Physics 5040
Spring 2009
Problem Set 1 Solutions
1. The laplacian in spherical coordinates is
1 ∂
2
2 ∂
∇ = 2
r
+ angle terms .
r ∂r
∂r
Then if r 6= 0 we have
∇2
1
1 ∂
= 2
r
r ∂r
∂(1/r)
1 ∂
r2
= 2 (−1) = 0 .
∂r
r ∂r
If S is a small sphere about the origin, then (where ∂S is the boundary of S)
Z
Z
Z
1
1
1
∇2 d3 r =
∇ · ∇ d3 r =
∇ · r̂ da .
r
r
r
S
S
∂S
∂
But the gradient in spherical coordinates is ∇ = ∂r
r̂ + angle terms, so we
have
Z
Z
Z
Z
1
∂(1/r)
1
− 2 da = − dΩ = −4π .
da =
∇2 d3 r =
r
∂r
r
∂S
∂S
S
So we have shown that for r 6= 0 we have ∇2 1r = 0, while its integral about
the origin is −4π. This is just what we mean by −4πδ(r).
R
Here is another way to look at this. Consider the integral
∇2 1r f (r) d3 r.
Using the elementary vector identity ∇ · (f v) = ∇f · v + f ∇ · v we may write
Z
Z 1
1
f d3 r = f ∇ · ∇ d3 r
∇2
r
r
Z
Z
1
1
3
= ∇ · f∇
d r − ∇f · ∇ d3 r
r
r
Z
Z
r̂
1
= f ∇ · r̂ da − ∇f · − 2 d3 r
r
r
Z
Z
∂f 1 2
da
r dr d cos θ dϕ
= −f 2 +
r
∂r r2
But if these integrals are over an infinitesimal sphere,
R ε then the first integral is
just −4πf (0) and the second one vanishes because 0 ∂f
∂r dr = f (ε)− f (0) → 0
21
as ε → 0. This shows that ∇ r really does behave like −4πδ(r).
1
2. (a) Using the change of variables formula we have
Z
Z
1
δ(ax)f (x) dx = δ(y)f (y/a) dy
|a|
1
=
f (0)
|a|
which implies that
δ(ax) =
1
δ(x) .
|a|
(b) First note that
Z
Z
2
2
δ(x − a )f (x) dx = δ[(x − a)(x + a)]f (x) dx .
As you integrate from x = −∞ to x = ∞, the delta function will pick
out the values of the integrand wherever its argument equals 0. In this
case we get contributions at x = −a and at x = a, so by part (a) we have
Z
1
1
1
δ(x2 − a2 )f (x) dx =
f (−a) +
f (a) =
[f (−a) + f (a)]
|−2a|
|2a|
2 |a|
which implies
δ(x2 − a2 ) =
1
[δ(x + a) + δ(x − a)] .
2 |a|
R
(c) First note that δ(f (x))g(x) dx will give contributions wherever f (x) =
0. Let xi be a zero of f , i.e., f (xi ) = 0. Expanding about xi we have
f (x) = f (xi ) + f ′ (xi )(x − xi ) +
1 ′′
f (xi )(x − xi )2 + · · ·
2!
= f ′ (xi )(x − xi ) + · · · .
If x − xi is very small, then (x − xi )2 is really negligible, so by part (a)
again we have, near x = xi ,
δ(f (x)) = δ(f ′ (xi )(x − xi )) =
1
δ(x − xi ) .
|f ′ (xi )|
And as in part (b) we then have in general
X
1
δ(f (x)) =
δ(x − xi ) .
′
|f (xi )|
i
(d) Simply integrate by parts:
Z
Z
dδ(x − y)
f (x) dx
δ ′ (x − y)f (x) dx =
dx
Z
d
=
(δ(x − y)f (x)) dx − δ(x − y)f ′ (x) dx
dx
= 0 − f ′ (y)
2
which implies that
δ ′ (x − y) = −δ(x − y)
d
.
dx
But be careful – you can also write
Z
Z
dδ(x − y)
d
d
f (y) dy =
f (x)
δ(x − y)f (y) dy =
dx
dx
dx
which implies that
δ ′ (x − y) = +δ(x − y)
d
.
dy
3. (a) From z = x + iy and z̄ = x − iy we have x = (z + z̄)/2 and y = (z − z̄)/2i
so that by the chain rule
∂x ∂
∂y ∂
1 ∂
∂
∂
=
+
=
−i
∂z
∂z ∂x ∂z ∂y
2 ∂x
∂y
∂
∂x ∂
∂y ∂
1 ∂
∂
=
+
=
+i
∂ z̄
∂ z̄ ∂x ∂ z̄ ∂y
2 ∂x
∂y
∂x
∂y
1 1
∂z
=
+i
= − =0
∂ z̄
∂ z̄
∂ z̄
2 2
∂ z̄
∂x
∂y
1 1
=
−i
= − =0
∂z
∂z
∂z
2 2
(b) Let f = u + iv be analytic so that u and v satisfy the Cauchy-Riemann
equations
∂u
∂v
∂u
∂v
=
and
=−
.
∂x
∂y
∂y
∂x
Then
∂u
∂v
∂u ∂x ∂u ∂y
∂v ∂x
∂v ∂y
∂f
=
+i
=
+
+i
+i
∂ z̄
∂ z̄
∂ z̄
∂x ∂ z̄
∂y ∂ z̄
∂x ∂ z̄
∂y ∂ z̄
1 ∂u
∂u
∂v
∂v
=
+i
+i
−
2 ∂x
∂y
∂x ∂y
= 0 by the C-R equations .
(c) Now suppose f = u + iv and ∂f /∂ z̄ = 0. Since we don’t know that
f is analytic, we don’t know that u and v satisfy the Cauchy-Riemann
equations. But from part (b) we have
0=
∂f
∂u
∂u
∂v
∂v
=
+i
+i
−
∂ z̄
∂x
∂y
∂x ∂y
so that equating the real and imaginary parts of this equation shows that
in fact u and v do satisfy the Cauchy-Riemann equations so that f is
indeed analytic.
