PHZ 3113 Fall 2010
Exam 3
Name:
This is a closed-book exam. Some possible useful information:
(1 + x)p = 1 + p x +
∞
X
1
p!
1
1
p(p − 1) x2 + p(p − 1)(p − 2) x3 + · · · =
xn
2
6
n! (p − n)!
n=0
ex = 1 + x +
∞
X
1 n
1 2 1 3
x + x +··· =
x
2
6
n!
n=0
sin a cos b =
1
[sin(a + b) + sin(a − b)]
2
cos a cos b =
1
[cos(a + b) + cos(a − b)]
2
sin2 ( 12 x) =
Z
f (x) =
dk
f̃ (k) eikx
2π
1
(1 − cos x)
2
Z
f̃ (k) =
dx f (x) e−ikx
1. The Pauli matrices σk , where k runs over 1, 2, 3 or x, y, z, are defined as
0 1
0 −i
1
0
σx =
,
σy =
,
σz =
.
1 0
i
0
0 −1
(a) Compute σk2 , for k = x, y, z. Compute the products σx σy , σy σz , σz σx , σy σx , σz σy ,
σx σz . Is there a pattern?
The squares are all the identity matrix,
σx2 = σy2 = σx2 =
1 0
0 1
= 1.
The products are
σx σy =
σy σz =
σz σx =
i
0
0
−i
0
i
i
0
0 1
−1 0
= i σz = −σy σx ,
= i σx = −σz σy ,
= i σy = −σx σz .
These can be summarized
σm σn = δmn 1 + kmn σk .
The objects i = i σx , j = i σy , k = i σz are a realization of Hamilton’s quaternion algebra
[http://mathworld.wolfram.com/Quaternion.html],
i2 = j2 = k2 = i j k = −1.
(b) What are the eigenvalues of σx , σy , σz ? What are the eigenvectors?
The matrix σz is already diagonal, with eigenvalues λ = ±1 and eigenvectors vz+ = (1, 0)
and vz− = (0, 1). We know that det M = λ1 λ2 and tr M = λ1 + λ2 . Since det σk = −1
and tr σk = 0 for all three matrices, all three have the same eigenvalues, λ = ±1. The
eigenvectors of σx and σy are
1
1
±1
±1
,
vy± = √
vx± = √
1
i
2
2
(normalization not required). Any eigenvectors can be multiplied by ±1 and by ±i to give
equivalent results.
(c) For any constant vectors a, b, show that
(σ · a)(σ · b) = (a · b) 1 + i (a × b) · σ ,
where 1 is the two-by-two identity matrix.
You can write out
(σ · a)(σ · b) = (ax σx + ay σy + az σz )(bx σx + by σy + bz σz )
and do out all the products, or use
(σ · a)(σ · b) = (am σm )(bn σn ) = am bn (δmn 1 + i kmn σk )
= (am bm ) 1 + i (a × b)k σk = (a · b) 1 + i (a × b) · σ.
(d) For fixed unit vector n̂, compute exp( 12 iθ n̂ · σ).
From (c),
(n̂ · σ)2 = (n̂ · n̂) 1 + i (n̂ × n̂) · σ = 1.
[Note that in part (a) you found that this holds for n̂ = x̂, n̂ = ŷ, and n̂ = ẑ.] So, any even
power is (n̂ · σ)2k = 1 and any odd power is (n̂ · σ)2k+1 = (n̂ · σ), and
h
i
1 ( 1 iθ)4 + · · · 1
exp( 12 iθ n̂ · σ) = 1 + 12 ( 12 iθ)2 + 24
2
h
i
1
1
1
1
1
3
5
+ ( 2 iθ) + 6 ( 2 iθ) + 120 ( 2 iθ) + · · · (n̂ · σ)
= cos( 12 θ) 1 + i sin( 12 θ) (n̂ · σ).
2. Three beads of equal mass m are constrained to lie on a circular hoop of radius a = 1.
Each bead is connected to its two neighbors by identical springs of spring constant k. The
resulting equations of motion for displacements θk are
m θ̈1 = k(θ2 + θ3 − 2θ1 ),
m θ̈2 = k(θ3 + θ1 − 2θ2 ),
m θ̈3 = k(θ1 + θ2 − 2θ3 ).
(a) Assume there are oscillatory solutions, θ1 (t) = c1 cos ωt, etc. What are the characteristic
frequencies of oscillations?
With θ1 = c1 cos(ωt), etc., θ̈k = −ω 2 θk , and the equations of
!
!
θ1
−2k
k
k
−mω 2 θ2 =
k −2k
k
θ3
k
k −2k
motion become
!
θ1
θ2 .
θ3
Write as a single matrix,
mω 2 − 2k
k
k
k
mω 2 − 2k
k
k
k
2
mω − 2k
!
θ1
θ2
θ3
!
= 0.
As usual, if the matrix is invertible the problem has only the trivial solution θ1 = θ2 = θ3 = 0.
So, the matrix must not be invertible, and thus has determinant zero,
det M = m3 ω 6 − 6km2 ω 4 + 9kmω 2 = mω 2 (mω 2 − 3k)2 = 0.
There is one mode with ω 2 = 0 and two degenerate modes with ω 2 = 3k/m.
(b) What are the oscillation patterns? Because of the high degree of symmetry of the problem,
you may be able to guess some of the characteristic modes, but show that you can choose
basis modes that are orthogonal.
