Theoretical Physics
Course codes: Phys2325/Phys3150
Course Homepage: http://bohr.physics.hku.hk/~phys2325/
Lecturer: Z.D.Wang,
Office: Rm528, Physics Building
Tel:
2859 1961
E-mail: zwang@hku.hk
Student Consultation hours: 3:00-5:00pm Monday
Tutor: Mr. Xianjia HUANG, Rm525
1
Text Book: Lecture Notes Selected from
Mathematical Methods for Physicists
International Edition (4th or 5th or 6th Edition)
by
George B. Arfken and Hans J. Weber
Main Contents:
Application of complex variables,
e.g. Cauchy's integral formula, calculus of residues.
Partial differential equations.
Properties of special functions
(e.g. Gamma functions, Bessel functions, etc.).
Fourier Series.
Assessment:
One 3-hour written examination (70% weighting)
and course assessment (30% weighting)
2
1 Functions of A Complex Variables I
Functions of a complex variable provide us some powerful and
widely useful tools in in theoretical physics.
• Some important physical quantities are complex variables (the
wave-function )
• Evaluating definite integrals.
• Obtaining asymptotic solutions of differentials
equations.
• Integral transforms
• Many Physical quantities that were originally real become complex
as simple theory is made more general. The energy E  E 0  i
n
n
( 1/   the finite life time).
3
1.1 Complex Algebra
We here go through the complex algebra briefly.
A complex number z = (x,y) = x + iy, Where. i   1
We will see that the ordering of two real numbers (x,y) is significant,
i.e. in general x + iy  y + ix
X: the real part, labeled by Re(z); y: the imaginary part, labeled by Im(z)
Three frequently used representations:
(1) Cartesian representation: x+iy
(2) polar representation, we may write
z=r(cos  + i sin) or z  r  e i
r – the modulus or magnitude of z
 - the argument or phase of z
4
r – the modulus or magnitude of z
 - the argument or phase of z
The relation between Cartesian
and polar representation:
r  z 
x
2
 y
2

1/ 2
  tan 1  y / x 
The choice of polar representation or Cartesian representation is a
matter of convenience. Addition and subtraction of complex variables
are easier in the Cartesian representation. Multiplication, division,
powers, roots are easier to handle in polar form,
z1  z2  ( x1  x2 )  i( y1  y2 )
z1 z2  ( x1 x2  y1 y2 )  i ( x1 y2  x2 y2 )
z1 z 2  r1r2 e i 1  2 
z / z  r / r e i 1  2 
1
2
1
2
z n  r n e in
5
Using the polar form,
z1 z 2  z1 z 2
arg( z1z2 )  arg( z1 )  arg( z2 )
From z, complex functions f(z) may be constructed. They
can be written
f(z) = u(x,y) + iv(x,y)
in which v and u are real functions.
For example if f ( z )  z 2 , we have


f z   x 2  y 2  i2xy
The relationship between z and f(z) is best pictured as a
mapping operation, we address it in detail later.
6
Function: Mapping operation
y
v
Z-plane
x
u
The function w(x,y)=u(x,y)+iv(x,y) maps points in the xy plane into points
in the uv plane.
Since
ei  cos   i sin 
ein  (cos   i sin  ) n
We get a not so obvious formula
cos n  i sin n  (cos   i sin  ) n
7
Complex Conjugation: replacing i by –i, which is denoted by (*),
*
z  x  iy
We then have
zz *  x 2  y 2  r 2
Hence
 
z  zz
* 12
Special features: single-valued function of a
real variable ---- multi-valued function
Note:
z  re i
rei   2n 
ln z  ln r  i
ln z  ln r  i  2n 
ln z is a multi-valued function. To avoid ambiguity, we usually set n=0
and limit the phase to an interval of length of 2. The value of lnz with
n=0 is called the principal value of lnz.
8
Another possibility
| sin x |, | cos x | 1 for a real x;
however, possibly | sin z |, | cos z | 1 and even  
Question:
Using the identities :
e iz  e iz
e iz  e iz
cosz 
; sinz 
2
2i
to show (a) sin( x  iy )  sin x cosh y  i cos x sinh y
cos( x  iy )  cos x cosh y  i sin x sinh y
(b) | sinz |2  sin 2 x  sinh 2 y
| cosz |2  cos 2 x  sinh 2 y
9
1.2 Cauchy – Riemann conditions
Having established complex functions, we now proceed to
differentiate them. The derivative of f(z), like that of a real function, is
defined by
f  z  z   f  z 
f z  df
 lim

