¯2 = j+1 j+1 j j (6.1) j 2 j 2 3 and hence the propagator

µ ¶N/2 Z N−1 P mx˙ 2 m Y i² N−1 j h¯ 2 −V (¯xj ) U = dxk e j=0 = 2πi²h¯ k=1 µ m ¶N/2 Z NY−1 = dxk 2πi²h¯ k=1 ( " #) NX−1 2 2 im (xj+1 − xj) ω ² 2 2 × exp − (xj+1 + xj+1xj + xj ) . (6.2) 2¯h j=0 ² 3

Once again we come across coupled integrals. The trick to use here is to express the sum in the exponent in terms of the row vector x ≡ (x0, x1, ...xN )

19 and the matrix M, such that   1 ω2²  M0,0 = MN,N = ² − 3    ³ ´  1 ω2²  Mk,k = 2 ² − 3 ³ ´ (6.3)  2  1 ω ²  Mk,k+1 = Mk−1,k = − ² + 6     all other elements = 0 for 1 ≤ k ≤ N − 1. With these deﬁnitions we write the sum in the exponent as " # NX−1 2 2 (xj+1 − xj) ω ² 2 2 − (xj+1 + xj+1xj + xj ) = j=0 ² 3

NX−1 t = xMx = xiMijxj, (6.4) i,j=0 which inserted in (6.2) yields

µ ¶N/2 Z NY−1 m i² xMxt U = dxk e h¯ . (6.5) 2πi²h¯ k=1 Now, let N be the (N − 1) × (N − 1) matrix one gets by removing all elements from M containing indices 0 and N, i.e. the ﬁrst and last row and the ﬁrst and last column (the reason will become clear later!). We then have

t 2 2 xMx = M00x0 + MNN xN + M01x0x1 + M10x1x0 +

NX−1 +MN−1,N xN−1xN + MN,N−1xN xN−1 + xjNjlxl = j,l=1

NX−1 2 2 = M00(x0 + xN ) + 2M01(x0x1 + xN−1xN ) + xjNjlxl, (6.6) j,l=1 where we have used that M is totally symmetric. We thus have that the propagator is µ m ¶N/2 Z NY−1 U = dxk 2πi²h¯ k=1 (   ) im NX−1  2 2  × exp M00(x0 + xN ) + 2M01(x0x1 + xN−1xN ) + xjNjlxl . (6.7) 2¯h j,l=1

20 The ﬁrst two terms in the exponent are independent of the integration variables, so we can take them out from the integral: ( ) µ m ¶N/2 im U = exp M (x2 + x2 ) 2πi²h¯ 2¯h 00 0 N (   ) Z NY−1 im NX−1   × dxk exp 2M01(x0x1 + xN−1xN ) + xjNjlxl . (6.8) k=1 2¯h j,l=1

The integral above contains terms both linear and quadratic in xj in the 0 exponent. It is therefore very useful to deﬁne x = (x1, x2, ...xN−1) and to change variables in the following way. We know that the matrix N is hermitian, so it must be diagonalizable with eigenvalues ni and (orthonormal) eigenvectors ni , 1 ≤ i ≤ N − 1. We can thus deﬁne a matrix O, which contains the eigenvectors of N;

O = (n1 n2 ... nN−1). (6.9)

Here ni are column vectors. We now switch variables (remember that x is deﬁned as a row vector!) to  0 t t 0  x = qO ⇔ qO O = x O = q (6.10)   x0t = Oqt ⇔ OtOqt = Otx0t = qt, where we have used that the eigenvectors ni are orthonormal, which implies OtO = 1. We have ( ) µ m ¶N/2 im U = exp M (x2 + x2 ) 2πi²h¯ 2¯h 00 0 N   (   ) Z N−1  N−1  Y im  X  × dx exp 2M (x x + x x ) + x N x  = k  01 0 1 N−1 N j jl l k=1 2¯h  j,l=1  | {z } x0Nx0t ( ) µ m ¶N/2 im Z NY−1 = | det(O)| exp M (x2 + x2 ) dq | {z } 00 0 N k 2πi²h¯ 2¯h k=1 =1 (   ) NX−1 im t t × exp 2M01 (x0qjO1j + qjON−1,jxN ) + qO NOq  . (6.11) 2¯h j=1

