Quantum Mechanics PHYS 212B
Problem Set 2
Due Thursday, January 21, 2016
Exercise 2.1 Consider a two state system governed by Hamiltonian H and with energy eigenstates |E1 i
and |E2 i where H|E1 i = E1 |E1 i and H|E2 i = E2 |E1 i. Two other states are:
|E1 i + |E2 i
√
2
|E1 i − |E2 i
√
|yi =
2
|xi =
At time t = 0 the system is in state |xi. At what subsequent times is the probability to find the system in
state |yi biggest and what is this probability?
Solution 2.1
Time evolution operator
u(t, t0 ) = e−iH(t−t0 )/~
The state of the system |ψ(t)i with |ψ(t = 0)i = |xi,
|ψ(t)i = e−iHt/~ |xi
Then
1
(hE1 | − hE2 |)e−iHt/~ (|E1 i + |E2 i)
2
1 −iE1 t/~
=
e
− e−iE2 t/~
2
hy|ψ(t)i =
The probability to find state |yi,
P (|yi) = |hy|ψ(t)i|2 =
(E1 − E2 )t
1
1 − cos
2
~
The biggest probability
Pmax (|yi) = 1
when
t=
(2n + 1)π~
.
E1 − E2
Exercise 2.2 The usual Hydrogen-like wave functions for single atoms are derived with the assumption
that finite size of the nucleus can be neglected. Model the nucleus as an uniformly charged spherical shell
of radius R0 . What is the first order shift in energy of the 2s state? Assume that R0 a0 , where
a0 = 4π0 (~c)2 /(me c2 Z 2 e2 ) is the appropriate Bohr radius for a single electron atom with nuclear charge Z.
The spatial wave function of the unperturbed state is
3/2 Z
Zr
u(r) = (32π)−1/2
2−
e−Zr/2a0 .
a0
a0
1
Solution 2.2
The perturbation is
H0 =
Ze2
, 0 < r < R0
4π0 r
Then the first order energy shift is
h2s|H 0 |2si = 4π
Z
R0
u(r)H 0 u(r)r2 dr
0
Z
R0
= 4π
0
Z
a0
Z 2 e2
32π0 a0
Z
1
32π
3 Zr
2−
a0
2
Zr
e− a0
Ze2 2
r dr
4π0 r
Let r0 = Zr/a0 ,
h2s|H 0 |2si =
2 2
0
(2 − r0 )2 r0 e−r dr0
0
Z e
32π0 a0
Z
Z 2 e2
≈
32π0 a0
Z 4 e2 R02
=
16π0 a30
Z
≈
ZR0
a0
ZR0
a0
(2 − r0 )2 r0 (1 − r0 )dr0
0
ZR0
a0
4r0 dr0
0
0
since R0 a0 , i.e. r0 1, e−r ≈ 1 − r0 and only the first-order terms of r0 are kept in the integral kernel.
2