 f z 
z  0
 z 0  z
z
dz
provided that the limit is independent of the particular approach to the
point z. For real variable, we require that lim  f x   lim  f x   f xo 
x xo
x xo
Now, with z (or zo) some point in a plane, our requirement that the
limit be independent of the direction of approach is very restrictive.
lim
,
Consider
z  x  iy
 f   u  i v
f
u  iv

z x  iy
10
Let us take limit by the two different approaches as in the figure. First,
with y = 0, we let x0,
f
v 
 u
 lim   i 
z 0 z x 0 x
x 
lim
u
v

i
x
x
Assuming the partial derivatives, exist. For a second approach, we set
x = 0 and then let y 0. This leads to
f
u v
lim
 i

z  0  z
y y
If we have a derivative, the above two results must be identical. So,
u
v

x
y
u
v

y
x
11
These are the famous Cauchy-Riemann conditions. These CauchyRiemann conditions are necessary for the existence of a derivative, that
is, if f  x exists, the C-R conditions must hold.

Conversely, if the C-R conditions are satisfied and the partial
derivatives of u(x,y) and v(x,y) are continuous, f  z exists. (see the proof
in the text book).

12
Analytic functions
If f(z) is differentiable at z  z0 and in some small region around z0 ,
we say that f(z) is analytic at z  z0
Differentiable: Cauthy-Riemann conditions are satisfied
the partial derivatives of u and v are continuous
Analytic function:
Property 1:
 u  v0
2
2
Property 2: established a relation between u and v
Example:
Find the analytic functions w( z )  u ( x, y )  iv ( x, y )
if (a) u ( x, y )  x 3  3xy2
(b) v( x, y )  e  y sin x
13
1.3 Cauchy’s integral Theorem
We now turn to integration.
in close analogy to the integral of a real function
The contour z0  z0' is divided into n intervals .Let n  
wiith z j  z j  z j 1  0
for j. Then
n
z0
j 1
z0
f  j z j   f z dz

n 
lim
provided that the limit exists and is
independen t of the details of
choosing the points z j and  j ,
where  j is a point on the curve bewteen
z j and z j 1.
The right-hand side of the above equation is called the contour (path)
integral of f(z)
14
As an alternative, the contour may be defined by
z2
x2 y2
c z1
c x1 y1
x2 y2
 f z dz   ux, y   iv x, y dx  idy 
x2 y2
 udx  vdy i  vdx  udy

c
x1 y1
c x1 y1
with the path C specified. This reduces the complex integral to the
complex sum of real integrals. It’s somewhat analogous to the case of
the vector integral.
An important example

z n dz
c
where C is a circle of radius r>0 around the origin z=0 in
the direction of counterclockwise.
15
In polar coordinates, we parameterize
i
and dz  ire d , and have
1
2i

c
z  re i
n 1 2
r
z dz 
2
n
0
{
1
 expin  1 d
0
for n  -1
for n  - 1
which is independent of r.
Cauchy’s integral theorem
– If a function f(z) is analytical (therefore single-valued) [and its
partial derivatives are continuous] through some simply connected
region R, for every closed path C in R,
 f z dz  0
c
16
Stokes’s theorem proof
Proof: (under relatively restrictive condition: the partial derivative of u, v
are continuous, which are actually not required but usually
satisfied in physical problems)
 f z dz   udx  vdy  i  vdx  udy 
c
c
c
These two line integrals can be converted to surface integrals by
Stokes’s theorem
 A dl     A  d s
c
 s
Using A  Ax x  A y y and ds  dxdyz
We have
 Ax dx  Ay dy   A  d l     A  d s
c
c
s
 A y A x 
dxdy
  

y 
 x
17
For the real part, If we let u = Ax, and v = -Ay, then
 v u 
udx  vdy     dxdy
x y 

c


v
u

[since C-R conditions x
y
=0
]
For the imaginary part, setting u = Ay and v = Ax, we have
 u v 


vdx

udy


   dxdy  0
 x
y 
 f z dz  0
As for a proof without using the continuity condition, see the text book.
The consequence of the theorem is that for analytic functions the line
integral is a function only of its end points, independent of the path of
integration,
z2
z1
z1
z2
 f z dz  F z 2   F z1     f z dz
18
•Multiply connected regions
The original statement of our theorem demanded a simply connected
region. This restriction may easily be relaxed by the creation of a
barrier, a contour line. Consider the multiply connected region of
Fig.1.6 In which f(z) is not defined for the interior R
1.6 Fig.
Cauchy’s integral theorem is not valid for the contour C, but we can
construct a C for which the theorem holds. If line segments DE and
GA arbitrarily close together, then
A