21 The proof of | det(O)| = 1 can be found in the appendix. Now because O contains the eigenvectors of N, we also have

NO = N(n1 n2 ... nN−1) = (n1n1 n2n2 ... nN−1nN−1). (6.12)

The diagonal matrix P contains the eigenvalues of N and is obtained from

  n1    n   2    t  .  P ≡ O NO =   (n1n1 n2n2 ... nN−1nN−1) =  .     .  nN−1   n1 0 ... 0    0 n ... 0   2  =  . . . .  . (6.13)  . . .. .  0 0 . . . nN−1

Inserting this expression into equation (6.11) yields ( ) µ m ¶N/2 im U = exp M (x2 + x2 ) 2πi²h¯ 2¯h 00 0 N ( ) Z NY−1 im NX−1 × dqk exp [2M01qj(O1jx0 + ON−1,jxN ) + qjPjlql] = k=1 2¯h j,l=1 ( ) µ m ¶N/2 im = exp M (x2 + x2 ) 2πi²h¯ 2¯h 00 0 N ( ) Z NY−1 NX−1 h i im 2 × dqk exp 2M01qj(O1jx0 + ON−1,jxN ) + njqj .(6.14) k=1 2¯h j=1

We have achieved to make the integrals uncouple, so now we can complete the square for each qj in the exponent! Rewriting the terms in the sum over j,

2 2M01qj(O1jx0 + ON−1,jxN ) + njqj =

Ã !  M 2 M 2  01 01 2 = nj qj + (O1jx0 + ON−1,jxN ) − 2 (O1jx0 + ON−1,jxN ) (6.15). nj nj

22 Hence, we have ( ) µ m ¶N/2 im U = exp M (x2 + x2 ) 2πi²h¯ 2¯h 00 0 N ( ) NY−1 2 −imM01 2 × exp (O1kx0 + ON−1,kxN ) k=1 2¯hnk ( ) Z µ ¶2 imnk M01 × dqk exp qk + (O1kx0 + ON−1,kxN ) (6.16) 2¯h nk

The product over k here concerns of course still the integrals on the last line. The integrals above are clearly a product of N −1 gaussian integrals which we can solve directly. Keeping in mind that x0 and xN are not integrated over, ( ) N−1 Z µ ¶2 Y imnk M01 dqk exp qk + (O1kx0 + ON−1,kxN ) = k=1 2¯h nk Ã ! Ã ! Ã ! NY−1 2πih¯ 1/2 2πih¯ (N−1)/2 1 1/2 = = QN−1 = k=1 mnk m k=1 nk Ã ! Ã ! 2πih¯ (N−1)/2 2πih¯ (N−1)/2 = (det(P))−1/2 = (det(N))−1/2 . (6.17) m m

It is fairly easy to show that det(P) = det(N). The proof of this relation is done in the appendix. What about the second exponential factor in (6.16)? It can be related to the inverse N−1 of the matrix N by using that

1 −1 OkjOlj = Nkl , (6.18) nj which is proved in the appendix. Thus we rewrite ( ) NX−1 2 −imM01 2 exp (O1kx0 + ON−1,kxN ) = k=1 2¯hnk ( ) NX−1 2 ³ ´ −imM01 2 2 2 2 = exp O1kx0 + ON−1,kxN + 2O1kON−1,kx0xN = k=1 2¯hnk ( ) −imM 2 ³ ´ = exp 01 N −1x2 + N −1 x2 + 2N −1 x x . (6.19) 2¯h 11 0 N−1,N−1 N 1,N−1 0 N

23 This is almost everything we need to do. To conclude, so far we have seen that ( ) µ m ¶N/2 im U = exp M (x2 + x2 ) 2πi²h¯ 2¯h 00 0 N ( ) −imM 2 ³ ´ × exp 01 N −1x2 + N −1 x2 + 2N −1 x x 2¯h 11 0 N−1,N−1 N 1,N−1 0 N Ã ! 2πih¯ (N−1)/2 × (det(N))−1/2 = m

µ m ¶1/2 µ T ¶−N/2 = (det(N))−1/2 2πih¯ N ( ) im h³ ´ ³ ´ i × exp x2 + x2 M − M 2 N −1 − 2x x M 2 N −1 , (6.20) 2¯h 0 N 00 01 11 0 N 01 1,N−1

