Physics 5040
Spring 2009
Problem Set 1
1. Show
∇2
1
= −4πδ(r) .
|r|
Hint : First consider the case where |r| =
6 0. Then look at the integral of
∇2 (1/ |r|) over a small sphere about the origin. Use the divergence theorem.
2. Show the following very useful properties of the Dirac delta function:
1
δ(x) .
(a) δ(ax) = |a|
1
2
2
(b) δ(x − a ) = 2|a|
[δ(x − a) + δ(x + a)] .
(c) Let f (x) have zeros at the points xi for i = 1, . . . , n. Then
δ(f (x)) =
n
X
δ(x − xi )
i=1
|f ′ (xi )|
.
(d) Let δ ′ (x − y) be the derivative of the delta function with respect to its
first argument. (But be careful – not everyone is consistent with this
convention. You need to work out each case individually to be sure that
you’re not off by a (−) sign.) Then
δ ′ (x − y) = −δ(x − y)
d
.
dx
3. (Complex delta function) In this problem you will show that
∂
1
= π δ(z − z0 )
∂ z̄ z − z0
where z̄ is the complex conjugate of z. You will also need to remember the
Cauchy-Riemann equations: if f (x, y) = u(x, y) + iv(x, y) is analytic (i.e.,
differentiable throughout some neighborhood), then
∂v
∂u
=
∂x
∂y
and
∂u
∂v
=−
.
∂y
∂x
(a) From the coordinate representation z = x + iy and z̄ = x − iy, use the
chain rule to write ∂/∂z and ∂/∂ z̄ in terms of ∂/∂x and ∂/∂y. Use these
to show that
∂z
∂ z̄
=
= 0.
∂ z̄
∂z
1
(b) Note that a function f (x, y) can be written as f (z, z̄). Show, using the
results of part (a) together with the Cauchy-Riemann equations, that if
f is analytic, then
∂f
= 0.
∂ z̄
(c) On the other hand, suppose we have a function f (z) = u + iv such that
∂f /∂ z̄ = 0. Show that
∂u ∂v
∂u ∂v
= 0.
−
+i
+
∂x ∂y
∂y
∂x
Therefore u and v satisfy the Cauchy-Riemann equations so that f (z) is
analytic.
(d) Now recall Green’s theorem
Z I
∂v
∂u
dx dy
(1)
−
(udx + vdy) =
∂x ∂y
R
C
where C is a simple closed curve that bounds the simply connected region R. (You should remember this from calculus. Alternatively, it is a
good exercise to derive this by applying the divergence theorem in two
dimensions to the vector (v, −u). You will have to rotate the tangent to
C to get the outward directed unit normal.) Show this can be written in
the form
Z
I
∂g
dA(z)
(2)
g dz = 2i
R ∂ z̄
C
where dA(z) = dx dy.
Hint : From part (a) you know that
∂g
1 ∂g
∂g
.
=
+i
∂ z̄
2 ∂x
∂y
Now, in (1), let u = g, v = ig.
(e) The “complex delta function” has the property
Z
δ(z − z0 )F (z) dA(z) = F (z0 )
R
or
Z
R
δ(x − x0 )δ(y − y0 )F (x, y) dA(x, y) = F (x0 , y0 )
where the point z0 = x0 + iy0 is contained within the region R. In
equation (2), let g(z) = F (z)/(z − z0 ) where F (z) is analytic throughout
R. Show that
Z
I
∂
1
F (z)
F (z)
dA
dz = 2i
∂ z̄ z − z0
R
C z − z0
2
where C is a simple closed curve bounding R. From this, conclude that
Z
1
1
∂
F (z0 ) =
dA
F (z)
π R
∂ z̄ z − z0
and therefore that the action of ∂/∂ z̄ (1/(z − z0 )) is that of a complex
delta function:
∂
1
= π δ(z − z0 ) .
∂ z̄ z − z0
4. Consider the integral
I± = lim
ε→0
Z
b
−a
f (x)
dx
x ± iε
where f (x) is continuous and where a and b are positive real numbers and
ε > 0.
(a) Show that I± can be written
Z
I± = lim lim
ε→0 δ→0
(b) Explain why
Z
lim lim
ε→0 δ→0
Z b
xf (x)
xf (x)
dx +
dx
2
2
2
2
−a x + ε
+δ x + ε
Z b
Z +δ
εf (x)
xf (x)
dx
∓
i
lim
dx .
+ lim lim
2
2
ε→0 −a x2 + ε2
ε→0 δ→0 −δ x + ε
−δ
−a
−δ
xf (x)
dx +
x2 + ε2
Z
b
+δ
Z b
xf (x)
f (x)
dx
=
P
dx .
2
2
x +ε
−a x
(c) Now explain why
lim lim
ε→0 δ→0
(d) Show
lim
ε→0
Z
b
−a
Z
+δ
−δ
xf (x)
dx = 0 .
x2 + ε2
εf (x)
dx = f (0)
x2 + ε2
Z
∞
−∞
du
= πf (0)
u2 + 1
and hence conclude that
Z b
Z b
f (x)
f (x)
lim
dx = P
dx ∓ iπf (0)
ε→0 −a x ± iε
−a x
Z b
Z b
f (x)
=P
dx ∓ iπ
δ(x)f (x) dx .
