PHY4604–Introduction to Quantum Mechanics
Fall 2004
Practice Test 1 solutions
October 6, 2004
These problems are similar but not identical to the actual test. One or two parts will
actually show up.
1. Short answer.
• Explain the 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.
• Explain the significance of h̄ in quantum mechanics, and give an example
of a place where it shows up.
Planck’s const. is a quantum (smallest possible unit) of angular momentum. It shows up, for example, in the orbits of the Bohr atom, where
an electron may not have any value of angular momentum, but only an
integer number of nh̄
• Discuss the uncertainty principle briefly
If you make a measurement of an object’s position and momentum simultaneously, you cannot measure them both arbitrarily precisely. The
product of the uncertainty in knowledge of the object’s position ∆x and
in momentum ∆p fulfills the Heisenberg relation
∆x∆p >
∼ h̄
(1)
• Explain the difference between the 2 versions of Schrödinger’s equation
h̄2 ∂ 2 ψ
∂ψ
=−
+Vψ
ih̄
∂t
2m ∂x2
and
−
h̄2 ∂ 2 ψ
+ V ψ = Eψ
2m ∂x2
The first is the general version, the second applies for ”stationary states”,
i.e. if ψ is an eigenfunction of the Hamiltonian. These special states change
in time in a trivial way, ψn (x, t) = ψ(x) exp(−iEn t/h̄).
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• What are the units of P (x, t), the probability density in 1 dimension?
Justify your answer.
Must be 1/volume, since
R
dxP (x, t) = 1.
• Calculate the commutator [px , x2 ]
[px , x2 ]ψ = (−ih̄∇)(x2 ψ) − x2 (−ih̄∇)ψ
= −ih̄2xψ − ix̄2 ψ 0 + ih̄x2 ψ 0
= −ih̄2xψ.
So [px , x2 ] = −ih̄2x.
• Calculate the expression for the Bohr levels of the hydrogen atom from the
Bohr-Ehrenfest quantization condition.
See notes.
2. Consider a wave packet defined by
ψ(x) =
Z ∞
−∞
dkf (k)ei(kx−ωt)
(2)
with ω = h̄k 2 /2m and f (k) given by


 0
k < −∆k/2
f (k) =  a −∆k/2 < k < ∆k/2

0 ∆k/2 < k
(a) Find the form of ψ(x) at t = 0.
Z
ψ(x) =
ikx
dkf (k)e
=a
Z a/2
−a/2
dkeikx
a
= = (eix∆k/2 − e−ix∆k/2 )
ix
∆kx
a
2i sin
.
=
ix
2
(b) Find the value of a for which ψ(x) is properly normalized.
Need
R
dx|ψ|2 = 1, so
Z
sin2 ∆kx
2
x2
= 4a2 π∆k/2
1 = 4a2
2
dx
(3)
q
so a = 1/(2π∆k). I would have given you this integral on the formula
sheet on a real test.
(c) How is this related to the choice of a for which
Z ∞
−∞
dk|f (k)|2 = 1?
(4)
√
From the definition of f , the integral means a = 1/ ∆k. Put another way,
the way the Fourier transform f is defined here, |f |2 is not normalized to
1 but to 1/2π.
(d) Show that for a reasonable definition of ∆x, the size of the packet given
by your answer in a), ∆k∆x > 1.
A good estimate of the size of the packet might be the first time sin ∆kx/2
goes to zero, which occurs at x ' ∆x/2 = 2π/∆k. So ∆x∆k ' 4π for this
packet.
3. A particle in an infinite square well (of width a) has as its initial wave function
an equal mixture of the first two stationary states:
Ψ(x, 0) = C[ψ1 (x) + ψ2 (x)]
(a) Normalise Ψ(x, 0). (That is, find C.)
First we’ll need to find ψ1 (x), ψ2 (x), the 1st 2 stationary states. From
notes or just from guessing with the standing wave boundary conditions
at x = ±a/2, they are
s
ψ1 =
s
ψ2 =
2
πx
cos
a
a
2
2πx
sin
a
a
corresponding to energy eigenvalues −h̄2 ∇2 /2mψα = Eα ψα
h̄2 π 2
2ma2
4h̄2 π 2
=
2ma2
E1 =
E2
Time dependence: we will need to know how these 2 states evolve in
time. Since they are eigenstates of H, we must have Hα ψα = Eα ψα =
ih̄(∂/∂t)ψα , which has the solution
ψα (x, t) = ψα (x)e−iEα t/h̄
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Now, on to business. The normalization condition is
Z
dx|Ψ|2
1 =
Z
= C2
dx(|ψ1 |2 + |ψ2 |2 + ψ1∗ ψ2 + ψ2∗ ψ1 )
= C 2 (1 + 1 + 0 + 0)
where the last 2 terms vanish because they are odd in x. So C 2 = 1/2.
(b) Find Ψ(x, t) and |Ψ(x, t)|2 . Express the latter in terms of sin and cos using
eiθ = cos θ + i sin θ. Use ω = π 2 h̄/2ma2 .
From above, Ψ(x, t) = C(ψ1 (x)e−iE1 t/h̄ + ψ2 (x)e−iE2 t/h̄ ), and
|Ψ(x, t)|2 = C 2 (|ψ1 |2 + |ψ2 |2 + ψ1 ψ2 (ei(E1 −E2 )t/h̄ + e−i(E1 −E2 )t/h̄ )
= C 2 (|ψ1 |2 + |ψ2 |2 + 2ψ1 ψ2 cos 3ωt,
where I used the info given, that E1 = h̄ω, E2 = 4h̄ω. Note the oscillation
in time.
(c) Compute < x >. Notice that it oscillates in time. What is the frequency
of the oscillation? What is the amplitude?
Z
hxi = C
=
2
dx(|ψ1 |2 + |ψ2 |2 + 2ψ1 ψ2 cos 3ωt)x
2 Z a/2
πx
2πx
16a
dx cos
sin
x = 2 cos 3ωt,
a −a/2
a
a
9π
where note the 1st two terms vanished because they were even in x, but
now last one wasn’t. So this is a particle which is sloshing back and forth
in the well. Again I would have given you this integral on a real test.
(d) Compute < p >.
Z
hpi = C
2
dx((ψ1∗ + ψ2∗ )(−ih̄∇)(ψ1 + ψ2 )
´
4 ³ i(E2 −E1 )t/h̄
ih̄ e
− e−i(E2 −E1 )t/h̄
3a
4h̄
sin 3ωt
=
3a
= −
(e) Find the expectation value of the Hamiltonian operator, H, in terms of E1
and E2 .
Z
(ψ1∗ + ψ2∗ )H(ψ1 + ψ2 )
hHi = C 2
Z
= =C
Z
= C2
2
(ψ1∗ + ψ2∗ )(E1 ψ1 + E2 ψ2 )
(E1 |ψ1 |2 + E2 |ψ2 |2 ) = (E1 + E2 )/2,
4
R
where in the 2nd to last step I used dxψ1∗ ψ2 =0, etc. from the properties
of sin, cos, or recall that eigenfunctions belonging to different eigenvalues
are orthogonal. In the last step I used the normalization of ψ1 , ψ2 .
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