Download Path Integrals in Quantum Mechanics

Path Integrals in Quantum Mechanics
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
1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Outline . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
2
2
2 Path Integrals from ordinary Quantum Mechanics
2.1 The Schrödinger equation and time evolution . . . . . . . . .
2.2 The propagator . . . . . . . . . . . . . . . . . . . . . . . . . .
4
4
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
8.1 Free particle the easy way . . . . . . . . . . . . . . .
8.2 Details on the computation of the harmonic oscillator
8.2.1 Why det(P) = det(N), | det(O)| = 1 . . . . .
8.2.2 Calculation of det(N) . . . . . . . . . . . . . .
8.2.3 Calculation of elements of N−1 . . . . . . . .
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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 explore 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:
Ĥ|ψ(t)i = ih̄
∂
|ψ(t)i
∂t
⇒ |ψ(t)i = e−iĤ(t−t0 )/h̄ |ψ(t0 )i
(2.1)
2
p̂
where the (time independent) Hamiltonian operator is Ĥ = 2m
+ V (x̂). 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; x̂ and p̂ are the
position and momentum operators, respectively. These have corresponding
eigenstates and eigenvalues;
x̂|xi = x|xi
(2.2)
p̂|pi = p|pi.
(2.3)
and
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),
4
(2.4)
Z
dx|xihx| = 1.
(2.5)
∂
For later purposes, let us also state that in the x basis, p̂ = −ih̄ ∂x
, which
together with the normalizing condition (2.4) for the momentum operator
implies
hx|pi = √
1
eipx/h̄ .
2πh̄
(2.6)
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
∂
ψ(x, t),
∂t
(2.8)
h̄2 ∂ 2
+ V (x).
2m ∂x2
(2.9)
Hψ(x, t) = ih̄
where
H=−
In this project work, we will only deal with such time independent Hamiltonians, 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|ψn i satisfy the equation
Ã
!
h̄2 ∂ 2
−
+ V (x) ψn (x) = En ψn (x),
2m ∂x2
(2.10)
or in ket notation
Ĥ|ψn i = En |ψn i,
ψn (x) = hx|ψn i.
(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 )/h̄ .
The general solution thus becomes
ψ(x, t) =
X
cn e−iEn (t−t0 )/h̄ ψn (x)
n
5
(2.12)
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:
ψ(x, t) = hx|ψ(t)i = hx|e−iĤ(t−t0 )/h̄ |ψ(t0 )i
(2.14)
We may also insert an identity operator (eq. (2.5)), yielding
Z
ψ(x, t) = hx|e
−iĤ(t−t0 )/h̄
|
Z
=
dx0 |x0 i hx0 |ψ(t0 )i =
Z
dx0 hx|e−iĤ(t−t0 )/h̄ |x0 iψ(x0 , t0 ) ≡
{z
}
ψ(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
U (x, t; x0 , t0 ) = hx|e−iĤ(t−t0 )/h̄ |x0 i
(2.16)
is the so called propagator. What is the meaning of this quantity? First of
all, notice that e−iĤ(t−t0 )/h̄ |x0 i is the state of some object at time t, supposed
