Solutions: Tentamen i Kvantmekanik IFA521
kl. 14-19 30 May 2012
Lecturer: Joseph A. Minahan (471-3291)
Test Problem #1 – True False
Please answer the following 20 questions as either True or False. If you don’t
know the answer, guess — no points are subtracted for incorrect answers.
1. The Schrödinger equation is a nonlinear equation. False
~ is a
2. Each component of the the orbital angular momentum operator L
Hermitian operator. True
3. It is possible to know one component of spin at the same time but not
two. True
4. Two bosons can be in the same single-particle state. True
5. Wave functions can be nonzero in classically forbidden regions. True
6. A spin-1/2 particle in an l = 2 orbital angular momentum state can have
total angular momentum j = 2. False
7. The expectation value of r2 is larger for the 2s state than the 2p state in
hydrogen. True
8. Since the hydrogenic potential is always less than 0, bound states must
have negative energy. True
9. The measurement of the energy for a system will always result in one of
the eigenvalues of the Hamiltonian. True
10. Immediately after the measurement described in the last question, the
system is in an eigenstate of the Hamiltonian with eigenvalue equaling the
result of the measurement. True
11. The size of a hydrogenic atom is proportional to the charge of the nucleus.
False
12. The energies of a hydrogenic atom are proportional to the charge of the
nucleus. False
13. The potential V (x) = ax3 + bx5 + cx7 will have energy eigenstates which
are odd under a parity transformation. False
14. The operator Ô1 Ô2 is Hermitian if Ô1 and Ô2 are both Hermitian. False
15. One can find Planck’s constant by measuring the maximum energy of
ejected electrons from a metal surface irradiated with light, at two different
frequencies. True
(continued on the next page)
Test Problem #1 – True False continued
16. The total spin of two spin-1/2 fermions in the same single particle position
state must be zero. True
17. Energy eigenstates are stationary states. True
18. An eigenstate of Lz has a probability density that is constant in the azimuthal angle φ. True
19. The expectation value of the position x obeys the classical equations of
motion. True
20. The total spin of a filled shell in an atom is always zero, but not necessarily
the total orbital angular momentum. False
Test Problem #2
Consider the following wave-function for a particle in a one dimensional box
h
i
ψ(x) = C 4 ψ1 (x) − 3 ψ2 (x) − ψ3 (x) − 3 ψ4 (x) + i ψ5 (x)
where ψn (x) are normalized eigenstates with energies En = E1 n2 (E1 is the
energy of the ground state.)
(a) Find C so that ψ(x) is properly normalized.
(b) Find the probability that a measurement of the particle’s energy will be
9E1
(c) Find the expectation value of the energy in terms of the ground state energy
E1 .
(d) Assuming that Ψ(x, 0) = ψ(x), find Ψ(x, t) in terms of E1 and the wavefunctions ψn (x).
(a) 1 = |C|2 (|4|2 + | − 3|2 + | − 1|2 + | − 3|2 + |i|2 ) = 36|C|2 ⇒ C = 1/6
(b) P3 = 1 × |C|2 = 1/36
(c) hEi =
1
2
2
2
2
2
36 (1×|4| +4×|−3| +9×|−1| +16×|−3| +25|i| )E1
=
115
18 E1
(d)
Ψ(x, t)
=
1 −iE1 t/~
4e
ψ1 (x) − 3e−4iE1 t/~ ψ2 (x) − e−9iE1 t/~ ψ3 (x)
36
−3e−16iE1 t/~ ψ4 (x) + i e−25iE1 t/~ ψ5 (x)
Test Problem #3
Consider the two-state system of a spin 1/2 particle, where the eigenstates of
Sz are given by
1
0
↑ ≡
↓ ≡
0
1
~ pointing in the z direction, B
~ = (0, 0, B),
Suppose there is a magnetic field B
and a corresponding Hamiltonian given by
e ~
~ ·µ
H = −B
~ , where µ
~ =−
S,
mc
~ = ~ ~σ ,
S
2
(the Pauli matrices ~σ can be found in the useful formulae.)
