3. the Wave Function

3. The wave function

[Last revised: Saturday 28th November, 2020, 12:53]

115

Continuous spectrum

So far we have assumed observables with a discrete spectrum of possible values. • This is in particular the case of Sz.

B But there are observables, like the position or the momentum of a free particle, that have a continuum of possible values. As a consequence of postulate III, we will need an operator with a continuous set of eigenvalues. Then the operator will be acting on a Hilbert space of inﬁnite dimension.

B As we will see in this chapter, most of the results that we have already obtained can be generalized to a continuous spectrum.

116 Continuous spectrum

Let us revise ﬁrst the concept of orthonormal basis. • B If A is an operator with a (non-degenerate) discrete spectrum, A a = a a . | i | i Then a is an orthonormal basis, {| i}

1,a = a0 a a0 = d (Krönecker delta), d = aa0 aa0 8 0,a = a0 ⌦ ↵ < 6 B Consider now a generic operator X with a continuous: spectrum, X x = x x . | i | i Then x is an orthonormal basis in the following sense, {| i}

• , x = x0 (!!) x x = d(x x ) (Dirac delta), d(x x )= 0 0 0 8 0,x = x0 ⌦ ↵ < 6 :

117

Continuous spectrum

B The Dirac delta is strictly not a function, it is a distribution that assigns a complex number to each smooth function in x = x0 constrained to satisfy

• x0+e dx d(x x0) f (x)= dx d(x x0) f (x)= f (x0), ˆ • ˆx e 0 equivalently, a distribution that fulﬁlls

• dx d(x x0)=1, d(x)=0, x = 0. ˆ • 8 6 From now on, deﬁnite integrals extend from • to • unless otherwise stated.

118 Continuous spectrum

The Dirac delta can be otained as the limit when L • or e 0+ of the B ! ! functions:

L 1 ikx sin Lx (a) dL(x)= dk e = 2p ˆ L px

1 x2/(2e2) (b) de(x)= e (Gaussian) (2pe2)1/2

e/p (c) de(x)= (Breit-Wigner) x2 + e2

e e q(x + 2 ) q(x 2 ) dq(x) (d) de(x)= e ! dx q(x)=Heaviside step function

119

Continuous spectrum

For continuous spectra the completeness relation is generalized as follows • A : Â a a = I, X : dx x x = I. a | ih | ˆ | ih |

B Any vector in the Hilbert space of an operator A can be expressed in the basis of eigenvectors,

A : y = Â a a y Â a ca | i a | ih | i ⌘ a | i where c is the component of y along a . a | i | i B In the case of X with a continuous spectrum,

X : y = dx x x y dx x y(x) | i ˆ | ih | i ⌘ ˆ | i where y(x)= x y is the wave function of y , analogous to the component of h | i | i y along x . | i | i 120 Continuous spectrum

Notice that the vectors in the Hilbert space of A must be normalized, • 2 y y = Â y a a y = Â c⇤a ca = Â ca = 1. h | i a h | ih | i a a | | Likewise,

2 y y = dx y x x y = dx y⇤(x)y(x)= dx y(x) = 1. h | i ˆ h | ih | i ˆ ˆ | |

This requires y(x) L2(R),asquare-integrable function in R. 2

121

Continuous spectrum

B The probability to obtain an eigenvalue a of the observable A on the state y is | i

p = y P y = y a a y a h | A,a | i h | Â | iih i| | i ai=a ! 2 = Â ai y i | h | i | for the general case of a degenerate eigenvalue.

B Analogously, the probability to obtain a value x of the observable X in the interval [x , x ] on the state y is 0 1 | i x1 p = y P y = y dx x x y [x0,x1] h | X,[x0,x1] | i h | ˆ | ih | | i ✓ x0 ◆ x1 = dx y(x) 2. ˆ x0 | |

Hence, y(x) 2 is the probability density to obtain a value x of X that gives the B | | probability to obtain x in the interval [x, x + dx].

122 Continuous spectrum

Finally, the braket of two states •

f y = dx f x x y = dx f⇤(x)y(x) h | i ˆ h | ih | i ˆ is the overlap between their wave functions.

B And for an arbitrary function F = F(X),

F(X) x = F(x) x , | i | i the matrix element

f F(X) y = dx f F(X) x x y = dxF(x) f x x y h | | i ˆ h | | ih | i ˆ h | ih | i = dxF(x)f (x)y(x). ˆ ⇤

123

Position representation

The position of a particle will be given by the eigenvalue of the position operator, •

~X =(X1, X2, X3)=(X, Y, Z).