−1 −1 where we have used that NN−1,N−1 = N11 (see the appendix). M00 and M01 are known, see equation (6.3), but we still have to determine −1 −1 −1 det(N) and the elements N11 and N1,N−1 of the inverse N of N. The detailed steps of these derivations are put in the appendix, and here we only write down the answers. For large N:s we ﬁnd Ã ! sin ωT N ω2T N−1 det(N) ≈ N − , (6.21) ωT T 3N

T µ ωT ¶ N −1 ≈ 1 − cot ωT (6.22) 11 N N and ωT 2 N −1 ≈ . (6.23) 1,N−1 N 2 sin ωT Finally, using (6.3) with ² = T/N, as N → ∞

U(xN ,T ; x0, 0) =

( ) µ −imω ¶1/2 imω ·³ ´ 2x x ¸ = exp x2 + x2 cot ωT − 0 N , (6.24) 2πh¯ sin ωT 2¯h 0 N sin ωT which is the desired result.

24 Chapter 7

Imaginary time and statistical mechanics

An interesting and useful ﬁeld of application for path integrals is statistical mechanics. This is based on the insight that the partition function Z can be written as a propagator with an imaginary time argument. Ability to calculate path integrals is then very proﬁtable since the partition function contains all thermodynamic information about the system, i.e. statistical features like the expectation value of the energy.

7.1 General considerations

For the case where the energy eigenvalues are known, i.e. when the Hamil- tonian has been diagonalized, one gets the partition function by summing the Boltzmann factors, e−βE, for all system states1. In general, given the Hamiltonian, Hˆ , of a system its partition function becomes

Z = Tr e−βHˆ . (7.1)

In particular, if there is a continuous parameter x that labels all states, we can write Z Z = dxhx|e−βHˆ |xi. (7.2)

Hence Z can be thought of as the sum of the diagonal elements of the operator U;

−βHˆ Uij = hxi|e |xji. (7.3)

1 Here β = 1/kBT where kB is the Boltzmann constant and T is the temperature.

25 That we called this operator U is of course no coincidence—by comparing with the definition of the propagator (2.16) we see that this is exactly the 2 same if we make the identification β = i(tf − t0)/h¯. This is very interesting because it allows us to think of the (classical) statistical mechanics system as a dynamical quantum mechanics system evolving in imaginary time! Clearly, its conversion holds as well. We can actually make this connection more precise by explicitly changing variables into imaginary time (this is called a Wick rotation): dx dx dt dx τ = i(t − t ) ⇒ = = −i . (7.4) 0 dτ dt dτ dt We then find that the action goes over into

Z Ã !2 Z Ã !2 tf m dx ¯hβ m dx − V (x(t)) dt → −i + V (x(τ)) dτ. (7.5) t0 2 dt 0 2 dτ

We note that going to imaginary time causes a relative sign shift (with respect to the kinetic term) of the potential. This provides a thrilling perspective on the imaginary time approach—the particles can be thought of as moving in an inverted potential −V (x) when propagating in imaginary time!

7.2 Application to the harmonic oscillator

Now we go on and use the above to write an expression for the partition function in terms of path integrals Z Z = dxhx|e−βHˆ |xi =

Z Z x(τ=β¯h)=x R hβ¯ 2 − 1 m ( dx ) +V (x(τ))dτ = dx Dx(τ)e h¯ 0 2 dτ . (7.6) x(τ=0)=x The endpoints x(τ = 0) and x(τ = βh¯) are equal according to (7.2)—here we denote them x. The partition function is thus a sum over all paths of period βh¯. We will now use this result to calculate the partition function, and consequently the energy eigenvalues of the harmonic oscillator. According to the above, Z Z = dxU(x, −ihβ¯ ; x, 0) =

2 Here we denote the end time as tf .

26 Ã ! ( ) Z mω 1/2 −mω h i = dx exp x2 cos(−iβhω¯ ) − x2 = 2πhi¯ sin(−iβhω¯ ) hi¯ sin(−iβhω¯ ) Ã ! ( ) mω 1/2 Z −mωx2 = dx exp [cosh(βhω¯ ) − 1] = 2πh¯ sinh(βhω¯ ) h¯ sinh(βhω¯ )

Ã !  1/2 mω 1/2 π   = mω = 2πh¯ sinh(βhω¯ ) ¯h sinh(β¯hω) [cosh(βhω¯ ) − 1]

= [2(cosh(βhω¯ )) − 1]−1/2 (7.7).