3
(d) As an interesting exercise itself, let me show that Green’s theorem is just
the divergence theorem in two dimensions. To see this, recall that the
divergence theorem says (in an obvious notation)
Z
Z
A · n̂ da .
∇ · A dV =
S
V
In two dimensions, the “volume” becomes an area, and the bounding
“surface” becomes a curve. We need the outward unit normal to the
curve.
n̂
T
To find n̂, let the curve be parametrized by t so it may be written as
x(t) = (x(t), y(t)). Then the tangent T to the curve is given by
T = ẋ(t) = (ẋ(t), ẏ(t)) .
The outward unit normal is constructed by rotating T clockwise by 90
degrees and then dividing by kTk. The standard rotation matrix is given
by
cos θ − sin θ
R(θ) =
sin θ
cos θ
where, by convention, the rotation angle θ is defined in a counterclockwise
direction. (If you don’t know this, we will show it when we cover linear
algebra.) In our case, this means we want θ = −π/2 so that
0 1
ẋ
ẏ
R(−π/2)T =
=
−1 0
ẏ
−ẋ
and therefore
n̂ =
R(−π/2)T
=
kR(−π/2)Tk
−ẋ
ẏ
p
,p
2
2
ẋ + ẏ
ẋ2 + ẏ 2
.
Writing A = (Ax , Ay ), the divergence theorem may be written
Z
I Ay ẋ
Ax ẏ
p
ds
−p
(∂x Ax + ∂y Ay ) da =
ẋ2 + ẏ 2
ẋ2 + ẏ 2
4
where the “volume” element in this case becomes the area element da =
dxdy, and where the “area” element becomes the line element
p
p
ds = dx2 + dy 2 = ẋ2 + ẏ 2 dt .
So now the divergence theorem becomes
I
Z ∂Ax
∂Ay
dxdy = (Ax dy − Ay dx) .
+
∂x
∂y
If we now let A = (Ax , Ay ) = (v, −u) we obtain
Z I
∂v
∂u
dxdy = (u dx + v dy)
−
∂x ∂y
which is just Green’s theorem.
To show the complex form of Green’s theorem, let u = g and v = ig.
Then on the one hand
I
I
I
u dx + v dy = g(dx + idy) = g dz
while on the other hand
Z Z Z ∂g
∂u
∂g
∂g
∂g
∂v
−
−
+i
dxdy =
i
dxdy = i
dxdy
∂x ∂y
∂x ∂y
∂x
∂y
Z
∂g
= 2i
dA(z) by part (a) .
∂ z̄
H
R
(e) From g dz = 2i (∂g/∂ z̄)dA(z) we let g(z) = F (z)/(z − z0 ) where F (z)
is analytic to obtain
Z
I
∂
F (z)
F (z)
dA(z)
dz = 2i
z − z0
∂ z̄ z − z0
Z
Z
∂F (z)/∂ z̄
1
∂
dA(z) + 2i
dA(z) .
= 2i F (z)
∂ z̄ z − z0
z − z0
But ∂F (z)/∂ z̄ = 0 by part (b), and since F (z) is analytic we may apply
the Cauchy integral formula to the left hand side to obtain
Z
1
∂
dA(z)
2πiF (z0 ) = 2i F (z)
∂ z̄ z − z0
or
1
F (z0 ) =
π
Z
∂
F (z)
∂ z̄
1
dA(z) .
z − z0
Note that for z 6= z0 we have (∂/∂ z̄)[1/(z − z0 )] = 0 because ∂z/∂ z̄ = 0.
But this is infinite for z = z0 just as we expect for a delta function.
5
4. (a)
I± = lim
ε→0
Z
b
−a
Z
f (x)
dx = lim
ε→0
x ± iε
Z
b
−a
f (x) x ∓ iε
dx
x ± iε x ∓ iε
b
xf (x) ∓ iεf (x)
dx
x2 + ε2
−a
Z b
Z b
xf (x)
εf (x)
= lim
dx
∓
i
dx
2
2
2
2
ε→0
−a x + ε
−a x + ε
Z −δ
Z b
xf (x)
xf (x)
= lim lim
dx +
dx
2
2
2
2
ε→0 δ→0
−a x + ε
δ x +ε
Z b
Z δ
εf (x)
xf (x)
dx
∓
i
lim
dx
+ lim lim
ε→0 −a x2 + ε2
ε→0 δ→0 −δ x2 + ε2
= lim
ε→0
(b) Assume that the order of taking limits can be interchanged. Then we
can first let ε → 0 in both integrals because neither integral includes the
origin. This leaves us with
Z −δ
Z b
xf (x)
xf (x)
dx +
dx
lim lim
2
2
2
2
ε→0 δ→0
δ x +ε
−a x + ε
Z −δ
Z b
f (x)
f (x)
= lim
dx +
dx
δ→0
x
x
δ
−a
Z b
f (x)
=P
dx .
−a x
(c) If we take the limit δ → 0 before letting ε → 0, then the integrand
xf (x)/(x2 + ε2 ) is continuous and non-singular. Then as δ → 0 we
Rδ
clearly have −δ → 0.
(d) Letting x = εu we have
Z b/ε
Z b
ε2 f (εu)
εf (x)
dx
=
lim
du
lim
ε→0 −a/ε ε2 (u2 + 1)
ε→0 −a x2 + ε2
= lim
ε→0
Now rearrange the expression
Z ∞
Z
=
−∞
to write
Z
+
−∞
b/ε
−a/ε
−a/ε
=
Z
∞
−∞
6
Z
b/ε
−a/ε
Z
b/ε
+
−a/ε
−
Z
f (εu)
du .
u2 + 1
−a/ε
−∞
−
Z
∞
b/ε
Z
∞
b/ε
.
Taking the limit ε → 0, the last two integrals vanish and we are left with
Z ∞
Z ∞
Z b/ε
f (εu)
f (εu)
f (εu)
du
=
lim
du
=
lim 2
du
lim
2
2
ε→0 −∞ u + 1
ε→0 −a/ε u + 1
−∞ ε→0 u + 1
Z ∞
du
= f (0)
.
