PHZ 3113 Fall 2010
Homework #7, Due Friday, November 12
1. Compute
Z ∞
−∞
dx
[1 + (x − a)2 ][1 + (x + a)2 ]
.
As a function of the complex variable z, the denominator can be factored,
[1 + (z − a)2 ][1 + (z + a)2 ] = (z − a − i)(z − a + i)(z + a − i)(z + a + i),
and vanishes at z = ±a±i. Thus, there are four poles, two in the lower half-plane (Re[z] < 0)
and two in the upper half-plane (Re[z] > 0).
The desired integral can be extended to a closed coutour integral in the complex plane in
either the upper or lower half-plane, either of which will wrap around two of the four poles.
Closed in the upper half plane the integral has contributions from poles at z± = ±a + i,
I
h
1
f (z) dz = 2πi
(−a + i − a − i)(−a + i − a + i)(−a + i + a + i)
i
1
+
(a + i − a + i)(a + i + a + i)(a + i + a + i)
h
i
h −i
i
1
1
π
= 2πi
.
+
= 2πi
=
2
8ia(i − a) 8ia(i + a)
4(1 + a )
2(1 + a2 )
The additional contribution from the arc with z = Reiθ , dz = iReiθ dθ vanishes,
Z π
iReiθ dθ
2
→0
→
iθ
2
iθ
2
3R3
R→∞ 0 [1 + (Re − a) ][1 + (Re + a) ]
lim
(∝ 1/R3 is enough).
Thus,
Z ∞
−∞
dx
[1 + (x − a)2 ][1 + (x + a)2 ]
=
π
.
2(1 + a2 )
2. Let
1
Q= √
2
0
1
0
1
0
1
0
1
0
!
.
(a) Compute Q2 . Compute Q3 . What is Q to any odd power, Q2k+1 .
By direct multiplication,
1
Q2 =
2
Q3 =
1
√
2 2
0
1
0
1
0
1
0
1
0
1
0
1
0
2
0
1
0
1
!
!
0
1
0
1
0
1
0
1
0
0
1
0
1
0
1
0
1
0
!
!
1
=
2
1
0
1
1
= √
2 2
0
2
0
0
2
0
1
0
1
2
0
2
!
.
0
2
0
!
= Q.
Since Q2 Q = Q, iteration then gives Q2k+1 = (Q2 )k Q for any k: all odd powers of Q are
just Q.
(b) What is sinh(ζQ)?
sinh(ζQ) =
∞
X
k=0
∞
X
1
1
ζ 2k+1 Q2k+1 = Q
ζ 2k+1 = Q sinh ζ.
(2k + 1)!
(2k + 1)!
k=0
3. Show that two successive rotations about a given axis is equivalent to a single rotation
about the same axis. Find the net rotation angle.
One way is to write out the net transformation. For rotations about a single axis we need
consider only 2 × 2 matrices. Rotations by θ and φ are
cos θ − sin θ
cos φ − sin φ
Rθ =
,
Rφ =
,
sin θ
cos θ
sin φ
cos φ
and the net transformation is
cos θ cos φ − sin θ sin φ − sin θ cos φ − cos θ sin φ
Rθ Rφ = Rφ Rθ =
sin θ cos φ + cos θ sin φ
cos θ cos φ − sin θ sin φ
This is
R=
cos(θ + φ)
sin(θ + φ)
− sin(θ + φ)
cos(θ + φ)
.
or a single rotation by angle θ + φ. What else would you expect?
Or, write abstractly
Rθ = exp(θJ),
Rφ = exp(φJ),
R = Rθ Rφ = Rφ Rθ = exp[(θ + φ)J].
This requires exp(A) exp(B) = exp(A + B), which follows from the power series only as long
as [A, B] = 0, AB = BA.
4. Let the three matrices
J1 =
0
0
0
0
0
0 −1
1
0
!
J2 =
0
0
−1
0
0
0
1
0
0
!
J3 =
0 −1
1
0
0
0
0
0
0
!
form a vector J = (J1 , J2 , J3 ).
(a) The commutator of two matrices is defined to be [A, B] = AB − BA. Compute the
commutators [Ji , Jj ]. There are nine ij combinations, but only three need to be multiplied
out in detail. See if you can encapsulate your results using ijk .
Direct multiplication (there are only a few nonzero terms) gives
[J1 , J2 ] = J3 ,
[J2 , J3 ] = J1 ,
[J3 , J1 ] = J2 .
Reversing the order reverses the sign, and the commutator of anything with itself vanishes;
this is summarized by
[Ji , Jj ] = ijk Jk .
(b) The action of a rotation by θ on a vector v is v 0 = exp(θ · J ) v. Show that for small ∆θ,
the change in v is ∆v = ∆θ × v.
This does not use the commutators. For small θ,
 



θ2
0 −θ3
v1
θ2 v3 − θ3 v2
v 0 − v = (θ · J )v =  θ3
0 −θ1   v2  =  θ3 v1 − θ1 v3  = θ × v.
−θ2
θ1
0
v3
θ1 v2 − θ2 v1
This is “easy” if you notice that the jk components of the matrix Ji are (Ji )jk = −ijk , so
(θ · J )jk vk = θi (Ji )jk vk = −ijk θi vk = +jik θi vk = (θ × v)j .
(c) The action of a rotation can also be considered as acting on J by conjugation, such that
J 0 = exp(θ · J ) J exp(−θ · J ). Show that for small ∆θ, the change in J is ∆J = −∆θ × J .
This does use the commutators. For small θ, the linear term in the expansion of the exponential is enough, and
∆J = J 0 − J = exp(θ · J ) J exp(−θ · J ) − J ≈ −J (θ · J ) + (θ · J )J ,
so in components
(∆J )i = −Ji (θj Jj ) + (θj Jj )Ji = −θj (Ji Jj − Jj Ji )
= −θj (ijk Jk ) = −ijk θj Jk = −(θ × J )i .