it was in the state |x0 i at time t0 . Hence, hx|e−iĤ(t−t0 )/h̄ |x0 i = 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 |x0 i 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
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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
U (x, t; x0 , t0 ) = hx|e−iĤ(t−t0 )/h̄
X
!
|ψn ihψn | |x0 i =
n
=
X
hx|ψn ie−iEn (t−t0 )/h̄ hψn |x0 i =
n
=
X
n
ψn (x)ψn∗ (x0 )e−iEn (t−t0 )/h̄
(2.17)
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)]/h̄ ,
(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 ;
U (x, T ; x0 , 0) = hx|e−iHT /h̄ |x0 i
(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).
U (x, T ; x0 , 0) = hx|e−iHT /h̄ |x0 i = hx|e−iH(T −t1 )/h̄ e−iHt1 /h̄ |x0 i =
Z
= hx|e−iH(T −t1 )/h̄
dx1 |x1 ihx1 |e−iHt1 /h̄ |x0 i =
Z
=
dx1 hx|e−iH(T −t1 )/h̄ |x1 ihx1 |e−iHt1 /h̄ |x0 i =
Z
=
dx1 U (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 somewhere 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 = N ²;
−iH²/h̄
U = hx|e−iHT /h̄ |x0 i = hx| e| −iH²/h̄ e−iH²/h̄
{z · · · e
} |x0 i
(3.3)
N f actors
Now between every factor of e−iH²/h̄ we insert an identity operator (N-1 in
total), in the same way we did above. We get
U = hx|e−iH²/h̄ e−iH²/h̄ · · · e−iH²/h̄ |x0 i =
Z
= hx|e−iH²/h̄
Z
dxN −1 |xN −1 ihxN −1 |e−iH²/h̄
dxN −2 |xN −2 ihxN −2 | · · ·
Z
···
Z
−iH²/h̄
dx1 |x1 ihx1 |e
|x0 i =
dx1 dx2 · · · dxN −1
×hx|e−iH²/h̄ |xN −1 ihxN −1 |e−iH²/h̄ |xN −2 i · · · hx1 |e−iH²/h̄ |x0 i ≡
Z
≡
dx1 dx2 · · · dxN −1 UxN ,xN −1 UxN −1 ,xN −2 · · · Ux1 ,x0 , (3.4)
where we have identified U (xj+1 , ²; xj , 0) = hxj+1 |eiH²/h̄ |xj i ≡ 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 intervals 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:
U=
X
Z
Upath →
paths
9
Upath ,
(3.5)
x
x
x
x
3
N-2
x
2
N-1
x
x
N
. . ..
1
x
0
(N-1)ε
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,
Uxj+1 ,xj = hxj+1 |e−iH²/h̄ |xj i → 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
Uxj+1 ,xj = hxj+1 |e−iH²/h̄ |xj i = hxj+1 |(1 − iH²/h̄ + O(²2 ))|xj i =
i²
= hxj+1 |xj i − hxj+1 |H|xj i +O(²2 ).
| {z } | h̄
{z
}
I
(3.6)
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
I = δ(xj+1 − xj ) =
dpj ipj (xj+1 −xj )/h̄
e
.
2πh̄
(3.7)
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
10
as
Ã
!Z
i²
i²
p̂2
II = − hxj+1 |H|xj i = − hxj+1 |
+ V (x̂)
dpj |pj ihpj |xj i =
h̄
h̄
2m
i² Z
=−
dpj
h̄
à 2
p
j
2m
i² Z dpj
=−
h̄ 2πh̄
!
+ V (xj+1 ) hxj+1 |pj ihpj |xj i =
à 2
p
j
2m
!
+ V (xj+1 )
eipj (xj+1 −xj )/h̄ . (3.8)
Here we have also used equation (2.6).
For simplicity, we switch to
V (x̄j ) = 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