(a) Find the normalized energy eigenstates and eigenvalues.
(b) Find the normalized eigenstates and eigenvalues of Sx in terms of the eigenstates of Sz .
(c) Assume that at time t = 0 the spin state is the negative eigenstate of Sx .
Find the state as a function of time, t.
(d) Find the expectation value hSz i as a function of t for the state in (c).
(e) Find the expectation value hSx i as a function of t for the state in (c).
~ is parallel to Sz , then the z spin states are the eigenstates.
(a) Since B
H| ↑i =
H| ↓i =
eB
eB~
Sz | ↑i =
| ↑i
mc
2mc
eB~
eB
Sz | ↓i = −
| ↓i
mc
2mc
(b)
1
|+ix = √ (| ↑i + | ↓i)
2
1
|−ix = √ (| ↑i − | ↓i)
2
(c)
1
|Ψ, 0i|−ix ⇒ |Ψ, ti = √ e−i(eB/2mc)t | ↑i − e+i(eB/2mc)t | ↓i
2
(d) Since [Sz , H] = 0, hSz i is constant in time.
hSz i = h−|Sz |−i = 1/2 − 1/2 = 0 .
(e)
e−i(eB/2mc)t | ↑i − e+i(eB/2mc)t | ↓i
1 −i(eB/2mc)t
=
e
− e+i(eB/2mc)t |+ix + e−i(eB/2mc)t + e+i(eB/2mc)t |−ix
2
eB
eB
t |+ix + cos
t |+ix
= −i sin
2mc
2mc
~
eB
~
eB
~
eB
⇒ hSx i = + sin2
t − cos2
t = − cos
t
2
2mc
2
2mc
2
mc
|Ψ, ti =
Test Problem #4
Suppose we have 5 noninteracting identical particles with mass m in a onedimensional harmonic oscillator with potential, V (x) = 12 mω 2 x2 .
(a) Assuming that the particles are bosons, find the ground-state energy of this
system.
(b) Assuming that the particles are fermions without spin, find the ground-state
energy of this system.
(c) Assuming that the particles are spin-1/2 fermions find the ground-state
energy of this system.
(d) Assuming that the particles are spin-1/2 fermions find the degeneracy of
the ground-state.
(e) Assuming that the particles are spin-1/2 fermions find the first two exited
energy levels and their degeneracies.
(a) If all particles are bosons, then they can all be in the same single particle
state. In this case the ground state is
1
5
Egs = 5 × ~ω = ~ω
2
2
(b) If the particles are fermions without spin then we can have only one particle
per single particle state. Therefore,
Egs = (1/2 + 3/2 + 5/2 + 7/2 + 9/2)~ω =
25
~ω
2
(c) If all particles are fermions with spin-1/2, then we can put up to two particles into each single particle state. Therefore,
Egs = (2 × 1/2 + 2 × 3/2 + 5/2)~ω =
13
~ω
2
(d) The particle in the highest level can either be spin up or down. Hence, the
degeneracy is 2.
(e) The first excited state is made from lifting the fermion at the E = 52 ~ω
energy level to the E = 72 ~ω level, or, taking a fermion in the 3/2 level and
moving it to the 5/2 level. Hence the energy of this state is 15
2 ~ω. In either
case there is one fermion in a half-filled level, so each of these configurations
has two possible states for the possible spin of the fermion. Therefore, the
total degeneracy is 4.
The second excited state can be made either by taking the fermion in the
5/2 level and moving it to the 9/2 level, or taking one of the fermions in the
3/2 level and moving it to the 7/2 level, or by taking one of the fermions
in the 1/2 level and moving it to the 5/2 level. The energy of this state is
then 17
2 ~ω. In the first case there are 2 possible states, in the second there
are 8 possible states (there are 3 half-filled levels) and in the last case there
are 2 states again. Hence the degeneracy is 12.
Test Problem #5
Assume that we have the Hamiltonian
H = H0 + H 0
where H0 is the Hamiltonian for the one-dimensional harmonic oscillator,
H0 = ~ω(a† a + 1/2) ,
and H 0 is a small perturbation given by
H 0 = λ (a† )3 + 3(a† )2 a + 3a† a2 + a3 .
