Quantum formula sheet

quantum formula shet

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QUANTUM MECHANICS FORMULA SHEET
Contents
1. Wave Particle Duality
1.1 De Broglie Wavelength
1.2 Heisenberg’s Uncertainty Principle
1.3 Group velocity and Phase velocity
1.4 Experimental evidence of wave particle duality
1.4.1 Wave nature of particle (Davisson-German experiment)
1.4.2 Particle nature of wave (Compton and Photoelectric Effect)

2. Mathematical Tools for Quantum Mechanics
2.1 Dimension and Basis of a Vector Space
2.2 Operators
2.3 Postulates of Quantum Mechanics
2.4 Commutator
2.5 Eigen value problem in Quantum Mechanics
2.6 Time evaluation of the expectation of A
2.7 Uncertainty relation related to operator
2.8 Change in basis in quantum mechanics
2.9 Expectation value and uncertainty principle

3. Schrödinger wave equation and Potential problems
3.1 Schrödinger wave equation
3.2 Property of bound state
3.3 Current density
3.4 The free particle in one dimension
3.5 The Step Potential
3.7 Potential Barrier
3.7.1 Tunnel Effect

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3.8 The Infinite Square Well Potential
3.7.1 Symmetric Potential
3.9 Finite Square Well Potential
3.10 One dimensional Harmonic Oscillator

4. Angular Momentum Problem
4.1 Angular Momentum
4.1.1 Eigen Values and Eigen Function
4.1.2 Ladder Operator
4.2 Spin Angular Momentum
4.2.1 Stern Gerlach experiment
4.2.2 Spin Algebra
4.2.3 Pauli Spin Matrices
4.3 Total Angular Momentum

5. Two Dimensional Problems in Quantum Mechanics
5.1 Free Particle
5.2 Square Well Potential
5.3 Harmonic oscillator

6. Three Dimensional Problems in Quantum Mechanics
6.1 Free Particle
6.2 Particle in Rectangular Box
6.2.1 Particle in Cubical Box
6.3 Harmonic Oscillator
6.3.1 An Anistropic Oscillator
6.3.2 The Isotropic Oscillator
6.4 Potential in Spherical Coordinate (Central Potential)
6.4.1 Hydrogen Atom Problem

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7. Perturbation Theory
7.1 Time Independent Perturbation Theory
7.1.1 Non-degenerate Theory
7.1.2 Degenerate Theory
7.2 Time Dependent Perturbation Theory

8. Variational Method

9. The Wentzel-Kramer-Brillouin (WKB) method
9.1 The WKB Method
9.1.1 Quantization of the Energy Level of Bound state
9.1.2 Transmission probability from WKB

10. Identical Particles
10.1 Exchange Operator
10.2 Particle with Integral Spins
10.3 Particle with Half-integral Spins

11. Scattering in Quantum Mechanics
11.1 Born Approximation
11.2 Partial Wave Analysis

12. Relativistic Quantum Mechanics
12.1 Klein Gordon equation
12.2 Dirac Equation

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1.Wave Particle Duality

1.1 De Broglie Wavelengths
The wavelength of the wave associated with a particle is given by the de Broglie relation

mv
h
p
h
 where h is Plank’s constant
For relativistic case, the mass becomes
2
2
0
1
c
v
m
m

 where m0 is rest mass and v is
velocity of body.

1.2 Heisenberg’s Uncertainty Principle
“It is impossible to determine two canonical variables simultaneously for microscopic
particle”. If q and
qp are two canonical variable then

2


qpq
where ∆q is the error in measurement of q and ∆pq is error in measurement of pq and h is
Planck’s constant ( / 2 )h .
Important uncertainty relations

2


xPx (x is position and
xp is momentum in x direction )

2

 tE (E is energy and t is time).

2

L (L is angular momentum, θ is angle measured)

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1.3 Group Velocity and Phase Velocity
According to de Broglie, matter waves are associated with every moving body. These
matter waves moves in a group of different waves having slightly different wavelengths.
The formation of group is due to superposition of individual wave.
Let If tx,
1
 and tx,
2
 are two waves of slightly different wavelength and frequency.
     tdxdkkAtkxA   sin,sin
21

21   tkx
tddk
A 







 sin
22
cos2
The velocity of individual wave is known as Phase
velocity which is given as
k
v
p

. The velocity of
amplitude is given by group velocity vg i.e.
dk
d
v
g


The relationship between group and phase velocity is given by



d
dv
vv
dk
dv
kv
dk
d
v
p
pg
p
pg  ;
Due to superposition of different wave of slightly different wavelength resultant wave
moves like a wave packet with velocity equal to group velocity.
1.4 Experimental evidence of wave particle duality
1.4.1 Wave nature of particle (Davisson-German experiment)
Electron strikes the crystals surface at an
angle. The detector, symmetrically located
from the source measure the number of
electrons scattered at angle θ where θ is the
angle between incident and scattered electron
beam.
The Maxima condition is given by
p
h
dnor
dn












where
2
cos2
sin2




S D
gv
phv
t

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1.4.2 Particle nature of wave (Compton and Photoelectric Effect)
Compton Effect
The Compton Effect is the result of scattering of a photon by an electron. Energy and
momentum are conserved in such an event and as a result the scattered photon has less
energy (longer wavelength) then the incident photon.
If λ is incoming wavelength and λ' is scattered
wavelength and  is the angle made by
scattered wave to the incident wave then
  cos1' 
cm
h
o

where
cm
h
o
known as
c which is Compton wavelength (
c= 2.426 x 10
-12
m) and mo is
rest mass of electron.
Photoelectric effect
When a metal is irradiated with light, electron may get emitted. Kinetic energy kof
electron leaving when irradiated with a light of frequency
o  , where
o is threshold
frequency. Kinetic energy is given by

max 0k h h  
Stopping potential
sV is potential required to stop electron which contain maximum
kinetic energy
maxk.

