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2
Theoretical Physics
Course codes: Phys2325
Course Homepage: http://bohr.physics.hku.hk/~phys2325/
Lecturer: Z.D.Wang,
Office: Rm528, Physics Building
Tel: 2859 1961
E-mail: zwang@hkucc.hku.hk
Student Consultation hours: 2:30-4:30pm Tuesday
Tutor: Miss Liu Jia, Rm525
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3
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 (80% weighting)
and course assessment (20% weighting)

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 Y)
E ®E 0
+ iG n n
• Evaluating definite integrals.
• Obtaining asymptotic solutions of differentials
equations.
• Integral transforms
• Many Physical quantities that were originally real become
complex
as 1/ simple G
theory is made more general. The energy
( ® the finite life time).
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We here go through the complex algebra briefly.
A complex number z = (x,y) = x + iy, Where.
We will see that the ordering of two real numbers (x,y) is significant,
i.e. in general x + iy ¹ 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 q + i sinq) or
r – the modulus or magnitude of z
q - the argument or phase of z
i = -1
5
z = r ×eiq

6
r – the modulus or magnitude of z
q - the argument or phase of z
The relation between Cartesian
and polar representation:
( )
( )
2 2 1/ 2
r = z = x +
y
q =
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,
z ± z = x ± x + i y ±
y
( ) ( )
1 2 1 2 1 2
z z = x x - y y + i x y +
x y
( ) ( )
1 2 1 2 1 2 1 2 2 2
( 1 2 ) 1 2 1 2
z z =r r eiq +q
( ) ( 1 2 ) 1 / 2 1 / 2 z z = r r ei q -q z n =r neinq

Using the polar form,
arg( ) arg( ) arg( ) 1 2 1 2 z z = z + z
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) = z2
, we have
The relationship between z and f(z) is best pictured as a
mapping operation, we address it in detail later.
7
z1z2 =z1 z2
f ( z) = (x2 - y2 )+ i2xy

8
Function: Mapping operation
x
y Z-plane
u
v
The function w(x,y)=u(x,y)+iv(x,y) maps points in the xy plane into points
in the uv plane.
e i
q
= +
i
e = +
i
q q
cos sin
in n
q q
(cos sin )
q
Since
We get a not so obvious formula
cos nq + i sin nq = (cosq + i sinq )n

Complex Conjugation: replacing i by –i, which is denoted by (*),
We then have
Hence
Note:
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 2p. The value of lnz
with
n=0 is called the principal value of lnz.
9
z* = x -iy
zz* = x2 + y2 = r 2
z = (zz* )1 2 Special features: single-valued function of a
real variable ---- multi-valued function
z = reiq
rei(q +2np )
ln z = ln r + iq
ln z = ln r + i(q + 2np )

10
Another possibility
> ®¥
£
x x
| sin |,| cos | 1 for a real x;
z z
however, possibly | sin |,| cos | 1 and even
Question:
Using the identities :
e - iz e -
iz
= + = -
i
iz iz
2
; sinz e
x iy x y i x y
2
+ = +
to show (a) sin( ) sin cosh cos sinh
x + iy = x y -
i x y
cos( ) cos cosh sin sinh
2 2 2
x y
= +
(b) | sinz | sin sinh
2 2 2
x y
| cosz | cos sinh
cosz e
= +

Having established complex functions, we now proceed to
differentiate them. The derivative of f(z), like that of a real function,
is
defined by
df
f z
f z z f z
+ d
- = = = ¢
lim lim
® ® d
provided that the limit is independent of the particular approach to
the
¢ = ¢ = ¢
f x f x f x
lim lim
® + ® -
point z. For real variable, we require that
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.
Consider
11
( ) ( ) ( ) f ( z)
dz
z
z
z z
d
d
d 0 d 0
( ) ( ) ( o )
x x x x
o o
dz =dx + idy
df =du +idv
,
u i v
= +
d d
x i y
f
d
z
d d
d
+