The (inﬁnite-dimensional) orthonormal basis of eigenvectors of ~X is ~x where {| i} ~x labels every point in the 3-dimensional space,

3 ~x ~x0 = d (~x ~x0)=d(x x0)d(y y0)d(z z0) (orthonormality) 3 ⌦ ↵ d x ~x ~x = I (closure). ˆ | ih |

B The wave function in this basis is called the position representation,

~x y y(~x)=y(x, y, z). h | i ⌘ The normalization of the physical states is expressed as

3 3 3 2 1 = y y = d x y ~x ~x y = d x y⇤(~x)y(~x)= d x y(~x) . h | i ˆ h | ih | i ˆ ˆ | |

124 Position representation

The Hilbert space of physical states is formed by the square-integrable functions • in R3, L2(R3). It is remarkable that, in particular, the wave function of ~x , | 0i 3 ~x ~x0 = d (~x ~x0) ⌦ ↵ is not a square-integrable functionand hence does not belong to = L2(R3). H In order to incorporate it, one must generalize to include distributions. B H This is the so called rigged or equipped Hilbert space, introduced to account for the continuous spectrum, as was done implicitly in the previous section.

B Furthermore, we should not be concerned that a state described by a delta function cannot be normalized, because it is not physical (but a limiting case), since one cannot measure the position ~x with inﬁnite precision. Let us discuss this below.

125

Position representation

B In practice, when we measure the position X of the state y , it collapses to a state | i that is actually a superposition of a continuum of eigenstates of X in the interval [x D/2, x + D/2] where D is a narrow (but non zero) range around x that our detector cannot resolve, x+D/2 y dx0 x0 x0 y . | i ! ˆx D/2 ↵⌦ ↵ These can be considered as eigenvectors of a common eigenvalue x, analogous to the case of a degenerate eigenvalue a of an operator A with a discrete spectrum,

y a a y p = a y 2, | i ! Â | iih i | i ) a Â | h i | i | ai i assuming that x y does not appreciably change within an inﬁnitesimal D h | i x+D/2 2 2 p(x)= dx0 x0 y = D x y . ) ˆx D/2 | | | h | i | ⌦ ↵ y(~x) is the amplitude and y(~x) 2 is the probability density to ﬁnd a particle at ~x. B | | 126 Position representation

Let us now introduce the operator that produces a space displacement • (translation) from the position ~x to ~x + x~0:

T(x~ ) ~x = ~x + x~ . 0 | i | 0i There is a continuous set of translations, ~x R3, that can be composed 0 2 (multiplied),

T(~x1)T(~x2)=T(~x1 + ~x2).

They have the mathematical structure of a Lie group called T3.

127

Position representation

Translations are represented by unitary transformations acting on , B H

~x T(~x ) ~x , ~x0 T(~x ) ~x0 | i ! 0 | i ! 0 † ↵ ↵ ~x T (~x0)T(~x0) ~x0 = ~x + ~x0 ~x0 + ~x0 h | 3 3 ↵ = ⌦d (~x + ~x ~x0 ↵ ~x )=d (~x ~x0)= ~x ~x0 0 0 † 1 T (~x )=T (~x )=T( ~x ). ⌦ ↵ ) 0 0 0 B The elements of a Lie group can be written in terms of the generators. Consider ﬁrst the generator of translations in R,

T(x ) x = x + x . 0 | i | 0i We deﬁne the generator K as the operator such that

T(dx) I idxK ⌘ for an inﬁnitesimal translation dx.

128 Position representation

B K is self-adjoint, 1 † † † T (dx)=I + idxK, T (dx)=I + idxK K = K, ) and satisﬁes the differential equation dT T(x + dx)=T(x)+ dx dx T(x + dx)=T(x)T(dx)=T(x) (I idxK) = T(x) idxT(x)K dT = iTK ) dx with the boundary condition T(0)=I. Hence,

iKx T(x) = e 1 = I iKx + ( iKx)2 + ... 2! x N = lim I iK . N • N ! ⇣ ⌘ 129

Position representation

A ﬁnite translation x is the composition of N • inﬁnitesimal translations x/N B ! generated by K. The displacement x is the parameter of this one-dimensional Lie group.

Next we will investigate how K commutes with the position operator in one • dimension:

XT(dx) x = X x + dx =(x + dx) x + dx , | i | i | i T(dx) X x = T(dx)x x = x x + dx , | i | i | i [X, T(dx)] x = dx x + dx = dxT(dx) x x ) | i | i | i 8 | i [X, T(dx)] = dxT(dx). )

130 Position representation

B Writing this expression in terms of the generator, to leading order in dx,

[X, T(dx)] = [X, I idxK]= idx [X, K] dxT(dx)=dx (I idxK) = dxI, we ﬁnd that

[X, K]=iI

[X,¯hK]=i¯hI.