Z
Uxj+1 ,xj =


dpj ipj (xj+1 −xj )/h̄ 
i²

e
1 −

2πh̄
h̄

|



+ V (x̄j ) +O(² )
=

2m

{z
}
à 2
p
!
j
2
≈e−i²H(pj ,x̄j )/h̄
Z
=
Z
≈
dpj ipj (xj+1 −xj )/h̄ −i²H(pj ,x̄j )/h̄
e
e
(1 + O(²2 )) ≈
2πh̄
dpj i²(pj ẋj −H(pj ,xj ))/h̄ Z dpj i²(pj ẋj − pj −V (x̄j ))/h̄
2m
=
e
=
e
2πh̄
2πh̄
2
Z
=e
−i²V (x̄j )/h̄
2
dpj i²(pj ẋj − pj )/h̄
2m
e
, (3.10)
2πh̄
where we have omitted terms of order ²2 and higher. Also we have approximated the derivative (xj+1 − xj )/² ≈ ẋj . 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
−ax2 +bx
e
dx =
1
π b2 /4a
e
a
(3.11)
This manipulation is not always allowed. For more complicated Hamiltonians the
choice of x̄j might not arbitrary.
11
r
−i²V (x̄j )/h̄
⇒ U (xj+1 , ²; xj , 0) = Uxj+1 ,xj = e
r
=
m
2
ei²mẋj /2h̄ =
2πi²h̄
m
2
ei²(mẋj /2−V (x̄j ))/h̄ .
2πi²h̄
(3.12)
As we argued earlier, the propagator for one "path" is a product of these
factors;
Upath =
NY
−1
Uxj+1 ,xj =
j=0
NY
−1
(r
j=0
)
m
2
ei²(mẋj /2−V (x̄j ))/h̄ =
2πi²h̄
µ
m
=
2πi²h̄
¶N/2
e
i²
h̄
PN −1
j=0
mx˙j 2
−V
2
(x̄j )
,
(3.13)
hence
Z
U=
µ
=
m
2πi²h̄
¶N/2 Z NY
−1
µ
=
dx1 · · · dxN −1 Upath =
i²
dxk e h̄
PN −1
j=0
mx˙j 2
−V
2
(x̄j )
=
k=1
m
2πi²h̄
¶N/2 Z NY
−1
i²
dxk e h̄
PN −1
j=0
L(ẋj ,x̄j )
(3.14)
k=1
2
where we have identified the Lagrangian L = m2ẋ − V (x).
PN −1
In
the
limit
N
→
∞,
²
→
0
the
sum
goes
over
to
an
integral
as
²
j=0 L(ẋj , x̄j ) →
R
L(ẋj , x̄j )dt = S[x(t)]. Hence, we may write
Z
U=
R
Dx(t)eiS[x(t)]/h̄ ,
³
(3.15)
´N/2 R Q
N −1
m
where the so called measure Dx(t) = 2πi²h̄
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/h̄ | ψ(t)i =
Z
dx0 hx| e−iH²/h̄ | x0 i hx0 | ψ(t)i =
=
|
Z
=
{z
}|
{z
U (x,t+²;x0 ,t)
}
ψ(x0 ,t)
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
U (x, t + ²; x0 , t) =
2πh̄i²
¶1/2
(
"
µ
i m(x − x0 )2
exp
− ²V
h̄
2²
µ
m
⇒ ψ(x, t + ²) =
2πh̄i²
"
¶1/2 Z
dx0 exp
·
i²
× exp − V
h̄
µ
x + x0
2
im(x − x0 )2
2²h̄
x + x0
2
¶# )
#
¶¸
ψ(x0 , t).
(3.17)
This is the time evolution according to the path integral theory.
Let us now make the substitution η = x − x0 , yielding
µ
ψ(x, t + ²) =
m
2πh̄i²
·
¶1/2 Z
h
i
dη exp imη 2 /2h̄²
µ
i
η
× exp − ²V x +
h̄
2
¶¸
ψ(x + η , t)
| {z }
(3.18)
x0
Since ² is small, so must η be, and we can expand
ψ(x + η, t) = ψ(x, t) + η
dψ η 2 d2 ψ
+
+ ...
dx
2 dx2
(3.19)
and
·
exp −
µ
η
i
²V x +
h̄
2
¶¸
µ
=1−
=1−
13
¶
i²
η
V x+
+ ... =
h̄
2
i²
V (x) + ...,
h̄
(3.20)
³
´
where we also have performed a Taylor expansion of V x + η2 and excluded
terms of order ²η and higher. Inserting the two expansions into equation
(3.18) we thus have
µ
ψ(x, t + ²) =
m
2πh̄i²
¶1/2 Z
h
i
exp imη 2 /2h̄²
"
i²
∂ψ η 2 ∂ 2 ψ
× ψ(x, t) − V (x)ψ(x, t) + η
+
h̄
∂x
2 ∂x2
#
dη.
The integral over η can easily be evaluated to give
ψ(x, t + ²) = ψ(x, t) −
h̄² ∂ 2 ψ i²
− V (x)ψ(x, t)
2im ∂x2
h̄
"
#
h̄2 ∂ 2
⇔ ih̄ [ψ(x, t + ²) − ψ(x, t)] /² = −
+ V (x) ψ(x, t).
2m ∂x2
(3.21)
As ² → 0 the left hand side turns into a time derivative and we finally have
"
#
∂
h̄2 ∂ 2
ih̄ ψ(x, t) = −
+ V (x) ψ(x, t).
∂t
2m ∂x2
(3.22)
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
m
U=
2πi²h̄
µ
m
=
2πi²h̄
¶N/2 Z
i²m
dx1 dx2 · · · dxN −1 e 2h̄
¶N/2 Z
im
dx1 dx2 · · · dxN −1 e 2²h̄
xj =
2²h̄
m
´1/2
µ
j=0
PN −1
j=0
Here it is convenient to change variables. Let yj =
³
PN −1
³
x˙j 2
=
(xj+1 −xj )2
m
2²h̄
´1/2
.
(4.1)
xj ⇔
yj , yielding