Let us take limit by the two different approaches as in the figure.
First,
with dy = 0, we let dx0,
i v
u
d
d
ö çè
f
d
lim lim
+ ¶
i v
= ¶
u
Assuming the partial derivatives exist. For a second approach, we set
dx = 0 and then let dy 0. This leads to
v
i u
f
+ ¶
=- ¶
d
lim
If we have a derivative, the above two results must be identical. So,
12
÷ø
= æ +
® z
® x
x
z d
x d
d
d 0 d 0
x
x
¶
¶
y
y
z
z ® d
¶
¶
d 0
v
=¶
y
u
¶
x
¶
¶
,
=-¶
v
x
¶
u
y
¶
¶

These are the famous Cauchy-Riemann conditions. These Cauchy-
Riemann conditions are necessary for the existence of a derivative,
that
is, if 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, exists. (see the
proof
in the text book).
13
f¢(x)
f¢(z)

14
Analytic functions
If f(z) is differentiable at and in some small region around ,
we say that f(z) is analytic at
Differentiable: Cauthy-Riemann conditions are satisfied
the partial derivatives of u and v are continuous
Analytic function:
Property 1:
Ñ2u = Ñ2v = 0
Property 2: established a relation between u and v
Example:
Find the analytic functions ( ) ( , ) ( , )
3 2
a u x y = x -
xy
if ( ) ( , ) 3
b v x y e x
w z u x y iv x y
( ) ( , ) = -
y sin
= +
0 z = z
0 z = z
0 z

We now turn to integration.
in close analogy to the integral of a real function
The contour is divided into n intervals .Let
wiith for j. Then
'0
z0 ® z
0 1 D = - ® j j j- z z z
15
¢
å ( ) ò ( )
=
®¥
D =
0
0 1
lim
z
z
n
j
j j
n
f z z f z dz
n®¥
provided that the limit exists and is
independent of the details of
z
z
choosing the points and ,
j
j
z
where is a point on the curve bewteen
j
z z
and .
j j-
1
The right-hand side of the above equation is called the contour (path)
integral of f(z)

As an alternative, the contour may be defined by
2 2
2 2
udx vdy i vdx udy
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
16
ò ( ) = ò[ ( ) + ( )][ + ]
1 1
2
1
, ,
x y
c x y
z
c z
f z dz u x y iv x y dx idy
= ò [ - ] + ò[ + ]
1 1
2 2
1 1
x y
x y
x y
c c x y
òc
z ndz
where C is a circle of radius r>0 around the origin z=0 in
the direction of counterclockwise.

In polar coordinates, we parameterize
and , and have
n
1 z dz r i n d
i
¹
0 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,
17
z = reiq
dz = ireiqdq
+ p
ò = ò [ ( + ) ]
q q
p p
2
0
1
exp 1
2 2
c
n
1 for n -1
=
ò f ( z)dz = 0
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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)
These two line integrals can be converted to surface integrals by
Stokes’s theorem
Using and
We have
18
ò f ( z ) dz = ò( udx - vdy ) + i ò( vdx + udy
)
c c c
ò × = òÑ´ ×
c s
A dl A d s
A Ax x Ay y
=  +  ds =dxdyz
ò( Axdx + Aydy ) = ò A × dl = òÑ ´ A ×
d s
c c s
A y ö
x
æ
¶
-
¶
A
ò ÷ ÷ø
ç çè
= dxdy
¶
¶
y
x

For the real part, If we let u = Ax, and v = -Ay, then
u
æ
udx vdy v
ö
+ ¶
¶
u
v
= -¶
¶
=0 [since C-R conditions ]
For the imaginary part, setting u = Ay and v = Ax, we have
ö
æ
v
vdx udy u
òf (z)dz = 0
+ = ¶ dxdy 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,
19
( ) dxdy
y
x
c ò
ò ÷ ÷ø
ç çè
¶
- = - ¶
y
x
¶
¶
( ) ò ò = ÷ ÷ø
ç çè
¶
y
x
1
ò ( ) = ( ) - ( ) = -ò ( )
2
2
1
2 1
z
z
z
z
f z dz F z F z f z dz

•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¢
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
20
1.6 Fig.
E
ò ( ) =-ò ( )
D
A
G
f z dz f z dz

ù
é
ò = ò + ò + ò + ò
¢
f z dz f ( z)dz
'2
é
= ò + ò
ò ( ) ò ( ) ¢ ¢
f z dz f z dz
'1
C
ù
ABD®C EFG®-C
21
( )
( )
ABD DE GA EFG
ABDEFGA
ú ú
û
ê ê
ë
f (z)dz 0
ABD EFG
= úû
êë
=
C1 C2

