Lecture Note
The Schrodinger Equation and Stationary States
-
University:
College -
Course:
College Mathematics Academic year:
2025
-
Views:
358
Pages:
142
Author:
Rajkumar Behera
M.SC. PHYSICS - II YEAR
DKP21 - QUANTUM MECHANICS
(From the academic year 2016-17)
Most Student friendly University - Strive to Study and Learn to Excel
V.RAVICHANDRAN
ASSOCIATE PROFESSOR OF PHYSICS
SRI K.G.S. ARTS COLLEGE
SRIVAIKUNTAM Syllabus
QUANTUM MECHANICS
UNIT I: The Schrodinger equation and Stationary states
Physical basis of quantum mechanics- wave particle duality- The Schrodinger equation (timedependent and time independent)- Physical interpretation and conditions on the wave
function- Expectation values and Ehrenfest theorem- Stationary states and energy spectraParticle in a Square well potential.
General formulation: The fundamental postulates of wave mechanics- Dirac notation: BRA
and KET-Hilbert space- Hermitian operators and their properties- commutator relation and
Heisenberg uncertainty principle
UNIT II: Exactly Solvable systems
Linear Harmonic oscillator- Solving the one dimensional Schrodinger equation- Abstract
operator method-Particle in a box- Square well potential- Rectangular barrier potentialRigid rotator- Hydrogen atom
UNIT III: Approximation Methods
Time independent problems: Nondegenerate case: First and second order perturbationsDegenerate case: Zeeman effect- Stark effect- Variation method- WKB quantization ruleMethods for time dependent problems: Time dependent perturbation theory- First order
perturbation- Harmonic perturbation- Transition probability- Fermi’s Golden rule- Adiabatic
approximation- Sudden approximation
UNIT IV: Angular, Spin angular Momentum and Scattering theory
Matrix representation of angular momentum and spin angular momentum- Commutation
relations- Eigen values- Addition of angular momenta- Clebsch-Gordan coefficients (basic
ideas only) -Identical particle with spin-Pauli’s exclusion principle- Spin statistics- Stern
Gerlach experiment and electron spin Scattreing theory: The Scattering cross section- Born approximation- Partial wave analysisDifferential and total cross sections- Phase shift
UNIT V: Relativistic Quantum Mechanics
Klein Gordan equation for a free particle and in an electro magnetic field- Dirac equation for
a free particle-Charge and current densities
Dirac matrices- Plane wave solution- Negative energy states- Zitterbewegung- Spin angular
momentum-Spin-orbit coupling- Fine structure
Books for study and reference:
1. L. Schiff,Quantum Mechanics (Tata McGraw Hill, New Delhi, 1968)
2. P.M. Mathews and K. Venkatesan, A text Book of Quantum Mechanics (Tata
McGraw Hill,New Delhi, 1987)
3. V.K. Thankappan, Quantum Mecahnics (Wiley-Eastern, New Delhi, 1985)
4. J. Singh, Quantum Mechanics- Fundamentals and Applications in Technology
(John Wiley, Newyork,1997)
5. Goswami, Quantum Mecahnics (W.C.Brown, Dubuque,1992)
6. Y.R. Waghmare, Foundations of Quantum Mecahnics (Wheeler, New Delhi,
1996)
7. S. Devanarayanan, Quantum Mecahnics (Sci. Tech Publications Pvt. Ltd.
Chennai,2005) 1
UNIT I
1.1 Matter Waves
According to Debroglie, the wavelength λ associated with a moving particle having
momentum p is given by
h
h
p mv
where h is Planck’s constant
This wave associated with a particle is called matter waves.
Consider a particle possessing wave length λ. The value of ψ at any point x at time t is given
by
or
0 sin t
0 sin 2t
where ν is the frequency of the particle. Let the particle moves with velocity v along positive
x-direction. Using Lorentz transformation theory, we can write
t '
t
vx '
c2
v2
1
c2
vx'
2 t '
c2
0 sin
v2
1
c2
(1.1)
But the standard equation of wave motion is
2
T'
0 sin
x'
t '
u '
(1.1A)
where 0 is the amplitude,
T'
is the period and u' is the phase velocity
Comparing eqns. (1.1) and (1.1A)
u '
c2
and
v
1
'
T'
v2
1
c2
According to Einstein’s mass-energy relation
E m0c h
2
Substituting this in eqn. (1.2)
or
m0c 2
h
(1.2) 2
v'
m0c 2 h
v2
1 2
c
mc 2
h
The wavelength of the particle
velocity u ' c
frequency v' mc 2 h
2
or
h
mv
(1.3)
This is the expression for the de-Broglie wavelength for a particle of mass m moving with
velocity v
1.2 Schroedinger time-independent wave equation.
The classical wave equation is written as,
1
2 =
2 dt 2
2
(1.4)
where is the displacement and v is the velocity of the wave. Let the wave amplitude is
periodic in time t as
(r, t)= (r)
(1.5)
Substituting eq. (1.5) in eq. (1.4), we get
= 2
2
2
2
Or 2 0
2
or
2
4
2
2
0 since
2
and being the frequency and wavelength of the de-Broglie waves. Rearranging the terms,
we get the wave equation in the form
2 + (42/2) = 0.
From the expression for de-Broglie wavelength =
(1.6)
h h
,
p m 3
4 2 m 2 v 2
+
= 0.
h2
2
Now,
E V
1
mv2 where E is the total energy and V is the potential energy of the
2
system from which,
mv = 2m( E V )
2 +
and the wave equation becomes
2m
( E V ) 0
2
(1.7)
This equation is known as Schroedinger time-independent wave equation.
1.3 Schroedinger Time-dependent equation
Schrodinger time independent wave equation can be written as
2
2m
( E V ) 0
2
it
Mulktiplying on the right by e and rearranging we get,
- ( r )e
2
i t
8 2 m
8 2 m
i t
V
e
E e i t
+
2
2
h
h
The right hand side can be written as
8 m
2
h
2
8 m
2
E(r , t )
h
2
E
(r, t )
(-i ) t
8 m ih (r , t )
2
h 2 t
2
=
since E=
= 2m i (r , t )
t
Substituting this in eqn. (1.8), we get
2 2
V i
t
2m
or
This is the required Schoedinger time – dependent wave equation
(a) Solution of the Time dependent Schroedinger equation
The Schroedinger Time-dependent wave equation is written as,
(1.8) 4
Now
2
( x, t )
2
x
2m
2
V ( x ) ( x, t )
( x, t )
i t
(1.9)
( x, t ) can be expressed as the product of two functions, one involving the time
coordinate alone and the other position coordinate alone. That is,
(x,t) = ψ(x) (t)
Substituting this in the eqn. (1.9) and simplifying we get,
2
1 2 d ( x)
1 d (t )
V ( x) ( x)
2
( x) 2m dx
i (t ) dt
The left hand side is a function of x while the right hand side is a function of t only. But x and
t are independent coordinates. This is possible only when they are separately equal to a
constant (E)
Hence,
2
1 2 d ( x)
V ( x) ( x) E
2
( x) 2m dx
and
1 d (t )
E
i (t ) dt
(1.9A)
From the first of equations (1.9A) ,
d 2 ( x) 2m
2 E V ( x) ( x) 0,
dx 2
(1.10)
which is Schroedinger time independent wave equation whose solution is given by (x).
Considering the second of equations (1.9A), we have
d (t )
i
E (t )
dt
d (t )
i
E dt.
(t )
or
log (t )
Integrating,
iEt
i Ent
(t ) exp
or
i En t
n (t ) exp
or in general for the nth state,
Hence the general solution is
(x,t) =
a
n
n
n ( x, t ) =
a
n
n
n
iEnt
( x) exp
(1.11) 5
(b) Stationary State Solutions
Consider the probability distribution function * for a system in the state represented by
the wave function
(x, y, z, t)
iE n t
= a n n ( x, y, z ) exp
n 1
(1.12)
Taking the complex conjugate
* (x, y, z ,t) =
a
m
m
iE t
* m * ( x, y, z ) exp m
(1.13)
Multiplying eqns. (1.12) and (1.13),
i( E E )t
* an am n ( x, y, z ) m ( x, y, z ) 'an am n ( x, y, z ) m ( x, y, z ) exp m n
m n
' indicates m n
If an values are zero for all values except for one value of En then * will be independent
of time and the state represented by wave function would be stationary. In this case the
wave function is represented by
i En t
n ( x, y, z, t ) an n ( x, y, z ) exp
(1.14)
1.4 Physical significance of
The wave function is large where the particle is most likely to be and small elsewhere. The
finite region in which the wave function is appreciably different from zero is called a wave
packet. According to Max Born, the absolute square of (r,t), that is 2 gives the
probability per unit volume of finding the particle at some position r at any time t. The
probability of finding the particle in a volume element d is P(r,t) d = (r,t) 2 d . The
total probability of finding the particle in the whole region is unity.
ie. 2 d = 1
satisfying this condition is said to be a normalised wave function.
(a) Limitations on :
a) must be finite for all values of x, y, z of the region
b) must be single valued
c) must be continuous in all regions
(1.15) 6
d) is analytical i.e. it possesses continuous first order derivative
e) vanishes at infinity space
1.5 Orthogonal, Normalised and Orthonormal wave functions
For any two wave function, 1 (x) and 2 (x) in the limit a to b, if the condition
a
*
2
( x) 1 ( x) dx 0.
b
is satisfied then, then 1 ( x) and 2 ( x) are said to be mutually orthogonal wave functions
We know that the probability of finding a particle in the volume element d is given by
ψψ* d or 2 d. The total probability of finding the particle in the entire space is unity.
i.e., (r , t ) 2 d 1
where the integration extends over the entire space.
This can also be written as,
(r , t ) (r , t ) d 1
Any wave function satisfying the above condition is said to be normalized to unity or simply
normalised. If is not a normalised wave function, then it can be made normalised by
multiplying it with a constant called normalisation constant (A).
To find the normalisation constant,
( A ) * Ad
A
2
d 1
or
A
1
2
*
d
(1.16)
The functions which are orthogonal as well as normalised are called ortho-normal functions.
1.6 Pobability current density
If P(r) is the probability of finding a particle at a given point, then the probability of finding
the particle in a volume V bounded by the surface area A is
P P(r )d *d
V
V
dP d
d * d
ie.
*d
dt dt V
dt
dt
V
From Time dependent Schrodinger eqn.
*d
(1.17) 7
2 2
V i
2m
t
(1.18)
The complex conjugate of the above eqn. is
(1.19)
2 2
*
* V * i
2m
t
Multiplying equations (1.18) by * and (1.19) by and subtracting the two we get
d * d
*
( * 2 2 *)
dt
dt
2im
Using this eqn. (1.17) can be written as
dP
( * 2 2 *)d
dt
2im V
ie.
dP
( * *)dA (volume integral is changed into
dt
2im A
surface integral)
If we define a new vector S such that,
S (r , t )
or
( * *) , then we can write
2im
dP
S .dA div S
dt
A
dP
div S 0
dt
This is analogous to the equation of continuity in hydrodynamics
(1.20)
div j 0 . Here is
t
the fluid density and j is the current density. Thus we can define S is the probability current
density
1.7 Expectation values of dynamical quantities
The average value or the expected value of any function f(x) is given by 8
f = * f ( x)dx
(1.21)
Thus the expectation value is the mathematical expectation for the result of a single
measurement or it is the average of the large number of measurements on independent
systems. The expectation value of position vector r is given by
r * (r, t ) r (r, t ) d
(1.22)
The expectation value of potential energy (V) is
V = * (r, t ) V (r, t ) d
(1.23)
We can prove the classical equations are valid for the expectation values of the dynamical
quantities, For example, consider the classical equation,
E=
p2
V
2m
Considering expectation values throughout we get,
E =
Operating on a wave function
p2
V
2m
gives
2
p
V
E =
2m
Using the operator formalism
Eˆ i , pˆ i , we can write the above eqn. as
t
2 2
i
V
t
2m
(1.24)
Multiply eqn. (1.24) on the left by *,
* i
2 2
*
* V
t
2m
(1.25)
This means the expectation value is obtained by corresponding operator acting on
and
multiplied on the left by * . Therefore
E = * i
d ;
t
p * ( i)
d
(1.26) 9
1.8 Ehrenfest Theorem (Newton’s Law of motion)
Statement:
The average motion of a wave packet described by a wave function
agrees with the
corresponding classical motion of the particle. That is,
(a)
dr p
dt m
(b)
dp
V
dt
(1.27)
Proof:
The Schroedinger time dependent wave equation is written as
i
2
2 V
t
2m
2
2m
2 V
t
2mi
or
(1.28)
Taking its complex conjugate,
2 * 2m
2 V
t 2mi
(1.29)
For a normalized wave function, the expectation value of x is given by
x = * x d
(1.30)
d x
dt
Differentiating w.r.t. t,
*
t x d * x t d
(1.31)
since dx/dt=0.
Substituting the value of
and
from equation (1.28) and (1.29) into equation (1.31),
t
t
we get,
d x
2
2m
*
2 V x d
dt
2mi
2mi
2
*
x d
2mi
*
x
2
2m
2 V d
2mi
x d
*
2
(1.32)
The first integral on R.H.S. can be integrated by parts by taking u x and dv 2 * 10
xd .x d .( x )d
2
2
(1.33)
Using Gauss divergence theorem, the first term of the integral of R.H.S can be transformed to
an integral over a surface at infinity and since the wave function vanishes at infinity, we
have
.x d x dA 0
(1.34)
Thus from equation (1.33), we get
xd ( x )d
2
2
(1.35)
Integrating again equation (1.35) by parts and again the surface integral vanishes at infinity,
we have
xd ( x )d
2
2
(1.36)
Using equation (1.36), equation (1.32) becomes
dx
dt
2mi
( x ) d 2mi
2
2mi
x ( ) d
2
[ ( x )
2
x 2 ] d
d
mi
x
1
m
(1.37)
i x d
1
p
x
m
Considering y and z components, we have
d r 1 p
dt
m
(1.38)
Differentiating once again the eqn. (1.37) and multiplying by m,
d p
d
x
dt
i dt
d
x
i
t x d i
*
d
x t 11
2
2m
* 2
2m
V
d
(
V
)
d
x
i 2mi
i 2mi x
2
2 x
2 2
2m V
2
d
( )
2m
x
x
2 x
(1.39)
The first two terms cancel out and we get
d p
x
dt
V
x
V
d
x
(1.40)
Considering all other components, we get
d p
V = F (force acting on the system)
dt
(1.41)
This eqn. shows that the rate of change of momentum is equal to force, which is Newton’s
second law of motion.
1.9 Postulates of wave mechanics
Postulate:1
Quantum system is characterised by a wave function (r , t ) . It is a complex function of
space and time and contains all possible informations about the system.
Postulate:2
The probability of observing a particle at time t within volume dτ is
dP (r , t )
1
2
(r , t ) d
N
(1.42)
Postulate:3
Any wave function (r , t ) can be expanded interms of complete set of orthonormal functions
as
(r , t )
C (t ) (r)
a
a
a
(1.43) 12
Postulate:4
Time evolution of a system described by the wave function (r , t ) is given by the Schrodinger
equation
i
2 2
V
t
2m
(1.44)
1.10 Dirac’s ‘BRA’ and ‘KET’ vectors : Dual Space
The quatum matrix theory can be put into compact form by making use of notations invented
by Dirac. Accordingly each dynamical state may be represented by a certain type of vector
known as ket vector or simply ket represented by the symbol |, In order to distinguish the
kets from each other, we complete each symbol by inserting a particular letter in the middle.
For example | a denotes the ket vector corresponding to state
a of the system
( a in old notation)
The kets form a linear vector space and any linear combination of several ket vectors is also a
ket vector. For example let us consider two kets | a and | b and two arbitrary complex
numbers c1 and c2 the linear combination.
| v c1 | a c2 | b
(1.45)
is also a vector of ket space.
Accordingly ket vectors may be multiplied by complex numbers and added together to give
other ket vectors. The vectors of dual space are called bra vectors or simply bras and denote
in general one of them by the symbol | , the mirror image of the symbol for a ket vector. If
one wants to specify a particular one of them by a label b, (say), it is written in the middle as
b| .
There exists one to one correspondence between the vectors of bra and ket spaces. Bra and
ket thus associated by this one to one correspondence are said to be conjugate of each other
and are labelled by the same indices. Thus the bra conjugate to ket | a is represented by the
symbol a | ; subject to the conditions
| a | b a | b |
and
c | a c a | 13
The connection between dual spaces is given by defining the scalar product of a bra vector
and a ket vector such that
a || b a , b ) a b dq
We have the rules that any complete bracket expression denotes a number and any
incomplete bracket expression denotes a vector of the bra or ket space according to whether it
contains the first or second part of bracket. The first three (bra) and the last three (ket) letters
of bracket provide the names for the two kinds of state vectors (bra and ket). The bra vector
| is analogous to the complex conjugate (or hermitian adjoint) of the wave function of the
system. Evidently we have
a|b b| a
a | | b | b | a | b a | b
a | c | b c a | b
where
The bra conjugate to ket
c is any number.
| c1 | a | c2 | b is
| c1 a | c2 | b |
Thus the correspondence between the kets and bras is analogous to the correspondence
between the wave functions of wave mechanics and their complex conjugates. We further
note that if a ket vanishes, its conjugate bra also vanishes and vice – versa. A bra and a ket
vectors are said to be orthogonal if their scalar product is zero. The length of a bra vector
a | or of its conjugate imaginary ket vector | a is defined as the square root of the possible
number a| a .
1.11 Hilbert Space
A set of n vectors ai is said to be linearly independent if there exists no relation between
them of the form
n
c a 0
i 1
i
i
except for c1 c2 ........ c n 0. Otherwise such a relation will reduce the number of
independent terms. If ai and a j are mutually orthogonal i.e., their scalar product denoted by
(ai , a j ) vanishes for i j, then ai and a j are linearly independent. In three dimensional space 14
the maximum number of linearly independent vectors is obviously three. In general, any set
of
n -linearly independent vectors spans an n -dimensional linear space denoted by Rn . This
purely abstract concept of
n -dimensional space ( n a finite real positive integer) indeed
becomes essential in many problems in modern physics. A linear operator Pˆ of Rn transforms
one vector u of Rn into another vector v of Rn
i.e.,
Pˆ u v.
A vector u i which is carried over by a given operator P̂ . An operator of
space Rn has in general
n -dimensional
n eigen values. If some of them coincide, the system is said to have
degeneracy. All these notions can be carried out into a (complex) space with (countable)
infinitely many dimensions (n ) . If one imposes restriction that the scalar product
|
u
2
(u, u) ui
2
i 1
exists, where u (u1, u2....) is a sequence of real or complex scalars and is called a vector i.e.,
if all infinite series which occur and convergent, then the space is said to be Hilbert space.
In brief the vector space is said to be Hilbert space if it is complex and of countable infinite
dimensions such that all infinite series occurring in it are convergent. In Quantum Mechanics
we often deal with complex functions and the corresponding Hilbert space is that of
quadratically integrable (complex) function in i variables (q1 , q2 ,.... qi )
Let us now consider some properties of Hilbert space and some operations in it. The scalar
product of two functions viz.,
f (q1 , q2 , .... qi ) and g (q1 , q2 , .... qi ) is defined by
( f (q1 , q2 ,.... qi ) g (q1 , q2 ,.... qi )
f
(q1 , q2 , ....qi ) g (q1 , q2 , ....qi ) d q1 , dq2 , ....dqi
The condition (1.69) then becomes
| f |
2
dq1 dq2 .... dqi C exists
which is true for wave function
Now the wave function are orthogonal when
f
(q1 , q2 .... qi ) g (q1 , q2 .... qi ) dq1 dq2 .... dqi 0 15
In Quantum Mechanics a state function contains all the physically relevant information
about a physical system at a given instant of time and is represented by a vector along a
direction in the Hilbert space. If represents a physical state, then and a constant
multiple of both represent this state. Therefore the arbitrary representative vector of the
ray is usually normalised to one.
The states specified by different functions, i (forming complete basic set) and represented
by vectors along mutually perpendicular axes form a complete orthogonal system of
coordinates axes in the Hilbert space. The completeness of the system means that any wave
function associated with vector in the Hilbert space can be represented as the vector-sum
of its components along these axes i.e.,
bi i
i
there being an axis for each i . Thus each bi corresponds to the (complex) component of
arbitrary vector along i -axis.
The multiplication of two finite matrices is defined as
( AB) mn Ami Bin
i
where
m denotes the row and i the column of matrix A. This law may be extended to the
matrices having infinite number of rows and columns in the Hilbert space. Then this
multiplication is same as that obeyed by two operators. Hence any operator  can be
associated a matrix A whose matrix elements are defined as
A m Amnm n
n
n
Anm
n
where Anm n A m dq ( n , A m )
with respect to a basic set of eigen functions n . As each integral has a certain numerical
value, therefore
Anm n A m dq ( n , A m )
gives a doubly infinite array of number i.e. a matrix. Now a matrix operating on a wave
function (vectors) can be represented as a linear transformation in the Hilbert space .For
example in the equation.
Af g. 16
‘A’ transforms f int o g where f and g are two wave functions (vectors) in the Hilbert
space. Unless specified we consider only those eigen functions which are located in the
Hilbert space. This imposes a restriction to discrete spectrum.