2+1
u
−∞
To do this integral, let u = tan θ so that u2 +1 = sec2 θ and du = sec2 θ dθ.
Then
Z tan−1 ∞
Z ∞
π
π
du
=π
=
dθ
=
−
−
2
2
2
tan−1 (−∞)
−∞ u + 1
and hence
lim
ε→0
Z
b/ε
−a/ε
f (εu)
du = πf (0) .
u2 + 1
Putting all of this together, we finally obtain
Z b
Z b
f (x)
f (x)
dx = P
dx ∓ iπf (0) .
I± = lim
ε→0 −a x ± iε
−a x
5. (a) Let C be the contour
2
x iπ/2
2
−2
−2
Then
I=
I
C
ez
dz = 2πiez = 2πieiπ/2 = 2πi(i) = −2π .
z − iπ/2
z=iπ/2
(b) Let C be the contour
1
×
7
√
i 2
i
Then
I
I=
C
dz
=
2
z +2
I
dz
√
√
C (z + i 2)(z − i 2)
π
2πi
1
√
= √ =√ .
= 2πi
√
z + i 2 z=i 2
2 2i
2
6. The easy way to do this is to simply use the derivative formula
I
f (z)
2πi (n)
dz =
f (z0 )
n+1
n!
C (z − z0 )
where the contour C encloses the point z0 and f (z) is analytic within and on
C. Using d cosh z/dz = sinh z and d sinh z/dz = cosh z we have

I
 2πi cosh 0 = 2πi for n even
cosh z
2πi dn
n!
n!
dz
=
cosh
z
=
n+1
n
2πi

z
n! dz
z=0
for n odd
n! sinh 0 = 0
We can combine both of these possibilities into the single equation
I
2πi 1 + (−1)n
cosh z
.
dz =
z n+1
n!
2
Now here is the hard way to do this. From
ez = 1 + z +
z3
z2
+
+ ···
2!
3!
we have
cosh z =
and e−z = 1 − z +
z2
z3
−
+ ···
2!
3!
z2
z4
ez + e−z
=1+
+
+ ···
2
2!
4!
and
cosh z
1
1
1
1
1
= n+1 + · n+1−2 + · n+1−4 + · · · .
z n+1
z
2! z
4! z
If n is even then we have
1
1 1
1
cosh z
= n+1 + · · · +
· +
z + higher powers of z .
z n+1
z
n! z
(n + 2)!
H
H
But dz/z k = 0 if k ≥ 2, and z k dz = 0 for all k ≥ 1. Therefore
I
2πi
cosh z
dz =
if n is even .
n+1
z
n!
If n is odd, we obtain
cosh z
1
1
1
1
1
= n+1 + · · · +
+
·
+
z2 + · · ·
z n+1
z
(n − 1)! z 2
(n + 1)! (n + 3)!
8
so that
I
cosh z
dz = 0
z n+1
if n is odd .
As above, we can combine both of these into the single equation
I
cosh z
2πi 1 + (−1)n
.
dz
=
z n+1
n!
2
7. If P (z) is a polynomial, then so is
P (z) − P (a)
z−a
Q(z) =
although this isn’t so obvious. In general, if you have a polynomial f (z)
and you divide by z − α, you obtain f (z) = (z − α)g(z) + r where g(z) is a
polynomial and either r = 0 or deg r < deg(z − α) = 1 so that r is a constant.
(This statement is a consequence of the division algorithm.) Clearly, f (α) = r
so that f (z) = (z − α)g(z) + f (α) or f (z) − f (α) = (z − α)g(z) and therefore
the polynomial g(z) is given by
f (z) − f (α)
.
z−α
g(z) =
(In particular, note that if α is a root of f (z), then f (α) = 0 and f (z) =
(z − α)g(z).)
In any case, this shows that Q(z) is in fact a polynomial and hence is differentiable, i.e., it is analytic and can be integrated. Alternatively, by definition
of derivative we have
lim Q(z) = lim
z→a
z→a
P (z) − P (a)
= P ′ (a)
z−a
so that Q(z) doesn’t have a pole at z = a and can be integrated. Either way,
by the Cauchy-Goursat theorem we have (where C encloses the point a)
I
I
I
I
P (z) − P (a)
P (z)
dz
0=
Q(z) dz =
dz =
dz − P (a)
z
−
a
z
−
a
z
−a
C
C
C
C
I
P (z)
dz − 2πiP (a)
=
C z−a
which implies that
1
P (a) =
2πi
9
I
C
P (z)
dz .
z−a
8. (a) Let z = eiθ so dz = izdθ or dθ = −iz −1 dz. Then cos θ = (z + z −1 )/2
and sin θ = (z − z −1 )/2i and we have
Z 2π
dθ
I=
3
−
2
cos
θ + sin θ
0
I
−iz −1 dz
=
−1 ) + (z − z −1 )/2i
|z|=1 3 − (z + z
I
2 dz
=
2 + 6iz − (1 + 2i)
(1
−
2i)z
|z|=1
Using the quadratic formula, the roots of the denominator are
√
−6i ± −36 + 20
2−i
= 2 − i,
:= α, β .
2(1 − 2i)
5
Only the root β = (2 − i)/5 lies inside the circle |z| = 1. But you have to
be careful factoring the denominator. Factoring it as (z − α)(z − β) only
applies to a monic polynomial (where the coefficient of z 2 is 1). Here
the factorization is actually (1 − 2i)(z − α)(z − β) because multiplying
a polynomial by a constant doesn’t change its roots. Alternatively, you
could have just factored out 1−2i before applying the quadratic formula,
I just thought the algebra would be worse. Either way, the pole is at
z = β and we have
I=
4πi
(2πi)2
=
= π.
(1 − 2i)(β − α)
(1 − 2i)(−4/5)(2 − i)
(b) Use the same substitution z = eiθ as in part (a). Then with a > b > 0
we have
I
Z 2π
−iz −1 dz
dθ
=
I=
−1 )/2i
a + b sin θ
|z|=1 a + b(z − z
0
I
I
2 dz
2/b dz
=
=
.