Cauchy’s integral formula: If f(z) is analytic on and within a closed
contour C then
f z dz
in which z0 is some point in the interior region bounded by C. Note
that
here z-z0 ¹0 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
22
( ) ( 0 )
0
2 if z
z z
C
= p
ò -
ò ( ) ò f ( z
) =
-
-
f z dz
-
C C
dz
z z
z z
2
0
0 0

Let , here r is small and will eventually be made to
approach zero
f z dz i
(r®0)
f z re
= ò q = p
if z d if z
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.
23
- = iq
0 z z re
( ) ( ) q q
q
q
rie d
re
dz
z z
C C
i
i
ò ò +
=
-
2 2
0
0
( 0 ) 2 ( 0 )
2
C

f z dz
0 0
i 0, z exterior
1
Derivatives
Cauchy’s integral formula may be used to obtain an expression for
the derivation of f(z)
( ) ( )
d f z dz f z
¢ = 0
ç Ñ
ò- ¸ è ø æ ö
1
2
dz p i z z
0 0
ö
æ
-
1 1
1
f z dz d
pi p
Moreover, for the n-th order of derivative
24
ò ( ) ( ) =
-
î í ì
C
f z
z z
, z interior
2
0
p 0
ò ÷ ( ) ( )
÷ø
= ò ( -
) ç çè
f z dz
= 2
0 0 2 0
2
z z
dz z z i
( ) ( ) ( )
f z dz
ò( - ) +
= 1
0
0 2
!
n
n
z z
i
f z n
p

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
ò f z dz
= C
and ( ) 0 for every closed C within R, then f(z) is
analytic throught R (see the text book).
25


26
Examples
1. If ( ) a is analytic on and within
0
n
a circle about the origin, find .
n
n
n
a
å³
f z =
z ³ - å
( ) ( ) { } n j
f j z j a a z -
j n
n j
= +
1
!
( ) ( ) j
f j 0 = j!a
( ) ( = 0
) 1
ò ( ) = !
2 + 1
f z dz
n
n
n n i
z
a f
p

2.In the above case, on a circle of radius r about the origin,
then (Cauchy’s inequality)
Proof:
a = f z dz £ M r p
r
£ +
ò +1 2 1
1
n n r
where
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,
27
f ( z) £ M
a r n M
n £
( ) ( ) n n
z r
M
r
z
=
2
2
p
p
M(r) =Max z =r f (r)
( ) ( )
M R
f z dz
£
( )2 2
0
0
2
1
2 2
R
z z
i
f z
R
p
p p
-
¢ = ò
=M
R

Since R is arbitrary, let , we have
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.
28
R ® ¥
f ¢(z) =0, i.e, f (z) =const.

Taylor Expansion
Suppose we are trying to expand f(z) about z=z0, i.e.,
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
From the Cauchy integral formula,
29
( ) ( ) å¥
= -
n -
0
n
n 0 f z a z z
z¢ - z0 < z1 - z0
( ) ( ) ( )
¢ ¢
¢ ¢
f z dz
1
f z dz
ò =
¢ -
p
ò ( ¢ - ) - ( - )
f z 1
p
=
z z
2 i
z z z z
C C 0 0 2 i
( )
¢ ¢
f z dz
ò ( ¢ - )[ - ( - ) ( ¢ - )]
1
p
=
C 0 0 0 z z 1 z z z z
2 i

Here z¢ is a point on C and z is any point interior to C. For |t| <1, we
note the identity
t t t n
1 2
So we may write
- ¢ ¢
z z f z dz
1
p
¢ ¢
1
z z f z dz
n
n
f z
which is our desired Taylor expansion, just as for real variable power
series, this expansion is unique for a given z0.
30
å¥
=
= + + + =
- 0
1
1
n
t

( ) ( ) ( )
( ) òå¥
=
¢- +
=
C n
n
n
z z
i
f z
0
1
0
0
2
å ¥
( ) ò( ( )
)
=
¢- +
= -
0
1
0
2 0
n C
n
n
z z
pi
( )
( ) ( ) å¥
=
= -
0
0
0 n
! n
z z