0seV h h   , which is known as Einstein equation







photonincident
photon scattered
ElectronTarget Electron Scattered



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2. Mathematical Tools for Quantum Mechanics
2.1 Dimension and Basis of a Vector Space
A set of N vectors
N,........,
21 is said to be linearly independent if and only if the
solution of equation 


N
i
iia
1
0 is
1 2 Na = a = ..... a =0
Ndimensional vector space can be represent as 


N
i
iia
1
0 where i = 1, 2, 3 … are
linearly independent function or vector.
Scalar Product: Scalar product of two functions is represented as,, which is
defined as
dx
*
. If the integral diverges scalar product is not defined.
Square Integrable: If the integration or scalar product 
2
, dx  
is finite then the
integration is known as square integrable.
Dirac Notation: Any state vector  can be represented as  which is termed as ket
and conjugate of  i.e. * is represented by  which is termed as bra.
The scalar product of  and ψ in Dirac Notation is represented by   (bra-ket). The
value of   is given by integral 
rdtrtr
3*
,, in three dimensions.
Properties of kets, bras and brakets:






*
**
aa

Orthogonality relation: If  and  are two ket and the value of bracket 0
then,  is orthogonal.
Orthonormality relation: If  and  are two ket and the value of bracket 0
and 1 1 then  and  are orthonormal.
Schwarz inequality:  
2

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2.2 Operators
An operator A is mathematical rule that when applied to a ket  transforms it into
another ket i.e.
A   
Different type of operator
Identity operator I 
Parity operator  rr 
For even parity  rr  , for odd parity  rr 
Momentum operator
x
P i
x



 


Energy operator H i
t






Laplacian operator
2
2
2
2
2
2
2
zyx 










Position operator  rxrX 
Linear operator
For
2211  aa  if an operator A
ˆ
applied on  results in
1 1 2 2a A a A 
then operator A
ˆ
is said to be linear operator.

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2.3 Postulates of Quantum Mechanics
Postulate 1: State of system
The state of any physical system is specified at each time t by a state vector twhich
contains all the information about the system. The state vector is also referred as wave
function. The wave function must be: Single valued, Continuous, Differentiable, Square
integrable (i.e. wave function have to converse at infinity).
Postulate 2: To every physically measurable quantity called as observable dynamical
variable. For every observable there corresponds a linear Hermitian operatorA
ˆ
. The
Eigen vector of A
ˆ
let say
n form complete basis. Completeness relation is given
by
1
n n
n
I 


 
Eigen value: The only possible result of measurement of a physical quantity
na is one of
the Eigen values of the corresponding observable.
Postulate 3: (Probabilistic outcome): When the physical quantity A is measured on a
system in the normalized state  the probability P(an) of obtaining the Eigen value an of
corresponding observable A is 
2
1
ng
i
n
i
n
a
P a

 


 

where gn is degeneracy of state and
nu is the Normalised Eigen vector of A
ˆ
associated with Eigen value an.
Postulate 4: Immediately after measurement.
If the measurement of physical quantity A on the system in the state  gives the result
an (an is Eigen value associated with Eigen vector
na), Then the state of the system
immediately after the measurement is the normalized projection


n
n
P
P
where Pn is
projection operator defined by
n n .

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Projection operatorP
ˆ
: An operator P
ˆ
is said to be a projector, or projection operator, if
it is Hermitian and equal to its own square i.e. PP
ˆˆ
2

The projection operator is represented by
n n
n
 
Postulate 5: The time evolution of the state vector t is governed by Schrodinger
equation:  ttHt
dt
d
i  , where H(t) is the observable associated with total
energy of system and popularly known as Hamiltonian of system. Some other operator
related to quantum mechanics:
2.4 Commutator
If Aand B are two operator then their commutator is defined as A,B AB-BA
Properties of commutators
  
  
  
†
† †
, , ; , , ,
, , , ; , ,
, , , , , 0 (Popularly known as Jacobi identit y).
C C
C C B C
C C C
             
            
 
             
     

, 0f   
 

If Xis position and
xP is conjugate momentum then

1
,
n n
x
X P nX i

  
 
 and 
1
,
n n
x x
X P nP i

 
 

If b is scalar and A is any operator then , 0b 
If [A, B] = 0 then it is said that A and B commutes to each other ie AB BA.
If two Hermition operators AandB, commute ie, 0       and if A has non
degenerate Eigen value, then each Eigen vector of A
ˆ
is also an Eigen vector ofB.
We can also construct the common orthonormal basis that will be joint Eigen state of
AandB.
The anti commutator is defined as ,     

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2.5 Eigen value problem in Quantum Mechanics
Eigen value problem in quantum mechanics is defined as

n n na  
where an is Eigen value and
n is Eigen vector.
 In quantum mechanics operator associated with observable is Hermitian, so its Eigen
values are real and Eigen vector corresponding to different Eigen values are
orthogonal.
 The Eigen state (Eigen vector) of Hamilton operator defines a complete set of
mutually orthonormal basis state. This basis will be unique if Eigen value are non
degenerate and not unique if there are degeneracy.
Completeness relation: the orthonormnal realtion and completeness relation is given by
I
n
nnmnmn  

1
, 
where I is unity operator.
2.6 Time evaluation of the expectation of A (Ehrenfest theorem)

1 A
A,H
d
A
dt i t

 

where ,A His commutation between operator A and
Hamiltonian H operator .Time evaluation of expectation of A gives rise to Ehrenfest
theorem .

1d
R P
dt m
 , ,
d
P V R t
dt
  
where R is position, P is momentum and ,V R t is potential operator.
2.7 Uncertainty relation related to operator
If A
ˆ
and B
ˆ
are two operator related to observable A and B then
B
ˆ
,A
ˆ
2
1
B
ˆ
A
ˆ

where
2
2
A
ˆ
A
ˆ
A
ˆ
 and
2
2
B
ˆ
B
ˆ
A
ˆ
 .

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2.8 Change in basis in quantum mechanics
If k are wave function is position representation and kare wave function in
momentum representation, one can change position to momentum basis via Fourier
transformation.

 
 
1
2
1
2
ikx
ikx
x k e dk
k x e dx
 

 












2.9 Expectation value and uncertainty principle
The expectation value A of Ain direction of  is given by
A
A
 
 
 or
A
n na P where
nP is probability to getting Eigen value an in state  .