m
U=
2πi²h̄
¶N/2 Ã
|
{z
2²h̄
m
!(N −1)/2 Z
dy1 dy2 · · · dyN −1 ei
PN −1
j=0
(yj+1 −yj )2
(4.2)
}
≡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
i[(y1 −y0 )2 +(y2 −y1 )2 ]
dy1 e
Z
=
2
2
2
dy1 ei[2y1 −(2y0 +2y2 )y1 +y0 +y2 ]
15
(4.3)
Completing the square with respect to y1 in the exponent yields
Z
2
2
Z
2
dy1 ei[2y1 −(2y0 +2y2 )y1 +y0 +y2 ] =
s
=
R
1
dy1 e− 2i [(y1 −
y 2 +y 2
y0 +y2 2
y +y
) +( 0 2 2 )2 + 0 2 2 ]=
2
πi − 2 [ y02 +y22 − (y02 +y22 +2y0 y2 ) ]
4
e i 2
=
2
s
πi i (y2 −y0 )2
e2
, (4.4)
2
q
2
where we used that e−ax dx = πa . Inserting this into the integral over dy2
and performing the integral in the same way as above, we get
s
i
πi Z
2
2
dy2 e 2 (y2 −y0 ) +i(y3 −y2 ) =
2
s
=
s
=
πi
2
s
s
y2
2y 2
3i 2 2
πi Z
[y2 − 3 (y0 +2y3 )y2 + 30 + 33 ]
2
dy2 e
=
2
2
2
2y3
1
1
3i
πi Z
2 y0
2
dy2 e 2 [(y2 − 3 (y0 +2y3 )) + 3 + 3 − 9 (y0 +2y3 ) ] =
2
s
2πi 3i [ y03 + 2y32 − 1 (y02 +4y32 +4y0 y3 )]
e2 2 3 9
=
3
(πi)2 i (y3 −y0 )2
e3
. (4.5)
2
We now realize that if we compute the N − 1 integrals in this way, we
will get
Z
U =A
dy1 dy2 · · · dyN −1 ei
Ã
(πi)N −1
=A·
N
µ
m
=
2πi²h̄
¶N/2 Ã
2²h̄
m
!(N −1)/2 Ã
(πi)N −1
N
PN −1
j=0
!1/2
(yj+1 −yj )2
i
=
2
e N (yN −y0 ) =
!1/2
2 /N
ei(yN −y0 )
µ
m
=
2πiN ²h̄
¶1/2
=
2 /N
ei(yN −y0 )
.
(4.6)
Changing back from yi to xi we finally arrive at
µ
m
U (xN , T ; x0 , 0) =
2πiN ²h̄
¶1/2
e
im
(xN −x0 )2
2N ²h̄
r
=
m
2
eim(xN −x0 ) /2T h̄ .(4.7)
2πh̄iT
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 N ² = 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)]/h̄
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)]/h̄ comes
with the very small number h̄ 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 eachother and cause destructive interference.
For macroscopic (classical) particles, S is very large compared to h̄. 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 R"principle of least action". The
principle states that the action S[x(t)] = tt12 L(x, ẋ)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 h̄ that the interference will be destructive instead. It follows that
for macroscopic particles only the classical trajectory will contribute to the
motion.
x
Paths interfere
constructively
x
cl
Paths interfere
destructively
t
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)]/h̄ .
For microscopic (quantum mechanical) particles on the other hand, the
action is typically of the order of h̄. Hence it takes a comparably large change
in the action to achieve a significant change in the phase S[x(t)]/h̄. 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 h̄ → 0, even very small particles will obviously behave like
classical particles, since their action will then become large compared to h̄.
The limit h̄ → 0 is also the limit where the quantization of energy etc disappears, 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
simpler, we define x̄j such that x̄2j =
tential
x2j+1 +xj+1 xj +x2j
.
3
We then have the po-
mw2 2 mw2 (x2j+1 + xj+1 xj + x2j )
V (x̄j ) =
x̄ =
2 j
2
3
(6.1)
and hence the propagator
µ
m
U=
2πi²h̄
¶N/2 Z NY
−1
dxk e
i²
h̄
PN −1
j=0
−V (x̄j )
=
k=1
µ
m
=
2πi²h̄
(
mx˙j 2
2
"
¶N/2 Z NY
−1
dxk
k=1
#)
−1
im NX
(xj+1 − xj )2 ω 2 ² 2
× exp
−
(xj+1 + xj+1 xj + xj2 )
2h̄ j=0
²
3
.
(6.2)
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