Schwarz reflection principle
From the binomial expansion of for integer n (as an
assignment), it is easy to see, for real x0
g* z = z - x = z - x = 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
We expand f(z) about some point (nonsingular) point x0 on the real
axis
because f(z) is analytic at z=x0.
f ( n
) n
( x
å¥ ) =
Since f(z) is real when z is real, the n-th derivate must be real.
31
( ) ( )n
g z = z - x0
( ) [( ) n ] * ( *
)n ( * )
0
0
f * ( z) = f (z* )
( ) = ( -
)
0
0
0 n
! n
f z z x
( ) å¥
=( - ) n ( n ) ( ) = ( *
) 0
0
0
* *
f x
!
f z
n
f z z x
n
=

Laurent Series
We frequently encounter functions that are analytic in
annular region
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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
¢ ¢
f z dz
ù
é
1
p
= ò - ò
We let r2 ®r and r1 ®R, so for C1, while for C2, .
We expand two denominators as we did before
f z dz
= ò ò
f z dz
2 0 1 0 0 0 1 0 0
¥
1
å¥
1
1
p p
( ) ( ) n ¢ - ¢ ¢
n
1
2
(Laurent Series)
33
( ) ( )
z z
i
f z
C C
¢-
ú ú ú
û
ê ê ê
ë
1 2
2
0 0 z¢ - z > z - z
0 0 z¢ - z < z - z
ì
( ) ( )
¢ ¢
( )[ ( ) ( )]
( )
ü
ïý
¢ ¢
( ) ( ) ( ) [ ]ïþ
ïî
ïí
- - ¢ - -
+
¢ - - - ¢ -
1 2
1
C C
z z z z z z
z z z z z z
i
f z
p
( ) ( )
z z f z dz
( ) ( )
( z z ) f ( z )dz
z z i z z
i
n
n C
n C
n
-
+
¢ -
¢ ¢
= å - ò å ò
=
+
¥
=
+
0
1 0
0 0
0
0
1 2
2
= -
n
=-¥
n
f z an z z0

where
1
p
a 1
n n z z
Here C may be any contour with the annular region r < |
z-z0| < 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.
34
ò( ( ¢ )
¢
¢- ) +
=
C
f z dz
i
0 2

Example:
(1) Find Taylor expansion ln(1+z) at point z
(2) find Laurent series of the function
z z
+ = - +
ln(1 ) ( 1) 1
- ¢ ¢ =
¢
z dz
dz
( ) ( ) ò òå¥
+ ¢ +
1 2 2
¥
i
rie d
åò ( )
1
If we employ the polar form
a = -
+ - + - q
n r e
2 2 2
=
0
1
m
n m i n m
i
q q
p
=
¢ ¢ -
¢
=
0
1
1
1
2
m
n
m
n i z
n z z i
z
a
p p
f (z) [z(z 1)] 1 = - -
35
å¥
1
i n m
=- × + -
2 2 ,1
=
0
2
m
i
p d
p
- ³
î í ì
1 for n -1
<
=
0 for n -1
a n
1 1 2 3
( ) å¥
z z z zn
= -
= - - - - - - = -
- 1
1
1
n
z z z

The Laurent expansion becomes
å¥
=
1
n
n
n
n

Analytic continuation
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
å å¥
=
¥
1 1
=
z i
- -
ö çè
÷ø
æ
+
+
= - =
( ) 1
+
=
0
2
0
1
n
n
n
n
z i
i
f z f
z
f z
36
f (z) =1 (1+ z)

z z z n
1 2
Suppose we expand it about z = i, so that
1
f z 1
é
+ -
- -
z i
1 z i
1
=  2
converges for (Fig.1.10)
The above three equations are different representations of the same
function. Each representation has its own domain of convergence.
37
( ) å¥
=
= - + + = -
+ 0
1
1
n
z

( ) =
1 + i + ( z -
i
) =
( 1 + i )[ 1 + ( z - i ) ( 1 +
i
)] ( )
( ) úû
ù
êë
+
+
+
+
2
1 i
1 i
1 i
z - i < 1+ i = 2
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
will coincide everywhere as loAngd ams thiseys aioren w.eell-ddehfioneled..com

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