2 + 2iaz − b
2 + (2ai/b)z − 1
bz
z
|z|=1
|z|=1
√
The poles are at (−i/b)(a ± a2 − b2 ). Since a > b we see that
√
a + a2 + b 2
a
> >1
b
b
while
a−
√
√
√
a2 − b 2
a − a2 − b 2 a + a2 − b 2
√
=
b
b
a + a2 − b 2
=
b
a2 − (a2 − b2 )
√
√
=
< 1.
2
2
b(a + a − b )
a + a2 − b 2
10
√
√
If we let α = (−i/b)(a + a2 − b2 ) and β = (−i/b)(a − a2 − b2 ) then
only the pole at z = β contributes to the integral so that
I=
(2πi)(2/b)
2π
(2πi)(2/b)
√
=√
.
=
2
2
2
β−α
(2i/b) a − b
a − b2
(c) Again use z = eiθ . Then
Z 2π
I
dθ
−iz −1 dz
I=
=
−1 )]2
(5 − 3 sin θ)2
0
|z|=1 [5 − (3/2i)(z − z
I
I
4iz −1 dz
4iz −1 dz
=
=
−1
2
−1
)
(10iz − 3z 2 + 3)]2
|z|=1 [z
|z|=1 (10i − 3z + 3z
I
I
4i
z dz
4iz dz
=
=
2
2
2
9 |z|=1 (z − 10iz/3 − 1)2
|z|=1 (3z − 10iz − 3)
The denominator has zeros at 5i/3 ± 4i/3 = 3i, i/3 where only i/3 lies
inside the unit circle so that
I
4i
z
2πi 4i d
z dz
I=
=
·
·
2
2
2
9 |z|=1 (z − 3i) (z − i/3)
1! 9 dz (z − 3i) z=i/3
2
4 5 3i
5π
= (−2π) ·
=
.
9 4 8
32
(d) First note that
Then we can write
dz 1 + z 4 → 0 as |z| → ∞ .
I=
Z
∞
0
1
dx
=
4
1+x
2
I
C
dz
1 + z4
where C is the contour shown below and the integral over the semi-circle
goes to zero.
The poles are at z 4 = −1 = eiπ = ei(π+2πn) or z = ei(π/4+nπ/2) for
n = 0, 1, 2, 3. These are
eiπ/4 , ei3π/4√
, ei5π/4 , √
ei7π/4 . Only
√ z = i3π/4
√ the points
iπ/4
= −1/ 2 + i/ 2 lie in the
the poles at e
= 1/ 2 + i/ 2 and e
upper half-plane:
11
√
√
−1/ 2 + i/ 2
×
×
√
√
−1/ 2 − i/ 2
×
× 1/
√
√
1/ 2 + i/ 2
√
√
2 − i/ 2
So we have
I
1
dz
I=
2 C (z − z1 )(z − z3 )(z − z5 )(z − z7 )
1
1
2πi
+
=
2 (z1 − z3 )(z1 − z5 )(z1 − z7 ) (z3 − z1 )(z3 − z5 )(z3 − z7 )
1
1
+ √
= πi √
2 2(−1 + i) 2 2(1 + i)
π
= √ .
2 2
(e) As in part (d), we may write
Z ∞
Z
I
x2
1 ∞
1
x2
z2
I=
dx
=
dx
=
dz
(x2 + a2 )3
2 −∞ (x2 + a2 )3
2 C (z 2 + a2 )3
0
where C is the same contour as in part (d). The poles are at z = ±ia,
but only the pole at z = +ia is included inside C. Noting that we can
write (z 2 + a2 )3 = (z − ia)3 (z + ia)3 we have
π
z2
1 2πi d2
=
.
· 2
I= ·
3
2 2! dz (z + ia) z=ia
16a3
(f) The integral
I=
Z
0
∞
1
sin x
dx =
x(x2 + a2 )
2
Z
∞
−∞
sin x
dx
x(x2 + a2 )
is well-defined (there is no pole on the real axis) but not easy to do. We
note that this is the imaginary part of
Z
1 ∞
eix
dx
2 −∞ x(x2 + a2 )
which, however, is not well-defined (it has a pole on the real axis) and
hence we must find its principal value. From the general formula derived
in class we have
Z ∞
X
X
eix
eiz
eiz
P
dx = 2πi
+ πi
.
Res
Res
2
2
2
2
2
z(z + a )
z(z + a2 )
−∞ x(x + a )
y>0
y=0
12
Since eiz = eix e−y , we close the contour C in the upper half-plane (which
is why the first sum is over residues for y > 0). From z(z 2 + a2 ) =
z(z − ia)(z + ia), the only pole inside C is at z = ia and we have
Z ∞
eix
eiz
eiz P
dx
=
2πi
+
πi
2
2
z(z + ia) z=ia
z 2 + a2 z=0
−∞ x(x + a )
= 2πi
Therefore
I = Im
e−a
1
πi
+ πi 2 = 2 (1 − e−a ) .
−2a2
a
a
iπ
π
(1 − e−a ) = 2 (1 − e−a ) .
2a2
2a
(g) We want to evaluate
Z
I=
∞
ln x
dx .
x2 + a2
∞
ln x
dx
x2 + a2
0
Consider the integral
Ie =
Z
−∞
which we evaluate by considering the contour integral
I
ln z
dz
C (z + ia)(z − ia)
where C is the contour
×
ia
Since the integrand goes to zero on the semicircle at infinity, we have
Z ∞
I
ln x
ln z
e
I=
dx =
dz
2 + a2
x
(z
+
ia)(z
− ia)
−∞
C
π
ln z = ln ia
= 2πi
z + ia z=ia
a
=
where we used ln i = iπ/2.