Let us rewrite this expression a little!

Z = [2(cosh(βhω¯ )) − 1]−1/2 =

h i−1/2 h i−1/2 = eβ¯hω + e−β¯hω − 2 = eβ¯hω(1 + e−2β¯hω − 2e−β¯hω) =

h i−1/2 e−β¯hω/2 = eβ¯hω(1 − e−β¯hω)2 = = 1 − e−β¯hω

X∞ X∞ = e−β¯hω/2 (e−β¯hω)j = e−β(j+1/2)¯hω. (7.8) j=0 j=0

Comparing this with the deﬁnition of the partition function,

X∞ Z = e−βEj (7.9) j=0

ﬁnally yields the familiar result

Ej = (j + 1/2)¯hω. (7.10)

27 Chapter 8

Appendix

8.1 Free particle the easy way

Here the free particle problem is solved using ordinary QM techniques. We want to ﬁnd the propagator directly from the general expression (2.16). Again we let t = T , t0 = 0. With V (ˆx) = 0 we get:

−iHT/ˆ ¯h U(x, T ; x0, 0) = hx, T | e |x0, 0i. (8.1)

Inserting the identity operator in terms of the momentum eigenkets yields

−ipˆ2T/2m¯h U = hx, T | e |x0, 0i = Z = dp hx, T | e−ipˆ2T/2m¯h|pi hp|x , 0i = | {z0 } √ 1 eipx0/h¯ 2πh¯ Z 1 − iT p2+ip(x −x)/¯h = dp e 2mh¯ 0 , (8.2) 2πh¯ where we have used equation (2.6). This kind of gaussian integral has ap- peared several times earlier in this work. Completing the square or simply using equation (3.11) yields s 2 r 1 π − (x0−x) m m im(x−x )2/2T ¯h U = e 2¯hiT = e 0 . (8.3) 2πh¯ iT/2mh¯ 2πhiT¯

28 8.2 Details on the computation of the harmonic oscillator

8.2.1 Why det(P) = det(N), | det(O)| = 1 The proof of det(P) = det(N) is very simple and straight-forward. Here it is: det(P) = det(OtNO) = det(Ot) det(N) det(O) =

t = det(N) det(|O{zO}) = det(N) (8.4) =1 We now use this result to show that | det(O)| = 1: det(N) = det(P) = det(Ot) det(N) det(O) =

= det(O) det(N) det(O) = (det(O))2 det(N). (8.5) We conclude that det(O) = ±1

⇒ | det(O)| = 1. (8.6)

8.2.2 Calculation of det(N) From equation (6.3) and the deﬁnition of N we have  ³ ´  N ω2t  Nk,k = 2 t − 3N    ³ ´ N ω2t (8.7)  Nk,k+1 = Nk−1,k = − t + 6N     all other elements = 0, ³ ´ 1 1+ω2t2/6N 2 or with a = 2 1−ω2t2/3N 2 :  ³ ´  N ω2t  Nk,k = 2 t − 3N    ³ ´ N ω2t (8.8)  Nk,k+1 = Nk−1,k = −2a t − 3N     all other elements = 0.

29 This implies   1 −a 0 0 ... 0 Ã !   2 N−1  −a 1 −a 0 ... 0  N ω t   det(N) = 2N−1 − det  . .  ,(8.9) N−1  .. .  t 3N  0 −a 1 −a .  ...... where detN−1 denotes that it is a determinant of a (N − 1) × (N − 1) matrix. Note the symmetry of the matrix above and let AN−1 be detN−1 of that kind of matrix, i.e. Ã ! N ω2t N−1 det(N) = 2N−1 − A . (8.10) t 3N N−1

By starting to evaluate the determinant, beginning with the ﬁrst column, we get   −a 0 0 0 ... 0    −a 1 −a 0 ... 0    A = A + a det  . .  = N−1 N−2 N−2  .. .   0 −a 1 −a .  ......