So, the generator of translations in one dimension, K, multiplied byh ¯ , has the same commutation relations as P,

[X, P]=i¯hI.

131

Position representation

B In R3 the Lie group of translations has 3 independent generators, K~ ,

T(d~x)=I i(dxK + dyK + dzK )=I id~x K~ . x y z · One can easily check that the commutation relations of ~X and K~ are

[X , K ]=id I [X , P ]=i¯hd I. r s rs , r s rs B This is a remarkable result:

In QM the momentum operator is the generator of translations,

~P = h¯ K~ ,

since it is represented by the same operator in the Hilbert space,

i ~P ~x T(~x)=e h¯ · .

132 Momentum representation

How does the momentum ~P act on the position representation? • Consider ﬁrst R,

T(dx) x = x + dx | i | i i i with T(dx) x = I Pdx x = x dxP x | i h¯ | i | i h¯ | i ✓ ◆ x + dx x P x = i¯h | i | i. ) | i dx And for the bra, x + dx x x P† = x P = i¯h h | h |. h | h | dx

133

Momentum representation

Let’s ﬁnd the wave function of the kets P y in the position representation: • | i

B We see that P is represented by a differential operator in the basis x , {| i} . y = x y = y(x), | i h | i . d dy P = i¯h x P y = i¯h . dx ( h | | i dx

134 Momentum representation

Proceeding in the same way, in R3 one obtains • . ∂ ∂ ∂ ~P = i¯h~ = i¯h , , . r ∂x ∂y ∂z ✓ ◆ Instead of the basis ~x we can use the basis of momentum eigenstates ~p , • {| i} {| i} ~P ~p = ~p ~p | i | i This is also an inﬁnite-dimensional orthonormal basis,

3 ~p ~p0 = d (~p ~p0)=d(p p0 )d(p p0 )d(p p0 ) 1 1 2 2 3 3 3 ⌦ ↵ d p ~p ~p = I. ˆ | ih |

135

Momentum representation

The wave function of a state y in the momentum representation is • | i yˆ(~p) ~p y . ⌘ h | i with

3 3 3 2 1 = y y = d p y ~p ~p y = d p yˆ⇤(~p)yˆ(~p)= d p yˆ(~p) . h | i ˆ h | ih | i ˆ ˆ | |

B Let us see how to express ~p in the position representation. Consider ﬁrst R, | i P p = p p x P p = p x p . | i | i ) h | | i h | i . d Using P = i¯h in the basis x , we ﬁnd dx {| i}

d i px i¯h x p = p x p x p = N e h¯ dx h | i h | i ) h | i where N is a normalization constant that we will determine next.

136 Momentum representation

1 B Writing the Dirac delta in the form (a) and using the property d(ax)= a d(x), | | 1 p p0 = d(p p0)=d(h¯ (k k0)) = d(k0 k) h¯ ⌦ ↵ 1 ix(k k) 1 i (p p)x = dx e 0 = dx e h¯ 0 2ph¯ ˆ 2ph¯ ˆ 2 i (p p)x = dx p x x p0 = N dx e h¯ 0 ˆ h | i | | ˆ ⌦1 ↵ N = . ) (2ph¯ )1/2 Hence, the wave function of p in the position representation is | i 1 i px x p = e h¯ h | i (2ph¯ )1/2 and in R3,

1 i ~p ~x ~x ~p = e h¯ · . h | i (2ph¯ )3/2

137

Momentum representation

B And then, the wave function of ~x in the momentum representation is | i 1 i ~p ~x ~p ~x = ~x ~p ⇤ = e h¯ · . h | i h | i (2ph¯ )3/2

Now we can change from one basis to the other. • Consider a state y in the one-dimensional momentum representation, | i 1 i px yˆ(p)= p y = dx p x x y = dx e h¯ y(x)= [y(x)]. h | i ˆ h | ih | i (2ph¯ )1/2 ˆ F

B We see that yˆ(p) is the Fourier transform of y(x). B Analogously, y(x) is the inverse Fourier transform of yˆ(p):

1 i px 1 y(x)= x y = dp x p p y = dp e h¯ yˆ(p)= [yˆ(p)]. h | i ˆ h | ih | i (2ph¯ )1/2 ˆ F The Fourier transform is deﬁned for square-integrable functions and distributions.