(a) Find the first-order correction to the ground-state energy.
(b) Find the first-order correction to the energy of the first excited state.
(c) Find the second-order correction to the ground-state energy.
(a) ∆(1) E0 = h0|H 0 |0i = 0 (every term has a different number of a and a†
operators.)
(b) ∆(1) E1 = h1|H 0 |1i = h0|aH 0 a† |0i = 0. (Same reason as in (a).)
(c)
∆(2) E0 =
∞
X
|h0|H 0 |ni|2
(0)
n=1
E 0 − En
The only nonzero contribution is from
1
√
λ
h0|H 0 |3i = λh0| (a† )3 +3(a† )2 a+3a† a2 +a3 √ (a† )3 |0i = √ h0|a3 (a† )3 |0i = 6λ
6
3!
Therefore,
∆(2) E0 =
6λ2
2λ2
=−
−3~ω
~ω
Test Problem #6
The potential energy of an electron around a spherically symmetric charge density is
e
Q(r)
(1)
U (r) = −
4π0 r
where Q(r) is the total charge inside the radius r. Now consider an electron in
the ground state of a hydrogenic potential
V (r) = −
Ze2
.
4π0 r
Normalized radial wavefunctions are
R10 (r)
=
R20 (r)
=
R21 (r)
2(Z/aµ )3/2 exp(−Zr/aµ )
2(Z/2aµ )3/2 (1 − Zr/2aµ ) exp(−Zr/2aµ )
1
= √ (Z/2aµ )3/2 (Zr/aµ ) exp(−Zr/2aµ )
3
(a) Find the probability that an electron in the ground state is within a distance
r from the nucleus.
(b) Now suppose that a second electron is added. It sees the potential from
the nucleus as well as a repulsive potential due to its interaction with the
first electron in the ground state. Assume that the second electron sees the
first electron as a charged cloud, where the charge density of the cloud is
proportional to the probability density. Using equation (1), find the effective
potential that the second electron sees due to the first electron.
(c) Treating your result from (b) as a first order perturbation, find the correction to the energy for the two-electron hydrogenic atom if one electron is in
the ground state and the second is in the 2p (` = 1) state.
(a) Since the radial wave functions are normalized, we have
Z ∞
r2 dr(Rnl (r))2 = 1.
0
The probability that the electron in the ground state is within a distance r
from the nucleus is then
3
Z r
Z r
Z
(r0 )2 dr0 (R10 (r0 ))2 =
(r0 )2 dr0 4
exp(−2Zr0 /aµ )
aµ
0
0
a 3 Z 3 Z 2Zr/aµ
µ
dzz 2 e−z
= 4
2Z
aµ
0
2Zr 2Z 2 r2
exp(−2Zr/aµ )
= 1− 1+
+
aµ
(aµ )2
(b) The charge of an electron is −e and since we are assuming that the charge
density is proportional to the probability density, the charge inside a radius
r is
2Zr 2Z 2 r2
exp(−2Zr/aµ )
Q(r) = −e 1 − 1 +
+
aµ
(aµ )2
and so the contribution to the potential is
2Zr 2Z 2 r2
e2
1− 1+
+
exp(−2Zr/aµ )
U (r) = +
4π0 r
aµ
(aµ )2
(c)
∆E
=
=
=
=
h21m|U (r)|21mi
3 2
Z ∞
2
Z
2Zr 2Z 2 r2
1
Zr
2
−Zr/aµ e
−2Zr/aµ
1− 1+
r dr
e
+
e
3
2aµ
aµ
4π0 r
aµ
(aµ )2
0
Z
Z
∞
∞
z3
2z 4
2z 5
e2 Z 1
e2 Z 1
+
+
z 3 e−z dz −
e−z dz
4π0 aµ 24 0
4π0 aµ 24 0
81 243 729
1
2
10
e2 Z
1
e2 Z 164
−
−
−
=
4π0 aµ 4 324 243 729
4π0 aµ 729