PHY4604–Introduction to Quantum Mechanics
Fall 2004
Test 3 SOLUTIONS
Dec. 3, 2004
No calculators or other materials. If you don’t recall a formula, ask and I might be
able to help. If you can’t do one part of a problem, solve subsequent parts in terms of
unknown answer–define clearly. Each part is worth 10 points, for max=70. Problem
1 required, attempt 1 of remaining 2 problems; circle which ones you want graded.
Possibly helpful formulae and constants
Pn = |hn|ψi|2
q
L± |`mi = h̄ `(` + 1) − m(m ± 1)|`m ± 1i
L± = Lx ± iLy
X
1=
|nihn|

n
"
2
m
e
En = −  2
2h̄ 4π²0
#2 
1

n2
Hψ = Eψ
∂
Hψ = ih̄ ψ
∂t
p̂ = −ih̄∇
ψ(x) = hx|ψi
Z ∞
0
dy y 2 e−y = 2
Z
|`1 − `2 | ≤ ` ≤ `1 + `2
a†
≡
a
s
k
h̄ ∂
x∓ √
2
2m ∂x
[L2 , Lα ] = 0
ψ=
X
c n ψn
n
−iHt/h̄
ψ(x, t) = e
ψ(x, 0)
2
R10 = q
e−r/a0
3
4πa0
[Li , Lj ] = ih̄²ijk Lk
s
Y10 =
3
cos θ
4π
1
dp
√
eipx/h̄ φ(p, t)
2πh̄
Z
dx
φ(p, t) = √
e−ipx/h̄ ψ(x, t)
2πh̄
h̄2 2
H=−
∇ + V (r)
2m
(Ô† φ, χ) = (φ, Ôχ)
ψ(x, t) =
a0 = h̄2 /(mZe2 )
ψn`m = Rn` (r)Y`m (θ, φ)
s
Y11 =
3
sin θ eiφ
8π
s
R21 =
µ
1 1
3 2a0
¶3/2
r −r/2a0
e
a0
1. Short answer. Must attempt (only) 3 of 5.
(a) An electron moving in the Coulomb field of a proton is in a state described
by the wave function
√
1
Ψ = [4ψ100 − 3ψ21 −1 − ψ210 + 10ψ211 ]
6
(1)
i. What is the expectation value of the energy?
1
(16h100|H|100i + 9h21 − 1|H|21 − 1i + h210|H|210i
36
+ 10h211|H|211i)
1
1
=
(16E1 + 9E2 + 1E2 + 10E2 ) = (16E1 + 20E2 )
36
36
7
= −
Ryd.
12
hΨ|H|Ψi =
P
Note this is an example of general rule hψ|O|ψi = n on |cn |2 , where
the on are the eigenvalues of the Hermitian operator O, and the cn the
expansion coefficients of ψ in the basis of O eigenstates.
ii. What is the expectation value of L̂z ?
hψ|Lz |ψi =
X
h̄m|cn`m |2 =
n`m
h̄
h̄
(16 · 0 + 9 · −1 + 1 · 0 + 10 · 1) =
36
36
iii. What is L̂+ ψ?
q
h̄
[0 − 1(1 + 1) − (−1)(−1 + 1)3ψ210
6q
− 1(1 + 1) − (0)(0 + 1)ψ211 + 0]
√
h̄ 2
= −
[3ψ210 + ψ211 ]
6
L+ ψ =
(b) The allowed eigenstates of a single particle Hamiltonian H0 (1) are ψn (r)χσ ,
where χσ is a spin wave function for σ =↑, ↓, and the eigenvalues of the
Hamiltonian are En , increasing monotonically with n starting from the
ground state energy E0 , and do not depend on σ. If a second particle
is now added to the system without interacting explicitly with the first,
i.e. H = H0 (1) + H0 (2), write down a) the ground and excited state
wave functions if both spins are up; b) the ground and excited state wave
functions for one spin up and one spin down. In both a) and b), state
whether the wave functions you write down are spin singlet or spin triplet.
2
Ground state wave function for two spins up will be Ψt0 (1, 2) = A(ψ0 (r1 )ψ1 (r2 )−
ψ0 (r2 )ψ1 (r1 ))χ↑↑ , due to Pauli principle, where A is normalization coefficient. Putting both particles in ψ0 would violate Pauli. 1st excited state
would be Ψ1 (1, 2)t = B(ψ0 (r1 )ψ2 (r2 ) − ψ0 (r2 )ψ2 (r1 ))χ↑↑ . Both are spin
triplet (S = 1). b) For opposite spins we can have either a symmetric spatial ground state, Ψs0 = Aψ0 (r1 )ψ0 (r2 ) √12 (χ↑↓ − χ↓↑ ), spin singlet
(S = 0), or an antisymmetric spatial gound state, Ψt0 = A0 (ψ0 (r1 )ψ1 (r2 ) −
ψ1 (r1 )ψ0 (r2 )) √12 (χ↑↓ + χ↓↑ ) (S = 1). The 1st excited states in this case
would be either Ψs1 = B(ψ0 (r1 )ψ1 (r2 )+ψ0 (r2 )ψ1 (r1 )) √12 (χ↑↓ −χ↓↑ ), spin singlet (S = 0), or an antisymmetric spatial gound state, Ψt1 = A0 (ψ0 (r1 )ψ2 (r2 )−
ψ2 (r1 )ψ0 (r2 )) √12 (χ↑↓ + χ↓↑ ) (S = 1).
(c) The electron in a hydrogen atom occupies a state (ignoring spin)
s
1
ψ = R21 
Y10 +
3
s

