Physics 827
Problem Set 9
Due Friday 12/04/2009
9.1 Shankar 10.1.3 (page 258) (Important)
This is straightforward algebra. Just observe that
Ã
p21
+
p22
=
pI + pII
√
2
!2
Ã
+
pI − pII
√
2
!2
= p2I + p2II
with a similar result for x21 + x22 . This is obvious since we are rotating by π/4 and the
norm of the vector is preserved. Lazy that I am I would have said this and skipped
the algebra. Promoting the classical coordinates and momenta to quantum operators
the Hamiltonian becomes
PI2
P2
1
2
].
+ II + mω 2 [XI2 + 3XII
2m
2m
2
The Schrödinger equation is
h̄2
−
2m
Ã
∂2
∂2
+
∂x2I
∂x2II
!
ψE (xI , xII ) +
1
mω 2 [x2I + 3x2II ] ψE (xI , xII ) = E ψE (xI , xII ) .
2
Since the Hamiltonian is separable (this was the point of finding the normal coordinates) the total energy is additive and we can write ψE (xI , xII ) = ψEI (xI ) ψEII (xII )
and E = EI + EII . We obtain two decoupled equations and we can write down the
energies.
(b) Elementary calculus yield
∂
∂xI ∂
∂xII ∂
1 ∂
1 ∂
=
+
= √
+ √
.
∂x1
∂x1 ∂xI
∂x1 ∂xII
2 ∂xI
2 ∂xII
∂
∂xI ∂
∂xII ∂
1 ∂
1 ∂
=
+
= √
− √
.
∂x2
∂x2 ∂xI
∂x2 ∂xII
2 ∂xI
2 ∂xII
Therefore,
∂2
∂2
∂2
∂2
+
=
+
.
∂x21
∂x22
∂x2I
∂x2II
This yields the kinetic energy operator. The potential energy arises as before
Ã
x21
+
x22
2
+ (x1 − x2 ) =
xI + xII
√
2
!2
Ã
+
1
xI − xII
√
2
!2
+
³√
2xII
´2
= x2I + 3x2II .
9.2 Shankar 11.4.2
Clearly, the infinitesimal generator of translations, PX does not commute with the
Hamiltonian. [PX , H] 6= 0 since ∂/∂x does not commute with V (x) = V0 sin(2πx/a).
The Hamiltonian is invariant under translation by integer multiples of the basic period
√
a, i.e., [Ta , H] = 0. Given a plane wave eipx/h̄ / 2πh̄, the Hamiltonian H has nonzero matrix elements between this state and only the states with momentum given by
p ± (2πh̄)/a. Thus, one can conclude that the momentum is conserved up to integer
multiples of 2πh̄/a.
9.3 Read section 12.3 (p313-315) and do Shankar 12.3.3 and 12.3.4 (These are
important)
Consider the matrix element
1 Z 2π
2
2
hm|ψi = √
dφ e−imφ cos2 φ A e−ρ /(2∆ ) ;
2π 0
|hm|ψi|2 × ρdρ yields the probability for the particle to be found with an Lz value of
mh̄ and the value of the radial co-ordinate between ρ and ρ + dρ. Note the factor
ρdρ in two dimensions. Thus the probability to be in the state |mi is given by the
integral over ρ since we do not care what the radial coordinate is. The wave function
is separable and therfore, the constant arising from the ρ integral is independent of m
and thus the ρ-dependence can be ignored. So we can focus on the angular dependence
only:
³
´2
ψ(φ) ∝ eiφ + e−iφ ∝ e2iφ + 2 + e−2iφ .
Thus, the probabilities to obtain the values 2h̄, 0, and −2h̄ are in the ratio 1 : 4 : 1.
Normalization yields the answer.
The wave function in 12.3.4 can be written as
·µ
¶
µ
¶
¸
ρ
ρ
−ρ2 /(2∆2 ) 1
iφ
−iφ
ψ(ρ, φ) = A e
−i e +
+i e
.
2
∆
∆
Once more the probability of obtaining Lz value of +h̄ and the radial co-ordinate lying
between ρ and ρ + dρ is given by
2
2
π
2
2 ρ + ∆
|A|2 e−ρ /∆
× ρdρ .
2
∆2
The corresponding probability for −h̄ is the same: the key observation is that |(ρ/∆) −
i| = |(ρ/∆) + i|. Thus the two probabilities are equal.
2