13
π
π
π
iπ 2
ln a + ln i = ln a +
a
a
a
2a
e
Now
R ∞ weR 0look at
R ∞the integral I from a different viewpoint. We write
=
+
and
let
x
→
−x in the first integral to obtain
−∞
−∞
0
Z
0
−∞
ln x
dx =
2
x + a2
Z
∞
0
ln(−x)
dx =
x2 + a2
Z
0
∞
ln x
dx +
2
x + a2
Z
0
∞
x2
iπ
dx
+ a2
where we used ln(−x) = ln x + ln(−1) = ln x + iπ. Combining this with
our previous evaluation of Ie we obtain
Z ∞
Z ∞
iπ 2
iπ
ln x
π
dx +
dx .
ln a +
=2
2 + a2
2 + a2
a
2a
x
x
0
0
Finally, equating the real parts of this equation yields
Z ∞
π
ln x
dx =
ln a .
I=
x2 + a2
2a
0
(h) We want to evaluate the principal value integral
Z ∞
dω ′
.
I =P
′
′
2
2
−∞ (ω − ω)[(ω − ω0 ) + a ]
This has a pole on the real axis at ω ′ = ω and poles in the complex ω ′
plane at ω ′ = ω0 ± ia. As we did in class, we consider the contour
ω0 + ia ×
×
ω
Using the same formula that we used in part (f) we have
1
1
+ iπ ′
I = 2πi ′
′
2
2
(ω − ω)[ω − (ω0 − ia)] ω′ =ω0 +ia
(ω − ω) + a ω′ =ω
=
2πi
iπ
+
(ω0 − ω + ia)(2ia) (ω − ω0 )2 + a2
=
(π/a)(ω0 − ω − ia) + iπ
(ω0 − ω)2 + a2
=
(π/a)(ω0 − ω)
.
(ω0 − ω)2 + a2
14
(i) To evaluate the integral
I=
Z
∞
0
xλ−1
dx,
x2 + 1
0<λ<2
we apply the formula derived in class:
Z ∞
X
−π
Res[z λ−1 P (z)]
xλ−1 P (x) dx = iπλ
e sin πλ
0
C
where C is the “keyhole” contour
C
The poles are at x = ±i so we have z 2 + 1 = (z + i)(z − i) and
λ−1
λ−1
−π
−π
i
(−i)λ−1
I = iπλ
=
i
− (−i)λ−1 .
+
iπλ
e sin πλ 2i
−2i
2ie sin πλ
For the term in brackets, write i = eiπ/2 and −i = ei3π/2 . (You can not
use −i = e−iπ/2 here because the phase of z around the contour C was
restricted to 0 ≤ θ ≤ 2π.) Then
I=
−π
2ieiπλ sin πλ
iπ(λ−1)/2
e
− ei3π(λ−1)/2
−πe−i3π/2 iπλ/2 iπ
e
e − ei3πλ/2
iπλ
2ie sin πλ
iπλ/2
+πi
e
+ ei3πλ/2
=
iπλ
2ie sin πλ
π −iπλ/2
e
+ eiπλ/2
=
2 sin πλ
=
=
π
πλ
π
πλ
cos
=
cos
πλ
sin πλ
2
2
cos
2 sin πλ
2
2
=
π
2 sin πλ
2
15
where we used the identity
sin A ± B = sin A cos B ± cos A sin B
which implies
sin 2A = 2 sin A cos A .
9. First we have
I
I
ez
ez
dz
=
dz
2
2
2
2
2
|z|=4 (z + π )
|z|=4 (z − iπ) (z + iπ)
ez
ez
d
2πi d
.
+
=
1! dz (z + iπ)2 z=iπ dz (z − iπ)2 z=−iπ
I=
Evaluating the derivatives yields
d z
−2 e (z + iπ) = ez (z + iπ)−2 − 2ez (z + iπ)−3 dz
z=iπ
z=iπ
eiπ
1
2eiπ
1
=
1
−
−
=
(2iπ)2
(2iπ)3
4π 2
iπ
and
d z
−2 e (z − iπ) = ez (z − iπ)−2 − 2ez (z − iπ)−3 dz
z=−iπ
z=−iπ
2e−iπ
1
e−iπ
1
.
−
=
=
1
+
(−2iπ)2
(−2iπ)3
4π 2
iπ
Therefore
I = 2πi ·
i
1
= .
2π 2
π
10. Our integral is
I=
=
1
2πi
I
1
2πi
I
|z|=3
|z|=3
z 2 (z 2
eλz
dz
+ 2z + 2)
eλz
dz
z 2 [z − (−1 + i)][z − (−1 − i)]
where the roots of the quadratic in the denominator are z = −1 ± i, and there
is the additional pole at z = 0. For the second order pole at z = 0 we have
d λz 2
−1 e (z + 2z + 2) dz
z=0
= λeλz (z 2 + 2z + 2)−1 − eλz (2z + 2)(z 2 + 2z + 2)−2 z=0
=
λ−1
λ 2
− =
.
2
4
2
16
The residue at z = −1 + i is
eλz
e−λ eλi
=
z 2 [z − (−1 − i)] z=−1+i
4
and the residue at z = −1 − i is
eλz
e−λ e−λi
.
=
z 2 [z − (−1 + i)] z=−1−i
4
Adding these up we have
1 λ − 1 e−λ iλ
−iλ
+
9e + e )
I = 2πi ·
2πi
2
4
λ − 1 1 −λ
+ e cos λ .
2
2
R∞ 2
H
2
11. To evaluate 0 sin x dx we consider C eiz dz where C is the contour
=
B
R
π/4
O
A
R
RA RB RO
H
2
By Cauchy’s theorem we know that C eiz dz = 0 so that 0 = O + A + B .
⌢
On OA we have z = x from x = 0 to x = R. On AB we have z = Reiθ from
θ = 0 to θ = π/4. And on BO we have z = reiπ/4 from r = R to r = 0.
First consider BO: z 2 = r2 eiπ/2 = ir2 and dz = eiπ/4 dr so that
Z 0
Z 0
Z R
2
2
2
eiz dz = eiπ/4
e−r dr = −eiπ/4
e−r dr
R
R
0
√
iπ/4 π
= as R → ∞ = −e
2
r r
i π
1 π
.
