Download PHY4604–Introduction to Quantum Mechanics Fall 2004 Practice

PHY4604–Introduction to Quantum Mechanics
Fall 2004
Practice Test 3
November 22, 2004
These problems are similar but not identical to the actual test. One or two parts will
actually show up.
1. Short answer.
(a) Recall the Bohr energy levels of the Hydrogen atom are
En = − 2h̄m2 n2
³
e2
4π²0
´2
(1)
Compare the wavelengths of the 2p → 1s transitions in
i. hydrogen
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
ii. deuterium (mass of nucleus 2× that of H).
The mass which appears in the Bohr formula is the electron mass m,
which is replaced by the reduced mass µ = mmp /(m+mp ) if one allows
for the proton motion. Changing mp → 2mp doesn’t change much,
so to a good approximation the photon wavelength doesn’t change,
λD ' λH .
iii. positronium.
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 .
(b) An electron is in the ground state of tritium, a form of heavy hydrogen
where the nucleus has two neutrons in addition to a proton. A nuclear
reaction now changes the nucleus instantaneously to 3 He, i.e. two protons
1
and a neutron. Calculate the probability that theqelectron remains in the
ground state of 3 He. You may need ψ100 = (2/ 4πa30 ) exp −r/a0 , with
a0 = h̄2 /(mZe2 ), where Z is the nuclear charge.
Z, the nuclear charge or number of protons, is 1 for tritium and 2 for 3 He.
Thus the Bohr radius for 3 He is twice as small as for tritium or hydrogen,
3
a0He ' 12 at0 . The electron in the 3 He ion is more tightly bound to the
nucleus. We’re told the system starts out in the tritium ground state,
2
t
= q
ψ100
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
He
He
t
He
+ ...
+ cψ210
ψ100
= aψ100
+ bψ200
meaning the electron in the tritium ground state will be found after a
measurement in the 3 He ground state with probability amplitude
3
t
He
a = hψ100
|ψ100
i
µ
Z
¶
1
+ 31
at
a He
0
0
4
1
q
d3 r e
4π (at )3 (a3 He )3
0
0
4 √ Z∞
t
8
dr r2 e−3r/a0
=
t 3
(a0 )
0
√
Z
√
16 2
2 −3y
=
= 0.838
= 4 8 dy y e
27
=
−r
and probability = |a|2 = 0.702.
(c) An electron moving in the Coulomb field of a proton is in a state described
by the wave function
√
1
Ψ = [4ψ100 + 3ψ211 − ψ210 + 10ψ21−1 ]
6
(2)
i. What is the expectation value of the energy?
1
(16h100|H|100i + 9h211|H|211i + h210|H|210i
36
+ 10h21 − 1|H|21 − 1i)
1
1
=
(16E1 + 9E2 + 1E2 + 10E2 ) = (16E1 + 20E2 ).
36
36
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.
2
ii. What is the expectation value of L̂2 ?
X 2
hψ|L2 |ψi =
h̄ `(` + 1)|cn`m |2 =
n`m
h̄2
10h̄2
(16 · 0 + 9 · 2 + 1 · 2 + 10 · 2) =
36
9
iii. What is the expectation value of L̂z ?
X
hψ|Lz |ψi =
h̄m|cn`m |2 =
n`m
h̄
h̄
(16 · 0 + 9 · 1 + 1 · 0 + 10 · (−1)) = −
36
36
(d) How large would a constant magnetic field have to be to split two H-atom
states which are degenerate in zero field by an amount so as to maximally
absorb light of wavelength λ?
e
H = −µ · B = S · B
m
e
e
= = Sz Bz = h̄ms Bz ,
m
m
where the last step where the operator is replaced by its eigenvalues holds
only when applied to Sz eigenstates, and where I have used ms for the Sz
quantum number. The two Sz states have a difference of ∆ms = 1, so the
energy of the photon produced must be me h̄Bz ≡ hc
, or λ = (2πmc/(eBz )).
λ
(e) For two particles a and b such that `a = 1 and `b = 1, argue that it must
be true that
|` = 2, m = 1i = α|ma = 1, mb = 0i + β|ma = 0, mb = 1i,
where `, m are the quantum numbers corresponding to total angular momentum Lq= La + Lb , and find the coefficients α and β. Hint: use
L± |`mi = `(` + 1) − m(m ± 1)|`m ± 1i.
a) states given are the only possible ones with m = ma + mb and ma ≤ `a ,
mb ≤ `b .
b) Start at top of angular momentum ladder, where we know there is only one
possible |`m = `i state, equal to the one possible |ma = `a , mb = `b i, then
apply lowering operator as in HW, remembering square root factors to keep
states normalized:
√
L− |22i`m = (La + Lb )|11ima mb
√
2 · 3 − 2 · 1|21i =
2(|01i + |10i)
3
So divide by
√
2 to get |21i =
√1 (|10i
2
+ |01i).
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?
L2 |1mi = 2h̄2
Lz |1mi = h̄m
(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 L̂x (Hint: you will need to calculate the matrix elements
hm|L̂x |m0 i.)
Write Lx = (L+ + L− )/2, find, e.g.
L+ + L−
|1i = 0
2
√
2
L+ + L−
h1|
|0i = h̄
2
2
L+ + L−
| − 1i = 0...
h1|
2
h1|
where I’ve used the effect of L± acting on |mi, and the orthonormality of
the |mi. It is a little tedious to do them all, but eventually we find

Lx

0 1 0
h̄ 

√
=
 1 0 1 
2 0 1 0
(c) Find the eigenvectors of L̂x .
Now we just have a matrix eigenvalue problem, and I assume you can find
the eigenvectors, then normalize them. The results are
√ 





1/2
1/2
−1/
2
√
√





0√ 
|1i =  1/ 2  |0i = 
 | − 1i =  −1/ 2 
1/2
1/2
1/ 2
4
(d) If a system is prepared in the state vector


1
1 
|ψi = √  4 
,
26 −3
(3)
what is the probability that a measurement of L̂x yields the value 0?
P = |h0|ψi|2 = 4/13
3. Pauli principle.
Consider two electrons described by the Hamiltonian
H=
where
p̂21
p̂2
+ 2 + V (x1 ) + V (x2 )
2m 2m


 ∞
x < −a/2
V (x) =  0 −a/2 ≤ x ≤ a/2

∞ x > a/2
Assume both electrons are in same spin state.
(4)
(5)
(a) What is the lowest (ground state) energy?
(b) What is the energy eigenfunction for this ground state?
Parts a) and b): Denote the single-particle eigenfunctions of the ordinary
infinite square well by ψ0 , ψ1 , .... The two-particle ground state must be
antisymmetric under exchange of all particle labels according to Pauli, but
we are told that the spins are both the same, e.g. up. The wave function
must be
Ψ0 (1, 2) = (ψ0 (x1 )ψ1 (x2 ) − ψ1 (x1 )ψ0 (x2 ))χ↑↑
From our theorem about additive Hamiltonians, the energy of this state is
the sum of the energies of the 2 single particle states 0 and 1:
Eground = E0 + E1 =
h̄2 (π/a)2 h̄2 (2π/a)2
5h̄2 (π/a)2
+
=
2m
2m
2m
(c) What is the energy and the wave function of the first excited state (still
with equal-spin condition!)?
Ψ1 = (ψ0 ψ2 − ψ2 ψ0 )χ↑↑
Eexc,1 = E0 + E2 =
5h̄2 (π/a)2
h̄2 (π/a)2 h̄2 (3π/a)2
+
=
2m
2m
m
5