PHY4604–Introduction to Quantum Mechanics
Fall 2004
Practice Test 2 solutions
October 4, 2004
These problems are similar but not identical to the actual test. One or two parts will
actually show up.
1. Short answer.
• You are told that a potential can be written V (x) =
10
X
Vn x2n . All Vn ≥ 0.
n=1
What can you say about the solutions ψ to Hψ = Eψ?
Since the potential is an even function of x, the parity operator commutes
with the Hamiltonian. This means the eigenstates can be classified according to parity.
• What is tunnelling? Draw a picture describing the process, and describe
explicitly a situation which distinguishes classical physics from quantum
mechanics.
When a particle approaches a potential barrier higher than its total energy,
in classical physics it must be reflected with 100% probability. In quantum, mechanics, however, it is transmitted to the other side with some
probability, provided the barrier is not infinite.
• Write the Dirac expression hα|H|αi as an explicit integral in three dimensions, assuming that hr|αi represents a wave function ψα (r). If |αi is an
eigenvector of H with eigenvalue Eα , evaluate the integral.
Z
hα|H|αi =
d3 rψα∗ (r)Hψα (r)
• Identify: Ehrenfest’s theorem.
Expectation values obey classical laws, e.g. hp/mi = dhxi/dt.
• If A and B are self-adjoint, is the combination A + iB self-adjoint, antiself-adjoint, unitary, antiunitary, none of the above, or more than one of
the above? Defend your answer.
(A + iB)† = A† − iB † = A − iB,
1
which is neither self-adjoint nor anti-self-adjoint. For it to be unitary we
would need (A + iB)(A − iB) = A2 + B 2 + i[A, B] = 1, which we don’t
know anything about, so we can’t say. None of the above is therefore the
best answer.
• Prove that if (ψ, Oψ) = (Oψ, ψ) for all ψ, then O is self-adjoint.
We did this in class. Expand ψ in an orthonormal basis: ψ =
Then the relation we are given means
X
n,m
c∗m cn (ψm , Oψn ) =
X
n,m
P
n c n ψn .
c∗m cn (O† ψm , ψn ),
which since the relation is true for all ψ, means the cn ’s are independent,
∗
i.e. the only way you can satisfy the equation is to have Omn = Onm
,
where Omn ≡ (ψm , Oψn ). To prove self-adjointness, we need (χ, Oφ) =
P
P
(Oχ, φ)) ∀χ, φ. If we put χ = n fn ψn and φ = m gm ψm , we can expand
(χ, Oφ) =
X
m,n
= =
fn∗ gm Onm
X
m,n
∗
fn∗ gm Omn
= (Oχ, φ))
QED
• Prove that eigenvectors of a Hermitian operator corresponding to distinct
eigenvalues are orthogonal.
Let |1i, |2i be eigenvectors of Hermitian O corresponding to eigenvalues o1
and o2 . Self-adjoint implies
h1|O|2i = hO1|2i, ,i.e.
o2 h1|2i = o1 h1|2i,
where I wrote o1 instead of o∗1 since we know eigenvalues of Hermitian
operators are real. Now if o1 6= o2 the two vectors must be orthogonal,
h1|2i = 0.
• Find the expansion coefficient of the first excited SHO state in the function
(x2 + x20 )−2
It’s zero, since the first excited SHO state is odd, but the function you are
trying to expand is even.
2
• Draw a picture of a finite attractive square well in 1D, V = −V0 inside and
V = 0 outside, and sketch the form of the ground and 1st excited states of
negative energy, assuming such states exist. Be sure to indicate how these
functions differ from the analogous eigenfunctions of the infinite square
well problem. What happens to the number of bound states (E < 0) when
you make the well in this problem deeper?
Main difference from infinite square well
is that wave functions leak out of well. If you make the well deeper, you will trap
more bound states, as we saw in the java applet I showed in class.
2. Electrons incident on metal surface. Consider a metal occupying a halfspace x > 0, and model it as a region of constant potential −V0 . On the left for
x < 0 is a true vacuum with V = 0. Electrons are incident on the metal surface
from the left with initial energy E.
