PHY4604–Introduction to Quantum Mechanics
Fall 2004
Test 2 Solutions
Nov. 5, 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
Pn = |hn|ψi|2
Hψ = Eψ
∂
Hψ = ih̄ ψ
∂t
p̂ = −ih̄∇
ψ(x) = hx|ψi
(P Q)† = Q† P †
det(H − E) = 0
X
1=
|nihn|
n
T = |jtrans |/|jinc |
R = |jref l |/|jinc |
Z
dp
eipx/h̄ φ(p, t)
ψ(x, t) = √
2πh̄
Z
dx
φ(p, t) = √
e−ipx/h̄ ψ(x, t)
2πh̄
h̄2 2
H=−
∇ + V (r)
2m
(O† φ, χ) = (φ, Oχ)
T +R=1
s
L± ≡
k
h̄ ∂
x∓ √
2
2m ∂x
hr|r0 i = δ(r − r0 )
ψ=
X
c n ψn
n
−iHt/h̄
ψ(x, t) = e
Z
ψ(x, 0)
p = h̄k
dxeikx = 2πδ(k)
j=−
Z
[x, p̂] = ih̄
1
δ(ax) = δ(x)
a
ih̄ ∗
(ψ ∇ψ − ψ∇ψ ∗ )
2m
dxf (x)δ(x − x0 ) = f (x0 )
1
1. Short answer. Must attempt (only) 3 of 5.
(a) A Hamiltonian for a 2-level system is written as
H = ((E1 |1ih1| + E2 |2ih2| − t|2ih1| − t|1ih2|)
where the vectors |1i and |2i represent two orthonormal states of the system which span the Hilbert space of the problem. Write down the matrix
H of the problem in the |1i, |2i basis, and find the eigenvalues Ea and Eb .
The matrix elements hα|H|βi, where α, β each can be 1 or 2, can be easily
calculated to find
"
H =
E1 −t
−t E2
#
The eigenvalue equation Hψ = Eψ can be most easily solved for the
eigenvalues by looking for the characteristic polynomial of the matrix
"
H =
E1 − E
−t
−t
E2 − E
#
,
which is found by taking the determinant. So the eigenvalues are the
solutions of (E1 − E)(E2 − E) − t2 = 0, i.e.
µ
Ea,b =
q
1
E1 + E2 ± (E1 − E2 )2 + 4t2
2
¶
(b) Prove that if (ψ, Oψ) = (Oψ, ψ) for all ψ, then O is self-adjoint.
The definition of the adjoint is (O† χ, φ) = (χ, Oφ), so an operator is selfadjoint if (Oχ, φ) = (χ, Oφ)∀ χ, φ. This is not what we are told, but
we can always expand ψ in a basis of eigenfunctions of the operator O:
P
ψ = n cn ψn . Then the equation we are given reads
X
m,n
c∗n cm (ψn , Oψm ) =
X
m,n
c∗n cm (Oψn , ψm ),
but this is true for any ψ, so for any coefficients cn . The only way this can
happen is if (ψn , Oψm ) = (Oψn , ψm ).
Now the condition for self-adjointness can be expressed in a similar way if
P
P
we expand χ = n dn ψn and φ = n fn ψn :
X
m,n
d∗n fm (ψn , Oψm ) =
X
m,n
d∗n fm (Oψn , ψm ),
and now we could prove self-adjointness if (ψn , Oψm ) = (Oψn , ψm ), which
we just happen to know from above QED.
2
(c) 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.
(d) Consider a δ-function potential V = λδ(x) of strength λ > 0. Find the
change in the derivative at x = 0 of the wave function ψ which solves the
Schrödinger equation with energy E, in terms of the value of ψ at x = 0.
Show how you got there.
Schrödinger equation is
Ã
!
h̄2 ∂ 2
−
+ λδ(x) ψ = Eψ.
2m ∂x2
. Integrate from −² to ²:
h̄2
−
2m
Ã
!
Z ²
∂
∂
dxψ(x),
ψ|² −
ψ|−² + λψ(0) = E
∂x
∂x
−²
so in the limit as ² → 0, we have
Ã
∂
∆
ψ
∂x
!
=
2mλ
ψ(0)
h̄2
(e) Two operators A and B satisfy A = B † B + 2 and A = BB † + 5. Suppose
ψ is an eigenfunction of A with eigenvalue α . Show that if Bψ 6= 0, then
Bψ is an eigenfunction of A, and find the eigenvalue.
3
We want to show A(Bψ) = λ(Bψ) and find λ. If we knew the commutator
[A, B] we could do so, so let’s try to calculate it. First, subtract the two
expressions for A to find [B, B † ] = −3. Now
[A, B] = [BB † + 5, B] = BB † B − BBB †
= −B[B, B † ] = 3B
So A(Bψ) = (3B + BA)ψ = (3B + Bα)ψ = (3 + α)(Bψ) QED
Attempt 1 of remaining 2 4-part problems!
2. Eigenvalues and Eigenvectors. You are given three normalized eigenfunctions of the Hermitian operator Q̂, ψ0 , ψ1 , and ψ2 . The ground state ψ0 is
nondegenerate, with eigenvalue q0 , while ψ1 and ψ2 are degenerate with eigenvalue q1
(a) Calculate (ψ0 , Q̂ψ1 ) and (ψ1 , Q̂ψ1 ).