1.12 Hermitian operator
Operator A is said to be Hermitian if for two square integrable functions and the
following condition is satisfied.
| A A |
Adjoint or Hermitian conjugate of an operator:
For an orbitrary operator A, we can define another operator A+ such that
+
| A A | . Then A is called the adjoint or Hermitian conjugate of A
(a) Properties of Hermitian operators:
(i)
Product of two Hermitian operators is Hermitian, if and only if, they
commute
Consider two Hermitian operators A and B. If they commute then we will show AB is
hermitian. Let (x) and (x) are two functions. If A is hermitian, then we can write,
| AB | A | B |
If B is also hermitian ,
A | B | BA |
| AB | BA |
(1.46)
IF C=AB is hermitian then,
| AB | | C | C | AB |
From eqns.(1.46) and (1.47), AB=BA. Thus AB is hermitian only if they commute.
(1.47) 17
(ii)
Eigen values of hermitian operator are real:
A |
Let
n
a |
n
n
Now, n | A | n an n | n
(1.48)
Also A n | n an n | n
(1.49)
*
If A is hermitian then n | A | n A n | n . Using (1.48) and (1.49), we get
an an *
(1.50)
Hence the eigen values of hermitian operators are real
(iii)
Eigen functions of a hermitian operator corresponding to different eigen
values are orthogonal to each other.
Let P be any Hermitian operator and 1 and 2 be any two eigen functions of P. If A and
B are the two distinct eigen values of P corresponding to 1 and 2 , then we can write
P 1 A 1
P 2 B 2
or
P * 1* A * 1* A 1 *
P * 2 * B * 2 * B 2 *
since A and B are real.
(1.51)
From the general characteristics of the Hermitian operator P,
2
* P 1d P * 2 * 1d
(1.52)
Using eqn. (1.51),
or
2
* A 1d B 2 * 1d
or
A 2 * 1d B 2 * 1d
( A B) 2 * 1d 0
or 2 * 1d 0 since
A B . Hence proved.
(1.53) 18
1.13 Proof of Uncertainty principle for one-dimension Wave packet
The uncertainty in position along X - axis, by analogy with the standard deviation of
statistics, may be defined as
x x x
2
where x is the expectation value of
(1.54)
x. The probability of finding the particle between co-
ordinates x and x dx is given by ( x) ( x) dx, so that expectation of average value of
x, for normalised function (x) , is written as
x ( x) x ( x) dx,
(1.55)
Similarly the uncertainty in momentum is represented as
p p p
2
(1.56)
where p is the expectation value of momentum defined as
p i dx
x
i
Let us choose
dx
x
(1.57)
x p 0.
Now consider the integral
i
d
( x )dx
dx
Integrating by pars,
i
d
d
( x )dx i x i x
dx
dx
dx
d
dx since ψ vanishes at infinity.
= 0 i x
dx
But dx 1 for normalised wave function.
or
i
i
d
d
x dx i
x dx i
dx
dx
d
d
x dx i
x dx i
dx
dx 19
As the right hand side is imaginary, therefore it must be equal to the imaginary part of left
i
hand side i.e., 2i imaginary part of
d
x dx i
dx
(1.58)
Taking modulus of both sides and then squaring we get
d
4 Imaginary Partof i
x dx
dx
2
2
(1.59)
But we have
i
d
x dx imaginary part of
dx
i
d
x dx
dx
Equation (1.59) can now be written as
d
4 i
x dx
dx
2
2
(1.60)
According to Schwarz inequality, we have
f φ dx
2
f f dx φ φ dx
(1.61)
So that the left hand side of equation (1.59) (using f i (d / dx) and x can be
written as
d
4 i
x dx
dx
2
4 i
d
d
i
dx x x dx.
dx
dx
Therefore equation (1.59) itself can be written as
4 i
d
d
2
2
i
dx x dx .
dx
dx
or
d
i dx
or
p2 x2
2
2
x dx
2
.
4
2
.
4
(1.62)
Since
x 2 x 2 dx
and
d
p i
dx.
dx
2
2
Applying conditions expressed by (1.58) to equations (1.54) and (1.56), we get 20
and
(x)
2
,(p)
2
2
x
2
p ,
(1.63)
Using (1.63), equation (1.62) becomes
(x) (p)
2
xp
or
2
2
4
2
(1.64)
(a) Commutation Relation between Position and Momentum :
(i)
Let us find the commutation relation between x and p x .
In operator formalism
x xˆ and p x pˆ
i x
Consider the operation of x, p x on a function (x, )
xˆ pˆ x ( x) ( xˆ pˆ x
i.e.,
We have
and
xˆ p x
pˆ x xˆ
pˆ x xˆ ) ( x)
(1.65)
i x
( x ) x
i x
i
x
Equation (1.65) gives
xˆ
pˆ x x
i.e.
x
i x i
x
x
x
i
i x
x
i
xˆ pˆ x i
(1.66) 21
UNIT II
2.1 Linear Harmonic oscillator
In the case of a liner harmonic oscillator the force F = - K x can be represented by the
potential energy function V ( x) 1 / 2Kx 2
From the Schroedinger equation,
d 2 2m
1
2 E K x 2 0.
2
dx
2
(2.1)
Introducing a variable = α x, where α is constant, we can write,
d d d d
dx d dx d
d 2 d d d d
. .
dx 2 dx d d dx
and
d 2 d 2
.
d 2
d
Therefore eq. (2.1) becomes
d
2
or
2
d
2
2mE mK 2
0.
2 2
2
d 2 2mE mK 2
d 2 2 2 4 2
Let us choose α such that
(2.2)
0.
mK
1
2 4
(2.3)
and
2mE
2 2
Therefore equation (2.3) assumes the dimensionless form
d 2
( 2 ) 0.
d 2
(2.4)
Let the solution is of the form
( ) H ( )e
2
2
(2.5)
where H() is a polynomial of finite order in . If we consider the positive sign in the
exponent, then will diverge as . From eqn. (2.5)
2
2
d
H ' ( ) exp
H
(
)
exp
d
2
2
and
2
2
d 2
''
H
(
)
exp
H
(
)
exp
2
d 2
2
2
2
2
2
H ( ) exp
H
(
)
exp
H
(
)
exp
2
2
2
22
2
2
2
2
H ( ) exp
2
H
(
)
exp
(
1
)
H
(
)
exp
2
2
2
2
Substituting the value of () and d in equation (2.4) we get
d 2
2
2
2
exp
H ( ) 2H ( ) H ( ) H ( ) ( ) H ( ) 0
2
H ( ) 2H ( ) ( 1) H ( ) 0
or
(2.6)
Energy Levels.
Using power series, we can solve equation (2.6)
Let
H ( ) s (a0 a1 a2 2 ....),
=
s
a
v 0
On differentiation,
v
v
a
v 0
v
a0 0, s 0
s v
dH
av ( s v) sv1
d v
d 2H
av ( s v) ( s v 1) sv2
2
d
v
and
Substituting these values in equation (2.6), we get,
a
v
( s v) ( s v 1) s v 2 2 av ( s v) s v 1 ( 1) av s v 0
v
or
v
av (s v) (s v 1)
s v2
v
2 av (s v)
v
s v
( 1) av
v
v
s v
0
(2.7)
For H() to be a solution for all values of ξ, the coefficient of the individual powers of
must vanish separately, i.e., equating to zero the coefficients of various powers of we get
s(s-1) a0 = 0
(s + 1) (s) a 1 = 0
(s+2) (s+1) a2 – (2s + 1- ) a 0 = 0
(s+3) (s+2) a3 – (2s + 3 - ) a 1 = 0
... ... ... .... .. .. ... .... .... ... ... .. .. ..
... ... ... ... ... ... ... ... ... .... ... ... ....
(s+ν + 2) (s + ν + 1) av + 2 – (2s + 2 ν +1 - ) av = 0
From these expressions we can write,
av+2 =
(2.9)
2s 2v 1
av
( s v 1) ( s v 2)
(2.8) 23
where v is an integer. Since a0 cannot be equal to zero, from the first of equations (2.8), s = 0
or 1. If a0 is equal to zero, only odd powers appear. With a1 equals to zero, the series
contains even powers only. If we examine the convergence of the power series solution
defined by equation (2.9), we find that as ν , (av+2)/av 2/v, so that the series converges
for all finite values of .
Considering the series expansion of e , we have
2
2
e
1
2
=
4
2!
6
3!
...
v
v2
(v / 2)! (v / 2 1)!
...
= b0 + b2 2 + b4 4 + bv v + bv + 2v+2 + ...
1
bv2 (v / 2 1 ) !
2
1
bv
2 v
(v / 2 ) !
Lim
v
or
bv2 2
bv
v
This shows that H() diverges approximately as e2 and the product H ( )e
2 / 2
like e
in this region which tends to
,
2
2
will behave
as .So this is not an acceptable
solution. This situation can be avoided by coosing in such a way that the power series for
H() cuts off at some term, making H () a polynomial.
From eqn (2.9), by making
= 2s + 2v + 1.
we can make the series cut off. The index s can still be either 0 or 1, and corresponding to
these values, is equal to 2v + 1 or 2v+3 where 2v is an even integer. Or,
= (2n+1).
m
( 2n 1)
K
Then
2 En
or
1
E n n c .
2
n = 0,1,2...
where c =
angular
n=0,1,2...
E
9/2 ωc
...(2.9A)
K
is the classical
m
frequency
of
7/2 ωc
the
5/2 ωc
oscillator.
3/2 ωc
½ ωc
Fig. 2.5 24
Zero point energy. The energy levels given by equations (2.9A) are discrete and have equal
spacings. When n=0, the finite value of lowest energy or ground state energy is (1/2) ћc.
Hence zero point energy is given by,
E0
1
c
2
and all higher energy levels are separated by an amount equal to ћc. Zero point energy is
characteristic of quantum mechanics and is related to the uncertainty principle.
2.2 Infinitely deep potential well or particle in one-dimensional box
Consider a square potential well with infinitely high sides, as indicated in figure (2.1). The
particle is bounded by impenetrable rigid walls of width 2a as shown.The potential of the
well is represented by,
V(x) = 0
for –a a ,
C=0
and for x<-a
D= 0
Thus we have wave functions in the three regions as,
1 = A cos α x +B sin α x
-a < x < a,
(2.19)
2 = C e x
x<-a
(2.20)
3 = D e- x
x >a
(2.21)
Considering the equations (2.19) and (2.21) and applying
(1)
xa
the boundary condition
( 3) xa
We have
A cos a + B sin a = D e - a
(2.22)
and applying the other boundary condition
d 3
d 1
dx x a dx x a
-A sin a + B α cos a = -D e-
a
(2.23)
Again at x = – a and by applying (1) x = -a = (2) x = -a we have
A cos a - B sin a = C e - a
(2.24)
d
d
And by applying 1
we have,
2
dx x a dx xa
A sin a B cos a Ce
Adding equations (2.22) and (2.24) we have,
a
(2.24A) 28
2 A cos a = (C + D) e - a
(2.25)
And subtracting eqns. (2.22) and (2.24) we have
2 B sin a = (D - C) e - a
(2.26)
Similarly from equation (2.23) and (2.24A) we have,
2 B cos a = (C - D) e - a
(2.27)
2 A sin a = (C + D) e - a
(2.28)
Dividing eqn. (2.28) by (2.25) gives
tan a = ,
(2.29)
and from eqns. (2.27) and (2.26)
cot a = - ,
(2.30)
A solution of the equations (2.29) and (2.30) is obtained graphically.
Now consider equation (2.29) and let = a and η= a , then
tan =
2mVa 2
and 2 + 2 = 2 a2 + 2 a2 = a2 [α2 + 2] = 2 = constant for a given system.
η
η
η=ξtanξ
η=-ξcotξ
c
c
b
b
a
a
ξ
ξ
Fig. 2.4
Since and are positive values, the energy levels are found from the intersection in the
1/ 2
2mVa 2
2
first quadrant of the curve of tan against with the circle of known radius 29
Curves for three values of Va2 are shown in figure (2.4).
The product Va2 is a measure of the strength of the potential well – which binds the particle
to the well. In other words, the greater the depth (V) and breath (a) of a well, the greater the
number of bound states and greater the probability of retaining a particle in the well.
The first three increasing values of Va2 give altogether one, two and three energy levels,
respectively. As a special case let V approaches infinity. For the first group of solutions,
tan a = /
a = (2n + 1) /2
The characteristic energy values are
2 2 2
E 2n 1
n
8m a2
with n = 0,1,2 ...
(2.31)
For the second group of solutions,
tan a = -
0.
a n
or
The characteristic energy values are
2 2 2
E 2n
n
8m a2
n = 0, 1, 2...
(2.31A)
2.4 Rectangular potential barrier
Let us consider the one-dimensional problem where the potential is defined as in fig.2.5.
V(x)=V0
V ( x) 0 for x 0
V0 for 0 x a
0 for x a
V0
(2.31)
I
Here we have a potential barrier
between
x 0 and
x a.
If
a
particle having energy less than
E
II
III
V(x)=0
V(x)=0
X=a
X=0
Fig. 2.5 30
st
V0 , i.e., E V0 , approaches this barrier from the left, i.e., from 1 region, classically the
particle will always be reflected and hence will not penetrate the barrier. However, wave
mechanics predicts that the particle has some probability of penetrating to region 3rd, the
probability of penetration being greater if E V0 classical mechanics predicts that the
particle will always be transmitted ; while according to wave-mechanics, the particle has a
finite probability of transmission and hence it is not certain that the particle will penetrate the
barrier.
The Schroedinger equation for region-I is
1 2m
2 E 1 0 (since V 0) .
2
x
2
(2.32)
The Schroedinger equation for II region is
2 2 2m
2 ( E V0 ) 2 0.
x 2
(2.33)
The Schroedinger equation for III region is
2 3 2m
2 E 3 0.
x 2
(2.34)
Here 1 , 2 and 3 are the wave-functions for I, II and III regions respectively.
The general solutions of equations (2.32) (2.33) and (2.34) may be written as
1 A1eip x / B1eip x /
(2.35)
2 A2eip
(2.36)
1
1
2x
B2eip 2 x /
/
3 A3eip x / B3eip x /
1
1
(2.37)
where p 1 and p 2 ,are the momenta of particle in the corresponding regions, which are given
by
p 2 2m(E - v 0 )
P1 2mE
(2.38) 31
A1 , B1 , A2 , B2 and B3 are constants to be determined by boundary conditions.
In equation (2.35) the first term represents the wave travelling along +ve
X -axis
in the I
region, i.e., the incident wave and second term represents the wave travelling along negative
X -axis i.e., wave reflected at x 0.
In equation (2.36), the first term represents the wave travelling along (+) ve X -axis in the II
region, i.e., the wave transmitted at x 0 and second term represents the wave travelling
along (-) ve X -axis in the II region, i.e., the wave reflected at x a.
In equation (2.37) the first term represents the wave travelling along (+)ve X -axis in the III
region, i.e., the wave transmitted x a and the second term represents the wave travelling
along (-)ve X -axis in the III region; but no wave travels back from infinity in III region.
Consequently B3 0, so that the solution of equation (2.34), i.e., equation (2.37) can be
written as
3 A3eip x /
1
(2.39)
For the evaluation the constants A1 , B1 , A2 , B2 and A3 we shall apply the conditions at the
two boundaries x 0 and x a.
One conditions is that must be continuous at the boundaries, i.e.,
1 2 at x 0 .........( A )
) 2 3 at x a .........( B
(2.40)
The other condition is / x must be continuous at the boundaries i.e.,
1 2
x x at x 0 .........( A)
3
2
at x a .........( B)
x
x
(2.41)
Applying boundary condition (2.40A) to equations (2.35) and (2.36), we have
A1 B1 A2 B2
Applying boundary condition (2.40B) to equations (2.36) and (2.39), we get
(2.42) 32
A2e
ip 2 a /
B2e
ip 2 a /
A 3e
ip1a /
(2.43)
Differentiating equations (2.35), (2.36) and (2.39), we get
1
ip
1 A1eip1 x / B1e ip1 x /
x
2
ip
2 A2eip2 x / B2eip2 x /
x
(2.44)
(2.45)
3
ip
1 A3eip1 x /
x
(2.46)
Applying boundary conditions (2.41) to these equations, we get
p1A1 B1 p2 A2 B2
p2 A2e
and
ip2 a /
B2e
A1 B1
or
A e
2
ip2a /
p A e
1
3 ip1a /
p2
A2 B2
p1
ip2 a /
B2 e ip2a /
(2.47)
p1
A3 e ip1a /
p2
(2.48)
Solving (2.42) and (2.47) for A1 and B1 , we get
A1
A2
2
p B
p
1 2 2 1 2
p1 2
p1
(2.49)
B1
A2
2
p B
p
1 2 2 1 2
p1 2
p1
(2.50)
Solving (2.43) and (2.48) for A2 and B2 , we get
A2
A3
p i( p p )a /
1 1 e 1 2
2 p2
(2.51)
B2
A3
p i( p p )a /
1 1 e 1 2
2 p2
(2.52)
Substituting values of A2 and B2 from these equations in (2.49) and (2.50), we get 33
A1
A3 ip1a / p2 p1 ip2a / p2 p1 p2a /
e
1 1 e
1 1 e
4
p
p
1
2
p1 p2
(2.53A)
B1
A3 ip1a / p2 p1 ip2a / p2 p1 p2a /
e
1 1 e
1 1 e
4
p
p
1
2
p1 p2
(2.53B)
Equation (2.53A) may be written as
ip a /
A3
4e 1
A1 p2 p1 ip2a / p2 p1 ip a /
1 1 e 2
1 1 e
p1 p2
p1 p2
4 p1 p2 e ip1a /
( p1 p 2 ) 2 e ip2a / ( p1 p 2 ) 2 e ip2a /
4 p1 p2 e ip1a /
2
2
( p p )(e
1
2
2
(p
1
2 e
p )(
2
ip2a /
e
ip2a /
) 2 p1 p2 (e
ip2a /
e
ip 2 a /
)
2 p1 p2 e ip1a /
e ip2a /
(e ip2a / e ip2a / )
) 2 p1 p2
2
2
ip2a /
Dividing equation (2.53B) by (2.53A), we get,
B1
A1
p2
p 1 ip a /
p2
p 1 ip 2 a /
1
1
e 2
1
1
e
p
p
p
p
1
2
1
2
p2
p 1 ip a /
p2
p 1 ip 2 a /
1
1
e 2
1
1
e
p
p
p
p
1
2
1
2
e
( p1 p2 ) (e
2
ip 2a /
Now we consider the following two cases :
ip 2 a /
ip 2 a /
e
e
ip 2 a /
( p1 p2 )
( p12 p22 )(e
ip 2 a /
( p p22 )(e
2
1
ip 2 a /
2
2
(p p )(
1
2
2
ip 2 a /
e
) 2 p1 p2 (e
e
)
ip 2 a /
)
ip 2 a /
ip 2 a /
e
) 34
Case (i) E V0 ; in this case p1 2mE, p2 2m ( E V0 ), both are real so
A3
A1
4 p1 p2e ip1a /
p a
pa
( p 2 p 2) 2i sin 2 2 p1 p2 2 cos 2
1
2
2 p1 p2
2
2
pa
pa
i ( p p ) sin 2 2 p1 p2 cos 2
1
2
A3
A1
e ip1 a /
ip a/
2 p1 p2 e 1
2
2
pa
pa
i( p p ) sin 2 2 p1 p2 cos 2
1
2
Transmission coefficient T
Transmitte d flux
Incident flux
A3 A3 p1 / m A3 A3
A1 A1 p1 / m A1 A1
4p2 p2
1
2
pa
pa
( p 2 p 2) 2 sin 2 2 4 p 2 p 2 cos 2 2
1
2
1 2
2 2
4p p
1 2
2
2 2
2 2
pa
p a
( p p ) sin 2 2 4 p p 1 sin 2 2
1
2
1 2
4 p 12 p 22
2
( p 12 p )
2
2
2
sin
p2 a
4 p 12 p 22
35
Reflection Coefficient
2
2
p a
( p p ) 2i sin 2
B1
1
2
2
2
p a
p a
A1
( p p ) 2i sin 2 2 p1 p2 2 cos 2
1
2
pa
i ( p 2 p 2) sin 2
1
2
p
a
pa
2
2
2
i ( p p ) sin
2 p1 p2 cos 2
1
2
2
2
pa
i ( p p ) sin 2
B1
1
2
2
2
pa
pa
A1
i ( p p ) sin 2 2 p1 p2 cos 2
1
2
p1
BB
Reflected flux
m 1 1
R
p
A1 A1
Incident flux
A1 A1 1
m
B1 B1
or
2
2
2 p a
( p p ) sin 2
B B
1
2
R 1 1
2
2
2
2 2
p
a
2
2 p a
A1 A1
2
( p p ) sin
4 p p cos 2
1
2
1 2
pa
( p 2 p 2) 2 sin 2 2
1
2
p
a
p a
( p 2 p 2) 2 sin 2 2 4 p 2 p 2 1 sin 2 2
1
2
1 2
pa
( p 2 p 2) 2 sin 2 2
1
2
2
2
2
2 p2 a
( p p ) sin
4p2 p2
1
2
1 2
(2.54)
Case (ii), E V0 , then p1 2mE, is real and p2 2m ( E V0 ), is imaginary, therefore
ip2 is real.
