Physics 264L: Assignment 10
Made available: Friday, November 13, 2015
Due:
Monday, November 23, 2015, by 7 pm
1. Some properties of the classical harmonic oscillator
The most general solution for the position x(t) of the classical harmonic oscillator for a mass m
connected to a spring with spring constant k is given by
x(t) = A cos(ωt) + B sin(ωt),
where the angular frequency ω =
(1)
√
k/m.
(a) Show that Eq. (1) can be written as a single cosine function with the same angular frequency like
this
x(t) = C cos(ωt + ϕ),
(2)
where you need to explain how to express the amplitude C and the angle ϕ in terms of the
amplitudes A and B. That is, adding cosine and sine functions of the same frequency but with
different amplitudes has the same mathematical form as a single cosine function of the same
frequency, but with a new amplitude and phase.
√
√
A hint: one approach is to multiply the right side of Eq. (1) by A2 + B 2 / A2 + B 2 = 1 and use
trigonometry. An alternative approach, more in the spirit of 264L, is to use complex exponentials,
by observing that A cos(ωt) + B sin(ωt) = Re[Aeiωt ] + Re[(−iB)eiωt ] and then convert a certain
complex number to its polar form, x + iy = ρeiθ .
(b) Given a periodic function f (t) with period T , one can define its time-averaged value ⟨f ⟩ to be its
average over one period:
∫
1 T
⟨f (t)⟩ =
f (t) dt.
(3)
T 0
Given this, show that the time-average ⟨K⟩ of the kinetic energy (1/2)mv 2 = p2 /(2m) and the
time-average ⟨V ⟩ of the harmonic oscillator potential (1/2)kx2 = (1/2)mω 2 x2 are equal and given
by
1
⟨K⟩ = ⟨V ⟩ = E,
(4)
2
where E = p2 /(2m)+(1/2)mω 2 x2 is the total energy of the harmonic oscillator. Eq. (4) also holds
for the quantum averages (expectation values) of the same quantities for the harmonic oscillator
stationary states ψn .
2. Problem 4-13 on pages 190-191 of French and Taylor.
3. Probability for a position measurement of the mass of the harmonic oscillator in its ground
state to be in a classically forbidden region
Given the ground state wave function ψ0 (x) for the quantum harmonic oscillator
(
ψ0 (x) =
1
√
πa
)1/2
[
]
1 ( x )2
exp −
,
2 a
(
a=
ℏ
mω
)1/2
,
E0 =
1
ℏω,
2
(5)
show that the probability of measuring the position of the point mass m to be in a classically forbidden
region is approximately equal to 0.16 (but you should give your answer to six significant figures).
1
Based on the qualitative forms of the quantum harmonic oscillator states ψn (x), would you expect
that the probability to find a particle in a classically forbidden region to increase, stay the same, or
decrease with increasing energy En of the state?
Test your thinking by using Table 4-1 on page 167 of French and Taylor and Mathematica to calculate
the approximate probability for the mass to be in a forbidden region for the first excited energy state,
ψ1 .
Hint: Your calculations will involve definite integrals for which the integrands do not have antiderivatives that can be expressed as a finite algebraic expression involving “elementary” functions (polynomials, sin, cos, log, exp, etc) so you will instead need to find an approximate numerical answer. As a
first step in obtaining a numerical approximation, change variables in your integrals to a dimensionless
length y (so the length a becomes the new unit of length)
y = x/a,
(6)
and you will then obtain integrals that are pure numbers (as must be the case for probabilities),
without any parameters left. Then use Mathematica’s help system to learn how to use the numerical
integration function NIntegrate to find approximations to your simplified definite integrals.
4. The quantum wizards strikes again, this time messing around with the ground state of a
quantum harmonic oscillator
A particle with mass m is in the ground state of a quantum harmonic oscillator with classical frequency ω when a quantum wizard instantly makes the spring constant k = mω 2 four times stronger
(so the new oscillator frequency is twice as big). What is the probability that an energy measurement
will now give the value E = ℏω/2 (the original ground-state energy value), and what is the probability
that an energy measurement will now give the value E = ℏω (the new ground-state energy value)?
5. Average quantities for the quantum harmonic oscillator
Using the method of ladder operators discussed in Example 2.5 on page 49 of the Griffiths text (posted
on Sakai), show that, for the nth stationary state ψn (x) of the quantum harmonic oscillator with n ≥ 0,
(
)
1
⟨p2 ⟩n = n +
(7)
mℏω,
2
ℏ
σx σp ≥ ,
(8)
2
√
√
where σx = ⟨x2 ⟩ − ⟨x⟩2 and σp = ⟨p2 ⟩ − ⟨p⟩2 , and equality holds only for the ground state n = 0.
