Nakahara M. Geometry, Topology and Physics
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Proof. The completeness relation for the momentum eigenvectors is inserted into
the LHS of (1.92) to yield
x |e
−i
ˆ
Hε
|y=
dkx |e
−iε
ˆ
H
|kk|y
=
dk
2π
e
−iky
e
−iε
ˆ
H
x
e
ikx
where
ˆ
H
x
=−
1
2m
d
2
dx
2
+ V (x).
Now we find from the commutation relation of ∂
x
≡ d/dx and e
ikx
that
∂
x
e
ikx
= ike
ikx
+ e
ikx
∂
x
= e
ikx
(ik + ∂
x
).
Repeated application of this commutation relation yields
∂
n
x
e
ikx
= e
ikx
(ik + ∂
x
)
n
(n = 0, 1, 2,...)
from which we obtain
e
−iε[−∂
2
x
/2m+V (x )]
e
ikx
= e
ikx
e
−iε[−(ik+∂
x
)
2
/2m+V (x )]
.
Therefore,
x |e
−i
ˆ
Hε
|y=
dk
2π
e
ik(x −y)
e
−iε[−(ik+∂
x
)
2
/2m+V (x )]
=
dk
2π
e
−i[εk
2
/2m−k(x−y)]
e
−iε[−ik∂
x
/m−∂
2
x
/2m+V (x )]
· 1
where the ‘1’ at the end of the last line is written explicitly to remind us of the
fact ∂
x
1 = 0. If we further put p =
√
ε/2mk and expands the last exponential
function in the last line, we obtain
x |e
−iε
ˆ
H
|y=
2m
ε
e
im(x −y)
2
/2ε
d p
2π
e
−i[p+
√
m/2ε(x−y)]
2
×
∞
n=0
(−iε)
n
n!
i
2
εm
p∂
x
−
∂
2
x
2m
+ V (x)
n
· 1.
If we put q = p +
√
m/2ε(x − y) and use lemma 1.2, we obtain:
x |e
−iε
ˆ
H
|y=
2m
ε
e
im(x −y)
2
/2ε
dq
2π
e
−iq
2
×
1 + (−iε)V (x) +
(−ε
2
)
2
(−i)
ε
(x − y)∂
x
V (x )
+
(ε
2
) + (ε|x − y|
2
)
=
m
2πiε
e
iε(m/2)[(x−y)/ε]
2
× exp
−iεV
x + y
2
+
(ε
2
) + (ε|x − y|
2
)
.
Thus, the proposition has been proved.
Note that the average value (x +y)/2 appeared as the variable of V in (1.92).
This prescription is often called the Weyl ordering.
It is found from (1.92) that the integrand oscillates very rapidly for |x −y| >
√
ε and it can be regarded as zero in the sense of distribution (the Riemann–
Lebesgue theorem). Therefore, as x − y <ε, the exponent of (1.92) approaches
the action for an infinitesimal time interval [0,ε],
S =
ε
0
dt
m
2
v
2
− V (x)
m
2
v
2
− V (x)
ε (1.93)
where v = (x − y)/ε is the average velocity and x is the average position.
Equation (1.92) also satisfies the boundary condition for ε → 0,
x |e
−i
ˆ
Hε
|y
ε→0
−→ x|y=δ(x − y). (1.94)
This can be shown by noting that
∞
−∞
dx
m
2πiε
e
im(x −y)
2
/2ε
= 1.
The transition amplitude (1.79) for a finite time interval is obtained by
infinitely repeating the transition amplitude for an infinitesimal time interval one
after another. Let us first divide the interval t
f
− t
i
into n equal intervals,
ε =
t
f
− t
i
n
.
Put t
0
= t
i
and t
k
= t
0
+εk (0 ≤ k ≤ n). Clearly t
n
= t
f
. Insert the completeness
relation
1 =
dx
k
|x
k
, t
k
x
k
, t
k
| (1 ≤ k ≤ n − 1)
for each instant of time t
k
into (1.79) to yield
x
f
, t
f
|x
i
, t
i
=x
f
, t
f
|
dx
n−1
|x
n−1
, t
n−1
x
n−1
, t
n−1
|
×
dx
n−2
|x
n−2
, t
n−2
...
dx
1
|x
1
, t
1
x
1
, t
1
|x
0
, t
0
.