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3. Schrödinger Wave Equation and Potential Problems

3.1 Schrödinger Wave Equation
Hamiltonian of the system is given by
2
2
P
H V
m
 
Time dependent Schrödinger wave equation is given by
t
iH





Time independent Schrödinger wave equation is given by EH
where H is Hamiltonian of system.
It is given that total energy E and potential energy of system is V.

3.2 Property of bound state
Bound state
If E > V and there is two classical turning point in which particle is classically trapped
then system is called bound state.
Property of Bound state
The energy Eigen value is discrete and in one dimensional system it is non degenerate.
The wave function x
n of one dimensional bound state system has n nodes if n = 0
corresponds to ground state and (n – 1) node if n = 1 corresponds to ground state.
Unbound states
If E > V and either there is one classical turning point or no turning point the energy
value is continuous. If there is one turning point the energy eigen value is non-
degenerate. If there is no turning point the energy eigen value is degenerate. The particle
is said to be unbounded.
If E < V then particle is said to be unbounded and wave function will decay at ± ∞. There
is finite probability to find particle in the region.

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3.3 Current density
If wave function in one Dimension is x then current density is given by















xxim
J
x
*
*
2






which satisfies the continually equation 0


J
t


Where 
*
 in general J v
2
J v where v is velocity of particle?
If Ji, Jr, Jt are incident, reflected and transmitted current density then reflection coefficient
R and transmission coefficient T is given by

r
i
J
R
J
 and
t
i
J
T
J

3.4 The free particle in one dimension
Hψ = Eψ xE
dx
d
m



2
22
2



ikxikx
AeeAx


Energy eigen value E
m
k
2
22

 where
2
2

mE
k the eigen values are doubly degenerate
3.5 The Step Potential
The potential step is defined as







0
00
xV
x
xV
o

Case I: E > Vo
0
0
22
11
2
1




xBeAe
xBeAe
xikxik
xikxik



Hence, a particle is coming from left so D = 0.
R = reflection coefficient =
incident
reflected
J
J
=
2
1 2
1 2
k k
R
k k
 
 

 

xo
oV

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T = transmitted coefficient =
incident
dtransmitte
J
J
=
 
1 2
2
1 2
4k k
T
k k



where
 
2221
22

oVEm
k
mE
k


Case II: E < Vo
21
2
0
11

mE
kxBeAe
xikxik
I 


 
22
2
0
2

EVm
kxce
oxk
II




1
r
t
J
R
J
  and 0
t
i
J
T
J
 
For case even there is Transmission coefficient is zero there is finite probability to find
the particle in x > 0 .
3.7 Potential Barrier
Potential barrier is shown in figure.




Potential barrier is given by 









ax
axV
x
xV
0
0
00
0
Case I: E > Vo


 0
0
0
1
22
11
33
22
11



xEexx
axDeCexx
xBeAexx
xik
xikxik
xikxik



Where
21
2

mE
k
 
22
2

oVEm
k


Transmission coefficient

 
1
2
1sin
14
1
1









 T
a
Energy
o
x
oV

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Reflection coefficient

 
1
2
1sin
14
1











R Where
2
2
,

o
o
mV
a
V
E
 
Case II: E < Vo
3.7.1 Tunnel Effect

1 1
2 2
1
0
0
0
ik x ik x
I
k x k x
II
ik x
III
Ae Be x
Ce Be x a
Fe x





  
   
 



 
21
sin 1
4 1
R h  
 
and

 
21
1 sin 1
4 1
T h   
 

Where
o
o
V
EmV
a  ,
2
2


For E << Vo

 
2
22
1
16

EVma
oo
o
e
V
E
V
E
T











Approximate transmission probability
ak
eT
22
 where
 
22
2

EVm
k
o

3.8 The Infinite Square Well Potential
The infinite square well potential is defined as as shown in the figure










ax
ax
x
xV 00
0


Since V(x) is infinite in the region 0x and x a so the wave function corresponding
to the particle will be zero.
The particle is confined only within region 0 ≤ x ≤ a.
Time independent wave Schrödinger wave equation is given by

xV
o a x

ao
oV

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

E
dx
d
m

2
22
2


sin cosA kx B kx 
B = 0 since wave function must be vanished at boundary ie 0xso sinA kx
Energy eigen value for bound state can be find by ka nwhere 1,2,3...n
The Normalized wave function is for
th
nstate is given by

2
sin
n
n x
x
a a

 
Which is energy Eigen value correspondence to
th
n

2
222
2ma
n
E
n

 where n = 1, 2, 3 .....
othonormality relation is given by

0
sin sin i.e.
2
0
1
2
a
mn
m x n x a
dx
L L
m n
a
m
 

 
 


If x is position operator
x P Px is momentum operator and 
nx is wave function of
particle in n
th
state in one dimensional potential box then 
2
sin
n
n x
x
a a

  then


 
 
*
2 2
2 * 2
2 2
*
2 2 2 2
2 *
2 2
2
2 2
0
2
n n
n n
x n n
x n n
a
x x x x
a a
x x x x
n
P x i x
x
n
P x x
m x a
 
 

 

 








 
  

  


  






 

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The uncertainty product is given by
22
2
1
12
1


n
nPxx  
The wave function and the probability density function of particle of mass m in one
dimensional potential box is given by








3.8.1 Symmetric Potential
The infinite symmetric well potential is given by

22
0
22
a
x
a
a
xor
a
xxV



Schrondinger wave function is given by

)(0
2
cos
2
sin
)(0
2
cos
2
sin
0,
2
2
cossin
22
;
2
2
22
ii
ka
B
ka
A
i
ka
B
ka
Aso
x
a
xat
mE
kwherekxBkxA
a
x
a
E
dx
d
m












Hence parity () commute with Hamiltonian( )H then parity must conserve
So wave function have to be either symmetric or anti symmetric
2
a

2
a
a
xV
m
En
2
1
22
1


12
42 EEn 
13
93 EEn 
a
x
a

 sin
2
1

a
x
a


2
sin
2
2

a
x
a


3
sin
2
3

2
1
2
2
2
3

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For even parity
cosB kx and Bound state energy is given by 0
2
cos
ka

,....5,3,10
2
 n
a
n
k
ka 

Wave function for even parity is given as
a
xn
a

cos
2

For odd parity
Ψ (x) = A sin kx and Bound state energy is given by
sin 0
2
ka
 , 0
2
ka n
k
a

  2,4,6,........n
For odd parity Wave function is given as
a
xn
a

sin
2


The energy eigen value is given by ,......3,2,1
2
2
222
 n
ma
n
E
n


First three wave function is given by




a
x
a
x
a
x
a
x
a
x
a
x






3
cos
2
2
sin
2
cos
2
3
2
1




where
a
2
is normalization constant.