M0,0 = MN,N =
Mk,k = 2
³
1
²
−
1
²
−
ω2 ²
3
ω2 ²
3
´
(6.3)
³
´



1
ω2 ²

M
=
M
=
−
+

k,k+1
k−1,k

²
6







all other elements = 0
for 1 ≤ k ≤ N − 1. With these definitions we write the sum in the exponent
as
N
−1
X
j=0
"
#
(xj+1 − xj )2 ω 2 ² 2
−
(xj+1 + xj+1 xj + x2j ) =
²
3
N
−1
X
= xMxt =
xi Mij xj ,
(6.4)
i,j=0
which inserted in (6.2) yields
µ
m
U=
2πi²h̄
¶N/2 Z NY
−1
i²
t
dxk e h̄ xMx .
(6.5)
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
xMxt = M00 x20 + MN N x2N + M01 x0 x1 + M10 x1 x0 +
+MN −1,N xN −1 xN + MN,N −1 xN xN −1 +
N
−1
X
xj Njl xl =
j,l=1
= M00 (x20 + x2N ) + 2M01 (x0 x1 + xN −1 xN ) +
N
−1
X
xj Njl xl ,
(6.6)
j,l=1
where we have used that M is totally symmetric. We thus have that the
propagator is
µ
U=
(

m
2πi²h̄
¶N/2 Z NY
−1
k=1
dxk
)
N
−1
X
im 
× exp
M00 (x20 + x2N ) + 2M01 (x0 x1 + xN −1 xN ) +
xj Njl xl  . (6.7)
2h̄
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
U=
2πi²h̄
×
(
)
im
exp
M00 (x20 + x2N )
2h̄

(
Z NY
−1
¶N/2
)
N
−1
X
im 
dxk exp
2M01 (x0 x1 + xN −1 xN ) +
xj Njl xl  .
2h̄
k=1
j,l=1
(6.8)
The integral above contains terms both linear and quadratic in xj in the
exponent. It is therefore very useful to define x0 = (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





x0 = qOt ⇔ qOt O = x0 O = q
(6.10)
0t
t
t
t
t 0t
t
x = Oq ⇔ O Oq = O x = q ,
where we have used that the eigenvectors ni are orthonormal, which implies
Ot O = 1. We have
µ
m
U=
2πi²h̄
×
(
Z NY
−1
¶N/2
(
im
exp
M00 (x20 + x2N )
2h̄



)
N
−1

X
im 


dxk exp
2M01 (x0 x1 + xN −1 xN ) +
x
j Njl xl 


2h̄ 
k=1
j,l=1

|
{z
)
=
}
x0 Nx0t
µ
m
= | det(O)|
| {z } 2πi²h̄
=1
(
¶N/2
(
) Z N −1
Y
im
2
2
exp
M00 (x0 + xN )
dqk
2h̄
k=1

)
N
−1
X
im 
× exp
2M01
(x0 qj O1j + qj ON −1,j xN ) + qOt NOqt  . (6.11)
2h̄
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 ) = (n1 n1 n2 n2 ... nN −1 nN −1 ).
(6.12)
The diagonal matrix P contains the eigenvalues of N and is obtained from





t
P ≡ O NO = 




n1
n2
.
.
.
nN −1





 (n1 n1 n2 n2 ... nN −1 nN −1 ) =







=


n1 0
0 n2
..
..
.
.
0 0
...
...
...
0
0
..
.



.