2 = AN−2 + a(−a)AN−3 = AN−2 − a AN−3, (8.11) where we in the last step evaluated the (N − 2)-determinant along the ﬁrst row. To conclude we have

2 AN−1 = AN−2 − a AN−3 (8.12) and we also require

A0 = A1 = 1. (8.13)

One can easily convince oneself that this recursion relation can be written

Ã ! Ã !N−1 Ã ! A 1 −a2 1 N = . (8.14) AN−1 1 0 1

This problem will become greatly simpliﬁed if we diagonalise the matrix. In the basis of the eigenvectors of the matrix, the equation looks like Ã ! Ã !Ã ! N−1 λ−AN − AN−1 λ+ 0 λ− − 1 = N−1 (8.15) AN−1 − λ+AN λ− 1 − λ+

30 where the eigenvalues √ 1 i 4a2 − 1 1 λ = ± ≈ e±iωt/N . (8.16) ± 2 2 2 Here we have performed a Taylor expansion for large N’s using the expression for a as stated in the beginning of this subsection. Equation (8.15) can now be solved for AN−1, yielding

N N λ− − λ+ N sin ωt AN−1 = ≈ N−1 . (8.17) λ− − λ+ 2 ωt In the last step we have used equation (8.16) and made a Taylor expansion for large values of N. Inserting this into (8.10) immediately yields equation (6.21).

8.2.3 Calculation of elements of N−1 First we prove equation(6.18): 1 1 Nmk OkjOlj = NmkOkj Olj = OmjOlj = δml, (8.18) nj nj | {z } nj Omj where we in both steps have used that O contains the orthonormal eigenvectors of N. To ﬁnd the elements of the inverse, we use the formula (−1)j+kD N −1 = jk , (8.19) jk det(N) where Djk is the determinant of the matrix you get after erasing the j:th row and the k:th column of N. We ﬁrst restate: Ã ! N ω2t N−1 det(N) = 2N−1 − A . (8.20) t 3N N−1

−1 To get D11 (and later N1,1 ) we erase the ﬁrst row and column from N. The determinant then becomes Ã ! N ω2t N−2 D = 2N−2 − A . (8.21) 11 t 3N N−2 Equation (8.19) consequently yields t A N −1 = ³ ´ N−2 . (8.22) 11 ω2t AN−1 2N 1 − 3N 2

31 Using (8.17), that sin(α+β) = cos α sin β +cos β sin α and Taylor expanding gives T µ ωT ¶ N −1 ≈ 1 − cot ωT . (8.23) 11 N N Looking at equation (8.9) again, one can see that Ã ! N ω2t N−2 D = 2N−2 − × 1,N−1 t 3N   −a 1 −a 0 0 ... 0    0 −a 1 −a 0 ... 0    ×det  . .  = N−2  .. .   0 0 −a 1 −a .  ...... Ã ! N ω2t N−2 = 2N−2 − (−a)N−2, (8.24) t 3N since the matrix is triangular. Thus,

(−1)1+N−1D N −1 = 1,N−1 = 1,N−1 det(N) ³ ´ 2 N−2 (−1)N 2N−2 N − ω t (−a)N−2 aN−2t t 3N ³ ´ = ³ ´N−1 = 2 2 . (8.25) N ω2t ω t N−1 2N 1 − 2 AN−1 2 t − 3N AN−1 3N Taylor expansion of this expression ﬁnally yields

ωT 2 N −1 ≈ . (8.26) 1,N−1 N 2 sin ωT

32 Bibliography

[1] R. MacKenzie, Path Integral Methods and Applications, quant- ph/0004090, (2000).

[2] J.J. Sakurai, Modern Quantum Mechanics, (1994).

[3] J.W. Negele and H. Orland, Quantum Many-Particle Systems, (1998).

[4] C. Itzykson and J.B. Zuber, Quantum Field Theory, (1980).

[5] F. Hassan, Notes for the Quant. Mech.-I reading project, unpublished.

[6] R. Shankar, Principles of Quantum Mechanics, (1994).

[7] R.P. Feynman and A.R. Hibbs, Quantum Mechanics and Path Integrals, (1965).