PHYS 212A: Homework 2
October 29, 2013
1.18
a
Sakurai gives the proof within the text in 1.4.56.
b
See Sakurai’s derivation of the general uncertainty principle in section 1.4. Since expectation values
are commutitive numbers (or by elementary calculation) we have
[A, B] = [∆A, ∆B].
(1)
hα| [∆A, ∆B] |αi = λ∗ hα| ∆B∆B |αi − λ hα| ∆B∆B |αi
(2)
hα| {∆A, ∆B} |αi = λ∗ hα| ∆B∆B |αi + λ hα| ∆B∆B |αi
(3)
Thus, we can evaluate
and
For λ purely imaginary, the two terms will cancel and the expectation of the anticommutator
becomes zero. Now look at line 1.4.63 in Sakurai’s derivation. Since the anticommutator term
vanishes, the equality is exact.
c
We’ll expand the expectation values over the complete set hx00 |, with the normalization condition
hx0 |x00 i = δ(x0 − x00 ). Now consider
Z
∂
0
(4)
hx | ∆p |αi = dx00 δ(x0 − x00 )(−i~ 00 − < p >) hx00 |αi
∂x
Taking advantage of the explicit form of hx00 |αi given in the problem, we can evaluate the derivative
term explicitly to find
hx0 | ∆p |αi
=
R
00
2(x −hxi)
dx00 δ(x0 − x00 )(−i~( ip
− < p >) hx00 |αi
~ −
4d2
R
i~
= 2d
dx00 δ(x0 − x00 )(x00 − hxi) hx00 |αi
2
=
i~
2d2
hx0 |∆x|αi
1
1.21
p
Recall that for the infinite square well, the wave functions are of the form (2/a) sin(nπx/a).
To evaluate the uncertainty product we’ll take advantage of the fact that h(∆x)2 i = hx2 i − hxi2 ,
and h(∆p)2 i = hp2 i − hpi2 . Since we know the explicit form of the wavefunctions we can evaluate
integrals to find expectation values. The results of this procedure are
hx2 i = a2 [1/3 − 1/2n2 π 2 ]
(5)
hxi = a/2
(6)
hp2 i = ~2 /a2(nπ)2
(7)
hpi = 0
(8)
Combining these results, we can calculate the uncertainty product with the result
h(∆x)2 i h(∆p)2 i =
~2
[(nπ)2 /6 − 1]
2
(9)
1.28
a
[x, F (px )]CL =
∂x ∂F
∂F
∂x ∂F
=
−
∂x ∂px ∂px ∂x
∂px
(10)
b
Using i~[x, F (px )]CL = [x, F (px )]QM we have
[x, exp(ipx a/~)]QM = i~
∂
exp(ipx a/~) = −a exp(ipx a/~)
∂px
(11)
c
From part b we have
[x, exp(ipx a/~)] |x0 i = −a exp(ipx a/~) |x0 i
(12)
which can be rearranged to yield
x[exp(ipx a/~) |x0 i] = (x0 − a)[exp(ipx a/~) |x0 i].
(13)
This last equality implies that exp(ipx a/~) |x0 i is an eigenstate of the coordinate operator x with
the corresponding eigen value (x0 − a).
1.30
The solution to this problem is nearly identical to the solution of 1.28 b and c.
2
1.33
a
We’ll start by expanind hp0 |x|p00 i over a complete set of x0 .
hp0 |x|p00 i
R
=
hp0 |x|x0 i hx0 |p00 i dx0
x0 hp0 |x0 i hx0 |p00 i dx0
R 0 0 −ix0 (p0 −p00 )/~
1
= 2π~
dx x e
=
R
Now we recognize
1
2π~
Z
0
0
dx0 e−ix (p −p
00 )/~
= δ(p0 − p00 )
(14)
which implies
hp0 |x|p00 i = i~
∂
δ(p0 − p00 )
∂p0
(15)
We now expand over a complete set of momentum states and can write
=
R
hp0 |x|αi
R
dp00 hp0 |x|p00 i hp00 |αi = dp00 i~ ∂ 0 δ(p0 − p00 ) hp00 |αi
∂p
∂
0
= i~ 0 hp |αi
∂p
For part ii we have
hβ|x|αi
=
R
R
dp0 hβ|p0 i hp0 |x|αi
dp0 hβ|p0 i i~ ∂ 0 hp0 |αi
∂p
R 0 ∗ 0
= dp φβ (p )i~ ∂ 0 φ∗α (p0 )
∂p
=
b
This portion of the problem is nearly identical to 1.28, only with the roles of position and momentum
switched. Evaluation of the commutator [x, exp(ixΞ/~)] yields the result,
p[exp(ixΞ/~) |p0 i] = (p0 + Ξ)[exp(ixΞ/~) |p0 i]
(16)
which shows that the operator exp(ixΞ/~) is the momentum translation operator and that x is the
generator of momentum translation.
3