138 Momentum representation

B For example, the wave function of a particle perfectly localized at x0 is y(x)= x x = d(x x ) h | 0i 0 i 1 px0 yˆ(p)= [d(x x0)] = e h¯ F (2ph¯ )1/2 whose distribution of momentum is constant (all momenta are equally probable), 1 yˆ(p) 2 = p ( •, •). | | 2ph¯ 8 2

B And the wave function of a particle with a well-deﬁned momentum p0 is yˆ(p)= p p = d(p p ) h 0 | i 0 i 1 1 p0x y(x)= [d(p p0)] = e h¯ (plane wave) F (2ph¯ )1/2 whose spatial distribution is constant (all positions are equally probable), 1 y(x) 2 = x ( •, •). | | 2ph¯ 8 2

139

Momentum representation

B These results are as expected because, using the general uncertainty relations 1 D A D B y [A, B] y y y 2 |h | | i| and the commutation relation of X and P

[X, P]=i¯hI

one has that the product of uncertainties in position and momentum is

D X D P h¯ . y y 2 Hence, if one of them is perfectly known the other must be totally uncertain.

140 Momentum representation

In practice, the wave function • 1 i px y(x)= dp e h¯ yˆ(p) (2ph¯ )1/2 ˆ is not a plane wave but a wave packet, a superposition of plane waves, with yˆ(p) a function peaking more or less sharply at p = p , not quite as d(p p ). 0 0 B It is instructive to calculate [exercise] the expectation values and uncertainties X , D X, P , D P for different wave packets and check that the Gaussian h iy y h iy y wave packet,

2 (x x )2 (p p0) 0 i s 2 p0x c0 2(h¯ /s)2 y(x)=c0 e 2s e h¯ , yˆ(p)= e h¯ 1/2 with normalization c2 = 1 , is a minimum uncertainty packet, that satisﬁes 0 pps2 h¯ s h¯ D X D P = D X = , D P = y y 2 y p y p ✓ 2 2s ◆

141

Probability density and probability current density

We can now ﬁnd the Schrödinger equation for the wave function • y(~x, t)= ~x y(t) of a particle of mass m moving in a potential. h | i From the Schrödinger equation for y(t) , | i ∂ i¯h x y(t) = x H y(t) ∂t h | i h | | i and taking the position representation of a time-independent Hamiltonian, H = H(t), 6 P2 . h¯ 2 H = + V(~X) = 2 + V(~x), 2m 2mr we get the Schrödinger wave equation:

∂ h¯ 2 i¯h y(~x, t)= 2 + V(~x) y(~x, t). ∂t 2mr !

142 Probability density and probability current density

B In particular, we conﬁrm that energy eigenstates are stationary,

x H y (t) = E x y (t) = E y (~x, t), h | | E i h | E i E ∂ i E(t t ) i¯h y (~x, t)=E y (~x, t) y (~x, t)=e h¯ 0 y (~x, t ) ∂t E E ) E E 0 and satisfy

2 h¯ 2 + V(~x) yE(~x, t)=E yE(~x, t). 2mr !

143

Probability density and probability current density

Let us see now the evolution with time of the probability density • 2 $(~x, t)= y(~x, t) = y⇤(~x, t)y(~x, t), | | ∂$ ∂y⇤ ∂y 1 = y + y⇤ = [(Hy)⇤y y⇤Hy] ∂t ∂t ∂t i¯h i¯h 2 2 i¯h = y y⇤ y⇤ y = ~ y~ y⇤ y⇤~ y . 2m r r r · 2m r r ⇣ ⌘  ⇣ ⌘ Hence, the probability density satisﬁes the continuity equation:

∂$(~x, t) + ~ ~J(~x, t)=0, ∂t r · i¯h h¯ ~J(~x, t) y~ y⇤ y⇤~ y = Im(y⇤~ y). ⌘ 2m r r m r ⇣ ⌘ where ~J is the probability current density.

144 Probability density and probability current density

B Integrating this equation over an arbitrary region V of R3, and applying the divergence (or Gauss’s) theorem, ∂$ dV + dV ~ ~J = 0 ˆV ∂t ˆ r · ∂ dV $ + d~S ~J = 0 ) ∂t ˆV ˛S · we ﬁnd that ﬂux of the current density through the surface S enclosing the region V gives the total probability that has escaped or entered that region per time unit. If V = R3 the total probability is constant.