The transmittance or the transmission coefficient is given by 36
2 2
4 p p sech 2(ip a / )
2
1 2
T
2
2 2 2
2 2
( p p ) tanh (ip a / )4 p p
2
1 2
1 2
(2.55)
Incident wave
Transmitted wave
V0
E
X=0
X=a
Fig. 2.7
And reflection coefficient is given by,
R
( p12 p22 ) tanh 2(ip2a / )
2
( p12 p22) 2 tanh (ip2a / )4 p12 p22
(2.56)
2.5 The Rigid rotatOr with free axis
The system, consisting of two spherical particles attached together, separated by finite fixed
distance and capable of rotating about an axis passing through the centre of mass and normal
to the plane containing the two particles, constitutes, a rigid rotator. If these two particles are
constrained, to remain in one plane, then the direction of the axis of rotation is fixed and so
the system is called the rigid rotator with fixed axis. If the plane of these two particles can
move, then the axis of rotation is free to take any position in space and so the system is called
the rigid rotator with free axis. In a diatomic molecule the atoms vibrate with respect to each
other and so the distance between atoms will not be always constant ; while the distance apart
of the equilibrium position is constant. Thus the system of diatomic molecules is not really 37
Axis of rotation
r1
r2
m1
m2
r0
Fig. 2.7
rigid; how ever; it may be treated, at least as a first approximation, as a rigid rotator with free
axis.
Energy for the rotator
The kinetic energy of a particle of mass m can be expressed as
T
1
m ( x
2
2
y z )
2
2
(2.57)
where x, y , z are the components of the velocity of a particle along X , Y and Z axes
respectively. The transformation between Cartesian co-ordinates ( x, y, z ) and spherical coordinates (r, θ, φ) are given by
x r sin cos
y r sin sin
(2.58)
z r cos
so that the kinetic energy in spherical co-ordinates is expressed as
1
2
2
T m ( r 2 r 2 r 2 sin 2 )
2
(2.59)
If the distance r of the particle from the origin is fixed, its derivative r will be zero ; then
from equation (2.59) the kinetic energy would be
T
1
mr 2 ( 2 sin 2 2 )
2
(2.60) 38
Taking O, the centre of mass of the rotator, as origin, the K.E. of the particle of mass m1 is
given by
T1
1
m1r 12 ( 2 sin 2 2 )
2
Similarly the K.E of the particle of mass m is
2
T2
1
m2 r 2 ( 2 sin 2 2 )
2
2
Hence the total kinetic energy of the rotator will be
T T1 T2
1
1
m1r 2 ( 2 sin 2 2 ) m2 r 2 ( 2 sin 2 2 )
1
2
2
2
1
1
m1r 12 m2 r 22
2
2
2
( sin 2 2 )
(2.61)
As there is no potential energy of the rotator, total energy is given by
E T V T
(since V 0)
1
1
m1r 12 m2 r 22
2
2
2
( sin 2 2 ) (2.62)
But m1r 2 1 m2 r 2 I , the moment of interia of the system about the axis passing through the
1
2
2
centre of mass and perpendicular to the line joining the two masses.
E T
1
I ( 2 sin 2 2 )
2
(2.63)
The moment of inertia of the rotator may be expressed in a more convenient form as follows :
According to definition of centre of mass, rcm
0
But
m1r1 m2 r2
, we have
m1 m2
m1r1 m2 r2
ie. m1r1 m2 r2
m1 m2
r0 r1 r2
or
Substituting this in eqn. (2.64) we get
r2 r0 r1
(2.64) 39
m1r1 m2 r0 r1
r1
m2
r0
m1 m2
(2.65)
m1
r0
m1 m2
(2.66)
r2
Similarly
Then the moment of intertia of the rotator may be expressed as
2
2
m
m2
2 1
2
1
I m1r1 m2 r2 m
r
m2
r
m1 m2 0
1 m1 m2 0
2
m1m2 2
r
m1 m2 0
2
since, I r 0
(2.67)
where m1m2
(2.68)
m1 m2
is called the reduced mass of the system.
From equations (2.60) and (2.63) it is evident that the rigid rotator behaves like a single
particle of mass given by eqn. (2.68) placed at a fixed distance, equal to unity (since
r 1) from the origin, which in this case is the centre of mass of the system.
Wave equation for the rotator : The Schroedinger wave equation in three dimensions in
spherical co-ordinates is given by
1 2
1
1
2 2m
( E V ) 0
r
2
sin
2 2
2
r r r r sin
r sin 2 2
(2.69)
For a rigid rotator we have seen that potential energy is zero. r 1 and the mass
m may be
replaced by the moment of inertia I . Therefore the Schroedinger wave equation for a rigid
rotator becomes
1
1 2 2 I
E 0.
sin
sin
sin 2 2
(2.70)
This equation consists of two variables and which represent the precessional motion of
the rotator’s free axis and the rotation of the system respectively. 40
Solution of wave equation : Eigen function for the Rotator
Equation (2.70) may be solved by the method of separation of variables, i.e., the wavefunction ( , ) may be represented by
( , ) ( ) ( )
(2.71)
Where ( ) is function alone and ( ) is the function of alone.
Substituting in equation (2.70) and dividing throughout by , we get
1
1
2 2 IE
2 2 0.
sin
2
sin
sin
(2.72)
Multiplying this equation by sin 2 we get
2
sin
1 2 IE sin 2
0
sin
2 2
or
sin
2 IE 2
1 2
sin
2 sin
2
(2.73)
In this equation L.H.S is a function of alone, while R.H.S is a function of alone.
Therefore if this equation is to be satisfied, both sides must be equal to the same constant, m 2
(say) i.e.,
sin
2 IE 2
sin m 2
sin
2
and
(2.74)
1 2
m 2
2
Equation (2.74) may be rewritten as
2
m2 0
2
(2.75)
The solution of above equation may be written as
Aeim
where
m 0, 1, 2, 3 etc. (2.76) 41
A is any arbitrary constant which may be chosen in such a way that the function is
normalised,
i.e.,
2
0
m m d 1
i.e.
A2
i.e.
A
2
0
or
d 1,
2
0
Ae im Ae im d 1,
A2 2 1
or
(2.77)
1
(2 )
Thus the normalised function is
im
1 e
m
(2 )
Multiplying equation (2.74) by
, we get
sin 2
1
sin
sin
Let us introduce a new variable
2 IE
m2
2
sin 2
x such that
0.
(2.78)
x cos
sin (1 x 2 )
so that
(2.79)
x
sin
x
x
Then
sin
x
and in general,
sin
(2.80)
(2.81)
sin 2
(1 x 2 )
x
x
(2.82)
Using equation (2.80) and (2.81) and (2.82), equation (2.78) can be written in terms of
x
m2
2
(
1
x
)
x
1 x2
where
0.
x as
(2.83)
2 IE
2
(2.84)
Equation (2.83) is known as Legendre’s equation. It has physical significance only for values
of
x between the limits of -1 and +1 since x is equal to
cos . 42
( ) (1 x 2 ) m / 2 X ( x)
Where X (x) is the function of
(2.85)
x only.
Equation (2.85) yields
(m / 2) 1
2 (m / 2) dX
mx (1 x 2 )
X (1 x )
x
dx
(2.86)
(1 x 2 ) m / 2 X (1 x 2 ) (m / 2) 1 dX
2
(1 x ) mx
x
x x
dx
m (1 x 2 ) m / 2 m2 x 2 (1 x 2 ) (m / 2)1 X
2 x(m 1) (1 x 2 ) m / 2 X (1 x 2 ) ( m / 2)1 X
X
where we have assumed
d2 X
dX
and X
dx
d x2
(2.86a)
(2.87)
Using equations (2.85) and (2.86a), equation (2.83) can be written as
2 m/2
m2 x 2
m (1 x )
X {2 x (m 1) (1 x 2 ) m / 2 X
m2
(1 x 2 ) m / 2 X 0.
(1 x 2 ) (m / 2) 1 X
2
1x
(1 x 2 )
(m / 2) 1
Dividing throughout by (1 x 2 ) m / 2 , we get
(1 x 2 ) X 2(m 1) xX m(m 1) X 0
or
(1 x 2 ) X 2xX X 0
(2.88)
where m
1
and
m (m 1)
(2.89)
Now let us assume that X (x) may be expressed as a power series, as,
X a0 a1 x 2 x 2 a3 x3 ..................
or
X a1 2 a2 x 3a3 x .......... ........
2
(2.90) 43
X 2 a2 6a3 x 12 a4 x 2 .......
and
(2.91)
substituting these values in equation (2.88) and simplifying, we get
(2 a a ) 6a ( 2 )a 4 x 12a ( 4 2)a x2 20a 4 6a x3 ....
2
0
3
4
2
5
3
n
(n 1)(n 2)an2 2n n(n 1)an x .... 0
In order that the series may be zero for all possible values of x , the coefficients of individual
powers of
x must vanish separately, i.e., in general
(n 1)(n 2)an2 2n n(n 1)an 0
Where n 0,1, 2, 3,...
an 2
Substituting values of
2n n(n 1)
an
(n 1)(n 2)
and in above equation, we get
an2 (n m)(n m 1)
an
(n 1)(n 2)
(2.92)
This is called recursion formula for the coefficients in power series for X (x).
In order to obain a satisfactory wave function , it is necessary that X (x) should be a
polynomial breaking off after a finite number of terms, as in the case of harmonic oscillator.
The series will break after nth term if the numerator of equation (2.92) is zero
(n m)(n m 1) 0
i.e.
(n m)(n m 1)
(2.93)
Therefore, the sum (n m) may be replaced by l , where l is also zero or an integer.
l (l 1),
l m n 0,1,2,3,...
(2.94)
Substituting this value of in equation (2.83), we get
m2
2
(1 x )
l (l 1)
0.
x
x
1 x2
(2.95)
The solution of above equation contains the factor called the associated Legendre function
Pi m (x) which may be defined as 44
Pi m ( x) (1 x 2 ) m / 2
d m Pl ( x)
dx m
where Pl (x) is Legendre polynomial of degree l.
The solution of equation (2.95) is written as
BPl m ( x) BPl m (cos )
(2.96)
[since x cos ]
Where B is normalizing constant.
From orthogonal properties of associated Legendre’s Polynomials,
1
1
2 (l m)!
m
m
P
(
x
)
P
(
x
)
dx
1 k l
(2l 1) (l m)!
1
and
Pkm ( x) Pl m ( x)dx 0
for k l
fork l
(2.97)
According to normalizing condition,
ml
ml * d 1
1
B 2 Pl m ( x) Pl m dx 1
i.e.,
1
B2
or
2 (l m)!
1
2l 1 (l m)!
(2l 1) (l m)!
B
2 (l m)!
(2.98)
Substituting the value of B in (2.96) the normalized wave function is given by
(2l 1) (l m)! m
Pl (cos )
( )
2
(
l
m
)!
(2.99)
The complete wave-function or eigen-function for the rigid rotator is given by
(2l 1) (l m)! m
1 im
Pl (cos )
( )( )
e
2
(
l
m
)!
2
(2.100) 45
or
1
(2 )
ml ( , )
(2l 1) (l m)! m
Pl (cos )e im
2 (l m)!
(2.101)
Eigen values or energy levels of the rigid rotator.
From equations (2.84) and (2.94), we have
2 IE
l (l 1)
2
E El
or
l (l 1) 2
2l
l 0,1, 2, 3....
(2.102)
This equation gives allowed values for the energy (i.e. eigen values) of a rigid rotator with
free axis.
2.6 Rigid Rotator in a Fixed Plane
If we consider the rotator to be only in XY plane, then 90 and hence the
Schroedinger’s equation, in this case may be written as
2 2 IE
0
2
2
(2.103)
In this case m ( ), so that,
1 2 2 IE
constant m 2
2
2
So that we have
1 2
m2 0
2
2
m2 0
2
i.e.
(2.104)
where
2 IE
2
m
2
Eigen functions: The solution of equation (2.104) can be written as
(2.105) 46
m Ae im
(2.106)
where A is arbitrary constant and m 0, 1, 2, ....
According to normalisation condition
2
0
or
2
0
A2 2 1
or
m m d 1
Ae im Ae im d 1,
or
A
1
(2 )
The eigen functions are given by
m ( ) Ae im
1
e im
2
(2.107)
m 0, 1, 2, .... .
Eigen values : From equation (2.105), we have
E Em
m2 2
2I
2.7 The Hydrogen atom
P
The Schroedinger wave equation is
written as
r
2
2
E V 0.
2
θ
z
y
In spherical polar coordinates (fig 2.8)
φ
x = r sin cos
x
y = r sin sin
y
z = r cos
x
The expression for 2 in spherical
Fig. 2.8 47
polar coordinates is expressed as,
2
1 2
1
1
2
r
2
sin
2 2
2
r r r r sin
r sin 2
Hence the Schroedinger wave equation in a spherically symmetric potential may be written in
the spherical coordinates as
1
r2
r 2 r r
1
1
1 2 2
E V (r ) 0.
sin
r 2 sin 2 2 2
r 2 sin
The solution of the above equation can be separated in different variables of r, and by
writing.
(r,,) = R(r) () ()
and differentiating, dividing throughout by = R and multiplying by r2 sin 2 we get,
sin 2 d 2 dR 1 d 2 sin d
d 2 r 2 sin 2
E V (r ) 0.
r
sin
R dr dr d 2
d
d
2
(2.107A)
The second term of this equation depends only on and the rest is independent of . Hence
the second term should be equal to a constant . Let the constant is equal to -m2
That is,
1 d 2
m2
2
d
d 2
m 2 =0.
2
d
(2.108)
With this value for the second term, equation (2.107a) now can be written as,
1 d 2 dR 1 1 d
d m 2 2 r 2
2 E V (r ) 0.
r
sin
R dr dr sin d
d sin 2
(2.109)
The I and IV terms of this equation depend only on r and the II and III terms only on ;
therefore each part must be equal to a constant, (say) λ.
m2
1 d 2 dR 2 r 2
1
d
d
(E V )
r
sin
.
R dr dr
d
2
sin 2 sin d
so that we have
1 d
d
sin
sin d
d
and
m2
sin 2
0.
1 d 2 dR 2
r
2 {E V (r )} 2
2
r dr dr
r
(2.110)
R 0.
(2.111) 48
Equation (2.108), (2.110) and (2.111) are known as , and r equations respectively.
(b) Solution of equation :
d 2
m 2 0
2
d
The equation is
This is a second order differential equation whose solution is given by
= A e i m ,
(2.112)
where A is an arbitrary constant. The constant may be evaluated by normalizing , i.e.,
2
2
d 1
0
0
A2 d 1
or
A
1
2
(2.113)
therefore the solution becomes
1
2
e i m
(2.114)
The single valuedness of the function indicates that it should have the same value at = 0
and = 2; hence
A A e 2 i m
e
2 i m
= cos 2 m i sin 2 m = 1
or
(2.115)
This is true only when m is zero or an integer (positive or negative). Thus we write
1
2
e im ,
m = 0, 1, 2, .......
The quantity m is called the magnetic quantum number .
Solution for equation :
1 d
d
m2
sin
2 0.
In the θ-equation
sin d
d
sin
Let
x = cos ,
we have
d
dx d
d
sin
d
d dx
dx
(2.116) 49
d
d
sin
d
dx
or
(2.117)
d
d
sin
sin 2
d
dx
sin
or
d
d
(1 x 2 )
d
dx
(2.118)
Using equations (2.117) and (2.118), the equation becomes
m 2
d
d
2
0.
(1 x )
2
dx
dx
x
1
If = l (l + 1), then
(2.119)
2
d
2 ) d l (l 1) m 0.
x
(
1
dx
dx
1 x 2
(2.120)
Comparing this with associated Legendre equation
d
d
m2
m
2
pl m ( x) 0.
(
1
x
)
p
(
x
)
l
(
l
1
)
l
2
dx
dx
1
x
gives the solution of equation (20) as
m
m
= B pl (x) = B pl (cos )
| m | < l.
(2.121)
B is normalization constant which is evaluated by normalising in the following way.
0
sin d 1
(2l 1) (l m)!
2 (l m)!
1
B P
2
l
m*
( x) Pl m ( x)dx 1
1
Hence
B=
(2.122)
Thus the solution is
( ) =
(2l 1) (l m)!
m
pl (cos ),
2 (l m)!
pl ( x) (1 x 2 ) m / 2
m
with
1/ l
and
Pl ( x) 2
d
dm
pl ( x)
dx m
l
dx
l
2
( x 1) l
where x = cos
Solution of r-equation (radial equation )
(2.123) 50
In this problem V(r) = -
Ze2
, where Z is the atomic number. Consider this problem to one –
r
electron or hydrogen – like atom.
Equation for radial part can be written as
1 d 2 dR 2
Ze2 l (l 1) 2
r
E
R 0.
r 2 dr dr h 2
r
2r 2
Let
= αr
or r = /α ;
Then r-equation becomes
1
d 2 dR 2
2 d 2 d 2
For bound states, ie. E < 0,
2
2
Z e2 2 l (l 1) 2
E
R 0.
2 2
(2.124)
1 d 2 dR 2
Ze2 2l (l 1) 2
|
E
|
R 0.
2 d d 2
2 2
or
Choosing α such that
1 d 2 dR
2 d d
2 | E | 1
,
2 2
4
2 | E | 2 Ze2 l (l 1)
2 2 2 2 R 0.
and
'
(2.125)
2 Ze2 a
h 2
The equation (2.125) can be written as
1 d 2 dR ' 1 l (l 1)
R 0.
d 4
2 d
2
where the particular choice of the number (1/4) for the eigen value term
For large , the equation (2.126) has the form
(2.126)
2 E
is arbitrary .
2 2
1 d 2 dR 1
R 0.
2 d d 4
(2.127)
The solutions of equation (2.127) are
R =e + /2 and R =e - /2
(2.128)
Only the second solution is satisfactory (vanishing for ). The exact solution of
equation (2.126) is of the form
R() = F()e - /2
Substituting this value in equation (2.127), we have
2
/ 2
1 2
2
F ' ' ( ) (2 ) F ' ( ) ' l (l 1 F ( ) e 0.
4
4
(2.129) 51
As e-ρ/2 is not zero, the expression within the square brackets is zero :
2 F ' ' ( ) (2 ) F ' ( ) ( ' 1) l (l 1)
F ( ) 0.
(2.130)
F() = s L()
Let
F() = ss-1 L + s L
F() = s (s-1) s-2 L+2 ss-1 L + s L
Substituting these values in equation (2.130),
2 L' ' (2s 2 2 ) L' s(s 1) 2s s (( ' 1) l (l 1) L 0.
(2.131)
Let
L()
=
a
r 0
i.e.,
F()
=
r
r
a
r 0
sr
r
(2.132)
Substituting the values of F(), F(), and F() in eq. (2.130) we have
2 ar ( s r ) ( s r 1) s r 2 (2 ) ar ( s r ) s r 1
r 0
( ' 1) l (lr 0 1) a r s r 0
r 0
or
a (s r ) (s r 1)
r 0
r
s r
2 a r ( s r ) sr
r 0
sr
a ( s r ) s r 1 ( ' 1) a s r 1 l (l 1) a
0
r
r
r
r 0
r 0
r 0
sr
( s r )( s r 1) 2( s r ) l (l 1) ar
or
(2.132a)
r 0
s r 1
( s r ) ( '1) a
0.
r
r0
Equating the coefficients of s in eq. (2.132a), we have
s(s l) 2s l(l 1)a0 0
s(s l) l(l 1) 0
sin ce, a0 0
It means that either s = l or s = -(l+1); the value s=-(l+1) does not satisfy the condition of
well – behaved function and so, the only accepted value of s is l. Comparing now the
coefficients of s+r+1 we can write,
(s r 1)( s r ) 2(s r 1) l (l 1)ar 1 (s r 1 ' )ar
or
a r 1
s r 1 '
a r ( s r 1) ( s r 2) l (l 1)
(2.133) 52
ρ
The series (2.132) represented by coefficients (2.133) behaves like e for large and
therefore R = e-/2 F() will diverge as e+/2, which is not an acceptable solution.. Therefore
the series must break off after certain number of terms. From eq (2.133) let,
l + nr + 1 - = 0
(putting s =l)
or
= nr + l + 1.
(2.134)
Replacing ’ by integer n,
n = nr + l + 1.
(2.135)
It is seen that n is a positive integer and is identified as total quantum number. Eq. (2.131)
may now be written as (putting s = l and replacing by n):
L' ' (2l 1) 1 L' (n l 1) L 0
(2.136)
Comparing it with the associated Laguerre equation
p
p
p
L' ' q ( p 1 ) L' q (q p) L q 0,
We obtain, p = 2l + 1, q = n+l and so we get the solution of r-equation as
R(r ) Ce / 2 l L 2nll1 ( );
Here, C is the normalisation constant given by,
C
(n l 1)!
3
( ) .
3
2n(n l )!
(e) Energy of atomic levels and degeneracy.
We have assumed that
Ze2
'
or
' 2
2| E |
1/ 2
Z 2e 4
2 2| E |
Replacing ’ by n
En | En |
Z 2e4
(2.137)
2 2 n 2
Equation (2.137) is the expression for energy of an atomic state of a hydrogen – like atom
defined by the principal quantum number n.