2
22
2ma

2
a
2
a

2
22
2
9
ma

2
22
2
4
ma

2
a

2
a
2
a

2
a
x
3
x
2
x
1

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3.9 Finite Square Well Potential
For the Bound state E < Vo
Again parity will commute to Hamiltonian
So wave function is either symmetric or
Anti symmetric
For even parity



2
22
cos
2
a
xAex
a
x
a
kxCx
a
xAex
x
III
II
x
I









wave function must be continuous and differentiable at boundaries so using boundary
condition at
2
a
 one will get
2
tan
ka
k
tan   where
2
a

2
ka

For odd parity



2
22
sin
2
2
1
a
xAex
a
x
a
kxDx
a
xAex
x
III
x









using boundary condition one can get
cot    where
 
2
2

EVm
o

2
2

mE
k
and
2
2
22
2
amV
n
o
 which is equation of circle.
The Bound state energy will be found by solving equation
tann for even
2
a

2
a x
oV
xV

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cotn for odd
2
22
2


amV
n
o

one can solve it by graphical method.







The intersection point of = tan    (solid curve) and
2
2 2
2
2
o
mV a
  

(circle) give
Eigen value for even state and intersection point at = - cot    (dotted curve) and
2
2 2
2
2
o
mV a
  

(circle) give Eigen value for odd state.
The table below shows the number of bound state for various range of
2
0V a where Voa
2

is strength of potential.
Voa
2
Even eigen function Odd eigen function No. of Bound
states
m2
22


1 0 1
m
aV
m 2
4
2
22
2
0
22
 

1 1 2
m
aV
m 2
9
2
22
2
0
22
 

2 1 3
m
aV
m 2
16
2
9
22
2
0
22
 

2 2 4

n
tann
o
2
 
2
3 2

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The three bound Eigen function for the square well








3.10 One dimensional Harmonic Oscillator
One dimensional Harmonic oscillator is given by
  xxmxV
22
2
1

The schrodinger equation is given by


Exm
dx
d
m


22
2
22
2
1
2


It is given that x
m


 and

E
k
2

The wave function of Harmonic oscillator is given by.
 
2
2/14/1
2
!2
1


















 eH
n
m
nnn


and energy eigen value is given by

nE = (n+1/2); n = 0, 1, 2, 3, ....
The wave function of Harmonic oscillator is shown
0H ( ) = 1,
1H ( ) = 2 ,
2
2H ( ) = 4 -2 
It
nand
m wave function of Harmonic oscillator then 



mnnm
dxxx 


x
2
a

2
a
x
2
x
2
a
2
a

x
1

x
x
3
1n
2
0

E
xV
0n
2n
2
3
1

E
2
5
2

E

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Number operator
The Hamiltonian of Harmonic operator is given by
22
2
2
1
2
Xm
m
P
H
x

Consider dimensioned operator as HPX
ˆ
,
ˆ
,
ˆ where
X
m
X
ˆ



ˆ
x xP m P  HH
ˆ
 so  
22ˆˆ
2
1
ˆ
XPH
x .
Consider lowering operator  
xiPXa 
ˆ
2
1
and raising
†1
ˆ
2
xa X iP  
 

 2/1
ˆ
NH
where
†
N a a and  1/ 2H N    N is known as number operator
n is eigen function of N with eigen value n.
N n n n and
1
2
H n N n
 
 
 
 

nnnH 






2
1
where n = 0, 1, 2, 3,………
Commutation of a and
†
a:
†
[a, a ] = 1,
†
[a , a] = -1 and [N, a] = -a,
† †
[N a ] = a
Action of a and
†
a operator on n
11
001



nnna
abutnnna

Expectation value of
22
,,,
XXPPXX in stationary states
 
 
  







mnPn
m
X
PX
m
i
aa
Paa
m
X
X
X
X












2
1
,12
2
0,0
2
,
2
22

2
;0,
2
1 
 






XX PXnfornPX

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4. Angular Momentum Problem
Angular momentum in Quantum mechanics is given kLjLiLL
zyx
ˆˆˆ 
Where
xyzzxYyzX YPXPLXPZPLZPYPL  ;;
and
x
iP
X


,
y
iP
Y


,
z
ip
Z



Commutation relation
,
x y zL L i L 
 
, ,
y z xL L i L 
 
  , ,
z x yL L i L
2 2 2 2
X Y ZL L L L   and
2
, 0
XL L  
 
,
2
, 0
YL L 
 
,
2
, 0
ZL L 
 

4.1.1 Eigen Values and Eigen Function











































iL
iL
SiniL
Z
Y
X
cotsincos
cotcos

Eigen function of
ZL is


im
e
2
1
.
and Eigen value of
ZL m where m = 0, ± 1, ± 2...
L
2
operator is given by



















2
2
2
22
sin
1
sin
sin
1



L
Eigen value of
2
L is
2
( 1)l l where l = 0, 1, 2, ...l
Eigen function of
2
L is 
m
lP where 
m
lP is associated Legendre function
L
2
commute with L z so both can have common set of Eigen function.
, ( )
m m im
l lY P e

   is common set of Eigen function which is known as spherical
harmonics .
The normalized spherical harmonics are given by

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
 

 
2 1 !
( , ) 1 cos
4 !
mm m im
l l
l l m
Y P e
l m

  

 
 

l m l  



4
1
,
0
0 Y





i
eY sin
8
3
,
1
1  , 
0
1
3
, cos
4
Y  

 , 
1
1
3
, sin
8
i
Y e

  

 

 ,1,
22 m
l
m
l YllYL  l = 0, 1, 2, ……. And
  ,,
m
l
m
lZ YmYL  m = -l ,(-l +1) ..0,….. (l – 1)( l ) there is 2 1l
Degeneracy of
2
L is2 1l.
Orthogonality Relation 
2
' '
0 0
, , sin
m m
l l ll mm
d Y Y d
 
         

4.1.2 Ladder Operator
Let
X YL L iL
  and
X YL L iL
 
Let us assume ml,is ket associated with
2
L and
ZL operator.