145

Probability density and probability current density

B The probability current density integrated over R3 is the average particle velocity in the state y , | i 1 1 3 ~v y = y ~P y = d x y x x ~P y h i m h | | i m ˆ h | ih | | i 1 3 3 i¯h = d x y⇤( i¯h~ y)= d x y⇤~ y m ˆ r ˆ m r ✓ ◆ 1 3 3 i¯h = d x (i¯h~ y⇤)y = d x y~ y⇤ m ˆ r ˆ m r ✓ ◆ 3 i¯h 3 d x~J = d x y~ y⇤ y⇤~ y = ~v y. ) ˆ 2m ˆ r r h i ⇣ ⌘ B For an energy eigenstate yE, that is stationary, the probability density to ﬁnd the particle at ~x does not change with time, since ∂$ $ = y (~x, t)) 2 = y (~x, t )) 2 = ~ ~J = 0. | E | | E 0 | ) ∂t r · B The time evolution of a generic wave function will be described later.

146 Ehrenfest’s theorem

B We have already seen how expectation values change with time: d i ∂A y(t) A y(t) = y [A, H] y + y y . dt h | | i h¯ h | | i h | ∂t | i It is remarkable that the expectation values of X and P for a particle of mass m moving in a potential V(x), d i 1 X y = [X, H] y = P y dth i h¯ h i mh i d i dV P y = [P, H] y = = F(x) dth i h¯ h i dx h iy ⌧ y verify similar equations of motion as the the classical variables x and p: dx ∂H p = = dt ∂p m dp ∂H dV = = = F(x). dt ∂x dx

147

Ehrenfest’s theorem

? To check the commutators [X, H] and [P, H] given above notice that

[X, P2]=[X, P]P + P[X, P]=2i¯hP [A, BC]=[A, B]C + C[A, B], ( . dV d d dV [P, V]=PV VP = i¯h i¯hV V i¯h = i¯h . dx dx dx dx ✓ ◆ B From previous coupled differential equations one ﬁnds the Ehrenfest’s theorem, d2 m X y = F(x) y dt2 h i h i stating that the center of the wave function X moves like a classical particle h iy under the average force F . h iy

148 Ehrenfest’s theorem

B Although, at ﬁrst glance, it might appear that Ehrenfest’s theorem is saying that the quantum mechanical expectation values obey Newton’s classical equations of motion, this is not actually the case, because in general

F(x) = F( X ). h iy 6 h iy B Exceptions to this inequality are potentials of the form

n n 1 V(x)= lx , F(x)=nlx for: n = 0 (free particle), n = 1 (constant force) or n = 2 (harmonic potential). [Exercise]

149

Propagator

The time evolution of the wave function can be expressed in terms of the • propagator.

B This is a general approach that allows a perturbative treatment. It is frequently applied in Quantum Field Theory, where the propagator is a fundamental concept. We will introduce it here although it will not be used in this course.

The propagator K(~x , t ;~x , t ) is an integral operator that acts on the initial wave • 2 2 1 1 function and transforms it to the ﬁnal one,

y(~x , t )= d3x K(~x , t ;~x , t ) y(~x , t ). 2 2 ˆ 1 2 2 1 1 1 1 Let us explore its meaning.

150 Propagator

Suppose a particle perfectly localized initially (t ) at ~x . Its wave function is just • 1 i y(~x , t )=d(~x ~x ). 1 1 1 i

According to the propagator deﬁnition, at a time t2 the wave function will be

y(~x , t )= d3x K(~x , t ;~x , t ) d(~x ~x )=K(~x , t ;~x , t ). 2 2 ˆ 1 2 2 1 1 1 i 2 2 i 1

B Therefore, the propagator provides the probability amplitude to ﬁnd a particle in ~x2 at t2 if the particle was in ~x1 at t1, namely, it is the probability amplitude for the “propagation” between those points in that interval.

151

Propagator

i H(t2 t1) B Remember that y(t2) = U(t2, t1) y(t1) = e h¯ y(t1) . | i | i | i Then, taking y(t ) = ~x we have | 1 i | 1i ~x y(t ) = K(~x , t ;~x , t )= ~x U(t , t ) ~x ~x , t ~x , t h 2 | 2 i 2 2 1 1 h 2| 2 1 | 1i ⌘ h 2 2 | 1 1i where ~x , t and ~x , t are position eigenstates in the Heisenberg picture. | 1 1i | 2 2i B And in terms of energy eigenstates: i i H(t2 t1) E(t2 t1) y(~x2, t2)= ~x2 y(t2) = ~x2 e h¯ y(t1) = Â ~x2 E E y(t1) e h¯ h | i h | | i E h | ih | i 3 i E(t t ) = d x ~x E E ~x ~x y(t ) e h¯ 2 1 ˆ 1 Â 2 1 1 1 E h | ih | ih | i 3 i E(t t ) = d x y (~x )y (~x )e h¯ 2 1 y(~x , t ) ˆ 1 Â E 2 E⇤ 1 1 1 E i ¯ E(t2 t1) K(~x2, t2;~x1, t1)=Â yE(~x2)yE⇤ (~x1)e h . ) E (replace E E with dE E E if the energy spectrum is continuous) Â ˆ E | ih | | ih | 152 Propagator