Now consider the equation
n = nr + l + 1
can be satisfied for a given n for several combinations of nr and l. This implies that there are
several possible wave functions for a given energy value (n fixed). When this happens the
state is said to be degenerate. This holds good for every value of n > 1. 53
To determine the degeneracy
From equation (2.137) it is clear that energy eigen –values depend upon n and so are
degenerate with respect to both l and m. Thus for each value of n, l can vary from 0 to n – 1
and for each of these l values, m can vary from – l to +l. So, the total degeneracy is
n 1
(2l 1)
l 0
2(n)(n 1)
n n2
2
Non Coulomb field
(2.138)
Non central
field
m=+2
m=+1
m=0
m=-1
m=-2
central field
l=2
m=+1
Coulomb field
l=1
m=0
n=3
m=-1
l=0
m=+1
l=1
m=0
n=2
n=1
m=-1
l=0
m=0
l=0
m=0
Fig. 2.9
In particular
m = 0, 1, 2, ..... l
l = 0, 1,2 ... (n-1) 54
so that for n=2, we have l=0, m=0, and l=1, m=0,+1,-1 giving four wave functions or
quantum states etc. Fig. (2.10) shows the different eigen – states in case of hydrogen – like
atom.
The solution for radial wave function is found as
l
2 z 3 (n l 1)!
zr 2 zr 2l 1 2 zr
R(r )
exp
3
na na Lnl na
na0 2n{(n l )!}
0
0
0
The complete solution for the problem of hydrogen atom is obtained by multiplying the
solutions R(r ), ( ) and ( )
2.8 Normal or Ground state of Hydrogen Atom
For the ground state of the atom,
n = 1, l=0, m=0, and hence the wave-function is
1 / 2
e where a0 is the radius of Bohr’s first orbit (2.139)
3
a 0
100 =
r
But
2Zr 2r
2
because
n
1
,
Z
1
and
a
0
na0
a0
e 2
Hence, 100 1 3 exp r
a 0
a0
100 100*
and
2r
1
exp
3
a0
a0
(2.140)
(2.141)
since this expression for probability density is independent of and , the normal hydrogen
atom is spherically symmetric. Consider a small value dV = r2 sin dr d d, then the
probability of finding the electron in this volume is given by
100 100*r 2 dr sin dd
2r 2
1
exp
r dr sin dd
3
a0
a0
(2.142)
For a spherical shell between radii r and r + dr, this value becomes
=
2
2r 2
2r 2
1
4
exp
r dr 0 sin d 0 d 3 exp
r dr
3
a0
a0
a0
a0
Thus the probability of finding the electron between the distance r and r + dr from the
nucleus is
P(r )dr
2r 2
4
exp
r dr
3
a0
a0
The radial distribution function P(r) is given by 55
P(r )
2r 2
4
exp
r
3
a0
a0
The probability is maximum when dP/dr = 0.
or
2r
2r
dP 4 2 2
2r exp
3 r exp
dr a0 a0
a0
a0
or
2r
2r 2
4
exp
3
2r a 0
a
a0
0
0
or
2r 2
2r
a0
or
ra
0
0
0
(2.143)
Thus in the normal state, the maximum probability of finding the electron is at a distance
equal to the radius of the first Bohr’s orbit.
UNIT III 56
31 Perturbation theory
Perturbation theory is of two types.
1. Time independent perturbation theory
2. Time dependent perturbation theory
In time independent or stationary perturbation theory, the Hamiltonian is separated into two
parts, one represents the characteristics of the system which can be solved exactly and the
other due to perturbation.
Perturbed Hamiltonian H H (0) H ' where H (0) represents the unperturbed part and H ' the
perturbed part.
Time independent perturbation theory for nondegenerate case
The schrodinger eqn. is written as
H E
The Hamiltonian H 2 V
2m
2
Expanding H interms of some parameter λ as,
H H ( 0) H (1) 2 H ( 2) ...
(3.1)
Let the energy eigen values are E1( 0) , E2( 0) ...En( 0) and the eigen functions are 1(0) , 2(0) ... n(0) . It
is possible to expand the eigen functions and eigen values as power series in λ as
En En(0) En(1) 2 En( 2) ...
n n(0) n(1) 2 n( 2) ...
Substituting these En and n values in the schrodinger eqn.
(3.2)
H n En n we get
( H (0) H (1) )( n(0) n(1) 2 n( 2) ...) ( En(0) En(1) 2 En( 2) ...)( n(0) n(1) 2 n( 2) ...)
Collecting the like coefficients of λ, 57
n( 0) ( H
(0)
H
n(1) H n( 0) ) ( H
(0)
(1)
n( 2) H n(1) ...) ...
(0)
2
2
(1)
( E n n(1) E n n( 0) ) ( E n n( 2 ) E n n( 0) E n n(1) ) ... E n n( 0)
(1)
(0)
(0)
( 2)
(1)
(0)
For this eqn. to be valid for all values of λ, the coefficient of equal powers of λ on both sides
must be equal.
Comparing the coefficient of λ0,
H ( 0) n( 0) En n( 0)
( 0)
Comparing the coefficient of λ1,
H
n(1) H (1) n(0) En (0) n(1) En (1) n(0)
( 0)
(3.3)
Comparison of coefficient of λ2 gives
( 0)
( 2)
(1)
H (0) n( 2) H (1) n(1) En n( 2) En n(0) En n(1) ) etc.
(3.3A)
These eqns. represent unperturbed, first order, second order etc. perturbation equations.
First order perturbation theory:
To find the first order correction to energy En(1)
Consider the first order eqn H
n(1) H (1) n(0) E n ( 0) n(1) En (1) n(0)
( 0)
n(1) can be expanded as a series by using expansion theorem as
n(1) Cm m
( 0)
(3.3B)
m0
Substituting this eqn in (3.3) we get,
C
m
H ( 0) m( 0) H (1) n( 0) Cm En m( 0) En n( 0)
(0)
(1)
by using H ( 0) m(0) Em( 0) m(0)
m
or Cm Em(0) En(0) m(0) H (1) n(0) En(1) n(0)
(3.4A)
Multiply the above on the left by n( 0) and integrating we get,
*
m
( 0 )*
n
Cm [ Em( 0) En( 0) ] m( 0) d n0* H
n( 0) d En(1) n( 0)* n( 0) d
(1)
Here we have used the kronecker delta function
(3.4B) 58
(j0) d ij 0 if i j
( 0 )*
i
1 if i j
(3.4C)
Therefore,
1
En(1) n( 0)* H (1) n( 0) d n | H 1 | n H nn
To find the first order correction to wave function n(1) :
Multiplying eqn (3.4A) by m( 0)* and integrating,
C [E
m
( 0)
m
m
En(0) ] m(0)* n(0) d m( 0)*H n(0) d En(1) m(0)* n(0) d
(1)
Applying the condition (3.4C),
Cm
( 0 )*
m
H (1) n( 0) d
Em( 0) En( 0)
,m n
Substituting this value of Cm in eqn (3.3B) we get the first order correction for the wave
function.
To find the second order correction to energy En(2)
The second order correction to wave function n( 2) can be expressed as
n( 2) g m m( 0)
(3.5)
m
Substituting (3.5) and (3.3B) in (3.3A), we get
g
or
m
H ( 0) m( 0) Cm H
g
m( 0) g m En (0) m(0) Cm En (1) m( 0) En( 2) n( 0)
(1)
[ Em(0) En(0) ] m( 0) Cm [ En H (1) ] m(0) En( 2) n( 0)
(1)
m
(3.6)
m
Here we have used H (0) m( 0) Em(0) m(0)
Multiplying this eqn with n( 0)* and integrating over the space variables and applying the
kronecker delta function we get,
En( 2) Cm n(0)* H (1) m(0) d
m
Substituting the value of Cm we have, 59
E
( 2)
n
( 0 )*
m
H (1) n( 0) d n( 0)* H (1) m( 0) d
E m( 0) E n( 0)
m
( 2)
or E n n
'
( m | H (1) | n ) 2
E n( 0) E m( 0)
m n if
, mn
H (1) is Hermitian.
To find the second order correction to wave function n( 2) :
Multiplying the eqn (3.6) by m( 0)* and integrating,
g [E
( 0)
m
m
En( 0) ] m( 0)* m( 0) d
m
Cm m( 0)*[ En1 H (1) ] m( 0) d En( 20 m( 0)* n( 0) d
m
or
gm
m
Cm
En(1) m(0)* H (1) m(0) d
( 0)
E En
( 0)
m
mn
Substituting this value of g m in eqn. (3.5) we get the second order perturbed wave function.
3.2 Zeeman Effect (without electron spin)
The change in the energy levels of an atom when it is placed in uniform external magnetic
field, is called the Zeeman effect. Let us consider that the field strength B is applied on a
hydrogen atom, so that an electron of reduced mass carrying the charge 'e' is moving in a
field whose vector potential is A. The magnetic induction B in terms of vector potential A
can be written as
B curl A x A
Then the constant magnetic potential A in terms of magnetic induction B can be expressed as
A
[since B x r
1
(B x r).
2
x A x r (for B x A)
( . r) A - (A.) r 3A - A 2A
A
1
(B x r )
2 60
The classical Hamiltonian of a particle of mass carrying charge
e and moving in field of
vector potential A may be expressed as
H (P, r )
where
1
( p eA) 2 V (r )
2
=
p2
e
e2 2
V (r )
(p A A p)
A
2
2
2
=
H 0 H H
P2
H
V (r ), is the unperturbed Hamiltonian
2
0
H
e
(p A A p)
2
e2 2
H
A
2
and
Ist Order Zeeman effect :
For weak fields and second order perturbation term H containing A 2 may be neglected and
hence the perturbed Hamiltonian takes the form
H H 0 H
p2
e
V (r )
(p A A p)
2
2
ˆ and keeping in mind the vector identity
PP
i
div (A ) div A A
(A ) ( A) A
or
A A
i
i
We note (p A A p)
( A )
A
i
i
( A) A A
i
i 61
( A) 2 A
i
A [
(3.7)
1
1
(B x r )] (B x r )
2
2
Using the vector identity
div ( A x B) B curl A A curl B, we get
x B 0 and x r 0
A ψ
and
1
1
(B x r) ψ B (r x )ψ
2
2
(3.8)
(3.9)
using (3.8) and (3.9) equation (3.7) gives
1
(p A A p) 0 2 B (r x )
i
2
=
B r x B (r x p)
i
= B L .
or
p A A p B. L
(3.10)
(3.11)
The energy eigen functions of the unperturbed H atom are usually chosen to be eigen state
of Lz with eigen values m , m being magnetic quantum number. It is customary for
convenience to choose magnetic field along Z axis then B L BL z
First order energy correction
E n | H | n n |
e
BLz | n
2
e
e
B m n| n m
B
2
2
(3.12) 62
can
take
(2l 1) values
for
since m varies form l to l
m=+2
m=+1
with a difference of unity
The selection rule permits only
those
transitions
in
l=2
m=0
(d-state)
m=-1
which
m=-2
magnetic quantum number m
changes
selection
by
0
rule
or
1. This
coupled
with
expression (3.12) is sufficient to
explain the spectrum of normal
Zeeman effect. For an example
we
consider
the
m=+1
l=1
m=0
(p-state)
Δm=0
transitions
between two states with l 2
Δm=-1
m=-1
Δm=+1
Fig. 3.1
and l 1, (i.e. between d and
p states). The transitions are shown in the fig. 3.1
3.3 First Order stark effect in Hydrogen atom
The effect of change in energy levels of an atom in the presence of an electric field is called
stark effect. The unperturbed Hamiltonian of Hydrogen is given by,
H
( 0)
2 2
1 e2
where r is the distance between the nucleus and the electron of an
2
4 0 r
hydrogen atom and µ-the reduced mass
Due to an external electric field of strength E, the perturbed Hamiltonian term is given as
H ' Ee r cos , where e is the charge of the electron
For the ground state of the atom, n=1, l=0, m=0
The non-degenerate wave function is
(3.13) 63
nlm 100 R10 (r )Y00 ( , ) R10 (r )
1
4
for the ground state of hydrogen, the first order energy correction due to perturbation is
H
'
100,100
* H
100
100d 0 since H’ is having odd parity
'
Thus there is no first order stark effect in hydrogen atom
First Excited State
Consider the first excited state of hydrogen. Here n=2,
l=0, 1 and m=1, 0, -1. The
corresponding wave functions are,
1
even parity
4
3
210 R21(r )Y10 ( , ) R21(r )
cos
odd parity
4
3
i
211 R21(r )Y11( , ) R21(r )
sin e
odd parity
8
200 R20 (r )Y00 ( , ) R20 (r )
21, 1 R21(r )Y1, 1 ( , ) R21(r )
3
sin e
8
i
odd
(3.14)
parity
Out of the four wave functions 200 is of even parity and all other wave functions are of odd
parity. The secular determinant is given as
' , 200 E
H 200
' , 200
H 211
' , 200
H 210
' , 1, 200
H 21
(1)
' , 211
H 200
(1)
' , 211 E
H 211
' , 211
H 210
' , 1, 211
H 21
' , 210
H 200
' , 210
H 211
' , 210 E
H 210
' , 1, 210
H 21
(1)
' , 21, 1
H 200
' , 211
H 211
' , 21, 1
H 210
' , 1, 21, 1 E
H 21
0
(3.15)
(1)
If the integrand value is of odd parity, then the integral value is zero. Hence,
'
*
(1)
(1)
H 200
is odd parity and the net
, 200 200H 200d 0 since 200 - even parity and H
integrand is odd parity. In the same manner we can show
(1)
(1)
(1)
(1)
(1)
(1)
(1)
(1)
(1)
H 211
, 211, H 210, 210, H 211, 211 , H 211, 210, H 211, 211 , H 210, 211, H 210, 211 , H 211, 211, H 211, 210 are all zeros 64
Consider
1
3
R20 (r )r cos
sin ei R21(r )r 2 dr sin dd
8
4
(1)
H 200
, 211
2
But,
e
i
(1)
d 0 . Hence H 200, 211 0
0
(1)
(1)
(1)
On the same ground, H 211
are zero
, 200, H 200, 211 , H 211, 200
(1)
(1)
So we are left with two elements namely H 200
, 210 and H 210, 200
(1)
First, let us evaluate H 200
, 210 :
(1)
*
(1)
H 200
, 210 200H 210d
200
Here,
210
1
4
1
2a0
3/ 2
r
r
2 exp
a0
2a0
1
3
cos
4
2a0
3/ 2
r
r
exp
3a0
2a0
where a0 is the Bohr’s first orbit radius
(1)
Using these values and H r cos and assuming r , we get the r-integral value as,
a0
1 a05
4
5
2
e
d
e
d
0
8 3a04 0
=
a0
25 6 3 3a0
8 3
The Θ-integral gives the value as 2/3
and
φ-integral gives 2π
3eEa0
0,0
3eEa0
unperturbed
Therefore,
(1)
H 200
, 210
1
4
3
2
(3 3a0 ) 2 3a0
4
3
(1)
Similarly we get H 210
, 200 3a0 eE
Substituting these values the secular determinant becomes
Fig.3.2 65
E (1)
0
0
E (1)
3a0
0
0
0
0
E (1)
0
0
E (1)
3a0
0
0
There are four roots for this determinant as E(1) =0, 0, 3 a0 , -3 a0
The corresponding perturbed energy values are
E (1) 0, 0, 3eEa0 , 3eEa0
(3.16)
Thus for the first excited state of hydrogen atom, one energy level is raised by 3eE a0 , and
one is decreased by the same amount. The other two levels remain unchanged.
3.4 The Variation Method
The expectation value of energy in normalized state is given by
E *H d
(3.17)
If we choose the wave function as variable function, then the integral (3.17) is known as
variation integral and gives an upper limit to the energy E0 of the lowest state of the system.
The function is the variation function and its choice may be quite arbitrary, but more
wisely, it is chosen such that E approaches more closely to E0. If the variation function
equals the function 0 of the lowest state, then energy E will be equal to E0, i.e.
E 0 H 0 d E0
*
(3.18)
If ψ ≠ ψ0, they by expansion theorem ψ may be expanded in terms of a complete set of
orthonormal functions 0, 1, 2…obtaining
an n with an an * 1 and Hn Enn
n
(3.19)
n
Substituting this in equation (3.17), we get
E an *am n * Hm d
n
But
Hm =Emm
(3.20) 66
we have
E an *am n * Emm d
n
an *am Em n * m d
n
=
a
n
* am Em nm
n
Therefore E a n * a n E n
n
Since mn 1 for m n
0 for m n
| an | En
2
n
Substracting the ground state energy E0 from both sides, we get
E E0 | a n |
2
E n E 0
(3.21)
n
As |an|2 is positive and En ≥ E0 (always) for all values of n; therefore right hand side is
positive or zero. Thus we have proved that < E > is always an upper limit to E0, i.e.
< E > ≥ E0
(3.22)
This theorem is the basis of the variation method for the calculation of the approximate eigen
value of the system. If we choose a number of wave-functions 1, 2,3… and calculate
the E1, E2, E3... coresponding to them, then each of these values of E is greater than the
energy E0, So that the lowest one is the nearest to E0. Often the functions 1, 2,3…are
only distinguished by having different values of some parameter the process of minimising
E with respect to this parameter may then be carried out in order to obtain the best
approximation to E0. Which from the trial function will follow.
3.5 Physical applications of Variation method
Ground State of Helium : We use the variation method with a simple trial function to obtain
an upper limit for the energy of the ground state of the helium atom. The helium atom
consists of a nucleus of charge + 2e and two electrons each of charge ‘- e’ If we consider the
nucleus at rest, the Hamiltonian will be 67
2 2
1 e2
2
21
H
1 2 2e
2m
r1 r2 r12
(3.23)
where 12 and 22 and Laplacian operators
-e
for the first and second electrons
r12
-e
r1
at a distance r1 and r2 from the nucleus,
r12 r2 r1 is the distance between two
r2
+Ze
e2
electrons. If the interaction energy
r12
between two electrons were not present,
Fig. 3.3
the ground state eigenfunction of He would
be product of two normalized hydrogen like wavefunctions u100 (r1 ) u100 (r2 ) given by
3
z / a0 r1 r2
z
r1 r2 u100 r1 u100 (r2 ) 3 e
a0
with
z 2 and a 0
2
me2
We shall use (r1, r2) as a trial function and treat z to be the variation parameter, so that it is not
necessarily equal to 2. The expectation value of Hamiltonian H is the sum of expectation
values of kinetic energy and potential energy individually.
H = K.E. + P.E.
e 2
of electrons.
= T + V + interaction energy
r
12
then
e2
= + +
r12
Now the expectation values of hydrogen-like atoms (having one electron) with z atomic
number in general are
z
2
2
2
e
2ze
1 z
T
,V
. Since .
2a 0
a0
r1 a 0
But helium atom in ground state has two electrons, so it will be twice of hydrogen-like atom,
2
2
2
2
2z e
z e
i.e.
=
=
(3.24)
2a 0
2a 0 68
and
=2×
Hence
=
2
2
2z e
a0
2
4z e
a0
2
2
(3.25)
2
z e
4z e
e
2a0
a0
r12
(3.26)
Electron Interaction Energy : The expectation value of the interaction energy between the
electron is
e2
<
>=
r12
*
r1 r2
2
z3
3 e2
a 0
e 2 z / a0 r1 r2 d 3 r1 d 3 r2
1
r
e2
r1 r2 d 3 r1 d 3 r2
r12
12
Substituting,
2z
2z
2z
r1 1 and
r2 2 ,
r12 12
ao
a0
a0
we get
e2
ze 2
r12
32 2 a0
e 1 2
12
d 3 1 d 3 2
Solving the spherically symmetric integral by knowledge of electrostatics as in perturbation
theory, we get
e2
ze2
202
2
r12
32 a 0
5ze2
8a 0
(3.27)
The expectation value of Hamiltonian (3.26) for the trial function is
e 2 z 2 4e 2 z 5e 2 z e 2 2 27
H
z z
a0
a0
8a0 a0
8
Differentiating with respect to z and for minimum , equating the differential to zero,
H e 2 2 27
z z 0,
z
z a0
8
This gives z
27
1.69. Thus the lowest upper limit for the ground state energy of helium
16
atom obtained with trial function 69
e2
a0
2 2
27 2 27 27
e2
27 e
.
2
.
85
8 16
a0
16 a0
16
(3.28)
By the help of perturbation method, the ground state energy of helium atom comes out
e2
2
2.85 , where a0
.
m0 e 2
a0
The hydrogenetic wavefunctions give the best energy value when z
27
16
rather than 2. It
indicates that each electron screens the nucleus from the other electron, therefore the effective
nuclear charge being reduced i.e. 2 27 5 by of an electronic charge.
16
16
Hence "effective charge" in the nucleus is less than 2.
3.6 Connection formulas for penetration of a barrier
Figure (3.4) represents potential function V(x) as a function of x. Let a particle has energy E.
Then the whole region is divided into three regions
Region I where V > E
Region II where E > V
Region III where V > E
There are two types of turning
points, specified by points x1 and
V(x)
x2. When we pass from region I
V
to II, then at the turning point x1,
the barrier is to the left. But if
pass from region II to region III,
the barrier is to the right.