2 
Y11
3
i. If you measured the orbital angular momentum squared (L2 ), what
values could you get and what is the probability of each?
Only h̄2 `(` + 1) = 2h̄2 corresponding to ` = 1, since this is the only
angular momentum present in the wave function.
ii. What is the expectation value for L2 ?
Again 2h̄2 , since the action of L2 on the normalized wave function
yields 2h̄2 with probablility 1.
iii. If you looked for the particle at r = 0, what is the probability density
for finding it there?
0, since R21 ∝ r.
(d) You are given two particles in angular momentum states |`1 = 1, m1 = 0i
and |`2 = 3, m2 = 0i.
i. What are the possible values of the total angular momentum quantum
number `? What are the allowed eigenvalues of the total angular
momentum L2 , where L = L1 + L2 ?
By the triangle rule, ` can be between |`1 −`2 | and `1 +`2 , i.e. between
2 and 4, i.e. 2,3,4 are allowed. The corresponding L2 eigenvalues are
6h̄2 , 12h̄2 , and 20h̄2 .
ii. What are the possible values of the total angular momentum quantum
number m? What are the allowed eigenvalues of Lz ?
3
Only zero for both the m value m = m1 +m2 = 0 and the Lz eigenvalue
h̄m=0.
(e) For addition of two angular momenta `1 = 2, `2 = 1, use the table of
Clebsch-Gordon coefficients provided, and state what a and b and c are in
the following:
|m1 = 1, m2 = 0i = a|` = 3, m = 1i + b|` = 2, m = 1i + c|` = 1, m = 1i
a=
q
8/15, b =
q
q
1/6, c = − 3/10.
Attempt 1 of remaining 2 4-part problems!
2. Angular momentum 1. Consider an angular momentum 1 system.
(a) What are the possible eigenvalues of L2 and Lz corresponding to the eigenvectors |` = 1, mi?
Eigenvalue of L2 is h̄2 `(` + 1) = 2h̄2 , eigenvalue of Lz is mh̄, m = −1, 0, 1.
(b) In the basis where the eigenvectors |` = 1, mi of the operator L̂z are given
by (1, 0, 0), (0, 1, 0) and (0, 0, 1), construct the matrix representation of
the operator L̂y (Hint: you will need to calculate the matrix elements
hm|L̂y |m0 i. Express Ly in terms of L± .)
L = (L+ − L− )/(2i), so we calculate the matrix elements using L± |`mi =
`(` + 1) − m(m ± 1)|`m ± 1i. Note only the matrix elements between
states one unit of angular momentum apart are nonzero. Result is
qy

Ly

0
1 0
h̄ 

= i √  −1 0 1 
2
0 −1 0
(c) Find the eigenvalues and eigenvectors of L̂y , and indicate which corresponds to which!
In order: eigenvalues are −h̄, 0, h̄. Corresponding (normalized) eigenvectors are