+
=−
2 2
2 2
⌢
For AB we have z 2 = R2 ei2θ = R2 cos 2θ + iR2 sin 2θ and hence
Z
π/4 2 Z π/4
iz
iR2 cos 2θ −R2 sin θ
iθ
e dz = e
e
iRe dθ
0
0
≤
Z
π/4
2
e−R
0
17
sin 2θ
R dθ .
But, as we saw in class, sin 2θ ≥ 4θ/π for 0 ≤ θ ≤ π/4 and we have
Z
π/4 2 Z π/4
2
2
π
(1 − e−R ) .
eiz dz ≤
e−4R θ/π R dθ =
0
4R
0
Then as R → ∞ this integral vanishes and we are left with
r r
Z ∞
1 π
i π
ix2
e dx −
0=
.
+
2 2
2 2
0
2
Using eix = cos x2 + i sin x2 and equating real and imaginary parts we obtain
r
Z ∞
Z ∞
1 π
.
cos x2 dx =
sin x2 dx =
2 2
0
0
H
12. We want to evaluate C eiz /z dz over the contour C given by
Γ
γ
−R
ε
−ε
R
Since there are no poles inside C we have
Z −ε ix
Z iz
Z R ix
Z iz
I iz
e
e
e
e
e
dz =
dx +
dz +
dx +
dz .
0=
z
x
z
x
γ
Γ z
−R
ε
C
Now let x → −x in the first integral and combine this with the third integral
to obtain
Z R ix
Z R ix
Z R
Z −ε ix
e
e − e−ix
sin x
e
dx +
dx =
dx = 2i
dx .
x
x
x
ε
ε
ε
−R x
R
Next, it is easy to see that Γ → 0 by Jordan’s lemma. Alternatively, just
note
iz iR cos θ −R sin θ e e
|dz| = e
iReiθ dθ
z iθ
Re
For
R
γ
= e−R sin θ dθ → 0
as R → ∞ .
, let z = εeiθ and do the integral:
Z
Z
Z iεeiθ
iθ
e
iθ
iεeiθ
iεe dθ = lim
e
i dθ =
lim eiεe i dθ
lim
ε→0 γ
ε→0
ε→0 γ εeiθ
γ
Z 0
=
i dθ = −iπ .
π
18
Putting all of this together and letting both ε → 0 and R → ∞ we obtain
Z ∞
sin x
2i
dx − iπ = 0
x
0
or
Z
∞
0
π
sin x
dx = .
x
2
13. The contour C in this case is the rectangle
×
×
×
n
×
×
...
×
×
−2 −1
×
×
1
2
...
×
N
×
where the shape really doesn’t matter as long as it encloses the poles of sin πz
along the real axis.
(a) By the Cauchy integral formula we have
I
X
π(cot πz)f (z) dz = 2πi
Res(all poles inside C) .
C
In class we showed that the expansion of sin πz about z = n is given by
π2
(−1)n
1
2
1+
=
(z − n) .
sin πz
π(z − n)
6
Then the residue of
π(cot πz)f (z) =
π(cos πz)
f (z)
sin πz
at z = m is
π(−1)m (−1)m
f (m) = f (m)
π
since cos mπ = (−1)m . Then we have
1
2πi
I
C
π(cot πz)f (z) dz =
N
X
f (m) +
n=m
19
X
poles of f (z)
inside C
Res[π(cot πz)f (z)]
or
N
X
f (m) =
n=m
1
2πi
I
π(cot πz)f (z) dz −
C
X
Res[π(cot πz)f (z)] .
poles of f (z)
inside C
(b) From part (a) with f (z) = 2x/(x2 + z 2 π 2 ) we have
N
X
m=−N
x2
2x
+ m2 π 2
1
=
2πi
I
C
π(cot πz)f (z) dz −
X
Res[π(cot πz)f (z)] .
poles of f (z)
inside C
As the sides of C go to infinity we see that
I
I
1
≤ π
π(cot
πz)f
(z)
dz
|cot πz| |f (z)| |dz|
2πi
2π
C
C
where |cot πz| ≤ 1 as we saw in class, and where |f (z)| ∼ 1/ z 2 → 0. This
shows that the integral vanishes as the rectangle becomes arbitrarily large.
Next, we see that the poles of f (z) are at z = ±ix/π so
X
Res[π(cotπz)f (z)]
poles of f (z)
inside C
π(cot πz)2x π(cot πz)2x +
= 2
π (z + ix/π) z=ix/π π 2 (z − ix/π) z=−ix/π
= −i cot(ix) + i cot(−ix) = −2i cot(ix)
= −2 coth x
where we used
cot ix =
cos ix
e−x + ex
= −i coth x .
= i −x
sin ix
e − ex
Lastly, we note that f (−m) = f (m) so that
∞
X
f (m) = 2
m=−∞
and we have
2
∞
X
f (m) + f (0)
m=1
∞
X
2x
2
+ = 2 coth x
2 + m2 π 2
x
x
m=1
20
or
∞
X
2x
1
= coth x − .
2 + m2 π 2
x
x
m=1
14. (a) The time-independent Schrödinger equation is
~2 2
∇ + V (r) ψ(r) = Eψ(r)
−
2m
or
2mE
2m
∇2 + 2 ψ(r) = 2 V (r)ψ(r) .
~
~
The Green’s function is defined by
(∇2 + k 2 )G(r, r′ ) = δ(r − r′ )
where k 2 = 2mE/~2 . Since
δ(r − r′ ) =
we write
1
(2π)3
1
G(r, r ) =
(2π)3
Z
(∇2 + k 2 )G(r, r′ ) =
1
(2π)3
Z
≡
1
(2π)3
Z
′
Then
which implies that
e
G(p)
=
and therefore
1
G(r, r ) =
(2π)3
′
Z
Z
′
d3 p eip·(r−r )
′
e
d3 p eip·(r−r ) G(p)
.