(6.13)
. . . nN −1
Inserting this expression into equation (6.11) yields
µ
m
U=
2πi²h̄
×
¶N/2
(
(
Z NY
−1
−1
im NX
dqk exp
[2M01 qj (O1j x0 + ON −1,j xN ) + qj Pjl ql ]
2h̄ j,l=1
k=1
µ
m
=
2πi²h̄
×
Z NY
−1
¶N/2
)
im
exp
M00 (x20 + x2N )
2h̄
(
)
=
)
im
exp
M00 (x20 + x2N )
2h̄
(
)
−1 h
i
im NX
dqk exp
2M01 qj (O1j x0 + ON −1,j xN ) + nj qj2 . (6.14)
2h̄ j=1
k=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,
2M01 qj (O1j x0 + ON −1,j xN ) + nj qj2 =
Ã
= n j  qj +
!2
M01
(O1j x0 + ON −1,j xN )
nj
22
−
2
M01
n2j

(O1j x0 + ON −1,j xN )2 (6.15)
.
Hence, we have
µ
m
U=
2πi²h̄
¶N/2
(
)
im
exp
M00 (x20 + x2N )
2h̄
(
NY
−1
2
−imM01
×
exp
(O1k x0 + ON −1,k xN )2
2h̄n
k
k=1
Z
×
(
)
¶2 )
µ
M01
imnk
qk +
(O1k x0 + ON −1,k xN )
dqk exp
2h̄
nk
(6.16)
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,
NY
−1 Z
k=1
(
=
NY
−1
Ã
k=1
Ã
=
µ
imnk
M01
dqk exp
qk +
(O1k x0 + ON −1,k xN )
2h̄
nk
2πih̄
m
2πih̄
mnk
Ã
!1/2
=
2πih̄
m
Ã
!(N −1)/2
(det(P))
−1/2
=
!(N −1)/2 Ã
k=1
2πih̄
m
=
!1/2
1
QN −1
¶2 )
=
nk
!(N −1)/2
(det(N))−1/2 .
(6.17)
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
Okj Olj = Nkl−1 ,
nj
(6.18)
which is proved in the appendix. Thus we rewrite
exp
( N −1
2
X −imM01
k=1
2h̄nk
)
2
(O1k x0 + ON −1,k xN )
=
)
( N −1
2 ³
´
X −imM01
2 2
2
2
=
= exp
O1k x0 + ON −1,k xN + 2O1k ON −1,k x0 xN
k=1
2h̄nk
)
2 ³
´
−imM01
−1
2
−1
−1 2
= exp
N11 x0 + NN −1,N −1 xN + 2N1,N −1 x0 xN . (6.19)
2h̄
(
23
This is almost everything we need to do. To conclude, so far we have seen
that
µ
m
U=
2πi²h̄
¶N/2
)
(
im
M00 (x20 + x2N )
exp
2h̄
(
)
2 ³
´
−imM01
−1 2
−1
× exp
N11
x0 + NN−1−1,N −1 x2N + 2N1,N
−1 x0 xN
2h̄
Ã
2πih̄
×
m
µ
m
=
2πih̄
!(N −1)/2
¶1/2 µ
(det(N))−1/2 =
T
N
¶−N/2
(
(det(N))−1/2
)
´³
´
i
im h³ 2
2
−1
2
−1
× exp
, (6.20)
x0 + x2N M00 − M01
N11
− 2x0 xN M01
N1,N
−1
2h̄
−1
where we have used that NN−1−1,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
of N. The
−1 of the inverse N
detailed steps of these derivations are put in the appendix, and here we only
write down the answers. For large N :s we find
sin ωT
det(N) ≈ N
ωT
Ã
N
ω2T
−
T
3N
µ
−1
N11
T
ωT
≈
1−
cot ωT
N
N
!N −1
,
(6.21)
¶
(6.22)
and
−1
N1,N
−1 ≈
ωT 2
.
N 2 sin ωT
(6.23)
Finally, using (6.3) with ² = T /N , as N → ∞
U (xN , T ; x0 , 0) =
µ
−imω
=
2πh̄ sin ωT
¶1/2
(
)
·
¸
´
2x0 xN
imω ³ 2
2
exp
x0 + xN cot ωT −
,
2h̄
sin ωT
which is the desired result.
24
(6.24)
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 Hamiltonian has been diagonalized, one gets the partition function by summing
the Boltzmann factors, e−βE , for all system states1 . In general, given the
Hamiltonian, Ĥ, of a system its partition function becomes
Z = Tr e−β Ĥ .
(7.1)
In particular, if there is a continuous parameter x that labels all states, we
can write
Z
Z=
dxhx|e−β Ĥ |xi.
(7.2)
Hence Z can be thought of as the sum of the diagonal elements of the operator
U;
Uij = hxi |e−β Ĥ |xj i.
1
(7.3)
Here β = 1/kB T 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
same if we make the identification2 β = i(tf − t0 )/h̄. 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):
τ = i(t − t0 ) ⇒
dx
dx dt
dx
=
= −i .
dτ
dt dτ
dt
(7.4)
We then find that the action goes over into
Ã
!2
Z tf
m dx
t0
2
dt
− V (x(t)) dt → −i
Ã
!2
Z h̄β
m dx
2
0
dτ
+ V (x(τ )) dτ.
(7.5)
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
dxhx|e−β Ĥ |xi =
Z=
Z
=
dx
Z x(τ =βh̄)=x
x(τ =0)=x
1
Dx(τ )e− h̄
R h̄β
0
m
2
2
( dx
dτ )
+V (x(τ ))dτ
.
(7.6)
The endpoints x(τ = 0) and x(τ = βh̄) are equal according to (7.2)—here
we denote them x. The partition function is thus a sum over all paths of
period βh̄.
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=
2
Here we denote the end time as tf .
26
dxU (x, −ih̄β; x, 0) =
Z
=
Ã
mω
dx
2πh̄i sin(−iβh̄ω)
Ã
=
!1/2
mω
2πh̄ sinh(βh̄ω)
Ã
=
(
h
i
−mω
exp
x2 cos(−iβh̄ω) − x2
h̄i sin(−iβh̄ω)
!1/2 Z
(
−mωx2
dx exp
[cosh(βh̄ω) − 1]
h̄ sinh(βh̄ω)
mω
2πh̄ sinh(βh̄ω)
!1/2 

