Physics 216
Solutions to Problem Set 1
Spring 2010
1. Prove that
(
x,t

T
[
X
(
t
1
)
X
(
t
2
)
···
X
(
t
n
)]

x
′
,t
′
)
=
integraldisplay
D
[
x
(
t
)]
x
(
t
1
)
x
(
t
2
)
···
x
(
t
n
)
e
iS
[
x
(
t
)]
/
planckover2pi1
,
where
T
is the timeordered product symbol,
S
[
x
(
t
)] is the action [which depends on
the path
x
(
t
)],
X
(
t
) is the position operator in the Heisenberg picture, and
x
(
t
) is the
eigenvalue of
X
(
t
) when acting on the position eigenstate

x,t
)
. Assume that
t
≥
t
i
and
t
′
≤
t
i
for all
i
= 1
,
2
,...,n
.
In general, the Heisenberg position operators
X
(
t
), evaluated at different times,
do not commute. Therefore, a product of Heisenberg position operators, where each
operator is evaluated at a different time, will depend on the order in which the
operators appears.
One particular order is one in which the largest time appears
to the left and all the other operators appear in time order (i.e., time decreases as
we move from left to right). This particular order corresponds to the timeordered
product,
T
[
X
(
t
1
)
X
(
t
2
)
···
X
(
t
n
)] =
X
(
t
i
1
)
X
(
t
i
2
)
···
X
(
t
i
n
)
,
where
t
i
1
>t
i
2
>
···
>t
i
n
.
Consider the derivation of the path integral for
(
x,t

T
[
X
(
t
1
)
X
(
t
2
)
···
X
(
t
n
)]

x
′
,t
′
)
=
(
x,t

X
(
t
i
1
)
X
(
t
i
2
)
···
X
(
t
i
n
)

x
′
,t
′
)
.
The problem states that
t
≥
t
i
1
and
t
′
≤
t
i
n
. We can begin the derivation of the path
integral in the same way as presented in the class lecture, by dividing up the time
interval between
t
′
and
t
into slices:
t
′
≤
t
i
n
≤
t
i
n
−
1
≤···≤
t
i
2
≤
t
i
1
≤
t.
In contrast to the derivation presented in class, we do not assume that
t
i
k
=
t
+
kǫ
(i.e., uniform time slices), since the
t
i
k
(
k
= 1
,
2
,...,n
) are fixed by the product of
the Heisenberg position operators.
Next, we insert a complete set of states,
integraldisplay
∞
−∞

x
i
k
,t
i
k
)(
x
i
k
,t
i
k

dx
i
k
,
immediately following each Heisenberg position operator (for
k
= 1
,
2
,...,n
). Using
X
(
t
i
k
)

x
i
k
,t
i
k
)
=
x
(
t
i
k
)

x
i
k
,t
i
k
)
,
where
x
(
t
i
k
) is the (
c
number) eigenvalue of
X
(
t
i
k
), it follows that:
T
[
X
(
t
1
)
X
(
t
2
)
···
X
(
t
n
)] =
integraldisplay
∞
−∞
dx
i
1
integraldisplay
∞
−∞
dx
i
2
···
integraldisplay
∞
−∞
dx
i
n
x
(
t
i
1
)
x
(
t
i
2
)
···
x
(
t
i
n
)
×(
x,t

x
i
1
,t
i
1
)(
x
i
1
,t
i
1

x
i
2
,t
i
2
)···(
x
i
n
,t
i
n

x
′
,t
′
)
.
(1)
1
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Since the
x
(
t
i
k
) are commuting
c
numbers, we can rewrite eq. (1) as:
T
[
X
(
t
1
)
X
(
t
2
)
···
X
(
t
n
)] =
integraldisplay
∞
−∞
dx
i
1
integraldisplay
∞
−∞
dx
i
2
···
integraldisplay
∞
−∞
dx
i
n
x
(
t
1
)
x
(
t
2
)
···
x
(
t
n
)
×(
x,t

x
i
1
,t
i
1
)(
x
i
1
,t
i
1

x
i
2
,t
i
2
)···(
x
i
n
,t
i
n

x
′
,t
′
)
.
To make further progress, we now divide up the entire interval from
t
′
to
t
into
N
equally spaced time intervals (each interval of size
ǫ
),
T
J
=
t
′
+
Jǫ,
where
J
= 1
,
2
,...,N
−
1
,
where
N
≫
n
. If
N
is large enough, then each one of the times (
t
1
,t
2
,...,t
n
) that
appears in the time ordered product of Heisenberg position operators will approxi
mately coincide with one of the
T
J
(up to an error that can be made arbitrarily small
in the
N
→∞
limit). Then insert the appropriate complete set of states,
integraldisplay
∞
−∞

X
J
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 Spring '09
 Physics, wkb approximation, EQ

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