Let us consider here the limit ε → 0, namely n →∞. Proposition 1.2 states that
for an infinitesimal ε,wehave
x
k
, t
k
|x
k−1
, t
k−1
m
2πiε
e
iS
k
where
S
k
= ε
m
2
x
k
− x
k−1
ε
2
− V
x
k−1
+ x
k
2
.
Therefore, we find
x
f
, t
f
|x
i
, t
i
= lim
n→∞
m
2πiε
n/2
n−1
j =1
dx
j
exp
i
n
k=1
S
k
. (1.95)
If n − 1 points x
1
, x
2
,...,x
n−1
are fixed, we obtain a piecewise linear path from
x
0
to x
n
via these points. Then we define S({x
k
}) =
k
S
k
, which in the limit
n →∞can be written as
S({x
k
})
n→∞
−→ S[x(t)]=
t
f
t
i
dt
m
2
v
2
− V (x)
. (1.96)
Note, however, that the S[x (t)] defined here is formal; the variables x
k
and x
k−1
need not be close to each other and hence v = (x
k
− x
k−1
)/ε may diverge. This
transition amplitude is written symbolically as
x
f
, t
f
|x
i
, t
i
=
x exp
i
t
f
t
i
dt
m
2
v
2
− V (x)
=
x exp
i
t
f
t
i
dtL(x , ˙x)
(1.97)
which is called the path integral representation of the transition amplitude. It
should be stressed again that the ‘v’ is not well defined and that this expression is
just a symbolic representation of the limit (1.95).
The integration measure is understood as
x = summation over all paths x (t) with x(t
i
) = x
i
, x (t
f
) = x
f
(1.98)
Figure 1.2. All the paths with fixed endpoints are considered in the path integral. The
integrand exp[iS({x
k
})] is integrated over these paths.
see figure 1.2. Although x or S({x
k
}) is ill defined in the limit n →∞,the
amplitude x
f
, t
f
|x
i
, t
i
constructed from x and S({x
k
}) together is well defined
and hence meaningful. This point is clarified in the following example.
Example 1.5. Let us work out the transition amplitude of a free particle moving
on the real axis with the Lagrangian
L =
1
2
m ˙x
2
. (1.99)
The canonical conjugate momentum is p = ∂ L/∂ ˙x = m ˙x and the Hamiltonian is
H = p ˙x − L =
p
2
2m
. (1.100)
The transition amplitude is calculated within the canonical quantum theory as
x
f
, t
f
|x
i
, t
i
=x
f
|e
−i
ˆ
HT
|x
i
=
d px
f
|e
−i
ˆ
HT
|pp|x
i
=
d p
2π
e
i p(x
f
−x
i
)
e
−iT (p
2
/2m)
=
m
2πiT
exp
im(x
f
− x
i
)
2
2T
(1.101)
where T = t
f
− t
i
.
This result is obtained using the path integral formalism next. The amplitude
is expressed as
x
f
, t
f
|x
i
, t
i
= lim
n→∞
m
2πiε
n/2
dx
1
...dx
n−1
exp
iε
n
k=1
m
2
x
k
− x
k−1
ε
2
(1.102)
where ε = T /n. After scaling the coordinates as
y
k
=
m
2ε
1/2
x
k
the amplitude becomes
x
f
, t
f
|x
i
, t
i
= lim
n→∞
m
2πiε
n/2
2ε
m
(n−1)/2
dy
1
...dy
n−1
exp
i
n
k=1
(y
k
− y
k−1
)
2
. (1.103)
It can be shown by induction (exercise) that
dy
1
...dy
n−1
exp
i
n
k=1
(y
k
− y
k−1
)
2
=
(iπ)
(n−1)
n
1/2
e
i(y
n
−y
0
)
2
/n
.
Taking the limit n →∞, we finally obtain
x
f
, t
f
|x
i
, t
i
= lim
n→∞
m
2πiε
n/2
2πiε
m
(n−1)/2
1
√
n
e
im(x
f
−x
i
)
2
/(2nε)
=
m
2πiT
exp
im(x
f
− x
i
)
2
2T
. (1.104)
It should be noted here that the exponent is the classical action. In fact, if we
note that the average velocity is v = (x
f
− x
i
)/(t
f
− t
i
), the classical action is
found to be
S
cl
=
t
f
t
i
dt
1
2
mv
2
=
m(x
f
− x
i
)
2
2(t
f
− t
i
)
.