(a) Write down the general solution to Schrödinger’s equation on the left and
the right. Take the amplitudes of the incident wave, reflected wave, and
transmitted wave to be A, B, and C.
On each side, the solution can be written as a sum of travelling waves.
We are told that electrons are coming from the left, so there will be for
x < 0 a reflected and an incident wave. Write these as Aeipx/h̄ + Be−ipx/h̄ ,
0
respectively. On the right, put Ceip x/h̄ for the transmitted
√ wave. Check
that
each satisfies the Schrödinger equation, with p = 2mE and p0 =
q
2m(E + V0 ).
(b) State the boundary conditions at the interface and use them to extract
relations between A, B, and C.
3
At the boundary, ψ and its first derivative must be continuous since V is
finite. So at x = 0 we have
A+B = C
(A − B)p = Cp0 .
(c) Calculate the probability current densities in the incident, transmitted,
and reflected waves, and verify probability conservation.
Probability current density is
j = −
ih̄ ∗
[ψ ∇ψ − ((∇ψ ∗ )ψ)]
2m
so substitute, find for incident current density jinc = p|A|2 /m, for reflected
jref l = −p|A|2 /m, & for transmitted jtrans = p0 |C|2 /m. For probability
to be conserved, we need jinc + jref l = jtrans , or p(|A|2 − |B|2 ) = p0 |C|2 .
We can always take A real without loss of generality. The first B.C. then
says that Im B=Im C, but this is consistent with the 2nd B.C. only if
Im B=Im C=0. In this case, multiplication of the two B.C. gives you the
probability conservation.
(d) Calculate the reflection probability of electrons if E = 0.1eV and V0 = 8eV .
Reflection coefficient is R = |jref l /jinc | = |B|2 /|A|2 , while transmission coefficient is T = |jtrans |/|jinc | = p0 |C|2 /(p|A|2 ). Solving boundary conditions
for |B|/|A|, |C|/|A| gives
Ã√
!
√
E − E + V0
p − p0
= √
= 0.64
R =
√
p + p0
E + E + V0
√ √
Ã
!
4 E E + V0
4pp0
T =
= √
= 0.36
√
(p + p0 )2
( E + E + V0 )2
Note classically all of the electrons would be transmitted! Quantummechanically, most are reflected.
3. 2-level system. A Hamiltonian for a 2-level system is written as
H = E0 ((|1ih1| − |2ih2| + |2ih1| + |1ih2|)
where the vectors |1i and |2i represent two orthonormal states of the system
which span the Hilbert space of the problem.
(a) Write down the matrix H of the problem in the |1i, |2i basis.
In the |1i, |2i basis, the Hamiltonian is
"
H = E0
4
1 1
1 −1
#
(b) Find the eigenvectors |ai, |bi and eigenvalues Ea and Eb of the Hamiltonian.
√
√
Ea =
2E0 , |ai = Aa (1, 2 − 1)
√
√
Eb = − 2E0 ,
|bi = Ab (1, − 2 − 1),
q
√
where Aa and Ab are normalization constants =1/ 4 ∓ 2 2
(c) The system is initially in state |1i. Find the probability that it is still in
state |1i after a time t.
Use time evolution operator: |ψ(t)i = eiHt/h̄ |1i. Now we can’t do much
with this until we express |1i in terms of the eigenvectors: |1i = (1, 0) =
ha|1i|ai + hb|1i|bi = Aa |ai + Ab |bi. So
|ψ(t)i = Aa |aie−iEa t/h̄ + Ab |bie−iEb t/h̄
Now the probability of finding it in state |1i is h1|ψ(t)i|2 , or
|A2a e−iEa t/h̄ + A2b e−iEb t/h̄ |2
(d) Suppose you are given another Hermitian operator Q = q1 |2ih2|. Determine whether it is possible to find a set of states which are simultaneous
eigenvectors of Q and H.
QH = E0 q1 (−|2ih2| + |2ih1|)
HQ = E0 q1 (−|2ih2| + |1ih2|)
So [H, Q] ∝ |1ih2| − |2ih1| =
6 0, in other words it is not possible.
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