PHY4604–Introduction to Quantum Mechanics
Fall 2004
Test 1 Solutions
October 8, 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
Z
√
2
dx e−x =
Z
2
dx x2 e−x =
Z a/2
√
π
π/2
a3
6
(1 − 2 )
24
π
−a/2
Z a/2
3
3
a
dx x2 sin2 2πx/a = (1 − 2 )
24
2π
−a/2
1
En = h̄ω(n + )
2
2 2
h̄ k
E=
2m
dx x2 cos2 πx/a =
s
L± ≡
Z ∞
0
Z
ψ(x, t) =
k
h̄ ∂
x∓ √
2
2m ∂x
n −x
dx x e
= n!
CV = dE/dT
dkf (k) exp(−i(kx − ωk t)
H = (L+ L− +
h̄ω
)
2
1
Hψ = Eψ
Hψ = ih̄
∂
ψ
∂t
p̂ = −ih̄∇
E=
h̄ω
−1
eβh̄ω
E = nh̄ω

En = − 
"
2
m
e
2
2h̄ 4π²0
#2 
1

[L+ , L− ] = −h̄ω
h̄2 2
H=−
∇ + V (r)
2m
L = nh̄
p = h̄k
[x, p̂] = ih̄
n2
1. Short answer. Must attempt (only) 3 of 5.
(a) When one attempts to observe through which slit an electron passes in
a Young’s double slit experiment with electrons, the interference pattern
observed on the screen disappears. Explain why this happens, according
to the Copenhagen interpretation of quantum mechanics. If the energy
hν of the photons used to detect which slit the electron passes through is
made very small, you might think the electrons themselves could hardly be
disturbed, so for hν → 0 the interference pattern must reappear. Explain
this apparent discrepancy.
The observation of which slit the photon passes through “collapses the
wave function”, which was originally a superposition of an electron going
through slit one and one going through slit 2. There can therefore be
no interference if every electron is observed. If the energy of the photons
is made very small, its wavelength becomes very large. Your ability to
determine which slit the electron goes through eventually goes away when
the wavelength is comparable to the interslit distance, and the interference
pattern therefore reappears.
(b) Explain why the measurement of a specific heat in simple solids varying
as T 3 was paradoxical in the context of the classical theory. What concept introduced by Einstein and Debye resolved the paradox? Define your
terms.
According to the classical equipartition theorem, the average energy per
atom was (1/2)kT for every quadratic degree of freedom in the Hamiltonian, i.e. for a simple solid 3 · (1/2 + 1/2)kT = 3kT , where there is a
kinetic term and a harmonic potential term for each of 3 Cartesian degrees
of freedom for each atom. The specific heat per atom should therefore be
cV = dE/dT = 3k, but a much smaller number was observed. Einstein and
Debye showed that the quantization of the normal modes of the system
gave a much lower average energy at low T .
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(c) Write down the second excited state (not the ground state, not the 1st
excited state, but the one above that!) wave function for a particle of mass
m in a 1D box of width a. Do not normalize. Find its energy eigenvalue.
The second excited state must have 2 nodes (points where ψ = 0), and go
to zero at the boundaries of the well. The solutions to the free Schrödinger
equation are trigonometric functions representing standing waves, so the
only choice is cos something. The ground state was cos πx/a (no nodes),
so choose cos 3πx/a. This is consistent with the boundary conditions and
has 2 nodes on [−a/2, a/2]. To find the energy eigenvalue, we can note
that the eigenfunction can be written cos kx, with k = 3Π/a. k is now the
wavevector corresponding to momentum h̄k. Since the particle is free, its
energy must be p2 /2m, or h̄2 (3π/a)2 /(2m).
(d) Evaluate the expression
[ar2 , px ],
(1)
where r is the magnitude of the displacement vector r in 3D, and p̂x is the
x-component of the momentum operator.
[ar2 , px ] = a[r2 , px ] = a[x2 + y 2 + z 2 , px ]
= a[x2 , px ]
by the linearity of commutators and the definition of r2 . The commutator
[x2 , px ]ψ is
³
´
−ih̄ x2 (∇x ψ) − ∇x (x2 ψ)
³
= −ih̄ x2 ψ 0 − 2xψ − x2 ψ 0
= 2ih̄xψ
´
So the desired commutator is [ar2 , px ] = 2iah̄x.
(e) Name three experiments which led to the questioning of classical theory
and the advent of quantum theory and describe briefly.
i. Specific Heat of Solids (see above)
ii. Photoelectric effect – Shine light on metal surface, only photons of
sufficiently high energy (frequency) able to kick an electron out. Increasing intensity of low-energy light doesn’t help. A single electron
can make a transition out of the solid only if a single photon carries
sufficient energy h̄ω.
iii. Bohr atom – Bohr’s calculation of energy levels of hydrogen atom
starting with angular momentum quantization. Explained why atoms
did not decay due to radiation by accelerating charges.
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iv. Blackbody radiation – Planck explained blackbody spectrum which
was very different from that predicted by Jeans (classical theory) by
assuming that the modes of a cavity were quantized.
2. SHO in gravitational field. Consider a simple harmonic oscillator transverse
to a gravitational field (no motion in vertical direction allowed at first). The
Hamiltonian is
h̄2 ∂ 2
1
(2)
H=−
+ mω 2 x2 + mgy.
2
2m ∂x
2
For y = 0 the lowest two SHO eigenfunctions are
³
ψ0 |y=0
ψ1 |y=0
where x0 =
q
´2
−1 x
1
√ 1/2 e 2 x0
= ψ0 (x) =
(x0 π)
³ ´2
√
µ ¶
− 12 xx
2
x
0
√
= ψ1 (x) =
e
(x0 π)1/2 x0
(3)
(4)
h̄/(mω), corresponding to E0 = h̄ω/2 and E1 = 3h̄ω/2.
(a) Find the expectation value of x in the state
√1 (ψ0
2
+ ψ1 ).
Z
hxi =
=
=
=
=
=
dx|Ψ|2
1Z
dx(ψ0 + ψ1 )∗ x(ψ0 + ψ1 )
2Z
1
dxx(ψ1∗ ψ0 + ψ0∗ ψ1 )
2
³ ´2
√
x
−
2 1 Z
√
dxx2 e x0
(x0 π) x0
√ Z
2
2
x0 √
dyy 2 e−y
( π) |
{z
}
√
π/2
√
x0 / 2
since the other terms are even functions times x which is odd, giving an
overall integrand over a symmetric interval, which vanishes.
(b) Find the eigenvalues E0 (y) and E1 (y), as well as ψ0 (x, y) and ψ1 (x, y).
Hint: this is not hard.
Use the theorem we proved on the homework for Hamiltonians which consist of additive independent terms. Here we have ψ(x, y) = ψ(x)ψ(y) and
E(y) = E + mgy, where E is the eigenvalue at y = 0. This works because
y is a constant as far as the x-motion is concerned. So we write down
immediately that E0 = h̄ω/2 + mgy and E1 = 3h̄ω/2 + mgy. The wave
functions are ψ0 (x, y) = ψ0 (x) and ψ1 (x, y) = ψ1 (x) also because the mgy
term is a constant, ψ(y) is also (check by substitution).
4
(c) Suppose one particle is in the state ψ0 at height y and a second particle
is in the state ψ1 at height 0. Find y such that the two particles have the
same energy.
E0 + mgy = E1
h̄ω/2 + mgy = 3h̄ω/2
⇒ y = h̄ω/mg
(d) Now remove the gravitational potential, and allow the particles to move
freely in the the y-direction (there’s no restoring term in the y direction).
For a particle with momentum py in the 1st excited state as far as its xmotion is concerned, what is the energy? What is its wavefunction (ignore
normalization)?
Hamiltonian is now
H=−
h̄2 ∂ 2
h̄2 ∂ 2
1
−
+ mω 2 x2 .
2
2
2m ∂x
2m ∂y
2
(5)
and the energy in 1st excited state is 3h̄ω/2 + h̄2 p2y /2m. The eigenfunction
is ψ1 (x)eipy y/h̄ .
3. Particle in Box. A particle of mass m is in a “box” of size a, i.e. it moves in
the 1D potential


