PHY4604–Introduction to Quantum Mechanics
Fall 2004
Practice Exam Solutions
Dec. 13, 2004
No materials allowed. If you can’t remember a formula, ask and I might help. If you
can’t do one part of a problem, solve subsequent parts in terms of unknown answer–
define clearly. All parts 10 pts., max=120. Problem 1 required, attempt 2 of remaining
3 problems; circle which ones you want graded.
Possibly helpful formulae and constants
q
L̂+ ψ`m = h̄ `(` + 1) − m(m + 1)ψ`m+1
q
L̂− ψ`m = h̄ `(` + 1) − m(m − 1)ψ`m−1
En =
me4 1
2h̄2µn2
¶
1
2
L̂+ L̂− + L̂2z − h̄L̂z
L̂− L̂+ + L̂2z + h̄L̂z
Eψ
∂ψ
ih̄
∂t
h̄2 2
−
∇ + V (r)
2m
En = h̄ω n +
L̂2 =
=
Hψ =
Hψ =
H =
Z
dx xn e−x = n!
Z a
0
a2 − cos(a2 ) sin(a2 )
2a
³ ´2
− 12 xx
1
0
ψ0 (x) = q
e
1/2
π x0
sin2 ax dx =
ψ1 (x) = q
1
2π 1/2 x0
ψ2 (x) = q
1
µ
à µ
x
4
x0
8π 1/2 x0
à µ
1
ψ3 (x) = q
¶
2x − 12
e
x0
48π 1/2 x0
1
³
x
x0
!
¶2
x
8
x0
´2
³
− 12
−2 e
¶3
!
x
x0
´2
−1
x
− 12
e 2
x0
³
x
x0
´2
1. Short answer. Must attempt (only) 4 of 6.
(a) Explain what is meant by a 2p state of an atomic electron.
The “2” refers to the principal quantum number n = 2, so state has energy E2 = −1 Ryd/4, and “p” means total angular momentum quantum
number ` = 1.
(b) What is the degeneracy of the 1st excited state (E = (5/2)h̄ω) of the
isotropic 3D simple harmonic oscillator?
The 3D simple harmonic oscillator has energies Enx + Eny + Enz , where
each En is h̄ω(n + 1/2), so Etot = h̄ω(nx + ny + nz + 3/2). The first excited
state has one quantum of excitation in either x, y, or z, so degeneracy =
3.
(c) Sketch the first 3 eigenfunctions of the 1D infinite square well with V = 0
for −a ≤ x ≤ a and V = ∞ otherwise. Label them according to their
parity.
Ground state ψ1 and 2nd excited state ψ3 are both even functions of x ⇒
even parity, in other words parity eigenvalue π = +1. 1st excited state ψ2
is odd function ⇒ odd parity, in other words parity eigenvalue π = −1 .
2
(d) State whether the following operators are self-adjoint, anti-self-adjoint,
unitary, or “none of the above” and why.
i. L̂x
Lx = (r × p)x is a product of self-adjoint operators representing observables, and therefore also self-adjoint. It is also an observable itself,
of course.
ii. xp̂ x
(xp̂ x)† = x† p̂† x† = xp̂ x. Self-adjoint.
iii.
d
dx
Operator is anti-self-adjoint as discussed in class. Proof: definition of
adjoint is (O† χ, ψ) = (χ, Oψ) ∀χ, ψ. For d/dx we have
(χ,
Z
d
dψ
ψ) =
χ∗ (x)
dx
dx
dx
!
Z Ã
d ∗
=
− χ (x) ψ(x)dx,
dx
where I’ve performed an integration by parts and assumed that χ, ψ →
0 at ∞. From defintition,
Ã
d
dx
!†
= −
d
dx
iv. a + a†
(a, a† are raising & lowering operators for the 1D simple harmonic
oscillator.)
Since (a)† = a† , this is self-adjoint.
(e) If a particle is in the state |ψi, and |ni is the nth eigenvector of Q̂ corresponding to eigenvalue qn , what is the probability of measuring q3 ?
|h3|ψi|2 .
3
(f) Identify:
i. Photoelectric effect
Photons of wavelength ν are shone on a metal surface, and found to
kick out an electron only if ν is greater than some threshold value,
independent of intensity of light. Evidence (Einstein) for quantum
nature of light, E = hν.
ii. Davisson-Germer effect
Davisson-Germer experiment (1927) proved wave nature of electron
hypothesized by de Broglie by diffracting electrons from a crystal.
iii. Ehrenfest theorem
says that expectation values of quantum mechanical observables obey
classical equations of motion. For example,
dhpi
i
i
dV
dhV i
= − h[H, p]i = − h−ih̄ i = −
,
dt
h̄
h̄
dx
dx
where the right hand side is now just the classical force on a particle,
i.e. this is Newton’s law on the average.
iv. Stefan-Boltzmann law
Total radiation energy density emitted from a blackbody at temperature T is ∝ T 4 .