Restricting ourselves to the case t > t we deﬁne the retarded propagator • 2 1

K+(~x , t ;~x , t ) q(t t ) ~x U(t , t ) ~x 2 2 1 1 ⌘ 2 1 h 2| 2 1 | 1i where q(t) is the Heaviside step function

1,t2 > t1 q(t2 t1)= 8 0,t < t < 2 1 whose derivative is the Dirac delta, :

∂q(t2 t1) = d(t2 t1). ∂t2

153

Propagator

∂ B The retarded propagator is the Green’s function of the operator H i¯h : ∂t

2 h¯ 2 ∂ i E(t t ) + (~ ) i¯ q( ) y (~ )y (~ )e h¯ 2 1 x2 V x2 h t2 t1 Â E x2 E⇤ x1 " 2mr ∂t2 #" E # i ¯ E(t2 t1) = q(t2 t1) Â E yE(~x2)yE⇤ (~x1)e h E i ¯ E(t2 t1) i¯h d(t2 t1) Â yE(~x2)yE⇤ (~x1)e h E

i i E(t t ) i¯h q(t t ) y (~x )y⇤ (~x ) E e h¯ 2 1 2 1 Â E 2 E 1 h¯ E ✓ ◆ = i¯h d(t t )d3(~x ~x ) 2 1 2 1 i E(t t ) where we have used H y = E y , d(t t )e h¯ 2 1 = 1 and E E 2 1 3 Â yE(~x2)yE⇤ (~x1)=Â ~x2 E E ~x1 = d (~x2 ~x1). E E h | ih | i

154 Propagator

B Finding K+ is difﬁcult in general, depending on the form of the potential V(~x). An easily solvable case is the free propagator, V(~x)=0,

3 i E(t t ) ~x U(t , t ) ~x = d p ~x ~p ~p ~x e h¯ 2 1 h 2| 2 1 | 1i ˆ h 2 | ih | 1i 2 1 i ~p ~x p with ~x ~p = e h¯ · , E = h | i (2ph¯ )3/2 2m 2 0 3 1 i ~p (~x x~ ) i p (t t ) K (~x , t ;~x , t )=q(t t ) d p e h¯ · 2 1 e h¯ 2m 2 1 ) + 2 2 1 1 2 1 ˆ (2ph¯ )3

3/2 x~ ~x 2 3p i m | 2 1| i m h¯ 2 t t = q(t t )e 4 e 2 1 2 1 2ph¯ (t t )  2 1 where we have used (to the third power in 3 dimensions),

• 2 i(ap2+2bp) p i p i b dp e = e 4 e a , a > 0. ˆ • a r A fully localized particle at t would get fully delocalized at t > t (instantly!!) ) 1 2 1 155

Propagator

B It is instructive to see [exercise] how a Gaussian wave packet evolves freely with time: its center moves with constant group velocity and the packet broadens being no longer minimal: Dx Dp > h¯ /2 (while Dp remains constant).

B The free propagator is used as a starting point to ﬁnd perturbative solutions at order Vn(~x). This is the usual approach in Quantum Field Theory.

156 Feynman formulation of Quantum Mechanics: path integral

In Classical Mechanics, dynamics is governed by Hamilton’s principle: • the trajectory of a system in the phase space (coordinates and velocities) is an extreme of the action (usually a minimum), x f = x(t f )

t f dS = d dtL(x, x˙)=0 ˆ ti

xi = x(ti) where the Lagrangian L(x, x˙) is a function which contains all physical information concerning the system. If it is conservative, 1 L(x, x˙)= mx˙2 V(x). 2 B From this variational principle one derives the Euler-Lagrange differential equations, whose solutions provide the equations of motion of the system.

157

Feynman formulation of Quantum Mechanics: path integral

In 1933, Dirac pointed out that, in contrast to Classical Mechanics, the action • seemed to play no relevant role in QM. He unsuccessfully speculated that the propagator might correspond to exp iS/¯h where S is the classical action { } evaluated along the classical trajectory.