(The value of ⟨x2 ⟩n = (n + 1/2)ℏ/(mω) was calcuated in class and was calculated in pages that were
posted from Griffiths’ book.) This inequality says that a position-momentum uncertainty principle
is satisfied for all stationary states of the quantum harmonic oscillator. The result for ⟨p2 ⟩n can be
rewritten as ⟨p2 /(2m)⟩n = (1/2)[(n + 1/2)ℏω] = En /2, which is the quantum version of the classical
result Eq. (4) you derived above.
Also use ladder operators to determine without any integration the value of
⟨x2 p2 ⟩n .
(9)
Based on what you you learn from the next problem about commutators, discuss whether would you
get the same average value Eq. (9) if you wrote the four operators in a different order: do the averages
⟨p2 x2 ⟩n , ⟨xp2 x⟩n , ⟨px2 p⟩n , ⟨pxpx⟩n , etc all give the same value? This is a concern since the operators
x and p = −iℏ∂x do not commute1 , [x, p] = xp − px = iℏ. If they don’t all give the same value, what
do we mean by Eq. (9)?
1 Which
is an occasion to mention an old and rather bad math joke: What is purple and commutes? An Abelian grape.
2
6. The commutator of two operators corresponding to observables
Operators play a huge role in quantum mechanics because they turn out to be the way that observables
are represented in the more general quantum theory2 . For example, the position x of a particle is
represented by the operator “multiplication of a function by the number x”, f (x) → xf (x), momentum
is represented by the operator p = −iℏ∂x so f → −iℏ∂x f , and the three components of the angular
momentum vector L, which classically are given by3
x̂ ŷ ẑ L = x × p = (ypz − zpy , zpx − xpz , xpy − ypx ) = x y z ,
(10)
px py pz are represented in quantum mechanics by three operators that act on a general scalar function f (x, y, z)
of three coordinate variables:
Lx = y (−iℏ∂z ) − z (−iℏ∂y ) ,
(11)
Ly = z (−iℏ∂x ) − x (−iℏ∂z ) ,
Lz = x (−iℏ∂y ) − y (−iℏ∂x ) .
(12)
(13)
Given two linear operators A and B that act on functions to give new functions, one defines their
commutator to be a new operator [A, B] that measures to what extent the order of the operators
matters
[A, B] = AB − BA
(14)
If [A, B] = 0 is the zero operator (multiplication of a function by zero), then the order doesn’t matter
and one says that “the operator A commutes with the operator B”.
The commutator of two operators plays a central role in quantum mechanics. If the commutator of two
operators is not zero, then an uncertainty principle holds between the observables that the operators
correspond to, one cannot measure in quick succession the precise values of two non-commuting operators. In the following, you will learn how to manipulate commutators algebraically and so discover
uncertainty principles that hold for the components of the angular momentum vector L.
(a) Show that for any operators A, B, and C and for any complex-valued constant c:
[A, A] = 0,
[A, B] = −[B, A],
[cA, B] = c[A, B],
[A + B, C] = [A, C] + [B, C],
[AB, C] = A[B, C] + [A, C]B.
(15)
(16)
(17)
(18)
(19)
(b) Show that for any function f (x)
df
.
(20)
dx
Hint: to show this result, show that both sides applied to some general function g(x) give the
same result.
[f (x), p] = iℏ
2 Only certain kinds of operators called “Hermitian operators” can correspond to physical quantities, such operators obey
the property that they can be moved from one side of an inner product to the other side without changing the value of the
inner product: A is an Hermitian operator if and only if ⟨ Ψ1 | AΨ2 ⟩ = ⟨ AΨ1 | Ψ2 ⟩ for all square-integrable wave functions
Ψ1 and Ψ2 . You might like to show that Hermitian operators have real-valued eigenvalues which is why they are good for
representing experimentally measurable quantities.’
3 Note how the second and third components are obtained by what is called “cyclic substitution” of the labels x → y → z → x
so you only need to remember the first component, from which you can quickly deduce the other two components.
3
(c) For the angular momentum operators Eqs. (11)-(13), show that
[Lx , Ly ] = iℏLz ,
[Ly , Lz ] = iℏLx ,
[Lz , Lx ] = iℏLy ,
(21)
where you only have to prove the first equality, the others follow by cyclic permutations of the
symbols x → y → z → x. The fact that these three commutators are non-zero means that one
cannot measure a precise value of the x-component of the momentum immediately after measuring
a precise value of the y component and vice versa, and similarly for pairs of other components of
the angular momentum.
Hint: apply both sides of the commutator relation [Lx , Ly ] = iℏLz to some general function f (x, y, z) and show that both sides lead to the same result.