2
2
, 1 ,L l m l l l m  0,1,2,.......l
,, , ...0,...
zL l m m l m m l l  
Action of L+ and L- on ml, basis
, 1 1 , 1L l m l l m m l m
      
, 1 1 , 1L l m l l m m l m
      
Expectation value of
XL and
YL in direction of ml,
0
xL, 0
yL 
2
2 2 2
1
2
X Y
L L l l m    
 



2
2
1
2
X Y
L L l l m     
 


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4.2 Spin Angular Momentum
4.2.1 Stern Gerlach experiment
When silver beam is passed to the inhomogeneous Magnetic field, two sharp trace found
on the screen which provides the experimental evidence of spin.
4.2.2 Spin Algebra
ˆˆ ˆ
X Y ZS S i S j S k   ,
2 2 2 2
X Y ZS S S S  
2
, 0
X
S S  
 
,
2
, 0
Y
S S 
 

2
, 0
Z
S S  
 
and
 ,
X Y ZS S iS  ,
Y Z XS S iS  ,
Z X YS S iS

2 2
, 1 ,
s sS s m s s s m   , ,
z s sS s m m s m where
ss m s   
X YS S iS
  and
X YS S iS
 
 , 1 1 , 1
s s s s
S s m s s m m s m

    


4.2.3 Pauli Spin Matrices
For Spin
1
2
Pauli matrix
1
2
s,
1 1
,
2 2
s
m  
Pauli Matrix is defined as
 e.anticommutMatrix Spin Pauli;0
10
01
0
0
01
10
222
kj
I
i
i
jkkj
zyx
zyx




































1 if is an even permutation of , ,
1 if is an odd permutation of , ,
0 if any two indices among , , are equal
jkl
jkl x y z
jkl x y z
j k l


  



zzyyxx SSS 
2
,
2
,
2


 , 1 1 , 1
s s s sS s m s s m m s m
     

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
















 










10
01
20
0
201
10
2

zyx S
i
i
SS



















01
00
00
10
 SS









10
01
4
3
2
2
S
For spin ½ the quantum number m takes only two values
2
1

sm and
2
1
. So that two
states are
2
1
,
2
1
and
2
1
,
2
1
, 
smS
2
1
,
2
1
4
3
2
1
,
2
1
2
2 
S ,
2
1
,
2
1
4
3
2
1
,
2
1
2
2


S
2
1
,
2
1
2
1
2
1
,
2
1

zS ,
2
1
,
2
1
2
1
2
1
,
2
1
 
zS
0
2
1
,
2
1

S ,
2
1
,
2
1
2
1
,
2
1

S
2
1
,
2
1
2
1
,
2
1

 S , 0
2
1
,
2
1

S
4.3 Total Angular Momentum
Total angular momentum J = L + S , kJjJiJJ
zyx
ˆˆˆ 
jmj,is the Eigen ket at J
2
and Jz and
x yJ J iJ
 
x yJ J iJ
 

jj mjjjmjJ ,1,
22
 ,
jjjz mjmmjJ ,, 
  1,11, 
 jjjj mjmmjjmjJ
  1,11, 
 jjjj
mjmmjjmjJ
J L S l s j l s     
jmjSLJ
jzzz  and
l s jm m m 

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5. Two Dimensional Problems in Quantum Mechanics
5.1 Free Particle
Hψ = Eψ


E
yxm















2
2
2
22
2


x and y are independent variable. Thus



1
,
2
1
2
yx
ik yik x
n
i k r
x y e e
e






 
Energy Eigen value  
2
2
22
2
22
k
m
kk
m
yx


As total orientation of k

which preserve its magnitude is infinite. So energy of free
particle is infinitely degenerate.
5.2 Square Well Potential
0V x x a   and 0y a 
otherewise
2 2 2
2 2
2
H E
m x y
 

  
    
  


The solution of Schrödinger wave equation is given by Wave function

,
2
4
sin sin
x y
yx
n n
n yn x
a aa


  
     
   

Correspondence to energy eigen value










2
2
2
222
,
2 a
n
a
n
m
E
yx
nynx

1,2,3...
xn and 1,2,3...
yn

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Energy of state (nx, ny) Degeneracy
Ground state
2
22
2
2
ma



(1, 1)

Non degenerate
First state
2
22
2
5
ma



(1, 2), (2,1)

2
Second state
2
22
2
8
ma



(2, 2)

Non-degenerate

5.3 Harmonic oscillator
Two dimensional isotropic Harmonic oscillators is given by
 
22
2 2 21
2 2 2
yx
pp
H m x y
m m
   
 1
x y
n n x yE n n     where 0,1,2,3...
xn 0,1,2,3...
yn
1
nE n  
degeneracy of the n
th
state is given by (n + 1) where n = nx + ny.

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6. Three Dimensional Problems in Quantum Mechanics
6.1 Free Particle
EH


E
zyxm


















2
2
2
2
2
22
2


Hence x, y, z are independent variable. Using separation of variable one can find the
 

zikyikxik
k
zyn
eeezyx
2/3
2
1
,,

  
3 / 2
.
2
ik r
e




Energy Eigen value  
2
2
222
2
22
k
m
kkk
m
zyx


As total orientation of kwhich preserve its magnitude is infinite, the energy of free
particle is infinitely degenerate.