G
f  z dz  
E

f  z dz
D
19


 f z dz
f z dz  





C
 ABD DE GA EFG 
 ABDEFGA 







      f z dz  0
ABD EFG 
 f z dz   f z dz
C1
ABD  C1'
C2
EFG  C2'
20
1.4 Cauchy’s Integral Formula
Cauchy’s integral formula: If f(z) is analytic on and within a closed
contour C then
f z dz
 2if z 0 
z  z0

C
in which z0 is some point in the interior region bounded by C. Note that
here z-z0 0 and the integral is well defined.
Although f(z) is assumed analytic, the integrand (f(z)/z-z0) is not
analytic at z=z0 unless f(z0)=0. If the contour is deformed as in Fig.1.8
Cauchy’s integral theorem applies.
So we have

C
f z dz

z  z0

C2
f z 
dz  0
z  z0
21
i
Let z  z 0  re , here r is small and will eventually be made to
approach zero

C2
f z dz
dz 
z  z0
(r0)

C2

f z 0  re i
re i
rie  d
i

 if z 0  d  2if z 0 
C2
Here is a remarkable result. The value of an analytic function is given at
an interior point at z=z0 once the values on the boundary C are
specified.
What happens if z0 is exterior to C?
In this case the entire integral is analytic on and within C, so the
integral vanishes.
22
1
2i

C
f z dz  f z 0 ,

z  z0
 0,
z 0 interior
z 0 exterior
Derivatives
Cauchy’s integral formula may be used to obtain an expression for
the derivation of f(z)
d  1
f   z0  

dz0  2 i
1

2i


f  z  dz 

z  z0 
d
f z dz
dz 0
 1 
1

 
 z  z 0  2i
f z dz
 z  z0 2
Moreover, for the n-th order of derivative
n!

n
f z 0  
2i
f z dz
 z  z 0 n1
23
We now see that, the requirement that f(z) be analytic not only
guarantees a first derivative but derivatives of all orders as well! The
derivatives of f(z) are automatically analytic. Here, it is worth to
indicate that the converse of Cauchy’s integral theorem holds as well
Morera’s theorem:
If a function f(z) is continuous in a simply connected region R
and  f ( z )dz  0 for every closed C within R, then f(z) is
C
analytic throught R (see the text book).

24
Examples
1. If f ( z )   a n z n is analytic on and within
n0
a circle about the origin, find an .
f  j  z   j! a j 

a n  z n j
n  j 1
f  j  0  j! a j
f n  0
1
an 

n!
2i

f z dz
z n 1
25
2.In the above case, f z   M on a circle of radius r about the origin,
then a n r n  M (Cauchy’s inequality)
Proof:
an 
where
1
2

f  z dz
z r
z n 1
M r   Max
 M r 
z r
2r
2r n 1

M
rn
f r 
3. Liouville’s theorem: If f(z) is analytic and bounded in the complex
plane, it is a constant.
Proof: For any z0, construct a circle of radius R around z0,
f  z 0  

1
2i
f  z dz
 z  z 0 2
R

M 2R
2 R 2
M
R
26
Since R is arbitrary, let R   , we have
f z  0, i.e, f (z)  const .
Conversely, the slightest deviation of an analytic function from a
constant value implies that there must be at least one singularity
somewhere in the infinite complex plane. Apart from the trivial constant
functions, then, singularities are a fact of life, and we must learn to live
with them, and to use them further.
27
1.5 Laurent Expansion
Taylor Expansion

n




f
z

a
z

z

n
0
Suppose we are trying to expand f(z) about z=z0, i.e.,
n 0
and we have z=z1 as the nearest point for which f(z) is not analytic. We
construct a circle C centered at z=z0 with radius z   z0  z1  z0
From the Cauchy integral formula,
f z  