−a x
−a
You will frequently see this written in the symbolic form
lim
ε→0
1
1
= P ∓ iπδ(x)
x ± iε
x
3
which obviously means to apply both sides of this to a function f (x) and
integrate over an interval that includes the origin.
Hint : Define a new integration variable u by x = εu and note that
Z b/ε
Z ∞ Z −a/ε Z ∞
=
−
−
.
−a/ε
−∞
−∞
b/ε
5. Using the Cauchy integral formula, compute the following integrals:
(a)
(b)
I
ez
dz
C z − iπ/2
where C is the boundary of a square with sides at x = ±2 and y = ±2.
I
dz
2+2
z
C
where C is a circle of radius 1 centered at i.
6. Let C be any closed contour containing the origin. Show
I
2πi [1 + (−1)n ]
cosh z
dz
=
.
n+1
n!
2
C z
7. Let P (z) be a polynomial, and let C enclose the point a. Using the CauchyGoursat theorem, prove
I
1
P (z)
dz = P (a) .
2πi C z − a
Hint : Consider integrating the polynomial
Q(z) =
Explain your reasoning.
P (z) − P (a)
.
z−a
8. Evaluate the following real integrals by contour integration:
Z 2π
Z 2π
dθ
dθ
.
(b)
,
a > b > 0.
(a)
3 − 2 cos θ + sin θ
a + b sin θ
0
0
Z ∞
Z 2π
dx
dθ
(d)
.
(c)
2
(5
−
3
sin
θ)
1
+
x4
0
0
Z ∞
Z ∞
x2
sin x
(e)
dx .
(f)
dx .
2 + a2 )3
2 + a2 )
(x
x(x
0
0
Z ∞
Z ∞
1
1
ln x
dx .
(h) P
(g)
dω ′ .
2
2
′
2
2
′
x +a
−∞ (ω − ω0 ) + a ω − ω
0
Z ∞ λ−1
x
dx ,
0 < λ < 2.
(i)
2+1
x
0
4
Hints: In part (c) you can get a quadratic in the denominator before you
square
it. Factor it and then square. For the integral in part (g), consider
R∞
and
close the contour. This gives an answer with real and imaginary
−∞
R∞
R0
R∞
R0
parts. Now write −∞ = −∞ + 0 and let x → −x in −∞ . Then you can
write this as the sum of real and imaginary integrals. And the answer for the
integral in part (i) can be put into a simple form in terms of sin(λπ/2).
9. Evaluate
I
C
ez
dz
(z 2 + π 2 )2
I
C
eλz
dz
z 2 (z 2 + 2z + 2)
where C is the circle |z| = 4.
10. Evaluate
1
2πi
where C is the circle |z| = 3.
11. Prove that
Z
∞
2
sin x dx =
0
Z
0
∞
1
cos x dx =
2
2
r
π
.
2
2
2
(Note this is sin x and not sin x.)
H
2
Hint : Consider the integral C eiz dz where C is the contour shown below:
B
R
π/4
O
R
A
You will also probably need to use the inequality sin ϕ ≥ 2ϕ/π for 0 ≤ ϕ ≤
π/2.
R∞
12. In class we showed that P −∞ (sin x)/x dx = π although the principal value
symbol P isn’t needed here since there is no pole at x = 0. Now I want you
to show that
Z ∞
sin x
π
dx =
x
2
0
H iz
by explicitly evaluating C e /z dz over all portions of the contour C shown
below:
5
−R
ε
−ε
R
13. This problem (easily) generalizes our treatment of summation of series discussed in class. Let f (z) be a meromorphic function (i.e., it has no essential
singularities, in other words, it has only poles of finite order), and let C be a
contour that encloses the zeros of sin πz that lie at z = n, n + 1, . . . , N where
n is some integer (either greater or less than 0). Furthermore, assume that
the poles of f (z) and sin πz are distinct.
(a) Prove
N
X
1
f (m) =
2πi
m=n
I
C
π(cot πz)f (z) dz −
X
Res [π(cot πz)f (z)] .
poles of f (z)
inside C
Hint : If you use what we already showed in class for the expansion of
sin πz, then this takes about two lines to prove.
(b) Use this result to show
∞
X
n=1
x2
1
2x
= coth x − .
2
2
+n π
x
The function coth x − 1/x is called the Langevin function and arises
in calculating the magnetization of a paramagnet.
Hint : Consider the function
f (z) =
x2
2x
+ z 2 π2
and let C be a rectangle that encloses the points z = −N, −N +1, . . . , N −
1, N . Letting the sides of the rectangle go to infinity, show that
I
π(cot πz)f (z) dz → 0 .
C
14. In this problem you will derive the basic equation for the nonrelativistic scattering of an incident plane wave.
(a) Find the Green’s function for the time-independent Schrödinger equation
(H0 + V )ψ = Eψ or
~2 2
−
∇ + V (r) ψ(r) = Eψ(r) .
