PHY{396 L. Solutions for homework set #6.
Problem 1(a):
Think of box's walls as mirrors. Every time the particle bounces o a wall, let us look
on its reection instead. The reected image y(t) of the particle's actual trajectory x(t) is
described by
dy = dx ( 1)# bounces thus far:
(S:1)
dt
dt
Thus, before the rst bounce y = x, after the rst bounce but before the second y = 2a x
or y = x (depending on which wall the particle bounces o of), after the second bounce but
before the third y = x or y = x 2a, etc., etc. At the end of the path, y0 x0 modulo 2a
for an even number of reections and y0 x0 modulo 2a for an odd number of reections.
Let 0 x0; x0 a; it is easy to see that for any imagined trajectory y(t) from y0 = x0
to y0 = 2na x0 there is precisely one real trajectory x(t) from x0 to x0 that is conned to
the box at all times. Therefore
0
Z
Zx
x0
D[x(t)]box =
and since
S [x(t)] =
0
+1 X 2naZZx
X
n= 1 Zt
t0
x0
D[y(t)]free
(S:2)
0
m x_ 2 dt
2
= S [y(t)];
(S:3)
it follows that
0
ZZx
Ubox (t0; x0; t0; x0) = D[x(t)]box eiS[x(t)]=h
=
x0
0
+1 X 2naZZx
X
D[y(t)]freeeiS[y(t)]=h
n= 1 x0
+1 X
X
=
Ufree t0; (2na x0); t0; x0 ;
n= 1 1
(S:4)
cf.
eq. (1), Q:E :D:
Problem 1(b):
As explained in class,
+1
X
n= 1
Ufree t0; (na + x0); t0; x0 =
+1
X
1 e2i`(x0 x0)=a e 22 ihT`2 =ma2
`= 1 a
(S:5)
where T = t0 t0. Let us apply this formula to eq. (1) by substituting a 7! 2a and x0 7! x0.
Thus,
+1
X X
1 e2i`(x0x0)=2a e 22ihT`2 =4ma2
0
0
Ubox (t ; x ; t0; x0) =
`= 1 2a
+1
2 X
1 = 2a + 2a e+2i`x0=2a + e 2i`x0=2a
`=1
e 22 ihT`2 =4ma2
=
+1
X
`=0
0
` (x0 ) ` (x ) e
e
2i`x0 =2a + e+2i`x0 =2a
iTE` =h
(S:6)
where
h `)2 ;
E` = (2ma
2
0 (x)
r
1 and
a
` (x)
=
r
2 cos `x for ` > 0:
a
a
(S:7)
In other words, the evolution kernel (1) describes a quantum particle in a box with Neumann
boundary conditions.
Problem 1(c):
We already mentioned that when the particle bounces o the walls an even number of times
2
y0 = 2na + x0 while for an odd number of bounce-os t0 = 2na x0. Therefore,
0
ZZx
0 (t0 ; x0 ; t0; x0 ) = D[x(t)]box eiS [x(t)]=h ( 1)#bounces
Ubox
=
=
x0
+1 X
X
2naZZx0
n= 1 +1 X
X
x0
()
D[y(t)]freeeiS[y(t)]=h
()Ufree t0; (2na x0 ); t0; x0 ;
n= 1 +1
X
X
1 e2i`(x0x0)=2a e 22ihT`2 =4ma2
= ()
`= 1 2a
+1 1 X
0
0 = 2a e+2i`x0=2a e 2i`x0 =2a e 2i`x =2a + e+2i`x =2a
`=1
e 22ihT`2 =4ma2
=
+1
X
`=1
0
`(x0 ) ` (x ) e
(S:8)
iTE` =h
where the energies E` are exactly as in eq. (S.7) | but excluding the E0 = 0 | and
` (x)
=
r
2 sin `x :
a
a
(S:9)
And this is of course precisely the spectrum of the box with Dirichle boundary conditions.
