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2
Theoretical Physics
Course codes: Phys2325Course Homepage: http://bohr.physics.hku.hk/~phys2325/
Lecturer: Z.D.Wang, Office: Rm528, Physics Building Tel: 2859 1961 E-mail: [email protected]
Student Consultation hours: 2:30-4:30pm Tuesday
Tutor: Miss Liu Jia, Rm525
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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. WeberMain 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)
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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 ( the finite life time).
iEE nn0
/1
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1.1 Complex Algebra
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 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
r – the modulus or magnitude of z - the argument or phase of z
1i
5
ierz
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r – the modulus or magnitude of z - the argument or phase of z
The relation between Cartesian
and polar representation:
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,
1/ 22 2
1tan /
r z x y
y x
212121
ierrzz
212121 // ierrzz innn erz
)()(
)()(
2221212121
212121
yxyxiyyxxzz
yyixxzz
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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 , we have
The relationship between z and f(z) is best pictured as a mapping operation, we address it in detail later.
)arg()arg()arg( 2121 zzzz
7
2121 zzzz
xyiyxzf 222
Using the polar form,
2)( zzf
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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 pointsin the uv plane.
nin
i
ie
ie
)sin(cos
sincos
We get a not so obvious formula
Since
ninin )sin(cossincos
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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 2. The value of lnz with
n=0 is called the principal value of lnz.
iyxz *
222* ryxzz
21*zzz Special features: single-valued function of a real variable ---- multi-valued function
irez
nire 2
irlnzln
nirz 2lnln
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Another possibility
even and 1|cos||,sin|possibly however,
x;real afor 1|cos||,sin|
zz
xx
Question:
yx
yx
yxiyxiyx
yxiyxiyxi
ee iziz
222
222
iziz
sinhcos|cosz |
sinhsin|sinz | (b)
sinhsincoshcos)cos(
sinhcoscoshsin)sin( (a) show to
2
esinz ;
2
ecosz
:identities theUsing
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1.2 Cauchy – Riemann conditions
11
Having established complex functions, we now proceed to
differentiate them. The derivative of f(z), like that of a real function, is
defined by
provided that the limit is independent of the particular approach to the
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
zfdz
df
z
zf
z
zfzzf
zz
00limlim
oxxxx
xfxfxfoo
limlim
yixz viuf
,
yix
viu
z
f
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Let us take limit by the two different approaches as in the figure. First,
with y = 0, we let x0,
Assuming the partial derivatives exist. For a second approach, we set
x = 0 and then let y 0. This leads to
If we have a derivative, the above two results must be identical. So,
x
vi
x
u
z
f
xz
00limlim
x
vi
x
u
y
v
y
ui
z
f
z
0lim
y
v
x
u
,
x
v
y
u
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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).
xf
zf
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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:
Property 2: established a relation between u and v
022 vu
Example:
xeyxvb
xyxyxua
yxivyxuzw
y sin),( )(
3),( )( if
),(),()( functions analytic theFind23
0zz
0zz 0z
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1.3 Cauchy’s integral Theorem
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
'00 zz 01 jjj zzz
15
0
01
lim
z
z
n
jjj
ndzzfzf
n
The right-hand side of the above equation is called the contour (path) integral of f(z)
. and
bewteen curve on thepoint a is where
, and points thechoosing
of details theoft independen
is and existslimit that theprovided
1
j
j
jj
j
zz
z
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As an alternative, the contour may be defined by
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
22
11
2
1
,,
yx
yxc
z
zc
idydxyxivyxudzzf
22
11
22
11
yx
yx
yx
yxcc
udyvdxivdyudx
c
ndzz
where C is a circle of radius r>0 around the origin z=0 in the direction of counterclockwise.
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In polar coordinates, we parameterize and , and have
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,
irez diredz i
2
0
1
1exp22
1dni
rdzz
i
n
c
n
1- n for 1
-1n for 0{
0 dzzf
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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
c c c
udyvdxivdyudxdzzf
c s
sdAldA
yAxAA yx zdxdyds
c c s
yx sdAldAdyAdxA
dxdy
y
A
x
Axy
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For the real part, If we let u = Ax, and v = -Ay, then
=0 [since C-R conditions ]
For the imaginary part, setting u = Ay and v = Ax, we have
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,
dxdyy
u
x
vvdyudx
c
y
u
x
v
0dxdyy
v
x
uudyvdx
0dzzf
1
2
2
1
12
z
z
z
z
dzzfzFzFdzzf
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•Multiply connected regionsThe original statement of our theorem demanded a simply
connectedregion. 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 canconstruct a C for which the theorem holds. If line segments DE and GA arbitrarily close together, then
1.6 Fig.
E
D
A
G
dzzfdzzf
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1.4 Cauchy’s Integral Formula
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Cauchy’s integral formula: If f(z) is analytic on and within a closed contour C then
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
00
2 zifzz
dzzf
C
C C
dzzz
zf
zz
dzzf
2
000
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Let , here r is small and will eventually be made to
approach zero
(r0)
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.