6.2 Particle in Rectangular Box
Spinless particle of mass m confined in a rectangular box of sides Lx, Ly, Lz
 , , 0 ,
xV x y z x L   0 ,
yy L  0 ,
zz L 
=  other wise .
The Schrodinger wave equation for three dimensional box is given by


E
zyxm
H 

















2
2
2
2
2
22
2


Solution of the Schrödinger is given by Eigen function
x y z
n n n and energy eigen value
is
x y z
n n nE is given by

























z
z
y
y
x
x
zyx
nnn
L
zn
L
yn
L
xn
LLL
zyx

 sinsinsin
8










2
2
2
2
2
222
2
z
z
y
y
x
x
nnn
L
n
L
n
L
n
m
E
zyx


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6.2.1 Particle in Cubical Box
For the simple case of cubic box of side a, the
i.e. Lx = Ly = Lz = a





















a
zn
a
yn
a
xn
a
zyx
nnn zyx

 sinsinsin
8
3

 
222
2
22
2
zyxnnn nnn
ma
E
zyx


1,2,3...
xn 1,2,3...
yn 1,2,3...
zn
Energy of state (nx, ny, nz) Degeneracy
E of ground state
2
22
2
3
ma



(1, 1, 1)

Non degenerate
E of first excited state
2
22
2
6
ma



(2, 1, 1) (1, 2, 1) (1, 1, 2

3
E of 2
nd
excited state
2
22
2
9
ma



(2, 2, 1) (2, 1, 2) (1, 2, 2)

3

6.3 Harmonic Oscillator
6.3.1 An Anistropic Oscillator
 
222222
2
1
2
1
2
1
,, ZmYmXmZYXV
ZYX  
zzyyxxnnn nnnE
zyx
  


















2
1
2
1
2
1
where
0,1,2,3...
xn 0,1,2,3...
yn 0,1,2,3...
xn
6.3.2 The Isotropic Oscillator
x y z     
3
2
x y zn n n x y z
E n n n 
 
   
 
 
 where
x y zn n n n   0,1,2,3...n
Degeneracy is given by = 21
2
1
nn

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6.4 Potential in Spherical Coordinate (Central Potential)
Hamiltonion in spherical polar co-ordinate

2 2
2
2 2 2 2 2
1 1 1
sin
2 sin sin
r V r E
m r rr r r
  
  
   
        
        
         



2 2
2
2 2
1
2 2
L
r V r
m r rr mr
 

    
   
  
   


L
2
is operator for orbital angular momentum square.
So ,
m
lY are the common Eigen state of and L
2
because [H, L
2
] = 0 in central force
problem and 
2 2
, 1 ( , )
m m
l lL Y l l Y     
So  ,,r can be separated as ,
m
lf r Y  
 
 
2 22
2 2
12
0
2 2
d f r l lf
f r V r E f r
m r rdr mr
  
      
 
 


To solve these equations 

r
ru
rf
So one can get


0
2
1
2
2
2
2
22









 uE
mr
ll
rV
dr
ud
m


Where

2
2
2
1
mr
ll
is centrifugal potential and 

2
2
2
1
mr
ll
rV

 is effective potential
The energy Eigen function in case of central potential is written as
 

, , , ,
m m
l l
u r
r f r Y Y
r
       
The normalization condition is
 
2
, , 1r d   



2
2
22
0 0 0
, sin 1
m
l
u r
r dr Y d d
r
 
    

  


2
0
1u r dr


or 
2
2
0
1R r r dr



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Hauz Khas, New Delhi-16
6.4.1 Hydrogen Atom Problem
Hydrogen atom is two body central force problem with central potential is given by

2
0
4
e
V r
r
 


Time-independent Schrondinger equation on centre of mass reference frame is given by
 
2 2
2 2
, ,
2 2
R r V r R r E R r
m
 

 
      
  
   
 

where R is position of c.m and r is distance between proton and electron.
R,rRr
The Schronginger equation is given by


h
2
2M
1
R

R
2
R









h
2
2
1
r

R
2
rVr








E
Separating R and r part
 RER
M
R
2
2
2


rrVr
r 


2
2
2


Total energy
R rE = E + E
ER is Energy of centre of mass and Er is Energy of reduce mass µ


3/ 2
1
2
i k R
R e


 

2 2
2
R
k
E
M



i.eCentre of mass moves with constant momentum so it is free particle.
Solution of radial part


h
2
2
d
2
ur
dr
2

ll1h
2
2r
2

e
2
4
0r





urEr
For Hydrogen atom
em

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Hauz Khas, New Delhi-16
The energy eigen value is given by
nE=
 
4
2
2 2
0
2 4
em e
n


eV
n
2
6.13
 1,2,3...n
And radius of
th
norbit is given by
2 2
0
2
4
n
n
r
m e



1,2,3...n where me is mass of
electron and e is electronic charge.
nis known as participle quantum number which varies as1, 2,...n l n   For the
given value of n the orbital quantum numbern can have value between 0and1n (i.e. l
= 0, 1, 2, 3, ….. n – 1) and for given value of l the Azimuthal quantum number m varies
from – l to l known as magnetic quantum mechanics .
Degeneracy of Hydrogen atom without spin =
2
n and if spin is included the degeneracy
is given by
1
2
0
2 2 1 2
n
n
l
g l n


  
For hydrogen like atom
2
2
6.13
n
z
E
n where z is atomic number of Hydrogen like atom.
Normalized wave function for Hydrogen atom i.e. Rnl (r) where n is principle quantum
number and l is orbital quantum.

n l E(eV) Rnl

1

0

2
-13.6z
0
3 / 2
/
0
2
zr az
e
a
 
 
 


2

0

2
- 3.4 z
0
3 / 2
/ 2
0 0
2
2
zr az zr
e
a a
   
   
   


2

0

2
- 3.4 z
0
3 / 2
/ 20
0 0
1
24
zr az zr
e
a a
   
   
   

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The radial wave function for hydrogen atom is Laguerre polynomials and angular part of
the wave function is associated Legendre polynomials .