In 1948, Feynman developed Dirac’s idea and accomplished a new formulation of • QM based on writing the propagator as the sum over all possible paths (not just the classical one) of exp iS/¯h between the initial and the ﬁnal state. Somehow, a { } quantum particle manages to take all paths and the probability amplitude of each one adds up according to the superposition principle of QM.a

a R. P. Feynman, Space-time approach to nonrelativistic Quantum Mechanics, Rev. Mod. Phys. 20 (1948) 367.

158 Feynman formulation of Quantum Mechanics: path integral

B We have seen that in the canonical formalism, the propagator is

iHDt/¯h x , t x , t = x U(t , t ) x = x e x . f f i i f f i | ii f | ii ⌦ ↵ ⌦ ⌦ Notice that at every ﬁxed time t, the states x, t form a complete set, {| i} I = dx x, t x, t . ˆ | ih | Let us choose a set of intermediate times t t , t ,...,t with n 2 { 0 1 N} t t < t < < t t that will be assumed equidistant to simplify, i ⌘ 0 1 ··· N ⌘ f t f ti t = t + ndt , dt = . n 0 N

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Feynman formulation of Quantum Mechanics: path integral

B Then the propagator reads

x , t x , t = dx x , t x , t x , t x , t f f i i ˆ 1 f f 1 1 h 1 1 | i ii ⌦ ↵ ⌦ ↵ = dx dx x , t x , t x , t x , t x , t x , t ˆ 1 2 f f 2 2 h 2 2 | 1 1ih 1 1 | i ii ⌦ N 1 ↵ = dx dx x , t x , t ˆ 1 N 1 ’ n+1 n+1 n n ··· n=0 h | i with the notation x x , x x . i ⌘ 0 f ⌘ N For small enough dt,

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B In the momentum representation and neglecting terms of order (dt)2, x , t x , t = x (1 iHdt/¯h) x h n+1 n+1 | n ni h n+1| | ni = dp x p p (1 iHdt/¯h) x ˆ n h n+1 | nih n| | ni = dp x p [1 iH(p , x )dt/¯h] p x ˆ n h n+1 | ni n n h n | ni dpn ip (x x )/¯h iH(p ,x )dt/¯h = e n n+1 n e n n ˆ 2ph¯ dp i x x = n exp p n+1 n H(p , x ) dt , ˆ 2ph¯ h¯ n dt n n ⇢  and substituting this in the limit of large N, we have

dp0 dpN 1 x f , t f xi, ti = lim dx1 dxN 1 N • ˆ ··· ˆ 2ph¯ ··· 2ph¯ ! ⌦ ↵ N 1 i x +1 xn exp p n H(p , x ) dt . ⇥ h¯ Â n dt n n ( n=0  )

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Feynman formulation of Quantum Mechanics: path integral

B The expression above is an integral over all the possible values of x1,...,xN 1. Every set of values deﬁnes a path, i.e. a function x(t) given by the interpolation of

x(t0)=x0,...,x(tN)=xN, with ﬁxed x0 and xN:

t0 t1 t2 t3 tN

162 Feynman formulation of Quantum Mechanics: path integral

B There is also an integral over N momenta p0,...,pN 1. Therefore, we can write the propagator as the discretized version of the functional integral

i t f x , t x , t = x(t) p(t) exp dt [px˙ H(p, x)] f f i i ˆ D D h¯ ˆ ⇢ ti ⌦ ↵ where the x(t) have ﬁxed boundary conditions and the momenta p(t) are unbounded.

B The potential of a conservative system does not depend on p, so V factors out and we can integrate over the momenta, with the help of

• 2 i(ap2+2bp) p i p i b dp e = e 4 e a , a > 0. ˆ • a r Then we have,

dp idt i(x x ) m 1/2 im(x x )2 n exp p2 + p n+1 n = exp n+1 n ˆ 2ph¯ n 2mh¯ n h¯ 2pi¯hdt 2¯hdt ⇢ ⇣ ⌘ ⇢ 163

Feynman formulation of Quantum Mechanics: path integral

B Hence, m N/2 x f , t f xi, ti = lim dx1 dxN 1 N • 2pi¯hdt ˆ ··· ! ⌦ ↵ ⇣ ⌘ N 1 2 i m x +1 xn exp n V(x ) dt ⇥ h¯ Â 2 dt n ( n=0 " ✓ ◆ # ) that is the discretized version of the functional integral:

t f i m 2 x f , t f xi, ti = x(t) exp dt x˙ V(x) ˆ D h¯ ˆt 2 ⇢ i h i ⌦ ↵ i = x(t) exp S[x(t)] ˆ D h¯ ⇢ where the action S[x(t)] is a functional of all possible paths, in terms of the Lagrangian,

t f m S[x(t)] = dtL(x, x˙) , L(x, x˙)= x˙2 V(x). ˆ ti 2

164 Feynman formulation of Quantum Mechanics: path integral

This is the result we had advertised. We may interpret x(t) as a sum over all B D paths that we usually call path integral. ´ This alternative formulation of QM provides an extremely interesting and intuitive view of quantum processes and allows to derive the classical limit in a very natural way. As an illustration, let us consider the famous double slit experiment.