E
Region II
Region I
Region III
In region I, the wave-function
decreases exponentially, for
x and p1 2m V E , then
is approximated as
X2
X1
Fig. 3.4 70
1
A
x p dx
exp 1
p1
x1
(3.28a)
In region II, the wave function is oscillatory given by
x pdx
x
B
C exp i pdx
2
exp i
p2
p2
In region III, the wave-function decreases exponentially for
3
(3.29)
x
D
x p dx
exp 1
p1
x2
(3.30)
The regions of validity of these wave-forms of wave-functions are separated by the classical
turning points near which W.K.B. approximation fails. As 1, 2, 3 are all the
approximations of the same function , therefore the constants A, B, C and D cannot all be
arbitrary.
To connect the wave-functions at the turning points we assume that the potential energy
function is approximately linear in the neighbourhood of turning points x1 and x2. Thus at x1
and x2, we write
E V( x) C1 ( x x1 )
E V( x) C1 ( x x1 )
or
V( x) E C2 ( x x2 )
E V( x) C2 ( x x2 )
(3.31)
The schroedinger wave-equation
d 2 2m
2 E V( x) 0
dx 2
In the neighbourhood of x = x1 and x = x2, this eqn. takes the form
and
d 2 2mC1
( x x1 ) 0
dx 2
2
(near x1)
(3.32)
d 2 2mC 2
( x x2 ) 0
dx 2
2
(near x2)
(3.33)
Now we change the variable in equation (3.32) by substituting
1/ 3
2mC
z 2 1 ( x x1 )
d d dz
dx
dz dx
(3.34) 71
d 2mC1
1
dz 2
1/ 3
=
d 2 d d d d dz
dx 2 dx dx dz dx dx
and
1/ 3
d d 2mC1
dz dz 2
2mC
2 1
2/3
1/ 3
2mC1
2
d 2
dz 2
Making these substitutions in (3.32), we get
d 2
z 0
dz 2
(3.35)
1
3
Similarly by substituting z 2mC 2 x x 2
2
d 2
z 0
dz 2
equation (3.33) becomes
(3.36)
Equation (3.35) and (3.36) are Airy functions. We require a function which vanishes
asymptotically for large positive z (z > 0) corresponds to x x2) such a function is
s 3
1
cos
Ai(z) = sz ds
0 3
(3.37)
For Large (z) it has asytnptotic form
Ai( z )
Ai( z )
1
2 z
1
2 z
1
4
1
4
2
exp z
3
3
2
3
2
sin z 2
4
3
z 0
( z 0)
(3.38)
(3.39)
If the energy E is sufficiently large, the de-Broglie wavelength associated with the particle is
extremely small so the regions of validity of linear approximations contain many
wavelengths. The function Ai(z) which passes smoothly through the turning point provides
the required connections among the approximate forms equations (3.28a), (3.29) and (3.30). 72
In the neighbourhood of x1, we have
p1 2mC1 x x1 2mC1 3 z
2
2
and
x
x1
1
p1 dx 2mC1 3
2
x
z dx
x1
z
= z dz
0
x
Similarly
x1
1
p 2 dx 2mC1 3
2
x
x1
2 32
z
3
(3.40)
z dz z dz
0
3
2
z 2
3
(3.41)
Comparison of (3.40) and (3.41) with equations (3.38) and (3.39) shows that the functions
approximated to the left of x1 by
x p dx
1
1
exp 1 ;
p1
x1
x x1
(3.42)
has on the right the approximation
2
p2
x p dx
sin x 2 ; x x1
4
1
(3.43)
A similar analysis in the neighbourhood of points x2 shows that the function approximated to
the right of x2 by
3
1
x p dx
exp x 1 ;
2
p1
( x x2 )
(3.44)
is approximated in region II by
2
WKB
p2
x2 p dx
sin x 2 ; ( x x 2 )
4
1
x p dx
exp x 1 ;
p1
2
2
x2 p dx
sin x 2 ;
4
p2
Again the solution of equation (3.33) is the Airy function
(3.45)
x x2
x x2 73
Bi(z) =
1
0
sz s3
3 sin
e
s3
sz ds
3
(3.46)
This has asymptotic forms
Bi(z) ~
1
exp
z1/ 4
2 3/ 2
z (z 0)
3
(3.47)
Bi(z) ~
1
2
cos (z)3 / 2 (z 0)
1/ 4
4
( z)
3
(3.48)
A similar argument leads to the connection formulae
WKB
1
x1 p dx
exp 1 ; ( x x1 )
p1
x
1
x p dx
cos 2 ; ( x x1 )
4
p2
x1
(3.49)
Hence, the connection formulae at turning point x = a may be expressed in terms of Barrier to
the right and Barrier to the left as follows.
Barrier to the Right : Let V > E to the right of x = a.
For decreasing exponential
1
2
x p dx
x p dx
exp 1
sin 2
4
p1
p2
a
a
V>E
1
x p dx 2
x p dx
exp 1
cos 2
4
p1
p2
a
a
V(x)
(3.50)
E
V E to the left of x = a.
For decreasing exponential
or
1
2
a p dx
x p dx
exp 1
sin 2
4
p1
p2
x
a
(3.52)
1
2
a p dx
x p dx
exp 1
cos 2
4
p1
p2
x
a
(3.53)
For increasing exponential
or
1
1
a p dx
x p dx
exp 1
cos 2
4
p1
p2
x
a
(3.54)
1
1
a p dx
x p dx
exp 1
sin 2
4
p1
p2
x
a
(3.54A)
It may be noted that the connection formulae enable us only to obtain the relation between the
solutions in a region at some distance to the right of the turning point x = a, with those in a
region some distance to the left. In order to obtain the form of the wave-function in the
intermediate region, we should consider the exact, solution, which involves Bessel functions
of order
1
.
3
For applying W.K.B. approximation, the following requirements must be satisfied.
1. On either side of the turning point, there exist regions when the potential function
changes slowly so that W.K.B. approximation is applicable.
2. In the region near the turning point x = a, where W.K.B. method becomes
inapplicable, the kinetic energy can be represented approximately by a straight line (E
– V) = C (x – a). In other words the potential should not undergo a large fractional 75
change in slope within this region. Inside the barrier W.K.B. approximation begins to
hold after
x
a
[2m(V E)]
dx
becomes appreciably greater than unity.
3.7 Time Dependent Perturbation Theory
Consider the wave eqn,
H
(0)
( 0)
(0)
i
t
The general solution of the above eqn is
( 0)
a e
n 0
( 0)
n
i ( 0 )
En t
n(0)
(3.55)
H H (0) H (1)
Consider the Hamiltonian
H ( 0) is the unperturbed Hamiltonian and H (1) is a small perturbation varying very slowly
w.r.t. time
Substituting H in the perturbed eqn H E we get
( H ( 0) H (1) ) i
t
The general solution of this eqn is
(r , t ) an (t )n( 0) (r , t )
n
( 0) ( 0)
Substituting (3.56) and using H n i
n( 0)
we get
t
an (t ) H (1) n(0) i n(0)
n
Multiply by m( 0)* and integrating
an (t )
t
(3.56) 76
an (t )m(0)* H (1) n(0) d i
n
n
an (t ) ( 0)* ( 0)
m n d
t
am (t )
1
am (t ) an (t ) m(0)* H (1) n(0) d , m 0,1,2...
t
i n
or
i
Em (t ) and using mn
multiplying and dividing by exp
am (t )
Em En we get
1
an (t ) H 'mn exp(imnt )
i n
(3.57)
(1)
where H mn
m( 0)* H (1) n( 0) d
the perturbation is so small such that an (t ) does not vary very much with time.to find the
value of an(t)
Let an (t ) an(0) a1n (t ) 2 an( 2) (t ) ...
(3.58)
Substituting (3.58) in (3.57) we get
i
( 0)
(1)
a m (t ) (a n
a 1n 2 a n2 ...) H mn
exp( i mn t )
n
(3.58a)
Zereo Order Calculation:
Equating the coefficient of 0 we get the zero order calculation as
am(0) (t )
0 or
t
am( 0) (t ) constant t
(3.59)
This gives the initial state of the system. Let us assume the system is in particular state ‘l’ at
t=0. That is al( 0) (0) 0 or
ai( 0) il
First order calculation:
Equating the coefficients of λ
in eqn (3.58a), 77
(1)
i
(1)
am (t ) an(0) (t ) H mn
exp(imnt )
t
n
i
(1)
a (t ) an( 0) (t ) H mn
exp(imnt ) dt
n
t
or
(1)
m
i t
(1)
nl H mn exp( i ml t ) dt
n
t
i
H ml(1) exp(iml t ) dt
Let the perturbation is turned on at time t=0 and turned off at time t=t. Now we can write
am(1) (t )
i (1) t
H ml exp(iml t )dt
0
i t
i (1) e ml 1
H ml
ml
(3.60)
Probability of finding a particle in mth state:
Probabilit y P a m(1) (t )a m(1)* (t )
P=
1
ml2
4
ml2
2(1 cos ml t )
sin 2 ( ml t / 2)
H ml(1)
2
1
1
H ml(1)
(e iml t 1)( e iml t 1)
2
2
ml
2
2
H ml(1)
2
(3.61)
2
4 H ml(1) sin 2 ( ml t / 2)
2
2 ml2
sin 2 (ml t / 2)
The fig.(3.6) shows the plot of
against ml
ml2
Physical interpretation of the curve:
The maximum value of
sin 2 (ml t / 2)
occurs when ml 0 . Let ml x
ml2 78
sin 2 ( xt / 2) 1
2
x2
x
1
2
x
xt
2
xt ( xt ) 3
2 3.23 ...
2
2
t2
4
sin
-6π/t -4π/t
-2π/t
2
ml t / 2
ml2
t2/4
0
2π/t
4π/t 6π/t
ωml
Fig. 3.6
2
This is the highest value on the y-axis of the graph. The value of sin ( xt2 / 2) is zero when
x
xt
2n
n or x ml
2
t
ml
2
4 6
,
,
etc.
t
t
t
for
n 1,2,3 etc. respectively.
2
From the figure the main peak is proportional to t and the breadth is inversely proportional
to t . Therefore the area under the curve is proportional to t. This indicates that the transition
probability per unit time is proportional to the ‘ON’ time of perturbation.
Fermi Golden Rule(Transition to continuum):
The probability of transition lth state to mth state is given by 79
2
W am2 (t )
1 (1) 2 4 sin 2 (ml t / 2)
H ml
2
ml2
(3.62)
Since ml appears in the denominator, transition probability is maximum when ml is
small or Em and El are closer. If the energy levels are closer, they form a cluster or
continuum states.
Now, the total probability of transition am (t ) 2 . If (m) is the density of final states
then (m)dE will be the number of such states in the energy interval Em and Em dE
am (t ) am (t ) (m)dE
2
The total probability of transition to these states
2
The main contribution for transition comes from the peak
4 sin 2 (ml t / 2)
1 (1) 2
dE
Therefore transition probability T 2 H ml (m)
ml2
But dE d since E
2
4 sin ml t / 2)
1
T H ml(1) (m)
d
ml2
The integral value is 2t . Therefore T
2
1
(1) 2
H ml
(m)2t
or transition probability per unit time is T
2
(1) 2
H ml
(m)
(3.63)
This is Fermi Golden Rule. It states that,
The transition probability per unit time to states of continuum is
i.
ii.
iii.
Non-zero between continuum states
2
(1)
Proportional to H ml perturbation
Proportional to density of final states 80
3.8 Harmonic perturbation
Let us consider a perturbation is harmonic of frequency
as
k | H ' (t ' ) | m
or
0 for t ' 0
H ' km (t ' ) 2 k | H 0' | m sin t ' for 0 t ' t
for
t' t
0
(3.64)
where k | H 0 ' | m is independent of time.
The first order amplitude a (1) k (t ' ) at t’ is
t
2 k | H '| m sin t ' e
ak(1) (t ' t ) (i) 1 k | H ' | m ei km t 'dt '
(i) 1
0
t
i
0
0
2 k | H 0 '| m
i
k | H 0 '| m
ak(1) (t ' )
t eit '
0
e
t
km t '
dt '
e it ' i km t '
e
dt '
2i
i ( km )t '
ei ( km )t ' dt '
0
k | H '0 | m exp(i(km )t 1 exp(i(km )t 1
i
km
km
This reveals the amplitude depends on perturbation duration interval t and independent of
instantaneous time t’.The amplitude is appreciable when one of the terms in the denominator
is zero.Thus the effect of harmonic perturbation is either transfer or receive energy from the
system.
The probability of finding the system in kth state is given by
4 k | H ' 0 | m sin
2
2
a k(1) (t ' t )
2
( km )t
2
( km ) 2
2
(3.65) 81
3.9 Adiabatic approximation:
This method is used to find approximate solutions to time dependent Schrodinger eqn for the
systems where the Hamiltonian changes slowly in comparison eith the internal frequency of
wave functions.
Consider the schrodinger eqn i
H (t ) .
t
(3.65A)
Here H (t ) varies slowly such that a good approximation is obtained by solving sch eqn at
each instant of time.We assume H is constant and equal to instantaneous value H (t ' ) where
t’ is the value of t at which we calculate H.Then the stationary state wave functionobtained by
setting t=t’=const would satisfy
H (t ' ) n ( x, t ' ) En (t ' ) n ( x, t ' )
Since H is a slowly varying fn of time t’,the approximate soln can be written as
i t
n n ( x, t ) exp En (t ' )dt '
0
(3.66)
When this wave functionis known at zero time then at later times we can write
i t
an (t ) n ( x, t ) exp En (t ' )dt '
n
0
(3.66A)
Substituting (3.66A) in (3.65A)
i t
i t
n
i a n n an
exp En (t ' )dt ' an n En exp En (t ' )dt '
t
n
0
n
0
i t
H (t ) an n En exp En (t ' )dt '
n
0
Here we have used H (t ) n En n
t
n i
or i an n an
exp En (t ' )dt ' =0
t 0
n
82
i t
Em (t ' )dt ' and integrating over all space we get
*
exp
Multiplying the above eqn by m
0
i t
i t
i t
n
an n m * exp 0 En (t ' )dt ' exp 0 Em (t ' )dt ' d n an m * t exp 0 ( En Em )dt ' d 0
Using orthonormal conditions
a m an m *
n
n
d
t
i t
exp ( En Em )dt' d
0
(3.67)
To solve the integral of the above expression we know H (t ) n (t ) En (t ) n (t )
Differentiating w.r.t. t,
n En
n
H
n H
n En
t
t
t
t
Multiply this equation by m* and integrating,
m
*
E
H
n d m * H n d n m* n d En m* n d
t
t
t
t
(3.67A)
0
or
m
*H
n
n
* n
d H m
d Em m*
d since, H is Hermitian.
t
t
t
Substituting in (3.67A) and simplifying,
m
*
n
d
t
*
H
n d
t
, m n
Em En
m
Substituting in (3.67)
am
m*
i t
( En Em )dt '
0
0
En Em
H
n d exp
t
m n
Now, let us suppose that the system starts with as 1 and
By successive approximation we can solve for am.
an 0
(3.68)
for n s . 83
H
i t
t ms
ams
exp ( Es Em )dt ' 0 ,
( Es Em )
o
H
= *
t ms m
where
H
s d
t
H
,
t ms
Neglecting the slow variation of
ams
i
i
H
exp( ( Es Em )t ) exp( ( Es Em )t 0 )
2
i( Es Em ) t ms
where we have neglected the slow change of H
t ms
(3.69)
and the exponential factor is of the
order of unity.
Total probability of transmission to mth state is
2
am
2
4
( Em E s )
4
H
t ms
2
(3.70)
3.10 Sudden Approximation
The sudden approximation occurs when the Hamiltonian changes appreciably during a very
short but finite interval of time t0 . The sudden approximation consists of the change in
Hamiltonian discontinuous on different times. Let t 0 to be duration and to consider sudden
approximations, we consider the equation
H
k
t
i
exp ( Ek E f ) t
Ek E f
f
af
In sudden approximation the variation of
af
(3.71)
H
can not be neglected ; so
t
H
k
i
t
exp ( Ek E f ) t dt
( Ek E f )
f
t0
0
f | H (t ) | k
i
t 0
exp ( Ek E f ) t
( Ek E f )
0
t0
0
f | H (t ) | k
exp
( Ek E f )
i
i
( Ek E f ) t ( Ek E f ) t dt
(3.72) 84
The condition of sudden approximation, by the help of uncertainly relation E t , is
expressed (since t t0 is very small) as
t0
E
(3.73)
The physical interpretation of this condition is that of the energy of the system changes by an
amount E . In a time t 0 which is much less than the characteristic time associated with this
energy change, then the state of the system remains unaltered (i.e. there is no transition) and
so f | H (t ) | k 0; so equation (3.72) becomes
a
f
i t0
i
( Ek E f ) t dt
0 f '| H (t ) | k exp
(3.74)
If fk is the angular frequency of the transition from initial state k to a final state f , then
fk
Ef Ek
; so that we may write
af
i
t0
0
i jk t
f | H (t ) | k e
dt
(3.75)
When perturbation is switched on suddenly, H (t ) changes instantaneously in time t which
is small compared to period ( jk ) 1 so that the factor e
i jk t
in the above integral changes a
little and hence can be taken outside the integral, then we get
af
i i jk t t 0
e 0 f | H (t ) | k dt
i i fk t
e
f | H (t0 ) H (0) | k t0
it 0 i jk t
e
f | H |k
(3.76)
where H H (t0 ) H (0) and may be taken as the maximum value of interaction during its
sudden switch on.
Therefore sudden probability of transition from state k to state f will be given by
| af |2
t 02
| f | H '| k | 2
2
|
f
|
H
'
|
k
|
2
2
2 fk
(3.77)
This is used to calculate the transition probability under the influence of sudden perturbation. 85
UNIT IV
4.1 Addition of angular momenta
Consider two non interacting systems (1) and (2) with angular momenta J1 and J2
respectively. Now, the total angular momentum J=J1+J2. The components as well as the total
angular momenta satisfy the fundamental commutation relations
j1 j1 ij1;
Let
j2 j2 ij2 ;
j j ij
1 ( j1 , m1 ) is the simultaneous eigen vector of J12
(4.1)
and
them. Similarly 2 ( j2 , m2 ) is the simultaneous eigen vector of J 22
J1z and J 2 has no effect on
and
J 2 z and J 1 has
no effect on them. Hence we can write,
J12 1 ( j1 , m1 ) j1 ( j1 1) 1 ( j1m1 )
J1z 1 ( j1 , m1 ) m1 1 ( j1 , m1 )
J 22 2 ( j2 , m2 ) j2 ( j2 1) 2 ( j2 m2 )
(4.2)
J 2 z 2 ( j2 , m2 ) m2 2 ( j2 , m2 )
( j1 j2 m1m2 ) 1 ( j1 , m1 ) 2 ( j2 , m2 ) form a complete set of simultaneous
And the products
eigen vectors of mutually commuting set of operators J 12 , J 1z , J 22 and J 2 z . Using unitary
transformation, vector form of ( j1 , j2 , m1 , m2 ) can be transformed into ( j , j , j, m) . Let
1 2
j1 j2 and
( j1 , j2 , m1 , m2 ) be the eigen vector of
J z belonging to eigen value
(m1 m2 )
J z ( j1, j2 , m1, m2 ) ( J1z J 2 z ) 1 ( j1m1 ) 2 ( j2 m2 )
(m1 m2 ) 1 ( j1m1 ) 2 ( j2 m2 )
(4.3)
(m1 m2 ) ( j1, j2 , m1 , m2 )
Since m1 j1 , m2 j2
because j1
and
then m j1 j2 . That is the largest value of m is ( j1 j2 ) j
j2 are the largest values of m1
For a given value of
1
j1 1 and j2
2
j1
and
and
m2 respectively.
j2 let us find the possible values of j. Let 86
There will be three possible values for m1 and two possible values for m2 . Thus there will be
six possible values. They are of the form ( j1 , j2 , m1 , m2 ) as
1
1 1
1, ,1, correspond ing to m1 1, m2
2
2 2
1
1
1
1, ,1, correspond ing to m1 1, m2
2
2
2
1
1 1
1, ,0, correspond ing to m1 0, m2
2
2 2
1
1
1
1, ,0, correspond ing to m1 0, m2
2
2
2
1
1
1
1, ,1, correspond ing to m1 1, m2
2
2
2
(4.4)
1
1
1
1, ,1, correspond ing to m1 1, m2
2
2
2
The corresponding m m1 m2 values are
3 1 1 1 1 3
, , , , , ,
2 2 2 2 2 2
3 1 1 3
We can say that there are four states with m , , , corresponding to the total
2 2 2 2
angular momentum j
3
1 1
1
and two states with m , corresponding to j . Thus there
2
2 2
2
are two possible values for j as 3/2 and ½. It is possible to define six angular momentum
vectors of the form ( j1 , j2 , j, m) as 87
3
3
1 3 3
j ,m
2
2
2 2 2
3
1
1 3 1
1, , , corresponding to j , m
2
2
2 2 2
3
1
1 3 1
1, , , corresponding to j , m
2
2
2 2 2
1, , , corresponding to
3
3
1 3 3
1, , , corresponding to j , m
2
2
2 2 2
1
1
1 1 1
1, , , corresponding to j , m
2
2
2 2 2
1 1
2 2
1
2
1, , , corresponding to
j1 and j2
are the largest values of
(4.5)
1
1
j ,m
2
2
m1 and m2 . Hence the largest value of
m m1 m2 j1 j2
For this value of m we can write one function as
( j1 , j2 , j1 j2 , j1 j2 ) ( j1 , j2 , j1 , j2 )
(4.6)
The next value of m is j1 j2 1 which has two possible values for m1 and m2 as
m1 j1 and m2 j2 1
m1 j1 1 and
m2 j2 . That is ( j1 , j2 , j1 , j2 1), ( j1 , j2 , j1 1, j2 )
Correspondingly we have two -values as,
( j1 , j2 , j1 j2 , j1 j2 1), ( j1 , j2 , j1 j2 1, j1 j2 1)
Extending the discussion for m j1 j2 n , we get a set of vectors for corresponding
to
-vectors as
( j1 , j 2 , j1 n, j 2 ),
( j1 , j 2 , j1 n 1, j 2 1)
.................................