)
=
)
=
1/2
π

mω
[cosh(βh̄ω)
−
1]
h̄ sinh(βh̄ω)
=
= [2(cosh(βh̄ω)) − 1]−1/2 (. 7.7)
Let us rewrite this expression a little!
Z = [2(cosh(βh̄ω)) − 1]−1/2 =
h
i−1/2
= eβh̄ω + e−βh̄ω − 2
h
i−1/2
= eβh̄ω (1 + e−2βh̄ω − 2e−βh̄ω )
h
= eβh̄ω (1 − e−βh̄ω )2
= e−βh̄ω/2
∞
X
i−1/2
(e−βh̄ω )j =
j=0
=
=
e−βh̄ω/2
=
1 − e−βh̄ω
∞
X
e−β(j+1/2)h̄ω .
(7.8)
j=0
Comparing this with the definition of the partition function,
Z=
∞
X
e−βEj
(7.9)
j=0
finally yields the familiar result
Ej = (j + 1/2)h̄ω.
27
(7.10)
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 (x̂) = 0 we get:
U (x, T ; x0 , 0) = hx, T | e−iĤT /h̄ |x0 , 0i.
(8.1)
Inserting the identity operator in terms of the momentum eigenkets yields
U = hx, T | e−ip̂
Z
=
2 T /2mh̄
dp hx, T | e−ip̂
2 T /2mh̄
|x0 , 0i =
|pi hp|x0 , 0i =
|
{z
}
√ 1 eipx0 /h̄
2πh̄
iT
1 Z
2
=
dp e− 2mh̄ p +ip(x0 −x)/h̄ ,
2πh̄
(8.2)
where we have used equation (2.6). This kind of gaussian integral has appeared several times earlier in this work. Completing the square or simply
using equation (3.11) yields
1
U=
2πh̄
s
(x0 −x)2 m
π
e− 2h̄iT =
iT /2mh̄
28
r
m
2
eim(x−x0 ) /2T h̄ .
2πh̄iT
(8.3)
8.2
8.2.1
Details on the computation of the harmonic
oscillator
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(Ot NO) = 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.
8.2.2
(8.6)
Calculation of det(N)
From equation (6.3) and the definition of N we have