Problem 2(a):
Let x(t) = xcl (t)+ y(t) where xcl (t) satises the Newton equation mxcl (t) = f (t) m!2xcl(t)
as well as the initial and nal conditions xcl (t0) = x0 and xcl (t0) = x0 . Then clearly
x(tZZ0 )=x0
D[x(t)] =
x(t0 )=x0
3
y(ZtZ0 )=0
D[y(t)] :
y(t0 )=0
(S:10)
At the same time,
Zt
0
1 2 2
2 m! x + fx
S [x(t)] = dt 12 mx_ 2
t0
Zt
(S:11)
0
= S [xcl(t)] + 0 + dt 21 my_ 2 12 m!2y2
t0
and hence
U (t0 ; x0; t0; x0) =
=
x(tZZ0 )=x0
D[x(t)] eiS[x(t)]=h
x(t0 )=x0
eiS[xcl (t)]=h eiS
cl
=h A(t0
y(tZZ0 )=0
t0
Z
im
D[y(t)] exp 2h dt (y_ 2 !2 y2)
(S:12)
t0
y(t0 )=0
t0);
eq. (4). Note that the factor A(t0 t) depends only on the elapsed time but not on the
external force f (t), the classical trajectory xcl(t) or even the initial and nal points x0 and
x0.
cf.
Problem 2(b):
According to eq. (S.12),
A(t0 t0) =
where
y(tZZ0 )=0
D[y(t)] eiS [y(t)]=h
0
(S:13)
y(t0 )=0
Zt
0
S0[y(t)] = m2 dt (y_ 2 !2 y2):
t0
(S:14)
is a quadratic functional of the function y(t). To diagonalize this quadratic functional, we
expand y(t) into a Fourier series; in light of the initial and nal conditions y(t0) = y(t0) = 0,
4
we write
y(t) =
1
X
n=1
yn sin n
(t t0)
(S:15)
where = =(t0 t0). Consequently,
1
0 t0 ) X
m
(
t
2 2 ! 2 )y 2
(
n
S0 =
n
4
n=1
(S:16)
and therefore
A(t0
t0) = J
= J
= J
Y
1Z
n=1
1 Z
Y
dyne
n=1 s
1
Y
n=1
dyn
1
0 t0 ) X
im
(
t
2 2 ! 2 )y 2
exp
(
n
n
4h n=1
im(t0 t0 ) (n2 2
4
h
m(t0
!2 )yn2
(S:17)
4ih
t0) (n2 2 !2 )
Q
where J is the Jacobian of the linear variable transform from D[y(t)] to dyn . Note that
both the Jacobian J and the innite product on the last line of eq. (S.17) are badly divergent,
so what we really have here is
N
Y
s
4ih
A(t0 t0) = Nlim
J
N
0
!1 n=1 m(t t0 ) (n2 2 ! 2 )
(S:18)
for some `partial' Jacobians JN . By construction of the path integral, these Jacobians may
depend on the elapsed time t0 t0 and the particle's mass m, but they cannot possibly
depend on the potential energy V (x) of the harmonic oscillator. Consequently, for a free
particle we would have
N
Y
s
4ih
Afree(t0 t0) = Nlim
J
N
0
!1 n=1 m(t t0 ) (n2 2 0)
5
(S:19)
with exactly same Jacobians JN and therefore
N
Aosc(t0 t0) = Y
Afree(t0 t0) n=1
s
n2
2 =
n2
2 !2
"
Y
1
!(t0 t0)
n
2 !
0
= sin!(!t(0 t t t)0)
0
#
1=2
:
(6)
Please note that this is the correct eq. (6); the printed homework version distributed in class
had this eq-n wrong!
p
On the other hand, for a free particle, A(t0 t0) = m=2ih (t0 t0), hence
Aosc(t0
t0) =
r
m!
2ih sin !(t0 t0) :
(7)
Q:E :D:
Problem 2(c):
Without an external force f (t), the classical trajectory of an un-forced oscillator is simply
xcl (t) = A cos !(t t0) + B sin !(t t0)
(S:20)
where the initial and the nal conditions impose
0 x0 cos ! (t0 t0 )
B = x sin
!(t0 t0) :
A = x0 ;
(S:21)
Consequently,
0
Zt
Scl = dt 12 mx_ 2cl
t0
1 m! 2 x2 cl
2
m!
= 2 sin
!T
2x0x1 + (x20 + x02) cos !T
(S:22)
where T = t0 t0 and hence
0
x U^ (T )
jx0i =
r
m! exp
2ih sin !T
im! 2x x
2h sin !T 0 1
6
(x20 + x02 ) cos !T
: (S:23)
Now consider the trace of the evolution operator:
Z
tr U^ (T ) = dx hxj U^ (T ) jxi
r
Z
m!
im!
2
= 2ih sin !T dx exp 2h sin !T 2x (1 cos !T )
s
1
i
=
= 2(1 cos
!T ) 2 sin( 21 !T )
=
1
X
n=0
e
i(n+ 12 )!T :
In light of eq. (8), this trace implies En = h !(n + 12 ), Q:E :D:
7
(S:24)