( j1 , j 2 , j1 , j 2 n)
and
( j1 , j 2 , j1 j 2 , j1 j 2 n)
( j1 , j 2 , j1 j 2 1, j1 j 2 n)
......................................................
( j1 , j 2 , j1 j 2 n, j1 j 2 n)
The smallest value of j will occur corresponding to
(4.7) 88
j2 n j2
or
n 2 j 2 . So the smallest value of m is j1 j 2 . So, the total
number of eigen states are given by,
j1 j 2
(2 j 1) (2 j
j1 j2
1
1)(2 j2 1) . It is very clear to express the vector ( j1 , j2 , j, m) as a linear
combination of the vector ( j1 , j2 , m1 , m2 )
( j1 , j2 , j, m) C j1,m2,m ,m ( j1, j2 , m1 , m2 )
j j
(4.8)
1 2
m1 m2
j j
The coefficient C j 1,m2,m1 ,m2 represents the unitary transformation between two angular
momentum representations known as Clebsch-Gordon coefficients. In Dirac notation it is
represented as C j ,1m2,m1 , m2 j1 j2 m1m2 | j1 j2 jm
j j
Problem: Find the C.G. coefficient for the coupling of two angular momenta
j1 1 and j 2
1
2
Solution:
For simplicity let us use the notations
( j1 j2 jm) jm ; ( j1 j2 m1m2 ) m m
1 2
The six independent vectors are
1,1 / 2 ; 1, 1 / 2 ; 0,1 / 2 ; 0, 1 / 2 ; 1,1 / 2 ; 1, 1 / 2
and
3 / 2,3 / 2 ; 3 / 2,1 / 2 ; 3 / 2, 1 / 2 ; 3 / 2, 3 / 2 ; 1 / 2,1 / 2 ; 1 / 2, 1 / 2
Obviously,
3 / 2,3 / 2 1,1 / 2
3 / 2, 3 / 2 1, 1 / 2
Similarly, for the remaining vectors we have
(4.9) 89
3 / 2,1 / 2 a1 1, 1 / 2 a2 0,1 / 2
3 / 2, 1 / 2 a3 0, 1 / 2 a4 1,1 / 2
1 / 2,1 / 2 a5 1, 1 / 2 a6 0,1 / 2
3 / 2, 1 / 2 a7 0, 1 / 2 a8 1,1 / 2
The coefficients a gives the unitary transformation matrix.
We know,
J jm ( j m)( j m 1) j , m 1 where J J x iJ y ( J1 ) ( J 2 )
Also,
(1 m1 )(2 m1 ) m11, m 2 if m1 1 j1
( J1 ) m1, m 2
0 if m1 1 j1
( J 2 ) m1, m 2
(1 / 2 m2 )(3 / 2 m2 ) m , m 2 1 if m2 1 j2
1
0 f m2 1 j2
Thus J 3 / 2,3 / 2 3 3 3 3 13 / 2,1/ 2 by using eqn4
2
3 / 2,1 / 2
2 2
2
1
J 1,1 / 2 by using eq(4.70a)
3
(1 1)(2 1)
1
3
1
2
1
2
1, 1 / 2 0,1 / 2 .which gives a1 ; a2
3
3
3
3
0,1 / 2
(1/ 2 1/ 2)(3 / 2 1/ 2) 1, 1 / 2 by using (4.70b)
Applying the lowering operator again on
3 / 2,1/ 2
(4.10) 90
J 3 / 2,1 / 2 2 23 / 2, 1 / 2
or 3 / 2, 1 / 2
1 1
2 3
or
1
( J1 ) ( J 2 ) 1 1, 1 / 2 2 0,1 / 2
2
3
3
[(1 1)(2 1)]
0 , 1 / 2
0
2
[(1 0)(2 0)] 1,1 / 2
3
3 / 2, 1 / 2 2 / 3 0, 1 / 2 1 / 3 1,1 / 2 and 3 / 2,3 / 2 1, 1 / 2
1 / 2,1 / 2 is a linear combination of 1,1/ 2 and 0,1 / 2 which is orthogonal to 3 / 2,1 / 2 . This
demands
a1*a5 a2*a6 0
or
a6
a*
1/ 3
1
1*
a5
a2
2 /3
2
1
hence , 1 / 2,1 / 2 a5 1, 1 / 2
0,1 / 2
2
From the condition for normalisation
1 / 2,1 / 2
1
2 1, 1 / 2 0,1 / 2
3
If the lowering operator J- is applied we get
1 / 2, 1 / 2
1
[ 0, 1 / 2 2 1,1 / 2 ]
3
Thus we have
0
0
0
0
0 1,1 / 2
3 / 2,3 / 2 1
0
0
0
0
1 1, 1 / 2
3 / 2, 3 / 2 0
0
1/ 3
2/3
0
0
0 0,1 / 2
3 / 2,1 / 2
3 / 2, 1 / 2 0
0
0
2/3
1 / 3 0 0, 1 / 2
2 / 3 1/ 3
0
0
0 1,1 / 2
1 / 2,1 / 2 0
0
0
0
1 / 3 2 / 3 0 1, 1 / 2
1 / 2, 1 / 2
C .G .coefficient
(1/ 2 1/ 2)(3 / 2 1/ 2) 0,1 / 2
91
Problem: Find the C.G. coefficient for the coupling of j1
1
1
and j2 .
2
2
For the given problem, the four independent -states are:
m ,m 1/ 2,1/ 2 ; 1/ 2,1/ 2 ; 1/ 2,1/ 2 ; 1/ 2,1/ 2 and the possible values of j are 1 and 0.
1
2
Hence the possible -states are
jm 11; 10; 00; 1,1
Obviously
11 1 / 2,1 / 2
1, 1 1 / 2, 1 / 2
The remaining states can be written as
10 a1 1 / 2,1 / 2 a2 1 / 2,1 / 2
J 1,1 21,0
hence , 10
1
[( J1 ) ( J 2 ) ] 1/ 2,1/ 2
2
Since 00 is orthogonal to 10 we have
00
1
[ 1 / 2, 1 / 2 1 / 2,1 / 2 ]
2
The matrix transformation is
0
0
0 1 / 2,1 / 2
11 1
10 0 1 / 2 1 / 2 0 1 / 2, 1 / 2
0 1/ 2 1/ 2 0
00
1 / 2,1 / 2
0
1
1, 1 00
1 / 2, 1 / 2
C .G .coefficient
4.2 Commutation rules for angular momentum:
The commutation rules can be combined symbolically as,
L L i L.
(4.11) 92
Two different components of the angular momentum do not commute. It will not be possible,
in general, to measure Lx , Ly , and Lz simultaneously, because product of the uncertainties of
two quantities is proportional to the mean value of their commutator. It is of interest to obtain
the commutation relations of L with the components, Lx , Ly , Lz . Let us consider, for
example,
L , L L
2
z
2
x
L L L
Ly Lz , Lz
2
= Lx 2 Lz
2
2
z
x
2
y
Lz Lz Lx ) ( Lz Lz Lz Lz )
2
L
as
hence
L , L (L L L
2
z
x
x
z
z
2
2
Lz Lz Lz
2
2
0,
Lz Lx Lx ) ( Ly Ly Lz Lz Ly Ly ).
Adding and subtracting Lx Lz Lx and Ly Lz Ly on R.H.S, we get
= ( Lx Lx Lz Lx Lz Lx Lx Lz Lx Lz Lx Lx ) ( Ly Ly Lz Ly Lz Ly Ly Lz Ly Lz Ly Ly )
= Lx ( Lx Lz Lz Lx ) ( Lx Lz Lz Lx ) Lx Ly ( Ly Lz Lz Ly ) ( Ly Lz Lz Ly ) Ly
= i Lx Ly i Ly Lx i Ly Lx i Lx Ly 0.
Thus we conclude that L2 commute
(4.12)
with Lz . By symmetry, we conclude that it also
commutes with Lx and Ly . In other words it is possible to measure simultaneously L2 and any
single component of L.
Eigenvalues of LZ
We now try to find the eigenfunctions of LZ that is, we want to satisfy the equation,
LZ c .
(4.13)
c .
i
or
The solution of eq. (4.13) is
f (r, ) exp (i c / ),
(4.14)
where f (r , ) is an arbitrary function of r and .
Now must be a single valued function of x, y, z. In this case increase in the angle by 2
should not change the wavefunction, so that
f (r , ) exp (i c / ) f (r, ) exp i c ( 2 ) /
or
exp (i c 2 / ) 1 93
2c
2m
c m ,
where
m is an integer. Thus eigenvalues of
Lz are
Lz m
(4.15)
ei m f (r, ).
(4.16)
The eigenfunction are
4.3 Angular momentum in General
The total angular momentum can be written as a sum of the orbital and spin angular
momenta:
J LS
(4.17)
where J has the components
J x Lx S x
J y Ly S y
J z Lz S z
J follows the same commutation rules as orbital angular momentum does, i.e.,
J
x
J
, J y i J z
y
J z , J x i J y
, J z i J x
(4.18)
or equivalently
J J i J
and
J x , J x J y , J y J z , J z 0.
J , J J , J
2
2
Further
x
x
=
J
x
x
Jy Jz
2
2
(4.19)
, J y J y J y J x , J y J x , J z J z J z J x J y
= i J z J y J y J z J y J z J z J y 0.
Thus the operator
J 2 J x J y J z commutes with J x , J y and J z .
2
2
2
Commutation relation of J z with ladder operators J ( J x i J y ) and J ( J x i J y ).
J
z,
J J z J J J z J z ( J x iJ y ) ( J x iJ y ) J z J z J x J x J z i( J y J z J z J y )
= J z , J x i J y , J z i J y , i(iJ x ) i J y J x, ( J x iJ y ) J
Similarly
J
z,
J J
Commutation relation of J and J 94
J , J J J J J ( J x i J y ) ( J x i J y ) ( J x i J y ) ( J x i J y )
= J x (J x i J y ) i J y (J x i J y ) J x (J x i J y ) i J y (J x i J y )
On simplification, we get
= 2i( J y J x J x J y ) 2iJ y , J x 2i (iJ z ) 2J z
Commutation of J 2 with J and J
J
2
, J J x , J y J z , J
2
2
2
Now
J J J x J y i( J x J y J y J x )
and
2
2
J J J x J y i( J x J y J y J x )
2
2
1
2
2
(J J J J
Jx J y
)
2
2
2
2 1
2
J x J y J z (J J J J ) J z
2
or
J 2 , J 12 ( J J J J ) J z 2 , J ]
So,
1
J J , J 1 J , J , J Jz 2 J
2
2
a, b, c ab, c a, cb.
We known that
Applying this we get
J 2 , J 12 J J , J 12 J , J J 12 J J , J 12 J , J J J z J z , J J z , J J z
=
1
1
J (2J z ) 0 0 (2J z ) J J z (J ) J J z 0.
2
2
Similarly J 2 , J 0.
Allowed values of Total angular momentum : J
Since J z and J 2 commute, they will have a set of simultaneous eigen vectors and therefore
can be simultaneously diagonalized. Let the basic vectors in the representation in which
J 2 and J z are simultaneously diagonal be (m). This will satisfy
J 2 (m) 2 (m)
and
J z (m) m (m)
(4.20A)
(4.20B)
where and m are real numbers ( J 2 and J z are hermitian) . . We will use commutation
relations to determine the eigenvalue and
m . Let,
J J x iJ y
(4.21) 95
J J x iJ y
(4.22)
(J † J )
(4.23)
which satisfy the commutation rules :
J
J
J
Similarly
and
, Jz J
(4.24)
2 J
, Jz J
, J
z
(4.25)
2
and the identities J J ( J x iJ y ) ( J x iJ y ) J x J y i J x , J y , J J z J z
2
2
J J J 2 J z J z
2
and
2
(4.26)
Now consider the scalar product
( , J 2 ) ( , J x ) ( , J y ) ( , J z ) ( J x , J x ) ( J y , J y ) ( J z , J z )
2
2
2
(because J x , J y and J z are hermitian)
because for any vector X , ( X , X ) 0
( J z , J z )
Substituting (m), we have
(m), J (m) J (m), J (m) .
2
z
z
Using equations (4.20A) and (4.20B) We have
(m), (m) m (m), m (m) .
2
m2
or
(4.27)
Now from equation (4.24)
J , J z ( J , J z J z , J ) J
Operating on the function (m) , we have
J , J z (m) J z , J (m) J (m)
or
J z J (m) J J z (m) J (m)
= J m (m) J (m) (m 1)J (m)
or
(4.27A)
J z J (m) J J z (m) J (m)
Since J , J z 0
J 2 J (m) J J 2 (m) 2 J (m)
or
J 2 J (m) 2 J (m)
It is clear that the vector J (m) is an eigenvector of J z belonging to the eigenvalue
(m 1) and of J 2 belonging to the same eigenvalue 2 of J 2 . Similarly, J J (m) is an 96
eigenvector of eigenvalue (m 2) and belonging to the same eigenvalue 2 of J 2 . This
process cannot continue indentfinitely otherwise it will contradict equation (4.27). Let be
the greatest eigenvalue of J z for the given eigenvalue 2 of J 2 , and ( ) be the
corresponding eigenvector. Now from equation (4.27A)
J z J ( ) ( 1) J (
J ( ) 0,
hence
Since ( 1) cannot be an eigenvalue of J x . Thus
J J ( ) 0,
or
( J 2 J z J z ) ( ) 0
or
( 2 ) 2 ( ) 0
2
But ( ) is not identically zero, hence is the root of a quadratic equation. The larger root
of this equation is
1 1
2 2
(1 4 )
In a similar way
J z J (m) J J z (m) J (m) (m 1) ( J (m)
and hence
J
2
(4.28)
, J 0
J 2 J (m) J J 2 (m) 2 J (m
(4.29)
It is clear from equations (4.28) and (4.29), that the vector J (m) is an eigenvector of J z
belonging to the eigenvalue (m 1) J 2 belonging to the same eigenvalues 2 of J 2 , unless,
of course, if J (m) 0. Repeated application shows that J J (m) is an eigenvector of
J z belonging to the eigenvalue (m 2) . This cannot continue indefinitely otherwise it will
again contradict equation (4.27). If 1 is the least eigenvalue of J z then
J J (1 ) 0
or
( J 2 J z J z ) ( 1 ) 0
or
( 1 1 ) 2 ( 1 ) 0
or
12 1 0
2
2
The smaller root of the above equation would be 97
1 1
2 2
1 (1 4 )
Thus
1
and the eigenvalues of J z are
, ( 1) ,...( 1) ,
It follows that 2 must be an integer, i.e.,
1
1 1
n (1 4 ).
2
2 2
or
Rearranging, we obtain
nn
22
1
j ( j 1)
Thus
with
1 1
j 0, ,1, , ....
2 2
2
We may therefore infer that the eigenvalues of the operator J are the number j ( j 1) 2
1 1
where j 0, ,1, , ....
2 2
and each of these is (2 j 1) fold degenerate, the eigenvalues of J z for a given value of j
being
m j, ( j 1) ,.............( j 1), j
4.4 Angular momentum matrices
The matrices of J 2 and J z are given
J z m
and
J 2 j ( j 1) 2
We know that m (magnetic quantum number) varies from j to j i.e., it has (2 j 1)
values. Hence the dimensions of these diagonal matrices will be (2 j 1) i.e., (2 j 1) rows and
(2 j 1) columns. The explicit forms of angular momentum matrices are shown below :
j
0
...
J z
...
0
0 ...
j 1 ...
... ...
... ...
...
...
...
...
0
0
...
...
0 ... ... j
(4.30) 98
j ( j 1) ... ...
j ( j 1) ...
0
2
2
...
... ...
J
...
... ...
0
...
...
0
...
0
...
...
... ...
... ... j ( j 1)
Examples : Here we shall consider the following examples :
(i)
(ii)
For j 0, J 2 and J z are represented by null matrices of unit rank: (0)
1
1 1
J . In this case m ,
2
2 2
2
Dimension of J matrix = (2 j 1) 2 X (1/ 2) 1 2
i.e., there will be two rows and two columns.
11
Now
J 2 2 j ( j 1) 2 1 (3 / 4) 2
2 2
J 2 2 3 / 4 0
0 3/ 4
1
1
1
1
Further J z m for m and for m
2
2
2
2
0
2 1 / 2
Jz
0 1/ 2
J ( j m) ( j m 1)
Also
Only upper diagonal appears and rest of elements are zero.
1 1 1 1
1
1
J 1 0
for j and m
2
2
2 2 2 2
1 1 1 1
1
1
J 1
for j and m
2
2
2 2 2 2
J 0 1
0 0
J ( j m) ( j m 1)
Only lower diagonal appears and rest of elements are zero.
For
1
j ,
2
1
and m ,
2
J
1
1
and for j , and m , J 0
2
2
0
0
J
1
0
j 1. In this case m 1, , 1
(iii)
2
Dimension of J matrix = 2 j 1 3
i.e., there will be three rows and three columns.
J 2 2 j ( j 1) 2 1 (11) 2 2
Now
(4.31) 99
2 0 0
J 2 2 0 2 0
0 0 2
Further J z m , 0, according to the values of m as 1,0,1.
1 0 0
J z 0 0 0
0 0 1
Now
J ( j m)( j m 1)
0
for j 1 and m 1
2
for j 1 and m 0
2
for j 1 and m 1
(4.32)
(4.33)
4.5 Physical meaning of Identity
The word identical in quantum mechanics is to describe the particles that can be substituted
for each other under the most general possible circumstances with no change in physical
situation of the system.
Symmetrical and anti symmetric wave functions
i)
Symmetric wave function s : A wave function is symmetric if the interchange of any
pair of particles among its arguments leave the wave function unchanged.
ii)
Antisymmetric wave function A : wave function is antisymmetric if the interchange
of any pair of particles among its arguments changes the sign of the wave function.
If P is an exchange operator, then we must have
Ps (1,2) = s (2, 1)
PA (1,2) = A (2, 1)
This symmetry property of the wave function has relationship with the spin of the particle.
This relationship is listed here in the following postulates.
1. The identical particles having an integral spin quantum number are described by
symmetric wave function, i.e.,
Ps (1,2,3,..r,…s..n) = +s (1,2,3,...s,...r,...n) 100
This class of particles, i.e. the particles described by symmetric wave functions are
known as Bosons and obey Bose-Einstein Statistics. The examples of Bosons are photons
(spin 1), netural helium atoms in normal state (s = 0) etc.
2. The identical particles having half odd integral spin quantum number are described by
antisymmetric wave functions, i.e.
PA (1,2,3,..r,…s..n) = -A (1,2,3,...s,...r,...n)
This class of particles i.e. the particles described by antisymmetric wave function obey
Fermi-Dirac statistics and the particles are known as Fermi-particles or Fermions. The
examples of Fermions are electrons, protons, neutrons, muons (all spin
1
)
2
4.6 Construction of Symmetric and antisymmetric wave functions
An antisymmetric unnormalised wave function can be constructed by adding togather all the
perumated wave functions that arise from the original solution by means of an even number
of interchanges of pairs of particles and subtracting the sum of all the permuted wave
functions that arise by means of an odd number of interchanges of pairs of particles in the
original solution.
In the case where the Hamiltonian does not depend upon time, stationary state solutions
1,2,..n 1,2,...n e iEnt /
(4.34)
can be found and the time independent Schroedinger’s eqn. can be written as
H (1,2,......n) (1,2..n) = E (1,2,..n)
(4.35)
There are n ! solutions of the eqn. (eigen functions) derived from (1,2..n) by means of
permutations of its arguments belonging to the same eigen value E. Any linear combination
of these eigen functions is also an eigen function belonging to eigen value E. Hence the
system is degenerate and this type of degeneracy is called exchange degeneracy.
H (1,2) (1,2) = E (1,2)
The 2 ! = 2 solutions of this equation are (1, 2) and (2,1). The solutions correspond to a
single energy state E.
The symmetric wave function can be written as 101
s = (1,2) + (2,1)
(4.36)
and the antisymmetric wave functions is written as
A = (1,2) - (2,1)
(4.37)
Similarly for a system of 3 particles, the Schroedinger’s equation is
H (1,2,3) (1,2,3) = E (1,2,3)
(4.38)
This equation has 3! = 6 solutions corresponding to the same eigen values E. The six possible
functions obtained by exchanging the indices of the particles are
(1,2,3), (2,3,1), (3,2,1), (1,3,2), (2,1,3), (3,1,2)
Out
of these six functions, those arising by an even number of interchanges of the pairs of
particles in original wave function (1,2,3), are
(1,2,3), (2,3,1), (3,1,2)
and the functions arising by an odd number of interchanges of pairs of particles in original
function (1,2,3) are
(1,3,2), (2,1,3), (3,2,1)
So the symmetric wave function can be written as
s = (1,2,3) + (2,3,1) + (3,1,2) + (1,3,2) + (2,1,3) + (3,2,1)
(4.39)
and the antisymmetric wave function is
A = (1,2,3) + (2,3,1) + (3,1,2) – { (1,3,2) + (2,1,3) + (3,2,1)}
(4.40)
Here s and A are unnormalised wave functions.