ψ−1

−1
√ 
1
=
 −i 2 
2
1
4
and

ψ0

1
1  
√
=
 0 
2 1
and

ψ1

−1
1 √ 
=
 i 2 
2
1
(d) If a system is prepared in the state vector


1
1 

|ψi = √  4  ,
26 −3
what is the probability that a measurement of L̂y yields the value 0?
√
hψ0 |ψi = −1/ 13,
so P0 = |hψ0 |ψi|2 = 1/13.
3. Nuclear transition
(a) Given a central potential V (r), and a 2-particle Hamiltonian, H = p21 /2m1 +
p22 /2m2 + V (|r1 − r2 |), discuss the reduction of the problem to the effective
1-particle problem in terms of the center of mass momentum P, center of
mass position R, relative momentum p, relative coordinate r, total mass
M , and reduced mass µ. No derivations are required.
The Hamiltonian may be expressed as
H =
P2
p2
+
+ V (r),
2M
2µ
meaning that if we work in the frame where the total momentum is zero,
the Hamiltonian is just that of a single particle with momentum p moving
in a potential V (r).
(b) Compare the wavelengths of the 2p → 1s transitions in
i. hydrogen
5
The energy which must be carried away by a photon in the transition
is E2 − E1 = hc/λ, so
λH = hc/(E2 − E1 ) =
8h̄3
3m
µ
4π²0
e2
¶2
.
Note I have used m instead of µ, because µ = mmp /(m + mp ) is equal
to m up to an error of order O(10−3 ).
ii. deuterium (mass of nucleus 2× that of H). Give result in terms of λH
of part (a).
Changing mp → 2mp for deuterium doesn’t change the fact that the
nucleus is still 103 times heavier than the electron, so to roughly the
same order of approximation as before, the photon wavelength doesn’t
change, λD ' λH .
iii. positronium (electron and positron (same mass as electron, opposite
charge) bound together). Again give result in terms of λH of part (a).
In this case the positron is not heavier than the electron, but in fact
has the same mass. The reduced mass of 2 identical particles of mass m
is m/2, so we obtain the correct answer from the Hydrogen expression
by substituting m → m/2, or λpos = 2λH .
(c) Consider now an electron in the ground state of tritium, a form of heavy
hydrogen where the nucleus has two neutrons in addition to a proton,
and compare with an electron in the ground state of ionized 3 He, i.e. an
electron orbiting a nucleus with two protons and one neutron. Compare the
ground state Bohr radii of tritium and ionized 3 He, neglecting changes of
O(m/mn ), where m is the electron mass and mn is the mass of a nucleon.
Sketch the ground state wavefunctions ψ100 (r) for the two isotopes as a
function of r, clearly indicating qualitative differences.
6
Z, the nuclear charge or number of protons, is 1 for tritium and 2 for 3 He.
Thus the Bohr radius a0 = h̄2 /(mZe2 ) for 3 He is twice as small as for
3
tritium or hydrogen, a0He ' 12 at0 . The electron in the 3 He ion is more
tightly bound to the nucleus.
(d) A nuclear reaction (β-decay) changes a tritium nucleus instantaneously to
3
He. If the electron orbiting the tritium was in its ground state when the
reaction took place, calculate the probability that it remains in the ground
state of 3 He.
We’re told the system starts out in the tritium ground state,
2
t
ψ100
= q
t
3
4π(aH
0 )
e−r/a0 ,
which can be expressed as a linear combination of any complete set of states
in Hilbert space, for example the eigenstates of the 3 He Hamiltonian,
3
3
3
t
He
He
He
ψ100
= aψ100
+ bψ200
+ cψ210
+ ...
meaning the electron in the tritium ground state will be found after a
measurement in the 3 He ground state with probability amplitude
3
He
t
i
|ψ100
a = hψ100
µ
Z
4
1
q
d3 r e
4π (at )3 (a3 He )3
0
0
Z ∞
√
4
t
8
dr r2 e−3r/a0
=
t 3
(a0 )
0
√
Z
√
16 2
2 −3y
=
= 4 8 dy y e
27
=
7
−r
¶
1
+ 31
at
a He
0
0
and probability = |a|2
8