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Feynman formulation of Quantum Mechanics: path integral

B The double-slit experiment was ﬁrst performed with (sun)light by Thomas Young in 1801 and was key to accept the wave theory of light: the beams passing (diffracting) through two closely separated slits pierced on a plate interfere in their way to a screen where they display a fringe pattern of dark and bright bands.

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B In 1927, Davisson and Germer demonstrated that electrons show the same behaviour (using a nickel crystal instead of two slits), which was later extended to atoms and molecules! Similar experiments with low intensity beams of single photons or single electrons sent one by one (using a biprism instead of slits) have been performed with identical results:

Experiment by A. Tonomura et al. [American Journal of Physics 57 (1989) 117]: (b) 200, (c) 6000, (d) 40000, (e) 140000 electron sent one by one through a double slit.

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Feynman formulation of Quantum Mechanics: path integral

B The details of the experimental setup are not relevant for us. We will take it as a thought experiment with profound consequences. Richard Feynman in his Feynman lectures on physics called it “a phenomenon which is impossible, absolutely impossible, to explain in any classical way, and which has in it the heart of quantum mechanics. In reality, it contains the only mystery”. https://www.youtube.com/watch?v=aAgcqgDc-YM B One may think that a ﬂux of electrons behaves like a ﬂuid of many particles interacting with each other and this may cause the accumulation of impacts in bands on the screen resembling an interference pattern. But when sending electrons one by one there is no doubt that somehow every electron “interferes” with itself.

B Even more striking is the fact that if one detects which of the slit each electron goes through, or one closes the other slit, the interference pattern disappears! But how can an electron “know” if the other slit is open? And when both slits are open, how can one it interfere with another one emitted before or afterwards?

168 Feynman formulation of Quantum Mechanics: path integral

B The interpretation of Feynman is perhaps the most satisfactory, though contrary to the common sense.a The only thing we know for sure about the electron is that it comes from the source and ends up on the screen, but we have no information about its intermediate positions. So, in between, one can not say if the electron is here or there. Actually it is in a cohererent superposition of all possible paths.

B Of course, classically this makes no sense: nothing can be in more than one state at the same time. To the resulting probability amplitude (path integral) contribute not only the classical path but every path compatible with the boundary conditions. If the electron is detected through one of the slits then the possible paths are restricted in such a way that the interference cancels.

a “[Quantum theory] describes nature as absurd from the point of view of common sense. And yet it fully agrees with experiment. So I hope you can accept nature as She is — absurd.” R. P. Feynman in QED: the strange theory of light and matter, Princeton University Press, 1985.

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Feynman formulation of Quantum Mechanics: path integral

In the classical limit (¯h 0) one can apply the stationary phase approximation to B ! the functional integral: the only contribution to this oscillatory integral, with a rapidly varying phase, is the one that is an extreme of the integrand (the others cancel out), d S[x(t)] = 0. dx(t) cl We recover Hamilton’s principle! The principle of minimal action is just a good approximation of our quantum world in the classical domain.

B When we deal with macroscopic systems the action along different paths is always much larger thanh ¯ so the system obeys the familiar classical rules.

B Things change when the difference of the action along the possible paths is comparable toh ¯ .

170 Feynman formulation of Quantum Mechanics: path integral

B For example, consider two electron paths with constant velocities v1 = D/t y v =(D + d)/t, and assume d D, so v = v v . 2 ⌧ 1 ⇡ 2

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Feynman formulation of Quantum Mechanics: path integral

B Then i i 2 exp S[x ] + exp S[x ] µ 1 + cos Dj, h¯ 1 h¯ 2 ⇢ ⇢ 1 mv2t mv2t mDd pd d Dj = 2 1 = 2p h¯ 2 2 ! ⇡ ht¯ ⇡ h¯ l

where we have introduced de Broglie’s relation p = h/l with p = mv. This is exactly the phase difference of two “waves” in a diffraction experiment.

B The de Broglie’s wavelength assigned to a particle of momentum p should not be taken literally. It is rather an equivalent way of viewing things in the microcosmos.

172 A quantum skier...

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J. Al-Khalili, Quantum: A guide for the perplexed, Orion Publishing, 2012