It happens in many exactly solvable systems that the transition amplitude takes
the form
x
f
, t
f
|x
i
, t
i
=Ae
iS
cl
, (1.105)
where all the effects of quantum fluctuation are taken into account in the prefactor
A.
1.3.2 Imaginary time and partition function
Suppose the spectrum of a Hamiltonian
ˆ
H is bounded from below. Then it is
always possible, by adding a postive constant to the Hamiltonian, to make
ˆ
H
positive definite;
Spec
ˆ
H =
{
0 < E
0
≤ E
1
≤ E
2
≤···
}
. (1.106)
It has been assumed for simplicity that the ground state is not degenerate. The
spectral decomposition of e
−i
ˆ
Ht
given by
e
−i
ˆ
Ht
=
n
e
−iE
n
t
|nn| (1.107)
is analytic in the lower half-plane of t,where
ˆ
H|n=E
n
|n. Introduce the Wick
rotation by the replacement
t =−iτ(τ∈
+
) (1.108)
where
+
is the set of positive real numbers. The variable τ is regarded as
imaginary time, which is also known as the Euclidean time since the world
distance changes from t
2
− x
2
to −(τ
2
+ x
2
). Physical quantities change under
this change of variable as
˙x =
dx
dt
= i
dx
dτ
e
−i
ˆ
Ht
= e
−
ˆ
Hτ
i
t
f
t
i
dt
1
2
m ˙x
2
− V (x)
= i(−i)
τ
f
τ
i
dτ
−
1
2
m
dx
dτ
2
− V (x)
=−
τ
f
τ
i
dτ
1
2
m
dx
dτ
2
+ V (x)
.
Accordingly, the path integral is expressed in terms of the new variable as
x
f
,τ
f
|x
i
,τ
i
=x
f
|e
−
ˆ
H(τ
f
−τ
i
)
|x
i
=
¯
x e
−
τ
f
τ
i
dτ
1
2
m
dx
dτ
2
+V (x )
, (1.109)
where
¯
is the integration measure in the imaginary time τ .
For a given Hamiltonian
ˆ
H ,thepartition function is defined as
Z(β) = Tr e
−β
ˆ
H
(β > 0), (1.110)
where the trace is over the Hilbert space associated with
ˆ
H .
Let us take the eigenstates {|E
n
} of
ˆ
H as the basis vectors of the Hilbert
space;
ˆ
H|E
n
=E
n
|E
n
, E
m
|E
n
=δ
mn
.
Then the partition function is expressed as
Z(β) =
n
E
n
|e
−β
ˆ
H
|E
n
=
n
E
n
|e
−β E
n
|E
n
=
n
e
−β E
n
. (1.111)
The partition function is also expressed in terms of the eigenvector |x of ˆx.
Namely
Z(β) =
dxx|e
−β
ˆ
H
|x. (1.112)
If β is identified with the Euclidean time by putting β = iT ,wefindthat
x
f
|e
−i
ˆ
HT
|x
i
=x
f
|e
−β
ˆ
H
|x
i
,
from which we obtain the path integral expression of the partition function
Z(β) =
dy
x(0)=x (β)=y
¯
x exp
−
β
0
dτ
1
2
m ˙x
2
+ V (x)
=
periodic
¯
x exp
−
β
0
dτ
1
2
m ˙x
2
+ V (x)
, (1.113)
where the integral in the last line is over all paths periodic in [0,β].
1.3.3 Time-ordered product and generating functional
Define the T-product of Heisenberg operators A(t) and B(t) by
T [A(t
1
)B(t
2
)]=A(t
1
)B(t
2
)θ(t
1
− t
2
) + B(t
2
)A(t
1
)θ(t
2
− t
1
) (1.114)
θ(t) being the Heaviside function.