4.7 Distinguishability of identical particles
The two identical particles can be distinguishable from each other if the sum of the
probabilities of the individual wave functions in two states is equal to the probability derived
by the symmetrised wave function i.e, if
| 1,2 |2 | 2,1 |2 | 1,2 2,1|2
| 1,2 |2 | 2,1 |2 2 Re 1,2* 2,1
(4.41) 102
where Re denotes the real part of 1, 2* 2,1
It is possible only when overlap of wave-functions (1,2) and (2,1) is zero or
2 Re 1, 2* 2,1 = 0. Thus when the co-ordinates (space and spin) of two particles are not
the
same
between
exchange
degenerate
functions,
the
interference
term,
i.e. 2 Re (1,2) * (2,1) becomes zero and particle co-ordinates do not overlap.
4.8 The Palui’s Exclusion principle:
A particle, during its motion in space reflects the properties of the statistics which it obeys.
Consider two particle system which contains electrons in indistinguishable positions.
Electrons are
1
spin particles and obey Fermi Dirac statistics. If they occupy the same
2
position in space and have the same z-component of spin, it can be seen that the eigen
function of exchange operator for a case will be
P12 A(r1,s1; r2,s2) = A(r2,s2, r1,s1)
= A(r1,s1; r2,s2)
(4.42)
r r2
0 if 1
s1 s 2
(4.43)
The non existence of the wave-function under these conditions implies that there is zero
probability that the particle will occupy the same point in space and have identical spin
orientations. Pauli exculsion principle which states that no two particles obeying Fermi
Statistics can exist in the same quantum state.
4.9 Pauli Spin Matrices for electron
Like orbital angular momentum operators Lx, Ly, Lz, the spin operators, Sx, Sy and Sz to be
associated with the components of spin angular momentum satisfy the commutation relations.
[S x, S y ] S x S y S y S x i S z
[S y, S z ] S y S z S z S y i S x
[ S z , S x ] S z S x S x S z i S z
(4.44) 103
1
i.e. electron, then according the Unlenbeck and
2
1
Goudsmit hypothesis each of operators Sx, Sy and Sz must have just two eigen values and
2
1
. Now we introduce new auxiliary operators x, y and z such that
2
If we consider the case with spin
1
x
2
1
Sy y
2
1
S z z
2
Sx
(4.45)
The following properties of ’s may be noted:
Since the eigen values of each S are to be just
1
1
and , the eigen values of each must
2
2
be +1 and 1. Each of the operators x2 , y2 , z2 must therefore have only the eigen value 1
and such operator is only unit operator, therefore.
x2 y2 z2 = 1.
(4.46)
According to (4.45) and (4.46), the commutation rules satisfied by s must be
[ x , y ] x y y x 2iz
[ y , z ] y z z y 2i x
[z , x ] z x x z 2i y
Now
(4.47)
2i x y y x 2i x y y 2i x
y z z y y y y z z y
Hence x y y x , so that x and y anticommute.
Similarly any two of the ’s anticommute in pairs
x y y x 0
y z z y 0
z x x z 0
(4.48) 104
Finally from (4.47) and (4.48) we have
x y i z
y z i x
z x i y
Since each has two eigen values, so (2 x 2) matrix may be expected to fulfil the purpose
and we begin by associating with z, the simplest (2 x 2) matrix having the eigen values 1
and 1.
1 0
z
0 1
(4.49)
Now we have
a b 1
0
a b
x z
c d 0 1 c d
(4.50)
1 0 a b a b
0 1 c d c d
z x
and
(4.51)
But x and z anticommute, we must have
b
a b a
x z z x
0
c d c d
0
2a
0 2d 0
This yields a = d = 0
So that every matrix that anticommutes with (4.49) as x and y do, accordingly x must have
the form
0 b
x
c 0
(4.52)
The eigen values of (4.52) are ± (bc) so that if they are to be 1 and 1, we must set bc = 1.
0 1
1 0
And simple possibility is to take b =c =1, so x 105
0 i
and hence
0
It then follows form (4.48) that the matrix to be associated with y is
i
complete list of ’s becomes
0 1
0 i
1 0
, y
, z
1 0
i 0
0 1
x
(4.53)
These matrices x, y and z are called Pauli spin matrices associated with the components of
spin angular momentum.
1 0 1
1 0 i
1 1 0
Sx
, Sy
, Sz
2 1 0
2 i 0
2 0 1
(4.54)
These are same as Jx, Jy and Jz in total angular momenta (Angular Momentum chapter) when
1
j
2
Thus we have
and
S
2
2
x2 y2 z2 3
S x2 S y2 S z2
x2 y2 z2 3
4
4
2
2
4.10 Electron spin hypothesis : Stern -Gerlach experiment
The spin phenomenon has no classical analogue and approaching its theory we shall begin the
recounting an empirical clue. In 1925, Uhlenbeck and Goudsmit proposed that each electron
spins while revolving about nucleus and has a quantized spin angular momentum and being
an electrically charged body possesses a spin magnetic moment. This proposal was successful
in correlating spectral data for both unpertubed and perturbed complex atoms and in
interpreting the results of the Stern-Gerlackh experiment concerned with the deflection of
atoms projected into an inhomogeneous magnetic field. 106
Spin up
S
Atomic beam
z
N
slit
Fig. 4.1
Spin down
A beam of neutral atoms or molecules, collimated by slits S1 and S 2 is allowed to pass
through a non-uniform magnetic field B. The atoms or molecules are defected by a force
which according to classical physics, is given by
F (m. B)
(4.55)
Where m is magnetic moment vector.
The arrangement is such that in the region through which the beam of atoms passes, the
direction of B varies slowly, but the magnitude of B is strongly dependent on position, thus if
thus if the projection of m along the direction of B is denoted by mB , then we have
approximately.
F m B B
By measuring the deflection on the screen, the force F and hence the magnetic moment
along B may be determined. Classically mB can have any value ranging from
m to m i.e.
we would expect a single continuous trace on the screen; but experimentally, instead of a
continuous trace, discrete equidistant traces were observed, giving clear proof of quantisation
(discrete values) of the magnetic moment. Since the magnetic moment vector m appeared to
assume certain discrete directions in space, it is said to have undergone space quantisation.
Stem and Gerlach also measured allowed values of mB to moderate accuracy and found that
the values of mB appeared in the range from minimum
m to maximum m . The value of 107
maximum projection of m (i.e.m) is conventionally regarded as the magnetic moment of the
particle. Now the magnetic moment of charged particle m is related to angular momentum L
by the classical relation.
m-
Where m 0 = mass of electron,
e
e
L
2m 0
(4.56)
its charge. This relation, being a simple proportionality
between m and L is expected to be true in quantum mechanics also. Since any component of
L has (2l 1) eigen values, we may expect the projection of m in a fixed direction, such as on
B, to possess also (2l 1) distinct eigen values and to be expressible as
mB
e
m Bm
2m 0
(4.57)
where
B
is called Bohr magneton and
m
e
9.2032 x 10 20 J / Tesla
2m 0
can assume the values from l to l with steps of unity, i.e.,
(2l 1) values. Since l is an integer, (2l 1) is an odd number, therefore we expect an old
number of traces (2l 1) in Stern Gerlach experiment. But a beam of silver atom yielded two
traces in this experiment; which is an even number and a value of
m
e
m B
2m 0
m equal of
(4.58)
The extra-ordinary implications of this experiment could not be explained immediately. Later
Goudsmit and Uhlenbeck hypothesis of existence of electron spin and intrinsic magnetic
moment provided an explanation for this. On the basis of this theory the silver atom in an Sstate has two projections possible in space, namely.
m
e
m B
2m 0
(4.59)
The positive and negative signs signify the orientations of the magnetic moment in space, i.e.,
up or down. Goudsmit and Uhlenbeck also assumed that the electron has an intrinsic (or
spin) angular momentum, but this is not easy to measure directly as the magnetic moment. 108
The existence of intrinsic angular momentum of electron seems correct by the following
arguments :
(i)
Electron has a magnetic moment which may be assumed to be due to some
internal circulating currents of charged matter, then the appearance of an intrinsic angular
momentum is expected together with a magnetic moment.
(ii)
Conservation of angular momentum for an isolated system such as an atom
can not be maintained unless the electron moving in an electric field of nucleus consists of an
intrinsic momentum.
Dirac’s relativistic theory given a deeper understanding of intrinsic angular momentum and
magnetic moment of the electron. It is conventional to associate spin S with the magnetic
moment as
e
S , m
2
2m0
e
S , m
2
2m0
(4.60)
Theory of Stern Gerlach Experiment : The separation of beam of silver atoms into two
components in Stern-Gerlach experiment may be explained as follows : Let an atom of
magnetic moment
m enter a non-uniform slowly varying magnetic field B.
Then the force
acting on the atom, according to classical physics, is given by
F (m. B)
If magnetic field B is assumed along Z-direction and is the angle between m and B, then the
force on atom,
F m cos
B
z
(4.61)
Classically can have all possible values ; but quantum mechanically, according to property
of space quantization, it can have only two discrete values. When atom enters the nonuniforms field, it experiences an acceleration along z - direction given by
az
F
M0 109
where M 0 is the mass of atom. If L the length of magnetic field and
v
the velocity of atom
along the directions of beam, then time taken by atom in magnetic field.
t
L
v
If we assume the acceleration along z - direction to be constant, then displacement of atom
along z - direction is given by
1
1 F L
z az t 2
2
2 M0 v
B
m cos
2
L2
1
z L 1
m cos
2
M0
v 2
M ov
2
2
B
z
(4.62)
This is classical expression for the displacement of atom in an non-uniform field. Quantum
mechanically, due to space quantization cos 1, therefore for spin half particles, we have
1 mL B
2 M 0v 2 z
2
z
(4.63)
This expression gives two directions for spin half particles, which is actually observed in
Stern – Gerlach experiment. Thus Stern – Gerlach experiment provides a satisfactory
experimental evidence of the Goudsmit and Uhlenbeck hypothesis of space quantization.
Limitations of Stern-Gerlach Experiment
1.
The experiment can be carried out only for neutral atoms or molecules ; it can not be
carried out for changed particles.
2.
The Stern-Gerlach experiment can not be performed even for those neutral atoms or
molecules which remain in the field for a duration shorter than
.
mB z
4.11 Scattering Cross – Section
The effective area presented by the target to the incident beam of particles in the process of
scattering is called scattering cross section. Let a beam of particular of flux N is incident on a
scattering centre; particles are scattered in all directions. If dN is the number of particles 110
scattered per unit time into the solid angle d located in the direction and with respect to
the bombarding direction , then dN will be proportional to the incident flux and the solid
angle taken. Thus we can write,
dN = (,) N d
(4.63A)
where (,) is the proportionality constant. (,) has the dimensions of an area. This can
be regarded as the cross – section of the incident beam scattered into the solid angle d about
and . The proportionality constant is known as the differential scattering cross – section.
The total scattering cross section (σt) is obtained by integrating eqn. (4.63A) over the entire
solid angle.
Nscatt
= dN ( , ) Nd N ( , ) d
= N t
since, t = ( , ) d
N scatt
N
The cross – sections are usually measured in barns (1barn = 10-24 cm2) or millibrans
Thus
t
dA
dω
r
eikz
scatterer
Fig. 4.2
4.12 Quantum mechanical discription of scattering process:
a scattering experiment, a beam of particles impinges on a target which scatters them in a
sphere around the scattering centre. Figure (4.2) explains the general behaviour of the
particles scattering. In the quantum mechanical description of scattering, the total 111
wavefunction can be regarded as consisting, of two parts, one representing the incoming part
and the other, representing the scattered part.
The complete wavefunction is given by
e f ( , )
ikz
e
ikr
r
(4.64)
The first term represents a particle moving in the positive z-direction. The second term
represents a particle moving radially outward; its amplitude depends upon and and is
inversely proportional to r.
The magnitude of the incident flux density along z-axis is
v inc v e
2
=
ikz 2
where v is the velocity.
(4.65)
v
the magnitude of scattered flux is
2
v scat
2
eikr
v
2
v f ( )
2 f ( )
r
r
(4.66)
Let the scattered particles are intercepted by a detector of area dA which is placed at a
distance r from the scatterer. The solid angle subtended by the detector at the scattering
centre is
d
dA
r2
or dA d r 2
(4.67)
Therefore, the number of particles crossing the area dA per sec. is
r
2
| f ( ) |2 dA | f ( ) |2
dA
| f ( ) |2 d
r2
(4.68)
Now the differential scattering cross – section () is defined by
() d = (no. of particles scattered into d per unit time) / (no. of particles incident per
unit time)
v f ( ) d
v
2
() f ( )
or
2
(4.69)
This is the expression for differential scattering cross – section. 112
The Born approximation
When the kinetic energy of colliding particles is large compared to the interaction energy,
then Born approximation is best applied.
4.13 Perturbation approximation We wish to solve the wave equation for a relative
motion, i.e.,
2
2
V (r ) E ,
2
where =
m1m2
m1 m2
(4.70)
Its solution in asymptotic form is given by
ik z
( r , , )
e f ( , )
e
ikr
r
(4.71)
Where eikz represents the part of the particles moving in the
z- direction and the second
term represents the part which is moving radially outward after scattering.
Equation (4.70) can be written as
2 k 2 U (r) ,
k2
where
2E
2
and
U(r)
2V (r )
2
ikz
Substituting e v(r ) where v(r ) represents scattered wave part which is a small
perturbation . We have,
2 e i k z (r) k 2 e i k z (r) U (r) e i k z (r)
e
2
or
ikz
e
(r) k
2
ikz
(r) U (r) e U (r) (r)
ikz
since (r) is small in comparison with e i k z we can neglect the term U(r) (r)
Hence ,
2 e i k z 2 (r) k 2 e i k z k 2 (r) U (r) e i k z
or 2 (r) k 2 (r) U (r) e i k z
since
2e i k z k 2 e i k z
(4.72)
Therefore we have to solve the inhomogeneous wave equation the right hand side is known.
Sufficient criterion for the validity of our solution is 113
|(r) | « | e i k z| = 1, for all r.
(4.73)
Scattering cross – section :
The solution of inhomogeneous equations (4.72) can be expressed as
(r) G (r, r' ) F (r' ) dr'
k'
where Gk ' is the Green’s function.
For a free particle, Green’s function is expressed as,
G (r, r' )
k'
1
i k r - r'
e
4 | r - r' |
e i k |(r r ')|
(r)
F (r' ) dr'
4 | r - r' |
e i k | ( r r ' )|
U (r' ) e i k z ' dr'
(4.74)
4 | r - r' |
We assume that U(r) falls off rapidly for large r so that there is an asymptotic region in
(r)
which r is large in comparison with those values of r’ For r ,
(r r ' ) 2 r 2 r ' 2rr ' cos
2
r 2 r ' 2rr '
2
where cos
r'
2rr ,
r 2 1 2
r 2
r
2
2r ,
r 2 1
r
or
as r
1/ 2
(r r )
'
r
2r '
r 1
r
r r '
Similarly,
Now eqn (4.74) becomes
v(r )
v(r ) |r
1
4
1 r ' ikr ikr '
'
ikz '
'
r r 2 e e U (r )e dr
'
'
1 ikr
e U (r ' )eik ( z r ) dr '
4r
Now, (r ) eikz v(r ) eikz
'
'
1 ikr
'
e U (r ' )eik ( z r ) dr
4r 114
ikr
Comparing this with (r ) eikz e f ( , ) ,we have
r
f ( , )
1
4
U (r )e
'
ik ( z r ' )
dr
'
Denoting the incoming wave by the vector K0 and the scattered wave by K, we can write
Kz ' Kn0 .r ' K0 .r '
Therefore, f ( , )
1
4
Let us define a vector
U (r )e
[ i ( k0 k )] r 'dr'
'
Ƙ as
Now, f ( , )
1
4
Ƙ=K0-K
U (r ' )e[i .r ]dr
'
'
The scattering cross section
( , ) f ( , )
2
=
2
2
2
1
16 2
' [ i .r ]
'
V (r )e dr
'
U (r ' )e[i .r ]dr '
'
2
2
(4.75)
This gives the total scattering cross section.
4.14 Scattering by Spherically Symmetric potentials : Partial wave analysis
Consider a particle of energy E which moves in a central, spherically symmetric force field .
Let the wave function is the solution of the Schroedinger equation
2m
2 2 E V (r ) 0,
h
.
(4.76)
We have assumed that V is a function of r only. In a special case of symmetric problem, let
the wave function is independent of the angle . Therefore the general solution of eq.(4.37) is
= Rl(r) Pl(cos),
(4.77) 115
where R1(r) =
l (r )
r
and Pl (cos ) is the Legendre polynomial of order l. l satisfies the
equation
d 2 l
dr 2
l (l 1)
k 2 U (r ) 2 l 0
r
(4.78)
1/ 2
2mE
2mV (r )
where
k= 2
and U(r) =
2
To study the nature of asymptotic behaviour, we consider r to be so large that the U and l
terms in equation (4.78) can be neglected. Then the solution of equation (4.78) is of the form
ei k r. Hence the solution is
l
e ikr
= pl (cos ) pl (cos )
r l
r
l
(4.79)
Therefore, the general solution is
ikr
e ikr
e
Dl
= pl (cos ) Cl
r
r
l
Here Cl and Dl are constants.
(4.80)
From the quantum mechanical description of scattering,
e
ik z
ei k z
f ( )
r
(4.81)
From eqns. (4.80) and (4.81) we have
ei k z f ( )
e
ei k z f ( )
or
ikr
e ikr
e
Pl (cos ) Cl
Dl
r
r
r
l
e
ik r
sin l / 2 l
Pl (cos ) bl
.
r
l
ik r
(4.82)
where l/2 has been written for convenience in place of /2. Other constants have been
generalised into the new constant bl. The quantity l is the phase shift of the lth partial wave. It
measures the amount by which the regular function sin ( - (1/2) l +l) is displaced in ,
relative to the free particle function fl(). Thus the effect of the scattering potential is to shift
the phase of each outgoing partial wave.
To evaluate the constant bl:
From Bauer’s formula, substituting the value of e i k r in eq. (4.82) we get,
sin ( (1 / 2)l l
sin l / 2
e
k f ( )
bl Pl (cos )
i (2l 1)Pl (cos )
i
l
(4.83) 116
or
i (2l 1) P (cos ) sin( (1/ 2)l ) kf ( )e b P (cos ) sin( (1/ 2)l )
l
i
l
l
l
or
l
l
i
i (2l 1) P (cos )
l
ei
( l / 2 )
l
l
e i ( 1 / 2)
kf ( ) ei
2i
ei
bl Pl (cos )
( (1 / 2 ) l j )
e
2i
i ( (1 / 2 ) l l
)
ei i l (2l 1) pl (cos )e il / 2 bl pl (cos ) e (1/ 2)il il 2ikf ( )
l
l
or
l
il / 2
= e i
pl (cos
i (2l 1)e
l
b P (cos ) e
l
(1 / 2 ) il i l
l
(4.84)
Since e+i and e-i are linearly independent, each of the quantities in square brackets must
vanish. Therefore from the coefficient of e+i, we have
(i ) (2l 1) b e P (cos ) e
i l
l
0
il / 2
l
l
i
bl i l (2l 1)eil
or
(4.85)
Using eqn. (4.85) into equation (4.84) and equating the coefficients of ei to zero.
2ikf ( ) i l (2l 1)e
2i l
e
il / 2
Pl (cos ) i l (2l 1) Pl (cos )e il / 2
l
(2l 1) Pl (cos ) e2i l ,
l
eil / 2 i 1
= (2l 1) Pl (cos )ei eil e il
l
(4.86)
l
f ( )
or f ( )
1
(2l 1) Pl (cos )e
k l 0
i l
1
(2l 1) Pl (cos )e
k l 0
i l
e
il
e
2i
i l
sin l ,
(4.87)
The differential cross –section is then () =| f() |2
1
2
[ 2l 1) Pl (cos )ei l sin l ]
=
k 2 l 0
() =
l 0
(2l 1) 2 Pl (cos ) sin 2 l
k2
2
(4.88) 117
The total scattering cross – section is the integral of the above eqn.
ie.
t =
( ) d 2
t =
2
1
1
1
1
2
k2
(2l 1)
2
k2
(2l 1) sin
l
1
(4.89)
| ( ) | d (cos ).
2
k2
1
| f ( )|2 d (cos ).
(2l 1) 2 P1 (cos ) sin 2 l d (cos )
2
2
1
sin 2 l Pl (cos ) d (cos )
2
1
l
2
l
2
2l 1
from the orthogonality relation
1
1
Pl (cos ) d (cos )
2
4
k2
Hence, t =
2
2l 1
(2l 1) sin
2
l
(4.90)
l
The eqn. (4.90) shows that different partial waves contribute independently to the total cross
– section. Thus the phase shifts completely determine the scattering, and scattering cross –
section vanishes when each of the l is 0 or 180. Cross section is maximum when
1 = /2, 3/2,...