i0 rezz
drie
re
rezfdz
zz
dzzf i
C Ci
i
2 2
0
0
00 2
2
zifdzif
C
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Derivatives
Cauchy’s integral formula may be used to obtain an expression for
the derivation of f(z)
Moreover, for the n-th order of derivative
0
0 0
1
2
f z dzdf z
dz i z z
24
C
zf
zz
dzzf
i exterior z ,0
interiorz ,
2
1
0
00
0
2
000 2
11
2
1
zz
dzzf
izzdz
ddzzf
i
1
00 2
!n
n
zz
dzzf
i
nzf
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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
book). text the(see R throught analytic
is f(z) then R, within C closedevery for 0)( and
Rregion connectedsimply ain continuous is f(z)function a If
C dzzf
25
Morera’s theorem:
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. find origin, about the circle a
within andon analytic is a)( If 1.
Examples
0n
n
n
n
a
zzf
jn
jnnj
j zaajzf
1
!
jj ajf !0
12
1
!
0n
n
nz
dzzf
in
fa
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2.In the above case, on a circle of radius r about the origin,
then (Cauchy’s inequality)
Proof:
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,
Mzf
Mra nn
nn
rznn
r
M
r
rrM
z
dzzfa
11 2
2
2
1
rfMaxrM rz
22
00
2
22
1
R
RM
zz
dzzf
izf
R
R
M
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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.
R
.const)z(f,e.i,0zf
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1.5 Laurent Expansion
29
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,
0n
n0n zzazf
010 zzzz
C 00C zzzz
zdzf
i2
1
zz
zdzf
i2
1zf
C 000 zzzz1zz
zdzf
i2
1
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Here z is a point on C and z is any point interior to C. For |t| <1, we
note the identity
So we may write
which is our desired Taylor expansion, just as for real variable power
series, this expansion is unique for a given z0.
0
211
1
n
ntttt
C nn
n
zz
zdzfzz
izf
01
0
0
2
1
01
002
1
n Cn
n
zz
zdzfzz
i
0
00 !
n
nn
n
zfzz
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Schwarz reflection principle
From the binomial expansion of for integer n (as an
assignment), it is easy to see, for real x0
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.
Since f(z) is real when z is real, the n-th derivate must be real.
n0xzzg
*n
0**n
0* zgxzxzzg
** zfzf
0
00 !
n
nn
n
xfxzzf
*
0
00
**
!zf
n
xfxzzf
n
nn
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Laurent SeriesWe 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
We let r2 r and r1 R, so for C1, while for C2, .
We expand two denominators as we did before
(Laurent Series)
zz
zdzf
izf
CC
21
2
1
00 zzzz 00 zzzz
21000000 112
1
CCzzzzzz
zdzf
zzzzzz
zdzf
izf
zdzfzzzzizz
zdzfzz
i
n
n Cn
n Cn
n
001
001
00
21
1
2
1
2
1
n
nn zzazf 0
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where
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.
Cnn
zz
zdzf
ia
10
2
1
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0
222
1
mmnimn
i
ner
drie
ia
021 2
1
1
1
2
1
mn
mnn
z
zdz
izz
zd
zia
11zzzf
35
01,22
2
1
mmni
i
1- n for 0
-1nfor 1a n
1
3211
1
1
n
nzzzzzzz
The Laurent expansion becomes
Example: (1) Find Taylor expansion ln(1+z) at point z
(2) find Laurent series of the function
If we employ the polar form
1
1)1()1ln(n
nn
n
zz
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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,
0
20
1 1
1 ;)( ;
1
1)(
n
n
n
n
iz
iz
ifzf
zzf
36
z11zf
Analytic continuation
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Suppose we expand it about z = i, so that
converges for (Fig.1.10)
The above three equations are different representations of the same
function. Each representation has its own domain of convergence.
0
211
1
n
nzzzz
i1iz1i1
1
izi1
1zf
2
2
i1
iz
i1
iz1
i1
1
2i1iz
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 theywill coincide everywhere as long as they are well-defined.Cssfounder.com