13
Generalization to the case with more than
three operators should be trivial; operators in the bracket are rearranged so that the
time parameters decrease from the left to the right. The T -product of n operators
is expanded into n! terms, each of which is proportional to the product of n − 1
Heaviside functions. An important quantity in quantum mechanics is the matrix
element of the T -product,
x
f
, t
f
|T [ˆx(t
1
) ˆx(t
f
) ···ˆx(t
n
)]|x
i
, t
i
,(t
i
< t
1
, t
2
,...,t
n
< t
f
). (1.115)
Suppose t
i
< t
1
≤ t
2
≤ ··· ≤ t
n
< t
f
in equation (1.115). By inserting the
completeness relation
1 =
∞
−∞
dx
k
|x
k
, t
k
x
k
, t
k
| (k = 1, 2,...,n)
into equation (1.115), we obtain
x
f
, t
f
|ˆx(t
n
) ···ˆx(t
1
)|x
i
, t
i
=x
f
, t
f
|ˆx(t
n
)
dx
n
|x
n
, t
n
x
n
, t
n
|···ˆx(t
1
)
dx
1
|x
1
, t
1
x
1
, t
1
|x
i
, t
i
=
dx
1
...dx
n
x
1
...x
n
x
f
, t
f
|x
n
, t
n
···x
1
, t
1
|x
i
, t
i
(1.116)
13
The Heaviside function is defined by
θ(x) =
0 x < 0
1 x ≥ 0.
where use has been made of the eigenvalue equation ˆx(t
k
)|x
k
, t
k
=x
k
|x
k
, t
k
.If
x
k
, t
k
|x
k−1
, t
k−1
in the last line is expressed in terms of a path integral, we find
x
f
, t
f
|ˆx(t
n
)... ˆx(t
1
)|x
i
, t
i
=
xx(t
1
)...x(t
n
)e
iS
. (1.117)
It is crucial to note that ˆx(t
k
) in the LHS is a Heisenberg operator, while
x(t
k
)(=x
k
) in the RHS is the real value of a classical path x(t) at time t
k
.
Accordingly, the RHS remains true for any ordering of the time parameters in
the LHS as long as the Heisenberg operators are arranged in a way defined by the
T -product. Thus, the path integral expression automatically takes the T -product
ordering into account to yield
x
f
, t
f
|T [ˆx(t
n
)... ˆx(t
1
)]|x
i
, t
i
=
xx(t
1
)...x(t
n
)e
iS
. (1.118)
The reader is encouraged to verify this result explicitly for n = 2.
It turns out to be convenient to define the generating functional Z[J ] to
obtain the matrix elements of the T -products efficiently. We couple an external
field J (t) (also called the source) with the coordinate x (t) as x (t)J (t) in the
Lagrangian, where J (t) is defined on the interval [t
i
, t
f
]. Define the action with
the source as
S[x (t), J (t)]=
t
f
t
i
dt [
1
2
m ˙x
2
− V (x) + xJ]. (1.119)
The transition amplitude in the presence of J (t) is then given by
x
f
, t
f
|x
i
, t
i
J
=
x exp
i
t
f
t
i
dt (
1
2
m ˙x
2
− V (x) + xJ)
. (1.120)
The functional derivative of this equation with respect to J (t)(t
i
< t < t
f
) yields
δ
δ J (t)
x
f
, t
f
|x
i
, t
i
J
=
x ix(t) exp
i
t
f
t
i
dt (
1
2
m ˙x
2
− V (x) + xJ)
.
(1.121)
Higher functional derivatives are easy to obtain; the factor ix(t
k
) appears in the
integrand of the path integral each time δ/δ J(t) acts on x
f
, t
f
|x
i
, t
i
J
.Thisis
nothing but the matrix element of the T -product of the Heisenberg operator ˆx(t)
in the presence of the source J (t). Accordingly, if we put J (t) = 0intheendof
the calculation, we obtain
x
f
, t
f
|T [x(t
n
)...x(t
1
)] |x
i
, t
i
= (−i)
n
δ
n
δ J (t
1
)...δJ (t
n
)
x e
iS[x(t ),J (t)]
J =0
. (1.122)
It often happens in physical applications that the transition probability
amplitude between general states, in particular the ground states, is required
rather than those between coordinate eigenstates. Suppose the system under
consideration is in the ground state |0at t
i
and calculate the probability amplitude
with which the system is also in the ground state at later time t
f
. Suppose
J (t) is non-vanishing only on an interval [a, b]⊂[t
i
, t
f
]. (The reason for this
assumption will become clear later.) The transition amplitude in the presence of
J (t) may be obtained from the Hamiltonian H
J
= H − x (t)J (t) and the unitary
operator U
J
(t
f
, t
i
) of the Hamiltonian. The transition probability amplitude
between the coordinate eigenstates is
x
f
, t
f
|x
i
, t
i
J
=x
f
|U
J
(t
f
, t
i
)|x
i
=x
f
|e
−iH(t
f
−b)
U
J
(b, a)e
−iH(a−t
i
)
|x
i
, (1.123)
where use has been made of the fact H
J
= H outside the interval [a, b].By
inserting the completeness relations of the energy eigenvectors
n
|nn|=1
into this equation, we obtain
x
f
, t
f
|x
i
, t
i
J
=
m,n
x
f
|e
−iH(t
f
−b)
|mm|U
J
(b, a)|nn|e
−iH(a−t
i
)
|x
i
=
m,n
e
−iE
m
(t
f
−b)
e
−iE
n
(a−t
i
)
x
f
|mn|x
i
m|U
J
(b, a)|n.