For ω 2 = 0, the displacement vector satisfies
2θ1 = θ2 + θ3 ,
2θ2 = θ3 + θ1 ,
2θ3 = θ1 + θ2 ,
easily seen to require θ1 = θ2 = θ3 or
θ0 = (1, 1, 1).
This mode does not excite any of the springs.
For ω 2 = 3k/m, all three equations of motion reduce to θ1 + θ2 + θ3 = 0. This leaves a
two-dimensional space of solutions, and lots of freedom. One choice is to keep any one of the
masses fixed; the other two then move in opposite directions. This gives the three symmetric
possibilities θ1,1 = (0, 1, −1), θ1,2 = (−1, 0, 1), and θ1,3 = (1, −1, 0), of which any two are
independent (and the third is the sum of the other two). Another possibility is to take any
two displacements to be equal; the third then moves twice as far in the opposite direction,
θ01,1 = (−2, 1, 1), θ01,2 = (1, −2, 1), θ01,3 = (1, 1, −2). Again, the three of these sum to zero;
any one is the sum of the other two.
For an orthogonal choice of modes, take one of the first form and one of the second form,
θ1 = (0, 1, −1),
θ01 = (−2, 1, 1)
Of course, (2, 1, −3) and (4, −5, 1) also work, as do an infinity of other choices.
3. Let
f (x) =
−1 (−1 < x < 0)
+1 ( 0 < x < 1)
(a) Take f (x) be periodic with period L = 2 (so that −1 < x < 1 fills one complete period).
Write f (x) as a sine/cosine Fourier series or as a complex exponential Fourier series.
The complex exponential series is of the form
f (x) =
∞
X
n=−∞
cn einπx
(the π is there to map the period into 2π in phase). If you multiply by e−inπx for some n
and integrate,
Z 1
f (x) e−inπx dx = 2 cn
−1
(all the terms for different values of n on the right vanish). For n = 0 the integral of f
vanishes and c0 = 0; for n 6= 0 the integral is
Z 1
0
−inπx
e
Z 0
dx−
−1
e−inπx dx =
Thus, the coefficients are
h e−inπx i1 h e−inπx i 0
i
i
−
=
[(−1)n −1]− [1−(−1)n ].
−inπ 0
−inπ −1 nπ
nπ

 2
cn = inπ

0
(n odd)
(n even)
and the complex exponential series is
f (x) =
∞
X
odd n=−∞
2 inπx
e
.
inπ
If you combine the +n and −n terms, you get
2 inπx
4
2 einπx 2 ei(−n)πx
+
=
(e
sin(nπx),
− e−inπx ) =
inπ
i(−n)π
inπ
nπ
and the sine series is
f (x) =
∞
X
4
sin(nπx).
nπ
odd n=1
You can also obtain this directly integrating over sines and cosines; the coefficients an all
vanish for any odd function, and the bn are as obtained from the cn .
(b) Take the given values of f to be the only ones on the real line that are nonzero; f (x) = 0
for −∞ < x < −1 and 1 < x < ∞. Find the corresponding Fourier amplitude f̃ (k).
Z ∞
−∞
f (x) e−ikx dx =
Z 1
f̃ (k) =
0
e−ikx dx −
Z 0
−1
e−ikx dx =
e−ik − 1 1 − eik
−
,
−ik
−ik
1
2(1 − cos k)
(2 − eik − e−ik ) =
.
ik
ik
This was supposed to be easy, the function f is piecewise constant, but the eik term gave
trouble. It has three minus signs, one from the value f = −1, one from the minus in the
exponent −ikx, and one from evaluation at the lower limit x = −1. The net is, both eikx
and e−ikx have the same (negative) sign.
(c) Compute either the value of the sum S0 or of the integral I0
Z ∞
sin3 ( 12 y) dy
I0 =
.
y
−∞
∞
X
(−1)k
,
S0 =
2k + 1
k=0
[There are many ways to do this, but you might think about f ( 12 ).]
As it says,
f ( 12 ) =
∞
X
odd n=−∞
2 in π
e 2 =
inπ
∞
X
odd n=1
So,
S0 =
∞
4
4 X (−1)k
π
sin(n 2 ) =
= 1.
nπ
π
2k + 1
k=0
π
.
4
Or,
f ( 12 ) =
Z
dk 2(1 − cos k) i 1 k
2
e2 =
2π
ik
iπ
Z
dk
sin2 ( 12 k)
[cos( 12 k) + i sin( 12 k)] = 1.
k
The imaginary part (the integral over cosine) vanishes as the integral of an overall odd
function (the 1/k is odd), and the real part gives
I0 =
π
.
2
(d) Compute either the value of the sum S1 or of the integral I1 ,
S1 =
∞
X
k=0
Z ∞
sin4 ( 12 y) dy
I1 =
.
y2
−∞
1
,
(2k + 1)2
These are variations of Parseval’s theorem. For the sum,
Z 1
2=
−1
So,
|f 2 (x)| dx = 2
∞
X
n=−∞
|cn |2 =
∞
X
odd n=−∞
4
n2 π 2
.
π2
S1 =
.
8
For the integral,
Z
2=
|f (x)|2 dx =
Z
Z
Z ∞
dk 4(1 − cos k)2
dk
16 ∞ sin4 ( 12 k)
2
|f̃ (k)| =
=
dk.
2π
2π −∞
k2
k2
−∞ 2π
So,
I1 =
π
.
4