Quantum Mechanics: An Experimentalist’s Approach (1st Edition, 2014) – Solutions – Commins

Chapters 2-24 & All Figures Covered

,Solutions to problems in Quantum Mechanics by Eugene D. Commins
Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014
Chapter 2
2.1a) Let A u = λ u . Then u A u = λ u u . Thus,
u A u = u A† u = λ * u u
If A† = −A ,
− u A u = λ* u u = −λ u u
Thus:
λ* = −λ
so λ is imaginary or zero.
2.1b) If A† = A, B† = B , then
[ A, B ]† = ( AB − BA )† = B† A† − A† B†
= BA − AB = − [ A, B ]
Thus [ A, B ] is skew-Hermitian or zero.
2.1c) Suppose that A, B, I are n × n matrices, where n is a finite positive integer. Then
if:
AB − BA = iqI
we have:
trace(AB − BA) = 0
but:
trace ( iqI ) = iqn
which is a contradiction.

2.2. Straightforward computation using the method described in Sec. 2.5 of the text yields
the following results:

n=0: φ0 (x) = 1
n = 1: φ1 (x) = 3x
⎛3 1⎞
n=2: φ2 (x) = 5 ⎜ x 2 − ⎟
⎝2 2⎠
⎛5 3 ⎞
n = 3: φ 3 (x) = 7 ⎜ x 3 − x ⎟
⎝2 2 ⎠
Apart from normalization, these are the Legendre polynomials:
P0 ( x ) = 1
P1 (x) = x
3 1
P2 (x) = x 2 −
2 2
5 3
P3 ( x ) = x 3 − x
2 2



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,Solutions to problems in Quantum Mechanics by Eugene D. Commins
Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014
2.3. Consider the double integral:
x2 x2
I= ∫ x1
dx ∫
x1
[ f (x)g(y) − f (y)g(x)]2 dy (1)
Obviously, I ≥ 0. Expanding the integrand in (1), we have:
x2 x2 x2 x2 x2 x2
∫ x1
f 2 (x)dx ∫ g 2 (y)dy + ∫ g 2 (x)dx ∫
x1 x1 x1
f 2 (y)dy ≥ 2 ∫
x1
f (x)g(x)dx ∫
x1
f (y)g(y)dy
which immediately yields:
2

∫
x2

x1
f (x)dx ∫ g (x)dx ≥
2
x2

x1
2
(∫ x2

x1
f (x)g(x)dx ) (2)

2.4a) Let the (orthonormal) eigenvectors of G be ui and the corresponding (positive)
eigenvalues be gi = γ i2 where the γ i are real numbers. Similarly, let the eigenvectors of H
be w j and the corresponding eigenvalues be h j = η 2j where the η j are real. We have:

G = ∑ γ i2 ui ui , H = ∑ η 2j w j w j
i j

Thus,
GH = ∑ γ i2η 2j ui ui w j wj (1)
i, j

Now tr(GH) can be evaluated in any basis, for example the ui basis. Thus,

tr(GH ) = ∑γ 2
i η 2j um ui ui w j w j um
i, j, m
2 (2)
= ∑ γ η ui w j2
i
2
j w j ui = ∑ γ η 2
i
2
j ui w j
i, j i, j

since um ui = δ mi . Each term in the double sum (2) is non-negative. Hence:
tr(GH ) ≥ 0 (3)
To have tr(GH)=0, each term in the double sum must vanish. Since the eigenvalues of G
and H are all positive, this implies that ui w j = 0 for all i, j ; hence from (1) that
GH = 0. However, if we assume that the eigenvectors of G and H each form a complete
set, it is impossible to have ui w j = 0 for all i, j .
2.4b) As in 2.4a) we have: H = ∑ η j w j w j . Let A = ∑ η j w j w j . Then:
2

j j

A = ∑ η jη k w j w j wk wk
2

j, k

= ∑ η jηk w j δ jk wk = ∑ η 2j w j w j = H
j, k j

Also, A is Hermitian since the η j are all real. Then:




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, Solutions to problems in Quantum Mechanics by Eugene D. Commins
Published by Cambridge University Press, ISBN 9781107063990. © Eugene D. Commins 2014
u H u = u AA u = u A† A u = x x
where x = A u . Similarly, vH v = yy where y =Av . From Schwarz’
2
inequality, x x y y ≥ x y . Thus:
2 2
u H u v H v ≥ u A† A v = uH v
Also tr(H ) = ∑ ηi2 > 0 , since otherwise all ηi = 0 in which case H = 0.

2.5. In general, A u = α u + β u⊥ where α , β are complex numbers and u u⊥ = 0 .
Thus,
u A u = α u u + β u u⊥ = α u u
Hence,
u Au
α=
uu
which is real if A is Hermitian. Now,
AA u = α A u + β A u⊥
Thus,
u AA u = α u A u + β u A u⊥
2
= α2 u u + β u⊥ u⊥
Hence:
2
u Au 2
u AA u − = β u⊥ u⊥
uu
Therefore
2
2 u⊥ u⊥
u A2 u u Au
β = − 2 = A2 − A 2
uu uu uu
We can choose the norm of u⊥ such that u⊥ u⊥ = u u , and we can choose the phase
of u⊥ so that β is real and positive. Then,


β = A2 − A 2

2.6. Let A n = λn n .
a) Since [ A, B ] = 0 , AB n = BA n = λn B n . Thus B n is an eigenvector of A with
eigenvalue λn . Therefore either B n = consti n = bn n ; or B n and n are linearly
independent, in which case λn is a degenerate eigenvalue. Similar remarks hold for C:
C n = cn n where cn is a constant, or λn is a degenerate eigenvalue. However, since



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