 ∞
x < −a/2
− a/2 ≤ x ≤ a/2
x > a/2
V (x) =  0

∞
(6)
(a) Draw a picture of the 4 eigenfunctions of lowest energy. Make boundary
conditions clear.
Prob. 3a. Note eigenfunctions ψ1..4 are not normalized.
5
(b) Write down the general form of all eigenfunctions in two classes: those 1)
even and 2) odd under parity (x → −x). Give the corresponding eigenvalues for the nth member of each class, and specify normalizations.
ψne = Aen cos kx,
ψno = Aon sin kx,
and normalization
R
k = (2n + 1)π/a, n = 0, 1, 2, ...
k = 2nπ/a, n = 1, 2, 3....
dx|ψ|2 = 1 gives An =
q
(c) Calculate the root-mean-square displacement
excited states in terms of a, h̄, and m.
2/a for both e and o.
q
hx2 i in the ground and first
Z
πx
2 a3
6
a2
6
=
(1 − 2 ) = (1 − 2 )
a
a 24
π
12
π
Z
3
2
2a
3
a
3
2πx
=
(1 − 2 ) = (1 − 2 )
= (Ao1 )2 dx x2 sin2
a
a 24
2π
12
2π
hx2 i1e = (Ae1 )2
hx2 i1o
dx x2 cos2
(d) Suppose now the particle is in a state Ψ(x, 0) = √12 (ψ1e (x) − ψ2e (x)), i.e.
a superposition of the two lowest even eigenstates. Find Ψ(x, t) in terms
of the eigenvalues
Ene found above, and calculate the expectation value
R
hx(t)i ≡ dx x|Ψ(x, t)|2 .
Ψ(x, t) = (ψ1e (x)e−iE1e /h̄ − ψ2e (x)e−iE2e /h̄ ) and
1Z
∗
∗
dx x(|ψ1e |2 + |ψ2e |2 − ψ2e
ψ2e − ψ1e
ψ2e ) = 0
2
since Ψ is even in x.
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