2. Hydrogenic orbitals. An electron moving in the Coulomb field of a proton is
in a state described by the wave function (ignoring spin)
1
Ψ(r, θ, φ) = √ [ψ100 (r) + 3ψ311 (r, θ, φ)]
10
(a) What is the expectation value of the energy?
hHi =
1 Ryd
1
1
(E1 + 9E3 ) = −
(1 + 9/32 ) = − Ryd
10
10
5
(b) What is the expectation value of L̂2 ?
hL2 i =
1
1
(0 + 2h̄2 ) = h̄2
10
5
4
(1)
(c) Is the wavefunction an eigenstate of parity? Yes or no? Explain either
answer.
The parity of the hydrogenic wavefunctions is given by (−1)` . Therefore
the wavefunction given is an admixture of an even parity and an odd parity
wavefunction–meaning it is not an eigenstate of the parity operator.
(d) What is the expectation value of the operator
∂
∂φ
in this state?
∂
Easy way: recall Lz = −ih̄ ∂φ
. The expectation value of Lz is
1
(ψ100 + 3ψ311 , Lz (ψ100 + 3ψ311 ))
10
= 9h̄/10,
hLz i =
∂
therefore h ∂φ
i = (9/10)i.
3. Electron in hydrogenic state. The electron in a hydrogen atom occupies a
state
s
1 0
ψ = R21 (r) 
Y +
3 1
s

2 1
Y
3 1
(2)
where
s
R21 (r) =
s
Y10
=
3
cos θ
4π
µ
1 1
3 2a0
Y11
,
¶3/2 µ
=−
s
¶
r
e−r/2a0
a0
,
3
sin θeiφ
8π
(3)
(4)
(a) What value(s) could a measurement of the z-component of the orbital
angular momentum, L̂z yield, and what is the probability of each? What
is the expectation value of L̂z in this state?
m is either 0 or 1, so Lz can be either 0 or h̄, with probability 1/3 or 2/3,
respectively.
´
√
√
1³
ψ210 + 2ψ211 , Lz (ψ210 + 2ψ211 )
3
= 2h̄/3,
hLz i =
5
(b) Calculate the average distance of the electron from the nucleus in this
state.
´
√
√
1³
ψ210 + 2ψ211 , r(ψ210 + 2ψ211 )
3
1
2
=
(ψ210 , r ψ210 ) + (ψ211 , r ψ211 )
3
3
hri =
s
2 Z∞
µ
1
1 1
r2 dr r 
= ( + )
3 3 0
3 2a0
Z ∞
a0
=
dy y 5 exp −y
24 | 0
{z
}
¶3/2 µ
¶
2
r
e−r/2a0 
a0
120
= 5a0
(c) What is the expectation value of L̂x in this state?
µ
µ
¶
¶
√
√
L+ + L−
1
ψ210 + 2ψ211 ,
(ψ210 + 2ψ211 )
3
2
³
´
√
√
√
√ √
h̄
=
ψ210 + 2ψ211 , 2ψ211 + 0 + 2ψ21 −1 + 2 2ψ210
6
h̄
(2 + 2) = 2h̄/3
=
6
hLx i =
(d) If you measured the z-component of the angular momentum and the distance of the electron from the origin r simultaneously, what is the probability density for finding L̂z with eigenvalue zero at a distance r?
1 2
P m=0 (r) = r2 R21
(r)
3
6
4. Scattering potential. For a 1D potential as shown in the figure and E < V0 ,
(a) write down the Schrödinger equation and its general solution in the regions
I,II, and III assuming the particle is incident from the left.
2
h̄
Hψ = Eψ, H = − 2m
∇2 .
ψI = Aeipx/h̄ + Be−ipx/h̄ ,
ψII = Ceqx/h̄ + De−qx/h̄
ψIII = F eipx/h̄
with
p =
√
2mE
q
q =
2m(V0 − E)
(b) Write down the matching conditions at the boundaries x = a, b.
Require continuity of ψ and its derivatives at the interfaces:
ψI (a)
ψII (b)
ψI0 (a)
0
(b)
ψII
=
=
=
=
ψII (a)
ψIII (b)
0
(a)
ψII
0
ψIII (b),
in other words
Aeipa/h̄ + Be−ipa/h̄
Ceqb/h̄ + De−qb/h̄
(Aeipa/h̄ − Be−ipa/h̄ )ip
q(Ceqb/h̄ − De−qb/h̄ )
7
=
=
=
=
Ceqa/h̄ + De−qa/h̄
F eipb/h̄
(Ceqa/h̄ − De−qa/h̄ )q
F ipeipb/h̄
(c) Do not solve for all coefficients, but reduce the problem to a single equation
determining the eigenvalues.
This is tedious, not a good exam question.
B/A = e2iap/h̄
p + iq
or
p − iq
F/A = −4eipa/h̄ e−ipb/h̄ eq(a+b)/h̄
(iq −
p)(ipe2qa/h̄
pq
− ipe2qb/h̄ + qe2qa/h̄ + qe2qb/h̄ )
(d) Sketch the probability of finding the electron in all three regions.
8