(ψ0 , Q̂ψ1 ) = (ψ0 , q1 ψ1 ) = q1 (ψ0 , ψ1 ) = 0
(ψ1 , Q̂ψ1 ) = (ψ1 , q1 ψ1 ) = q1 (ψ1 , ψ1 ) = q1
(b) Assume ψ1 and ψ2 are linearly independent, but s ≡ (ψ1 , ψ2 ) 6= 0. Construct two new eigenfunctions ψ10 , ψ20 which are normalized and orthogonal,
both to each other and to ψ0 . Verify explicitly. Are ψ10 and ψ20 degenerate?
(Note: prime (0) does not mean derivative!).
Use Gram-Schmidt procedure:
ψ10 = ψ1
ψ20 = A(ψ2 + aψ1 ),
A is a normalization const., and we need (ψ10 , ψ20 ) = 0, (ψ10 , ψ0 ) = 0, and
(ψ20 , ψ0 ) = 0:
0 = (ψ10 , ψ20 ) = A(s + a),
so a = −s, ψ20 = A(ψ2 − sψ1 ), and requiring (ψ20 , ψ20 ) = 1 implies that
A = (1−|s|2 )−1 . The other orthogonality relations now follow immediately
because ψ0 is orthogonal to both ψ1 and ψ2 (eigenfunctions corresponding
to distinct eigenvalues are orthogonal). The new primed eigenfunctions
ψ10 , ψ20 of course have the same eigenvalues as the original ones, since they
are linear combinations of two eigenfunctions corresponding to the same
eigenvalue; they are still therefore degenerate.
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(c) Suppose the system is now prepared at time t = 0 in the state
ψ(x) = c(ψ0 (x) + ψ10 (x) + ψ20 (x)).
(1)
What are the probabilities that a particular measurement of Q in the state
ψ yields the result q0 ? q1 ?
√
Normalization of ψ implies that c = 1/ 3. The probability amplitude for
measuring q0 is the coefficient of ψ0 squared, i.e. 1/3. Both the other two
terms are eigenfunctions corresponding to q1 , so the probability is 2/3.
(d) Suppose Q̂ = H; find ψ(x, t) in terms of the eigenvalues E0 and E1 .
ψ(x, t) = e−iHt/h̄ ψ
1
= e−iHt/h̄ √ (ψ0 + ψ10 + ψ20 )
3
´
1 ³ −iE0 t/h̄
= √ e
ψ0 + e−iE1 t/h̄ ψ10 + e−iE1 t/h̄ ψ20
3
3. Finite step. A particle of mass m moves in the 1D potential


 0
x<0
(I)
0≤x≤L
(II)
x>L
(III)
V (x) =  V0

0
(2)
Assume that an electron is approaching this potential from the left (x < 0),
with 0 < E < V0 .
(a) Write down the solutions to the Schrödinger equation in the three regions
(I, II, III) as linear combinations of (possibly complex) exponential functions of x. Sketch the form |ψ| must have.
ψI = Aeipx/h̄ + Be−ipx/h̄ ,
ψII = Ceqx/h̄ + De−qx/h̄
ψIII = F eipx/h̄
(b) Determine the parameters which appear in the exponents in regions (I, II, III).
p =
q =
5
√
2mE
q
2m(V0 − E)
Probability density |ψ|2 . Note the incident probability density oscillates due to interference
between A and B terms. Within barrier, |ψ|2 decays monotonically. Finally, after barrier
there is no reflected wave so |ψ|2 is constant.
(c) State the boundary conditions on the wave function at x = 0 and x = L,
and use them to determine the coefficients of the exponential terms in the
wave function.
Require continuity of ψ and its derivatives at the interfaces:
ψI (0)
ψII (L)
ψI0 (0)
0
ψII
(L)
=
=
=
=
ψII (0)
ψIII (L)
0
ψII
(0)
0
ψIII (L),
in other words
A+B
Ce
+ De−qL/h̄
(A − B)ip
q(CeqL/h̄ − De−qL/h̄ )
qL/h̄
=
=
=
=
C +D
F eipL/h̄
(C − D)q
F ipeipL/h̄
(d) Write down expressions for the incident and reflected probability current
density in each of the three regions, in terms of the unknown parameters of
part (c). Give the definitions of the reflection and transmission coefficients
for the entire potential barrier and state the condition involving them
which expresses conservation of probability.
j = −
∂ψ ∗
ih̄ ∗ ∂ψ
(ψ
−ψ
)
2m
∂x
∂x
so
p
|A|2
m
p
= − |B|2
m
κ ∗
=
(C D − CD∗ )
m
p
=
|F |2
m
I
jinc
=
I
jref
l
II
jtot
III
jtrans
6
and
III
I
T = |jtrans
/|jinc
| = |F |2 /|A|2
I
I
2
2
R = |jref
l /|jinc | = |B| //|A|
and we must have T +R = 1 to conserve probability. Note there is no reason
to leave out the D term within the barrier for any finite L, because it poses
no physical problems. In fact, you can see that without D there is no way
to have a probability current density within the barrier, so you couldn’t
conserve probability from one side to the other! For a more complete
solution to this problem, see e.g. Gasiorowicz, Quantum Physics.
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