For the n
th
state there is n – l – 1 node
If Rnl is represented by ln, then

2
0
1
3 1
2
nl r nl n l l a   
 


2 2 2 2
0
1
5 1 3 1
2
nl r nl n n l l a    
 

1
2
0
1
nl r nl
n a


 
2
3 2
0
2
2 1
nl r nl
n l a














rR
10
0/ar
r
rR
20
0/ar
r
21R
0/ar

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7. Perturbation Theory
7.1 Time Independent Perturbation Theory
7.1.1 Non-degenerate Theory
For approximation methods of stationary states

o PH H H 
Where H Hamiltonian can be divided into two parts in that
oH can be solved exactly
known as unperturbed Hamiltonian and
pH is perturbation in the system eigen value
of
oHis non degenerate
It is known
n
o
nno EH  and1
pH W where 1
Now  
o n n nH W E    where
nis eigen function corresponds to eigen value
nE
for the Hamiltonian H
Using Taylor expansion
........
2210

nnnn EEEE  and .........
221

nnnn 
First order Energy correction
1
nEis given by
1
n n nE W 
And energy correction up to order inis given by
1
n n n nE E W   
First order Eigen function correction
1
0 0
m n
n m
m n n m
W
E E
 
 




And wave function up to order correction in 
0 0
m n
n n m
m n n m
W
E E
 
   

 


Second order energy correction

 
2
2
0 0
m n
n
m n
n m
W
E
E E
 




Energy correction up to order of
2

 
2
2
0 0 0
m n
n n n
m n
n m
W
E E W
E E
 
   

  



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7.1.2 Degenerate Theory
( )
n o p n n nH H H E    
0
nE is f fold degenerate  fEH
nnno ,.....3,2,1
0
 

To determine the Eigen values to first-order and Eigen state to zeroth order for an f-fold
degenerate level one can proceed as follows
First for each f-fold degenerate level, determine f x f matrix of the perturbation
pH
ˆ



















ff
f
f
ppfpf
ppp
ppp
p
HHH
HHH
HHH
H
.........
.
.
.........
.........
ˆ
21
22221
11211

where
p n p nH H

  
then diagonalised
pH and find Eigen value and Eigen vector of diagonalized
pH which
are
nE and
n respectively.
10
 nnn EEE  and 


f
nn q
1
 
7.2 Time Dependent Perturbation Theory
The transition probability corresponding to a transition from an initial unperturbed
state
ito perturbed
f is obtained as
 
1
'
0
'
fi
iw t
if f i
i
P t V t e dt  


Where
f i
fi
E E
w



and
 f o f i o i
fi
H H
w
   



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Hauz Khas, New Delhi-16
8. Variational Method
Variational method is based on energy optimization and parameter variation on the basis
of choosing trial wave function.
1. On the basis of physical intuition guess the trial wave function. Say
 
noo  ,.....,,
321 where
321,, are parameter.
2. Find  
   
  
nn
nn
n
H
E



,.....,,,.....,,
,.....,,,.....,,
,.....,,
32103210
32103210
3210 
3. Find  0,.....,,
321 


n
i
oE



Find the value of   ,.....,,
321 so that it minimize E0.
4. Substitute the value of   ,.....,,
321 in   ,.....,,
3210E one get minimum value
of E0 for given trial wave function.
5. One can find the upper level of
1 on the basis that it must be orthogonal to
0i.e.
1 0 0 
Once
1 can be selected the 2, 3, 4 step can be repeated to find energy the first Eigen
state.

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Hauz Khas, New Delhi-16
9. The Wentzel-Kramer-Brillouin (WKB) method
WKB method is approximation method popularly derived from semi classical theory
For the case1
d
dx

where ( )
2 ( ( )
x
m E V x




If potential is given as V(x) then and there is three region
WKB wave function in the first region i.e
1x x


1
1 1
exp ' '
x
I
x
c
P x dx
P x

 
   
  



WKB wave function in region II: i.e
1 2x x x 




' "
2 2
exp ' exp ' '
II
x
c c i i
P x dx P x dx
P x P x

  
     
   
 
 

WKB wave function in region III: i.e
2x x


2
3 1
exp ' '
x
III
x
c
P x dx
P x

 
   
  



9.1.1 Quantization of the Energy Level of Bound state
Case I: When both the boundary is smooth
1
( ) ( )
2
p x dx n 

where 0,1,2...n
 
2
1
1
2 2
2
x
n
x
m E V x dx n
 
  
 
 

 where 0, 1, 2,......n and
1xand
2x are turning
point
Case II: When one the boundary is smooth and other is rigid

2
1
3
( ) ( )
4
x
x
p x dx n 
 where 0,1,2...n
 
2
1
3
2
4
x
n
x
m E V x dx n 
 
  
 
 

 where 0, 1, 2,......n and
1xand
2x are turning
points
xV
I II III

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Case III: When both boundary of potential is rigid

2
1
( ) ( 1)
x
x
p x dx n 
 where 0,1,2...n
  
2
1
2 1
x
n
x
m E V x dx n   
 where 0, 1, 2,......n and
1xand
2x are turning
points .
9.1.2 Transmission probability from WKB
Tis defined as transmission probability through potential barrier V is given by
exp 2T   where  
2
1
1
2
x
n
x
m E V x dx 

and
1xand
2x are turning points.

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10. Identical Particles
Identical particle in classical mechanics are distinguishable but identical particle in
quantum mechanics are Indistinguishable.
Total wave function of particles are either totally symmetric or totally anti-symmetric.
11.1 Exchange Operator
Exchange operator
ijPas an operator as that when acting on an N-particle wave
function
1 2 3( .... ... ... )
i j N       interchanges i and j.
i.e
1 2 3 1 2 3( .... ... ... ) ( .... ... ... )
ij i j N j i NP              
sign is for symmetric wave function
s and sign for anti symmetric wave
function
a.
11.2 Particle with Integral Spins
Particle with integral spins or boson has symmetric states.
   
1 2 1 2 2 1
1
, ,
2
s          
 

For three identical particle:
 
   
   









123213312
132231321
321
,,,,,,
,,,,,,
6
1



s
For boson total wave function(space and spin) is symmetric i.e if space part is symmetric
spin part will also symmetric and if space part is ant symmetric space part will also also
anti symmetric.
11.3 Particle with Half-integral Spins
Particle with half-odd-integral spins or fermions have anti-symmetric.
For two identical particle:
  
122121 ,,
2
1
 
a
For three identical particle:

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 
   
   
1 2 3 1 3 2 2 3 1
1 2 3
2 1 3 3 1 2 3 2 1
, , , , , ,1
, , , , , ,6
a
           
   
           
   
 
    

For fermions total wave function(space and spin) is anti symmetric .ie if space part is
symmetric spin part will anti symmetric and if space part is ant symmetric space part will
also symmetric.