(1.124)
Now let us Wick rotate the time variable t →−iτ under which the exponential
function changes as e
−iEt
→ e
−Eτ
. Then the limit τ
f
→∞,τ
i
→−∞
picks up only the ground states m = n = 0. Alternatively, we may introduce a
small imaginary term −iεx
2
in the Hamiltonian so that the eigenvalue has a small
negative imaginary part. Then only the ground state survives in the summations
over m and n under τ
f
→∞,τ
i
→−∞.
After all we have proved that
lim
t
f
→∞
t
i
→−∞
x
f
, t
f
|x
i
, t
i
J
=x
f
|00|x
i
Z [J] (1.125)
wherewehavedefinedthegenerating functional
Z[J ]=0|U
J
(b, a)|0= lim
t
f
→∞
t
i
→−∞
0|U
J
(t
f
, t
i
)|0. (1.126)
The generating functional may be also expressed as
Z[J ]= lim
t
f
→∞
t
i
→−∞
x
f
, t
f
|x
i
, t
i
J
x
f
|00|x
i
. (1.127)
Note that the denominator is just a constant independent of Z [J ].Nowwehave
found the path integral representation for Z [J ],
Z[J ]=
x e
iS[x, J ]
(1.128)
where the path integral is over paths with arbitrarily fixed x
i
and x
f
.The
normalization constant
is chosen so that Z[0]=1, namely
−1
=
x e
iS[x,0]
.
It is readily shown that Z[J ] generates the matrix elements of the T -product
between the ground states:
0|T [x(t
1
) ···x(t
n
)] |0=(−i)
n
δ
n
δ J (t
1
) ···δ J(t
n
)
Z[J ]
J =0
. (1.129)
1.4 Harmonic oscillator
We work out the path integral quantization of a harmonic oscillator, which is an
example of systems for which the path integral may be evaluated exactly. We also
introduce the zeta function regularization, which is a useful tool in many areas of
theoretical physics.
1.4.1 Transition amplitude
The Lagrangian of a one-dimensional harmonic oscillator is
L =
1
2
m ˙x
2
−
1
2
mω
2
x
2
. (1.130)
The transition amplitude is given by
x
f
, t
f
|x
i
, t
i
=
x e
iS[x(t )]
, (1.131)
where S[x(t)]=
t
f
t
i
L dt is the action.
Let us expand S[x] around its extremum x
c
(t) satisfying
δS[x]
δx
x=x
c
(t)
= 0. (1.132)
Clearly x
c
(t) is the classical path connecting (x
i
, t
i
) and (x
f
, t
f
) and satifies the
Euler–Lagrange equation
¨x
c
+ ω
2
x
c
= 0. (1.133)
The solution of equation (1.133) satifying x
c
(t
i
) = x
i
and x
c
(t
f
) = x
f
is easily
obtained as
x
c
(t) =
1
sin ωT
[x
f
sin ω(t − t
i
) + x
i
sin ω(t
f
− t)] (1.134)
where T = t
f
−t
i
. Substituting this solution into the action, we obtain (exercise)
S
c
≡ S[x
c
]
=
mω
2sinωT
[(x
2
f
+ x
2
i
) cos ωT − 2x
f
x
i
]. (1.135)