Nk,k = 2
³
N
t
−
Nk,k+1 = Nk−1,k = −








or with a =
1
2
³
1+ω 2 t2 /6N 2
1−ω 2 t2 /3N 2









ω2 t
3N
³
+
ω2 t
6N
´
(8.7)
all other elements = 0,
´
:
Nk,k = 2
³
N
t
−
ω2 t
3N
Nk,k+1 = Nk−1,k = −2a








N
t
´
³
´
N
t
−
all other elements = 0.
29
ω2 t
3N
´
(8.8)
This implies

Ã
det(N) = 2
N −1
ω2t
N
−
t
3N
1 −a 0

−a
1 −a

!N −1
...
...
.
−a 1 −a . .
.. .. .. ..
.
.
.
.

detN −1 
 0

..
.
0
0
0
0
..
.




 ,(8.9)


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
det(N) = 2
N −1
N
ω2t
−
t
3N
!N −1
AN −1 .
(8.10)
By starting to evaluate the determinant, beginning with the first column, we
get

−a

 −a

AN −1 = AN −2 + a detN −2 
 0

..
.
0
1
0
−a
0
0
...
...
.
−a 1 −a . .
.. .. .. ..
.
.
.
.
0
0
..
.




=


= AN −2 + a(−a)AN −3 = AN −2 − a2 AN −3 ,
(8.11)
where we in the last step evaluated the (N − 2)-determinant along the first
row. To conclude we have
AN −1 = AN −2 − a2 AN −3
(8.12)
A0 = A1 = 1.
(8.13)
and we also require
One can easily convince oneself that this recursion relation can be written
Ã
AN
AN −1
!
Ã
=
1 −a2
1 0
!N −1 Ã
1
1
!
.
(8.14)
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
λ− AN − AN −1
AN −1 − λ+ AN
!
Ã
=
−1
λN
+
30
0
−1
λN
−
!Ã
λ− − 1
1 − λ+
!
(8.15)
where the eigenvalues
√
1
1 i 4a2 − 1
λ± = ±
≈ e±iωt/N .
2
2
2
(8.16)
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
AN −1 =
N
λN
N sin ωt
− − λ+
≈ N −1
.
λ− − λ+
2
ωt
(8.17)
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):
Nmk
1
1
Okj Olj =
Nmk Okj Olj = Omj Olj = δml ,
nj
nj | {z }
(8.18)
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
Njk
=
(−1)j+k Djk
,
det(N)
(8.19)
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
det(N) = 2
N −1
N
ω2t
−
t
3N
!N −1
AN −1 .
(8.20)
−1
) we erase the first row and column from N . The
To get D11 (and later N1,1
determinant then becomes
Ã
D11 = 2
N −2
N
ω2t
−
t
3N
!N −2
AN −2 .
(8.21)
Equation (8.19) consequently yields
−1
=
N11
³
t
2N 1 −
ω2 t
3N 2
31
´
AN −2
.
AN −1
(8.22)
Using (8.17), that sin(α + β) = cos α sin β + cos β sin α and Taylor expanding
gives
µ
−1
N11
≈
¶
T
ωT
1−
cot ωT .
N
N
(8.23)
Looking at equation (8.9) again, one can see that
Ã
ω2t
3N
...
...
.
0 −a 1 −a . .
... ... ... ... ...
N
D1,N −1 = 2
−
t

−a 1 −a 0
0

 0 −a 1 −a 0
!N −2
N −2

×detN −2 
 0

..
.
Ã
=2
N −2
N
ω2t
−
t
3N
×
0
0
..
.




=


!N −2
(−a)N −2 ,
(8.24)
since the matrix is triangular. Thus,
−1
N1,N
−1
=
(−1)N 2N −2
2N −1
³
³
N
t
N
t
´N −2
ω2 t
(−a)N −2
3N
´N −1
ω2 t
AN −1
3N
−
−
(−1)1+N −1 D1,N −1
=
=
det(N)
=
³
aN −2 t
2N 1 −
ω 2 t2
3N 2
´
AN −1
.
(8.25)
Taylor expansion of this expression finally yields
−1
N1,N
−1 ≈
ωT 2
.
N 2 sin ωT
32
(8.26)
Bibliography
[1] R. MacKenzie, Path Integral Methods and Applications, quantph/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).
33