(3)
2m
6
To do this, write (3) in the form
2mE
2m
2
∇ + 2
ψ(r) = 2 V (r)ψ(r)
~
~
and define the Green’s function by
(∇2 + k 2 )G(r, r′ ) = δ(r − r′ )
where k 2 = 2mE/~2 . (This is then the Green’s function for the Helmholtz
equation.) Write the Fourier transform as an integral over the dummy
variable p. In this case you want the solution to represent an outgoing
wave. Since the time dependence of the full Schrödinger equation goes
like e−iEt/~ , the overall wave must go like ei(k·r−Et/~) (why?). This tells
you how to pick the poles in the integral over p (in other words, how to
include the iε term). The answer you should get is
′
1 eik|r−r |
G(r, r ) = −
.
4π |r − r′ |
′
(4)
(b) Now write down the complete solution to (3).
(c) In a physical scattering experiment, the detectors are far from the scattering center so that |r| ≫ |r′ |. Show that in this limit we have
′
eikr −ik′ ·r′
eik|r−r |
=
e
|r − r′ |
r
where k′ = kr̂. Hence the solution for the scattered wave is given by
ψk (r) = eik·r + f (Ω)
eikr
r
(5)
where
Z
′ ′
m
f (Ω) = −
d3 r′ e−ik ·r V (r′ )ψk (r′ )
2
2π~
is called the scattering amplitude.
Remarks: In a more abstract formulation of quantum mechanics, the Schrödinger
equation is written as (H0 + V )|ψi = E|ψi or (E − H0 )|ψi = V |ψi. Then
denoting the inverse of the operator E − H0 by (E − H0 )−1 we have
|ψi = |ϕi + (E − H0 )−1 V |ψi
where |ϕi is a solution of the free particle equation H0 |ϕi = E|ϕi. The
problem now is how to define the inverse operator. Note that (E − H0 )−1
is nothing more than the Green’s function for the Schrödinger equation. To
solve for it, we add a small imaginary part to E in such a way that we get
outgoing (or incoming) waves. In other words, we write
|ψ (±) i = |ϕi +
1
V |ψ (±) i .
E − H0 ± iε
7
(6)
Equation (6) is known as the Lippmann-Schwinger equation. In coordinate space, this equation becomes
Z
1
(±)
hx|ψ i = hx|ϕi + d3 x′ hx|
|x′ ihx′ |V |ψ (±) i
E − H0 ± iε
which is just a more abstract form of the equation you wrote down in part
(b) for the particular case of an outgoing wave.
15. Consider the Klein-Gordon equation
( + m2 )ϕ = (∂µ ∂ µ + m2 )ϕ = 0
where ∂µ = ∂/∂xµ and x0 = t. Use the metric gµν = diag(1, −1, −1, −1) so
that = ∂02 − ∇2 = ∂ 2 /∂t2 − ∇2 .
(a) Define a Green’s function ∆F (x − y) for the Klein-Gordon equation by
( + m2 )∆F (x − y) = −δ(x − y) .
Show that ∆F (x − y) may be written as
∆F (x − y) = θ(x0 − y 0 )∆+ (x − y) − θ(y 0 − x0 )∆− (x − y)
√
where (with ωk = k2 + m2 and k(x − y) = kµ (x − y)µ )
Z
d3 k
+
∆ (x − y) = −i
e−ik(x−y)
(2π)3 2ωk
(7)
and −∆− (x) = ∆+ (−x). (The Green’s function ∆F is called the Feynman propagator for the Klein-Gordon equation, hence the notation.)
e
Hint : The method of adding −iε to k0 in the denominator of G(k)
as
we did in class for the wave equation Green’s function won’t work here
because we need nonzero results when the k0 contour is closed in both
the upper and lower half-planes. In the present case you will find that
e
the denominator of G(k)
is k02 − ωk2 . Now replace ωk by ωk − iε. This
gives a denominator that is commonly written in the form k 2 − m2 + iε
where you have to realize that ε is an infinitesimal quantity so that ωk ε is
essentially the same as ε etc. While we could, in principle, replace ωk by
ωk + iε, we would not arrive at the physical results required by quantum
field theory to describe the creation and annihilation of particles and
antiparticles.
(b) Show by direct calculation that ∆F as defined by equation (7) is indeed
a Green’s function for the Klein-Gordon equation.
Hints: (1) You will need to use the delta function as the derivative of
the step function. (2) You will need to use Problem (2d). (3) Evaluate
(∂ 2 /∂x0 2 )∆F (x − y) in terms of ∆± without using the explicit integral
representations of ∆± . (4) Now evaluate ( + m2 )∆F in terms of ∆± ,
again without the specific integrals. (5) Explain why ( + m2 )∆± = 0.
(6) Finally, you will have to evaluate (∂/∂x0 )(∆+ + ∆− ) at x0 = y 0 . For
this you need to use the explicit integral representations for ∆± .
8