(d) Using the identity Eq. (19) and the relations Eqs. (11)-(13), show that the operator L2 = L2x +
L2y + L2z corresponding to the magnitude squared of the angular momentum vector, does commute
with the Lx (and by symmetry also with Ly and with Lz ), i.e., show that
[ 2
]
L , Lx = 0.
(22)
This means that one can measure quickly in succession the the precise value of the magnitude
of the angular momentum and the precise value of exactly one of its components. (Note that
[L2 , Lx ] = 0 and [L2 , Ly ] = 0 does not imply [Lx , Ly ] = 0.)
Hints: Use the commutator properties that you derived above, Eqs. (15)-(19). As an example,
part of your calculation will involve an expression like [L2y , Lx ] which by Eq. (19) can be written
as [Ly Ly , Lx ] = Ly [Ly , Lx ] + [Ly , Lx ]Ly , but [Ly , Lx ] = −[Lx , Ly ] = −iℏLz , and so on.
We will stop at this point, but I would like to point out that you are very close to being able to use the
above results to deduce what possible values an angular momentum measurement will produce. The
idea is to introduce raising and lowering operators L± = Lx ± iLy and then follow an argument similar
to what we did for the harmonic oscillator. You can see the details on pages 160-166 of the Griffiths
book “Introduction to Quantum Mechanics, second edition”, and the conclusion is that the possible
experimental values of length-squared of the angular momentum are
L2 values: l(l + 1) ℏ2 ,
l = 0, 1/2, 1, , 3/2, · · · ,
(23)
where the quantum number l can have half-integer values (a surprise that emerges from the discussion
but in agreement with experiment). Possible simultaneous values of the z component Lz that can be
measured for a given angular momentum state l are:
Lz values: mℏ,
m = −l, −l + 1, · · · , l − 1, l,
(24)
i.e., for a given angular momentum state with quantum value l, there are 2l + 1 values of Lz . These are
remarkable results in that they follow just from the algebraic structure of commutators, without having
to refer directly to the Schrodinger equation. These results are also fundamental for understanding
properties of atoms and molecules since, when a photon strikes an atom, angular momentum is conserved and only certain transitions to new levels can occur that are consistent with angular momentum
of the atom and photon.
7. Consider a particle of mass m in the infinitely strong cubic well of length L = Lx = Ly = Lz described
in Section 5-4 of French and Taylor, with energy levels given by Eq. (5-12) on page 199, where the
integers nx , ny , nz satisfy nx ≥ 1, ny ≥ 1, and nz ≥ 1.
(a) Write down the general fully normalized wave function ψnx ,ny ,nz (x, y, z) corresponding to the
quantum numbers nx , ny , and nz
4
(b) Let the distinct energy values in increasing order be E1 , E2 , E3 , etc. Find the first six lowest
energy values and also give their degeneracies (the number of distinct quantum numbers that have
the same energy).
Note: To help you get started, the lowest energy or ground state E1 corresponds to the quantum
numbers nx = ny = nz = 1 and has degeneracy d = 1, and one of the wave functions corresponding
to the second energy level E2 has the quantum numbers nx = 2, ny = 1 and nz = 1. Two- and
three-dimensional symmetric potentials often have energy levels with degeneracies greater than
one, while one can show that the degeneracy of all bound-state energy levels in a one-dimensional
potential V (x) is always one. (An exception is a particle in a circular ring, because the wave
function doesn’t have to decay to zero at x = ±∞ in that case.)
(c) What is the degeneracy of level E14 ? Discuss briefly why this state has an unusual degeneracy.
8. Problem 5-6(a) on page 224 of French and Taylor. (Just do part (a).)
9. Problem 5-9, parts (a) through (c) only on pages 225-226 of French and Taylor.
10. Consider a spherically symmetric finite spherical well of radius R with potential V (r) = 0 for r < R
and V (r) = V0 for r ≥ R.
(a) From pages 200-201 of French and Taylor and from Eq. (4-4b) on page 158 of French and Taylor,
explain why no radially symmetric stationary bound state ψ(r) exists unless the depth of the
well V0 of the state satisfies
π 2 ℏ2
.
(25)
V0 ≥
8mR2
(b) Use Eq. (25) to estimate how strong is the nuclear force for a deuteron, which is a hydrogen
nucleus consisting of a proton bound to a neutron. Experiments show that the deuteron has
only a ground state with binding energy 2.23 MeV (this is the minimum energy needed to break
the deuteron apart into a neutron and proton) and no bound excited states. Assuming the two
particles act as if they are in a spherically symmetric finite well of radius R ≈ 2 × 10−15 m, and
that the ground state wave function depends on radius only (is radially symmetric), estimate the
height V0 in units of MeV of the spherical potential, which in turn gives you some insight about
how strong is the strong interaction compared to say the electrical Coulombic interaction.
5