11. Scattering in Quantum Mechanics





Incident wave is given by 
.
o
ik r
incr Ae 

. If particle is scattered with angle θ which is
angle between incident and scattered wave vector
ok

and k

Scattered wave is given by
 
.
,
ik r
sc
e
r Af
r
  

, where ,f is called scattering amplitude wave function.
is superposition of incident and scattered wave 







r
e
feA
rik
rik
o
o
 ,
differential scattering cross section is given by 
2
,
o
d k
f
d k

 

where  is solid angle
For elastic collision 
2
,
d
f
d

 


If potential is given by V and reduce mass of system is µ then




k
ok
 
2
2
2 ' 3
2
, ' ' '
4
ikrd
f e V r r d r
d
 
  


 




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11.1 Born Approximation

Born approximation is valid for weak potential V(r)
 
2
0
2
, ' ' sin ' 'f r V r qr dr
q

 

 


Where 2 sin
2
o
q k k k

  
 
for and for
o
k k
 

2
2
4 2
0
4
' ' sin ' '
d
r V r qr dr
d q
 







11.2 Partial Wave Analysis
Partial wave analysis for elastic scattering
For spherically symmetric potential one can assume that the incident plane wave is in z-
direction hence   exp cos
incr ikr  . So it can be expressed in term of a superposition of
angular momentum Eigen state, each with definite angular momentum number l
  
cos
0
2 cos
ik r ikr l
l l
l
e e i l l J kr p





   , where Jl is Bessel’s polynomial function and Pl
is Legendre polynomial.
   
0
, 2 1 cos
ikr
l
l l
l
e
r i l J kr P f
r
   


  
    
1
2 1 sin cos
li
l lf l e P
k

   
Total cross section is given by
 
2
2
0 0
4
2 1 sin
l l
l l
l
k

  
 
 
   
Where σl is called the partial cross section corresponding to the scattering of particles in
various angular momentum states and
lis phase shift .

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12. Relativistic Quantum Mechanics
12.1 Klein Gordon equation
The non-relativistic equation for the energy of a free particle is

2
2
p
E
m
 and
2
2
p
E
m
  
2
2
p
i
m t



 

 where
p i  is the momentum operator ( being the del operator).
The Schrödinger equation suffers from not being relativistic ally covariant, meaning it
does not take into account Einstein's special relativity .It is natural to try to use the
identity from special relativity describing the energy:
2 2 2 4
p c m c E  Then, just
inserting the quantum mechanical operators for momentum and energy yields the
equation
2 2 2 2 4
c m c i
t



   

 
This, however, is a cumbersome expression to work with because the differential operator
cannot be evaluated while under the square root sign.
which simplifies to
2
2 2 2 2 4
2
c m c
t



    


Rearranging terms yields
2 2 2
2
2 2 2
1 m c
E
c t
 
 

  
 

Since all reference to imaginary numbers has been eliminated from this equation, it can
be applied to fields that are real valued as well as those that have complex values.
Using the inverse of the Murkowski metric we ge
2 2
2
0
m c


     

where
2
( ) 0  

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In covariant notation. This is often abbreviated as
2
( ) 0   where
mc


and
2
2
2 2
1
c t

 


This operator is called the d’Alembert operator . Today this form is interpreted as the
relativistic field for a scalar (i.e. spin -0) particle. Furthermore, any solution to the Dirac
equation (for a spin-one-half particle) is automatically a solution to the Klein–Gordon
equation, though not all solutions of the Klein–Gordon equation are solutions of the Dirac
equation.
The Klein–Gordon equation for a free particle and dispersion relation
Klein –Gordon relation for free particle is given by
2
2
2 2
1
E
c t

 

  


with the same solution as in the non-relativistic case:
dispersion relation from free wave equation ( , ) exp ( . )r t i k r t   which can be
obtained by putting the value of in
2
2
2 2
1
E
c t

 

  

equation we will get
dispersion relation which is given by
2 2 2
2
2 2
m c
k
k

  

.
12.2 Dirac Equation
searches for an alternative relativistic equation starting from the generic form describing
evolution of wave function:



H
t
i
ˆ

If one keeps first order derivative of time, then to preserve Lorentz invariance, the space
coordinate derivatives must be of the first order as well. Having all energy-related
operators (E, p, m) of the same first order:

t
iE



ˆ
and
z
ip
y
ip
x
ip
zyx








 ˆ,ˆ,ˆ

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  mpppE
zyx  ˆˆˆˆ
321
By acting with left-and right-hand operators twice, we get
  
2
1 2 3 1 2 3
ˆ ˆ ˆ ˆ ˆ ˆ ˆ
x y z x y z
E p p p m p p p m               
which must be compatible with the Klein-Gordon equation
 
22222
ˆˆˆˆ
mpppE
zyx
This implies that
,0
ijji  for ji
0
ii
1
2

i
1
2

Therefore, parameters α and β cannot be numbers. However, it may and does work if they
are matrices, the lowest order being 4×4. Therefore, ψ must be 4-component vectors.
Popular representations are










0
0
i
i
i


 and










10
01

where
iare 2  2 Pauli matrices:





























10
01
0
0
01
10
321 
i
i

The equation is usually written using γµ-matrices, where
i i  for
The equation is usually written using γµ-matrices, where
i i  for 1,2,3i and
0  (just multiply the above equation with matrix β and move all terms on one side of
the equation):
0











m
x
i

 where










0
0
i
i
i


 and










10
01
0
Find solution for particles at rest, i.e. p=0:

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0
4
3
2
1
0 





























 m
t
i
































B
A
B
A
m
t
i
10
01

It has two positive energy solutions that correspond to two spin states of spin-½
electrons:











0
1
imt
Ae and










1
0
imt
Ae
and two symmetrical negative-energy solutions











0
1
imt
Be and










1
0
imt
Be