′
e
d3 p (−p2 + k 2 )eip·(r−r ) G(p)
′
d3 p eip·(r−r )
1
−p2 + k 2
′
−1
eip·(r−r )
=
d p
−p2 + k 2
(2π)3
3
Z
′
eip·(r−r )
.
d p 2
p − k2
3
Choose the pz -axis along r − r′ and do the angle integrals (where we let
R = |r − r′ |):
Z ∞
Z 1
Z 2π
−1
eipR cos θ
2
G(r, r′ ) =
p
dp
d
cos
θ
dϕ 2
3
(2π) 0
p − k2
−1
0
Z ∞
1 eipR − e−ipR
−1
2
p
dp
=
(2π)3 0
ipR p2 − k 2
Z ∞
p eipR
−1
.
dp 2
=
2
(2π) iR −∞
p − k2
21
To do this integral we invoke Jordan’s lemma. The contour must be closed
in the upper half-plane because R > 0 and i(pRe + ipIm )R = ipRe − pIm R so
eipR = eipRe e−pIm R . We have two choices for rewriting the denominator to
move the poles off the real axis:
p2 − (k ± iε)2 = p2 − (k 2 ± iε) = p2 − k 2 ∓ iε .
If we use p2 − (k + iε)2 , the poles are at p = ±(k + iε):
×
×
And if we use p2 − (k − iε)2 then the poles are at p = ±(k − iε):
×
×
In the first case, the residue at p = k + iε gives a term eikR , and in the second
case the residue at p = −(k − ive) gives a term e−ikR .
To choose between these two cases, observe that we want an outgoing wave,
which is of the form ei(k·r−Et/~) and not e−i(k·r+Et/~) . To see that ei(k·r−Et/~)
represents the outgoing wave, look at a point of constant phase k · r − Et/~.
As t increases, r must also increase to keep k · r − Et/~ constant. If we have
k · r + Et/~, then as t increases we must have r decrease to keep k · r + Et/~
constant, and hence this represents an incoming wave.
Therefore we choose to write p2 − (k + iε)2 (the first case above) so that
I
−1
peipR
′
G(r, r ) =
dp
(2π)2 iR
[p − (k + iε)][p + (k + iε)]
peipR −1 ikR
−1
·
2πi
·
e
=
=
2
(2π) iR
p + k + iε p=k+iε
4πR
′
=−
eik|r−r |
.
4π |r − r′ |
(b) The homogeneous equation (∇2 + k 2 )ϕ = 0 has the plane wave solutions
eik·r , and hence the general solution to the time-independent Schrödinger
22
equation is
ψk (r) = eik·r +
Z
d3 r′ G(r, r′ )
or
ψk (r) = e
ik·r
m
−
2π~2
Z
′
eik|r−r |
V (r′ )ψ(r′ ) .
d r
|r − r′ |
3 ′
(c) If |r| ≫ |r′ |, then
′
′2
2
2m
V (r′ )ψ(r′ )
~2
′ 1/2
|r − r | = (r + r − 2r · r )
2
r′
2r · r′
=r 1+ 2 −
r
r2
1/2
r · r′
≈r 1− 2
= r − r̂ · r′
r
′
2
where we used r′ /r2 ≈ 0 and (1 − ε)1/2 ≈ 1 − ε/2. So now eik|r−r | ≈
′
′ ′
eikr e−ikr̂·r := eikr e−ik ·r where we defined k′ = kr̂. In the denominator just
write |r − r′ | ≈ r since r ≫ r′ . Then
G(r, r′ ) = −
eikr −ik′ ·r′
e
4πr
m eikr
−
2π~2 r
Z
so that
ψk (r) = e
ik·r
:= eik·r + f (Ω)
′
′
d3 r′ e−ik ·r V (r′ )ψ(r′ )
eikr
r
where the scattering amplitude f (Ω) is defined by
Z
′
m
f (Ω) = −
d3 r′ e−ik·r V (r′ )ψ(r′ ) .
2π~2
15. (a) The defining equation for the Feynman propagator is
2
∂
2
2
∆F (x − y) = −δ 4 (x − y) .
−
∇
+
m
( + m2 )∆F (x − y) =
∂t2
As usual, write both ∆F (x − y) and δ 4 (x − y) in terms of their Fourier transforms:
Z
1
e F (k)
∆F (x − y) =
d4 k eik(x−y) ∆
(2π)4
Z
−1
−δ 4 (x − y) =
d4 k eik(x−y)
(2π)4
23
Acting on ∆F (x − y) with the Klein-Gordon operator x + m2 we have
Z
1
e F (k)
( + m2 )∆F (x − y) =
d4 k (−k 2 + m2 )eik(x−y) ∆
(2π)4
which implies that
e F (k) =
∆
Then
1
∆F (x − y) =
(2π)4
Z
1
=
(2π)4
Z
1
(2π)4
Z
=
+1
.
k 2 − m2
d4 k
eik(x−y)
k 2 − m2
0
0
eik0 (x −y ) e−ik·(x−y)
d k dk0
k02 − k2 − m2
3
0
d3 k dk0 e−ik·(x−y)
0
eik0 (x −y )
k02 − ωk2
where we have defined ωk2 = k2 + m2 .
To take into account both x0 > y 0 and x0 < y 0 , we add a small imaginary
part to ωk by writing ωk → ωk − iε. Then the poles occur at k0 = ±(ωk − iε).