1 f zdz
1
f zdz

2i C z  z
2i C z  z 0   z  z 0 
1
f zdz
2i C z  z 0 1  z  z 0  z  z 0 
28
Here z is a point on C and z is any point interior to C. For |t| <1, we
note the identity
1
 1 t  t2  
1 t


tn
n 0
So we may write
1
f z  
2i
1

2i
z  z 0 n f z dz 

  z 0 n 1

z
n

0
C


  z  z 0 n 
  z 0 n1

z
C
n 0


 z  z 0 
n 0
f z dz 
n
f n   z 0 
n!
which is our desired Taylor expansion, just as for real variable power
series, this expansion is unique for a given z0.
29
Schwarz reflection principle
n
From the binomial expansion of gz   z  x 0  for integer n (as an
assignment), it is easy to see, for real x0

g z   z  x 0 
*
  z
n *
*
 x0

n
 
 g z*
Schwarz reflection principle:
If a function f(z) is (1) analytic over some region including the real axis
and (2) real when z is real, then
f * z   f z *
We expand f(z) about some point (nonsingular) point x0 on the real axis
because f(z) is analytic at z=x0.

n   x 
n f
0
z  x0 
f z  
n!
n 0
 

Since f(z) is real when z is real, the n-th derivate must be real.

f
*
z    z
n 0
*
 x0

n
f n   x 0 
 f z*
n!
 
30
Laurent Series
We frequently encounter functions that are analytic in
annular region
31
Drawing an imaginary contour line to convert our region into a simply
connected region, we apply Cauchy’s integral formula for C2 and C1,
with radii r2 and r1, and obtain

1 
f z  


2i
C2
C1




 f  z dz 
 z  z


We let r2 r and r1 R, so for C1, z  z 0  z  z 0 while for C2,
We expand two denominators as we did before
z  z 0  z  z.0


1 
f z dz 
f z dz 

f z  




2i  z   z 0 1  z  z 0  z   z 0 
z  z 0 1  z   z 0  z  z 0 
C2
C1





1
z  z 0 n

2i n0
f z  


f  z dz 
 z   z0 n1
C1
a n  z  z 0 n


1
1

2i n0  z  z 0 n1
 z   z 0 
n
f  z dz 
C2
(Laurent Series)
n  
32
where
1
an 
2i
f  z dz 
 z   z 0 n1
C
Here C may be any contour with the annular region r < |zz0| < R encircling z0 once in a counterclockwise sense.
Laurent Series need not to come from evaluation of contour
integrals. Other techniques such as ordinary series
expansion may provide the coefficients.
Numerous examples of Laurent series appear in the next
chapter.
33

Example:
(1) Find Taylor expansion ln(1+z) at point z
(2) find Laurent series of the function
1
an 
2i
 z  n1
1
dz 
1

z z   1 2i
ln( 1  z )   (1)
n 1
n 1
zn
n
f z   zz  1
1

 m0
 z  m
dz 
z  n2
If we employ the polar form

1
rie i d
an  
2i m0 r n 2m e i n 2m 



1

 2i
 n  2m,1
2i
m 0
 1
an  
0
The Laurent expansion becomes
for n  -1
for n  - 1


1
1
  1 z  z2  z3    zn
z z  1
z
n1
34
Analytic continuation
f z   1 1  z 
For example
which has a simple pole at z = -1 and is analytic
elsewhere. For |z| < 1, the geometric series expansion f1,
while expanding it about z=i leads to f2,
1
f ( z) 
;
1 z

f1   ( z ) n ;
n 0
1   z i 
f2 



1  i n 0  z  i 
n
35
1
 1 z  z2  
1 z

n



z

n 0
Suppose we expand it about z = i, so that
f z  
1
1

1  i  z  i  1  i 1  z  i  1  i 
1

1 i
2



zi
z  i
1






1 i
1  i 2


converges for
z  i  1  i  2 (Fig.1.10)
The above three equations are different representations of the same
function. Each representation has its own domain of convergence.
A beautiful theory:
If two analytic functions coincide in any region, such as the overlap of s1 and s2,
of coincide on any line segment, they are the same function in the sense that they
36
will coincide everywhere as long as they are well-defined.