x0 > y 0
×
×
y 0 > x0
We have ik0 (x0 − y 0 ) = ik0Re (x0 − y 0 ) − k0Im (x0 − y 0 ), so for x0 > y 0 we must
close the contour in the upper half-plane, and for y 0 > x0 we must close in the
lower half-plane (where there is also an extra (−) sign because the contour is
taken in the clockwise direction). This gives us
∆F (x − y)
1
=
(2π)4
Z
0
3
d k dk0 e
−ik·(x−y)
24
0
eik0 (x −y )
[k0 − (ω − iε)][k0 + (ω − iε)]
ik0 (x0 −y 0 ) d3 k
−ik·(x−y) e
= θ(x − y )
(2πi)e
(2π)4
k0 − (ω − iε) k0 =−(ω−iε)
Z
ik0 (x0 −y 0 ) d3 k
−ik·(x−y) e
0
0
(−2πi)e
+ θ(y − x )
(2π)4
k0 + (ω − iε) k0 =ω−iε
Z
0
0
d3 k
0
0
= θ(x − y )(−i)
e−iωk (x −y ) e−ik·(x−y)
(2π)3 2ωk
Z
0
0
d3 k
0
0
+ θ(y − x )(−i)
eiωk (x −y ) e−ik·(x−y)
(2π)3 2ωk
0
0
Z
Now let k → −k in the first integral to write (remember that ωk = k0 )
Z
d3 k
0
0
e−ik(x−y)
∆F (x − y) = θ(x − y )(−i)
(2π)3 2ωk
Z
d3 k
0
0
eik(x−y)
+ θ(y − x )(−i)
(2π)3 2ωk
= θ(x0 − y 0 )∆+ (x − y) − θ(y 0 − x0 )∆+ (−(x − y))
or
∆F (x − y) = θ(x0 − y 0 )∆+ (x − y) − θ(y 0 − x0 )∆− (x − y)
where
∆+ (x − y) := −i
Z
d3 k
e−ik(x−y)
(2π)3 2ωk
and
∆− (x − y) := −∆+ (−(x − y)) = i
Z
d3 k
eik(x−y) .
(2π)3 2ωk
(b) Recall that
∂
θ(x0 − y 0 ) = δ(x0 − y 0 )
∂x0
and therefore also (by the chain rule)
∂
∂
∂
θ(x0 − y 0 ) =
θ(x0 − y 0 ) 0 (x0 − y 0 ) = −δ(x0 − y 0 ) .
∂y 0
∂x0
∂y
Furthermore, recall from Problem (2d) that
∂
∂
δ(x0 − y 0 ) = −δ(x0 − y 0 ) 0 .
0
∂x
∂x
We first evaluate ∂x20 ∆F (x − y). We have
∂x0 ∆F (x − y) = ∂x0 [θ(x0 − y 0 )∆+ (x − y) − θ(y 0 − x0 )∆− (x − y)] .
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I will write ∆± instead of ∆± (x − y) for simplicity. Then
∂x0 ∆F (x − y) = ∂x0 θ(x0 − y 0 )∆+ + θ(x0 − y 0 )∂x0 ∆+
− ∂x0 θ(y 0 − x0 )∆− − θ(y 0 − y 0 )∂x0 ∆−
= δ(x0 − y 0 )∆+ + θ(x0 − y 0 )∂x0 ∆+
+ δ(y 0 − x0 )∆− − θ(y 0 − x0 )∂x0 ∆−
= δ(x0 − y 0 )(∆+ + ∆− ) + θ(x0 − y 0 )∂x0 ∆+ − θ(y 0 − x0 )∂x0 ∆−
where we used the fact that δ(y 0 − x0 ) = δ(x0 − y 0 ).
Next we have
∂x20 ∆F (x − y) = ∂x0 δ(x0 − y 0 )(∆+ + ∆− ) + δ(x0 − y 0 )∂x0 (∆+ + ∆− )
+ δ(x0 − y 0 )∂x0 ∆+ + θ(x0 − y 0 )∂x20 ∆+
+ δ(x0 − y 0 )∂x0 ∆− − θ(y 0 − x0 )∂x20 ∆−
= −δ(x0 − y 0 )∂x0 (∆+ + ∆− ) + 2δ(x0 − y 0 )∂x0 (∆+ + ∆− )
+ θ(x0 − y 0 )∂x20 ∆+ − θ(y 0 − x0 )∂x20 ∆−
= δ(x0 − y 0 )∂x0 (∆+ + ∆− ) + θ(x0 − y 0 )∂x20 ∆+
− θ(y 0 − x0 )∂x20 ∆− .
Now note that
∇2 ∆F (x − y) = ∇2 [θ(x0 − y 0 )∆+ − θ(y 0 − x0 )∆− ]
= θ(x0 − y 0 )∇2 ∆+ − θ(y 0 − x0 )∇2 ∆−
because ∇2 only acts on spatial coordinates and the theta function depends
only on time coordinates. Therefore
( + m2 )∆F (x − y) = (∂x20 − ∇2x + m2 )∆F (x − y)
= ∂x20 ∆F (x − y)
+ (−∇2x + m2 )[θ(x0 − y 0 )∆+ − θ(y 0 − x0 )∆− ]
= δ(x0 − y 0 )∂x0 (∆+ + ∆− )
+ θ(x0 − y 0 )∂x20 ∆+ − θ(y 0 − x0 )∂x20 ∆−
+ θ(x0 − y 0 )(−∇2x + m2 )∆+
− θ(y 0 − x0 )(−∇2x + m2 )∆−
= δ(x0 − y 0 )∂x0 (∆+ + ∆− ) + θ(x0 − y 0 )( + m2 )∆+
− θ(y 0 − x0 )( + m2 )∆−
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But ∆± are integrals over d3 k where the 4-vector k is constrained by k 2 =
k02 − k2 = m2 . Therefore
(x + m2 )e±ik(x−y) = (−k 2 + m2 )e±ik(x−y) = 0
so that
( + m2 )∆+ = ( + m2 )∆− = 0 .
Finally, we must evaluate δ(x0 − y 0 )∂x0 (∆+ + ∆− ). Because of the δ(x0 − y 0 )
we actually have (with k0 = ωk )
{∂x0 (∆+ + ∆− )}x0 =y0
Z
(−i)∂x0
d3 k −ik(x−y)
ik(x−y)
e
−
e
0 0
(2π)3 2ωk
x =y
Z
−ik(x−y)
d3 k
ik(x−y)
(−iω
)
e
+
e
= −i
k
0 0
(2π)3 2ωk
x =y
Z
3
1
d k ik·(x−y)
=−
e
+ e−ik·(x−y)
3
2
(2π)
Z
d3 k ik·(x−y)
e
(let k → −k in second integral)
=−
(2π)3
=
= −δ 3 (x − y) .
So we are left with
( + m2 )∆F (x − y) = −δ(x0 − y 0 )δ 3 (x − y) = −δ 4 (x − y) .
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