Phase shifts are related to the potential energy function by the relation
Sin l =
-
a
0
V
Fl ( ) f l ( )d
E
(4.91)
Where fl() and Fl() are the functions in the differential equations
d 2 fl
d 2
and
l (l 1)
1 2 f l 0
(4.92)
2
d Fl V l (l 1)
1
Fl 0
2
d
2
E
(4.93)
To find the Phase shifts (l ):
Multiplying equation (4.93) by fl and equation (4.92) by Fl and subtracting, we get
d
d
When
f l ' Fl f l Fl ' V Fl f l 0
E
(4.94) 118
f l ' Fl f l Fl ' cos 1 l sin 1 l l
2
2
-sin 1 l cos 1 l l sin l
2
2
Hence from equation (4.94), we get
d
V
sin l Fl f l 0
d
E
V
Fl f l d
E
when the phase shifts are small, i.e., Fl() fl().
sin l
or
l
a
o
V
| f l ( ) |2 d
E
(4.95)
where a is the range of potential energy function.
At low energies approaching zero, only phase shift for l=0 partial waves are important and
scattering becomes isotropic.
4.15 Optical theorem. The optical theorem relates the total scattering cross – section to the
scattering amplitude in the forward direction (the scattering amplitude for = 0). Substituting
= 0 in eqN. (4.87) we get,
f (0)
1
(2l 1)Pl (cos(0))e
k l 0
i
l
sin l
Pl (cos 0) Pl (1) 1
But
f (0)
1
(2l 1)(cos l i sin l ) sin l
k l 0
The imaginary part of f(0) i.e., Im f(0) is given by
Im f (0) =
1
2
(2l 1) sin l
k l 0
(4.96)
According to eq. (4.90), we have
t
or
4
4 1
2
(2l 1) sin l .
2
k k
k l 0
t
4
. Im f (0)
k
(2l 1) sin
l 0
2
l
(4.97)
Thus the imaginary part of fprward scattering amplitude measures the suffering in intensity of
incident beam due to scattering. 119
UNIT V
5.1 Schrodinger relativistic equation (K.G.Equation)
Classical Hamiltonian function is given by
H= 1 ( p x2 p y2 p z2 ) V
2m
(5.1)
Schrodinger equation in Hamiltonian form is
i
(r , t )
2 2
(r , t ) V (r , t )
t
2m
(5.2)
Let the particle of rest mass m0 is moving in free space (V=0)
Now, the velocity of the particle is given by,
v2 x 2 y 2 z 2
Energy E=mc2=
m0c 2
(1
2
(5.3)
, where v
(5.4)
c
1
m02 ( x 2 y 2 z)2
m02v 2
m02 2c 2
p x p p (1 2 ) (1 2 ) (1 2 ) m02c2 1 2 1
2
2
y
2
z
2
= E2 m0 2c 2
c
p
2
x
p y2 pz2 m0 2 c 2 - E2 =0
2
c
Using the operators,
E i
h etc., we get
, px
t
2i x
2
2 2 4 2 2 2 1 2
m0 c 2 2 0
x 2 y 2 z 2 h2
c t
BY using eqn.(5.4)
(5.5) 120
or
or
2
2
2
4 2
1 2
2 2 2 2 m02c 2 2 2 0
y
z
h
c t
x
2 1 2
m 2c 2
2 2 0 2
c t
m02 c 2
where □2 = 2 12
□
2
c
2
2
t 2
is called D’Alembertian operator
(5.6)
(5.7)
The above equation is called Klein-Gordan equation for free particle.
5.2 K.G.equation in the presence of electromagnetic field
Consider a particle of charge ‘-e’ with rest mass m0 in an e.m. field.
The equation of motion is written as
m0 v
d
eE e (v H )
dt (1 v 2 c 2 )
c
(5.8)
where, v-velocity of the particle, c-velocity of light, E-electric field intensity, H-magnetic
field. E and H can be expressed as
E
1 A
c t
and H A .
Here, A and are the vector and scalar potentials.
Substituting in eqn. (5.8) we have
m0 v
d
e A e
e
[v ( A )]
dt (1 v 2 c 2 )
c t c
Consider the x-component value,
d
m0 v x
e e Ax e [v y (curl A) z v z (curl A) y ]
2
dt (1 v x c 2 )
x c t
c
(5.9) 121
e
e A
e dAx Ax
A
A
A
A A
A
vx x v y x vz v y y x vz z x x
x c dt
x
y
z
y
z t
x
c x
e
A
e dAx e Ax
A
vx y v y z vz
x c dt
c x
x
x
e
e dAx e
Ax vx Ay v y Az vz
x c dt
c x
e
e dAx e
A.v
x c dt
c x
m0 v x
d
e
e
Ax e ( A.v)
2
2
dt (1 v c ) c
x
c
x
or
(5.10)
This eqn. can also be written as
d
e
e
2
2
2
2
2
2
m0 c (1 v c e ( A.v)
m0 c (1 v c e ( A.v) (5.11)
dt vx
c
c
x
This is the Lagrangean equation
ie. d L L , where we have assumed L= m0c 2 (1 v 2 c 2 e e ( A.v)
dt vx
x
c
Canonical momentum px L 1
vx
2
m0c 2
2vx e
Ax
2
c
(1 v c ) c
2
2
or in general,
m0v
p
(1 v 2 c 2 )
e
A
c
E p j q j L p.v L
Total energy
=
m0v 2
(1 v c )
2
2
e
e
( A.v) m0c 2 (1 v 2 c 2 e ( A.v)
c
c
(5.12) 122
=
E=
m0
(1 v 2 c 2 )
m0 c 2
(1 v 2 c 2 )
v
2
c 2 (1 v 2 c 2 ) e
e
(5.13)
From eqn. (5.12)
p
or
m0v
e
A
c
(1 v 2 c 2 )
e
( p A)
c
2
2
m c
= ( p e A) 2
c
2
0
2
m0 v
(1 v
m02 c 2
= 1
(1 v 2 c 2 ) c 2
2
2
2
c )
m0 c 2
2
2
(1 v c )
2
= 12 ( E e ) 2 ,
c
since E=
c2 ( p
or
m0 c 2
(1 v 2 c 2 )
e
e 2
A) m02 c 4 ( E e ) 2 0
c
(cp eA) 2 m02c 4 ( E e ) 2 0
(5.14)
5.3 Dirac’s relativistic equation for a free electron
Dirac modified the Hamiltonian which is linear in momentum and mass as
H c . p m0c 2
(5.15)
We know the eigen value eqn. H E which can be written as
( E c . p m0c 2 ) 0
Using operators for E and p
(i
ic . m0 c 2 ) 0
t
(5.16) 123
Any solution of the above eqn. must be the solution of Schrodinger relativistic eqn
( E 2 p 2c 2 m02c 4 ) 0
eqn
commute with E
and
(5.16)
( E c . p m0c 2 )
Multiplying
by
and
(5.17)
(keeping
px, p y and pz commute;
p , while and do not commute ), we
can show that this agrees with schrodinger relativistic eqn (5.17) if and satisfy the
following relations
x2 y2 z2 1
x y y x y z z y z x x z 0
2 1
x x y y z z 0
(5.18)
and can be represented as
1
0
0
;
0 1
with
0
x
0
x
x etc.
0
Here x 0 1 ; y 0 i ; z 1
1 0
All
and
i
0
0
0 1
(5.19)
(5.20)
matrices called Dirac operators are of 4X4 dimension
5.4 Dirac’s Free particle solution or plane wave solution
From the equation
H i
t
Hamiltonian H c( x px y p y z pz ) m0c 2
and
(5.20a) 124
0
0
x px
0
p
x
0
0
0
px
px
0
0
0
px
0
etc. Substituting these x px ; y p y ; z pz values in eq.(5.20a)
0
0
we get
1
t
2
0 m0c 2 2 c( px ip y ) 3 cp z 4 i
t
cp z 1 c( px ip y ) 2 m0c 2 3 0 i 3
t
4
c( px ip y ) 1 cp z 2 0 m0c 2 pz 4 i
t
m0c 2 1 0 cp z 3 c( px ip y ) 4 i
(5.21)
Using the operator forms, the above eqn becomes
( E m0c 2 ) 1 cp z 3 c( px ip y ) 4 0
( E m0c 2 ) 2 c( px ip y ) 3 cp z 4 0
( E m0c 2 ) 3 cp z 1 c( px ip y ) 2 0
(5.22)
( E m0c 2 ) 4 c( px ip y ) 1 cp z 3 0
The above set of eqns have solutions if the determinant of the coefficients of is zero.
E m0 c 2
0
cp z
c( p x ip y )
2
0
E m0c
c( p x ip y )
cp z
0
cp z
c( p x ip y )
E m0 c 2
0
c( p x ip y )
cp z
0
E m0c 2
(5.23)
The determinant value gives
( E 2 m02c 4 c 2 p 2 ) 2 0
or
E ( p 2c 2 m02c 4
(5.24) 125
Taking 1 =1, 2 =0 we get, 3
Taking 1 =0, 2 =1 we get, 3
c( px ip y )
cp z
2 , 4
E m0c
E m0c 2
c( px ip y )
E m0c
2
, 4
(5.25)
cp z
.
E m0c 2
(5.26)
If we consider the negative roots E ,we get two more solutions as
1
cp z
E m0c 2
1
2
c( px ip y )
E m0c 2
c( px ip y )
E m0c 2
2
cp z
E m0c 2
3 =1
3 =0
4 =0
(5.27)
4 =1
(5.28)
If we write j (r, t ) u j ei ( k .r t ) with j=1,2,3... where u j are numbers, we can rewrite the
solution as
1
0
cp
z
u 1 ( p)
2 ;
E m0 c
c( p x ip y )
E m c2
0
0
1
c( p x ip y )
2
u ( p)
;
E m0 c 2
cp z
E m c2
0
cp z
2
E m0 c
1
1
c
(
p
ip
)
;
x
y
v ( p)
E m c2
0
1
0
c( p x ip y )
2
E m0 c
cp z
2
2
v ( p)
E m0 c
0
1
(5.29)
u’s and v’s are called Dirac spinors. u represents positive energy spinors with spin up and v
represents negative energy spinors with spin down. Solution for Dirac equation for a free
particle is thus obtained as the product of Dirac spinors multiplied by e
i ( k .r )t
. 126
5.5 Charge and current densities:
S(r, t )
P
(r, t ) 0,
t
(5.30)
Which is well known equation of continuity. The current density expression S(r, t ) has the
same form as in non-relativistic case, but the inspection of expression P(r, t ) indicates that it
can not be interoperated as position probability density in analogy with non-relativistic case
in which P(r, t ) due to following reason, :
The expression P(r, t ) may be expressed as
P(r, t )
2
t
2imc t
1
2mc 2
i
i
t
t
(5.31)
Now using Schroedinger equation in operator form and keeping in mind that the
Hamiltonian operator associated with dynamic observable energy E is Hermitian, we have
H i
i.e., i
E
t
t
and
H i
i.e., i
E
t
t
So equation (5.31) may be expressed as
1
E
P(r, t )
2
2mc
i.e.,
P(r, t )
1
2E
2
2mc
E
.
mc 2
(5.32)
From the expression E ( p 2 c 2 m 2 c 4 ) , e note that the energy of a particle can be either
positive or negative. Thus it follows that the expression for P(r, t ) is not definitely positive an
hence it can not be regarded as conventional position probability density. Thus it is necessary
to reinterpret
if Klein-Gordan equation is to be used. This was done by Pauli and
Wesskopf in 1934. According to them P multiplied by (e P) can be interpreted as charge
density which may be positive and negative since charge can have either sign ; then e S will
be corresponding current density. 127
5.6 Matrics for α and
The squares of all the four matrics are unity; so that their eigen-values are +1 and 1. Let us
arbitratily choose as the matrix that is to be diagonal and we rearrange its rows and
columns so that all the +1 eigen values are grouped together in the matrix of rank n and all
the 1 eigen values are grouped together in a matrix of rank m.
The matrix can be expressed as
1
0
=
1 1
(5.33)
which is an abbreviation of
1
0
=
0
0
0
0
0 1 0
0 0 1
0
1
0
0
(5.34)
All the four matrices x, y, z and are such that their squares are unity and they
anticommute with one another in pairs. We already have three well known 2×2 matrices x,
y and z called Pauli spin matrices; which satisfy the above properties, given by
0 1
0 i
1 0
x
,
,
y i 0 z 0 1
1 0
(5.35)
Since a 2 × 2 matrix has four elements, there are four and only four, independent 2 × 2
matrices, three of these x, y, z. The only other matrix linearly independent of these three
is
1 0
I=
0 1
which is a unit matrix and therefore commutes rather than anticommutes with every . Hence
we can not have fourth 2 × 2 matrix which satisfies both properties of Dirac Matrices. Now
we show that the Dirac Matrices must be even-dimensional. 128
Let us choose a representation in which is diagonal N × N matrix i.e.
O
b
= 1 bi
b N
O
(5.36)
As 2 = 1, b i2 1 and bi = 1 (I = 1, 2, . . . N)
Since anticommutes with each component of , we have
k + k = 0 (k = x, y, z)
This relation may be expressed as
k = k
k1 k k1 k
or
k1 k , as k1 k 1 ; we have
Taking trace of both sides, we get
Trace ( k1 k ) = Trace
or
Trace ( k k1 ) = Trace [Since Trace (ABC) = Trace (CAB)]
Trace () = Trace ()
(Since k k1 I)
(5.37)
This gives 2(Trace ) = or Trace () = 0
Similarly Trace (k) = 0
Thus
Trace () = Trace (k) = 0
This equation shows that the trace of each of matrices k and must be zero :
In matrix (13) let r of the bi’s are + 1 and the rest s of bi’s are 1 i.e.
b1 = b2 = br = 1
and
(5.38) 129
br+1 = br+2 = . . . = b3 = 1
so that r + s = N
(5.39)
But the condition that Trace () = 0 requires that
N
b
i l
i
r s 0 i.e., r = s
In view of this equation (16) shows
N = 2r
(540)
Thus Dirac matrices and must be even dimensional. Therefore we can not use 3 × 3
matrices. The next simplest choice is 4 × 4 matrices. As eigen values of all the four matrices
are +1 and 1. Let us arbitrarily choose as the matrix which is to be diagonal and we
arrange its rows and columns so that all the +1 eigen values are grouped together and all the
1 eigen values are grouped together in a matrix as
I
0
=
0 I
1
0
=
0
0
or
0
0
0 1 0
0 0 1
0
1
0
0
Therefore the matrix for k may be expressed as
0
k =
x 2
x1
0
(5.41)
where x1 has n rows and columns and x2 has m rows and columns. Since the square of
(5.41) is a unit matrix, we note that
x1 x2 = (1) n × n
x2 x1 = (1) m × m
(5.42)
(5.42A)
The unit matrix appearing on R.H.S of (5.42) has n rows and n columns while the unit matrix
on R.H.S. of (5.42A) has m rows and m columns. But no two matrices exist that satisfy (5.42) 130
and (5.742A) simultaneously if m n. Therefore we must have m = n = 2 for
4 × 4
matrices. It is apparent that y and z can be put in a form similar to (5.41).
Using Pauli spin matrices x, y and z and choosing
x1= x2= σx; then
0
x =
x
0
x 0
0 0
1
0 0 1
0 1 0
1 0 0
0 0 0
0
y =
y
0 0 0 i
y 0 0 i 0
0 0 i 0 0
i 0 0 0
0
z =
z
0 0
z 0 0
0 1 0
0 1
Similarly
1 0
0 1
0 0
0 0
(5.43)
We already have
1
0
=
0
0
0
0
0 1 0
0 0 1
0
1
0
0
These 4 × 4 matrices are evidently Hermitian and in abbreviated form may be expressed as
1 0 0
=
; 0
0 1
where each element is a matrix with two rows and two columns.
(5.44) 131
Negative Energy states
Now it is evident that there are two continuous energy spectrum E m02c 4 c 2 p 2 and
E
m02c 4 c 2 p 2
.When the Dirac electrons are at rest (p=0) then
E and E are
separated by
an amount of 2m0c 2 . Dirac suggested that all negative energy states are completely filled.
When electron from negative energy state picks up some energy and goes to positive state, a
vacancy called Dirac hole is created in negative energy states. This Dirac hole has all
similarities with an ordinary electron except a charge of ‘+e’. This is the antiparticle of
electron called positron.
5.7 Zitterbewegung Method
Consider the motion of an electron according to Dirac equation in Heisenberg representation.
TheHamiltonian is time dependent. The relativistic Hamiltonian in electromagnetic field
described by scalar potentials and vector potential A is
H (cp eA) βmc eφ
2
(5.45)
Now according to Heisenberg representation, the equation of motion for operator
x
1
x, H 1 x, (cp eA) βmc 2 eφ
i
i
x
(5.46)
Omitting the terms which commute with x, we get
x
1
x, x cp x c x x, p x c x i c x
i
i
i
(5.47)
But the velocity operator r v is given by
vc
(5.48)
The probability density function according to Dirac equation is † ; thereby given
momentum density for Dirac particle as † p and he velocity density for Dirac particle
appears to be
†v † (c ) c †
(5.49) 132
The eigen-values of each
are
1 therefore the observed value of any component of velocity
is c this is peculiar result, since according to relativistic mechanics speed of light is the
upper limit for the speed of material particle. To find the significance of the this result let us
investigate the motion of an electron under no field (i.e. A = 0 and 0). the Hamiltonian
then is expressed as
H c p βmc2
(5.50)
Writing the equation of motion for operator x , we have
x
But
1
1
[ x , H ]
( x , H H x )
i
i
(5.51)
x H H x x (c p mc2 ) (c p mc2 ) x
= c x ( x px y pz z pz mc) c ( x px y pz z pz mc) x 2 cp x
H x 2cpx H x H .
(5.52)
In view of this equation (5.51) gives
x
1
(2 x H 2cp x )
i
(5.53)
As for a free particle the energy and momentum are conserved, we have
i p x [p x H] 0 and
i H [H, H] 0
(5.54)
Therefore H and p x are independent of time. Keeping this in mind, the differential of equation
(5.53) with respect to time gives
x
1
(2 x H )
i
(5.55)
which may be expressed as
x 2i
H
x
(5.56) 133
Integrating w.r. t. time t , we get,
log ( x )
2i
Ht K
(5.57)
K being a constant of integration.
If at
t 0, x ( x ) t 0; we have
K log ( x )t 0
(5.58)
Equation (5.57) gives
log ( x )
2i
Ht log ( x ) t 0
i.e.
x
2i
log
Ht
(
)
x t 0
or
2i
x ( x ) t 0 exp Ht
(5.59)
Substituting this in (5.53), we get
2i 1
( x ) t 0 exp Ht (2 x H 2cp x )
i
(5.59A)
We have H 2 p 2c2 m2 c4 E 2 , therefore
H ( p 2c 2 m2 c 4 ) H 1
This implies that H is the reciprocal of H 1 with eigen value E 1
From equation (5.59A), we have
x cpx H 1 1 i ( x ) t 0 exp 2i Ht H
2
Therefore
1
2i
x c x c 2px H 1 i c ( x ) t 0 exp Ht H
2
1
1
(5.60) 134
Integrating,
1
2i
x c 2 p x H 1t c 2 ( x ) t 0 exp Ht H
4
2
x0
(5.60A)
where x0 is constant of integration
1
1
1
Since x †
[ x , H ]† [ H , x ] [ , H ] x
i
i
i
Also p x and H are Hermitian, therefore right hand side of (5.60A) is Hermitian if we ignore
the constant of integration which would obviously represent the initial position of particle.
Then x would be an observable quantity and its expectation value is given by
2
2
c pxt 1 c
2i
x
( x ) t 0 exp Et
2
E
4 E
(5.61)
Where ( x ) t 0 is the eigen value of x at t 0 .
Now we have
E ( p2c2 m 2 c4 ) 1 / 2 mc2
and writing 2 E , equation (5.61) may be expressed as
2
x
c pxt
E
2
1 c
( x ) t 0 e
4 E 2
it
(5.62)
In this equation the first term represents usual term
c 2 p x t c 2 p x t m t
x
xt
2
E
m
mc
(5.63)
of classical mechanics and the second term because of the exponential factor represents the
motion of particle, oscillating with angular frequency
.
This trembling motion of the
electron was first observed by Schroedinger and is called the Zitterbewgung and it imparts 135
the value to the velocity of electron. That is the electron’s motion is something like the
superposition of classical motion and electromagnetic wave motion
However the frequency
2 E 2mc2
is so high that the departure from the classical mechanics term x t is undetectable. The
Zitterbewgung did not appear in out no relativistic theory, the reason that this phenomenon is
due to the rest energy of the electron which remains unaccounted in classical mechanics.
5.8 SPIN-ORBIT ENERGY
The spin orbit coupling energy follows as a result to Dirac equation in a central field. The
term is however of order v 2 / c 2 and in order to obtain a consistent approximation we proceed
by two – component reduction of Dirac equation in the central field V (r ).
[c p mc2 V (r )] E .
(5.64)
Writing ψ 1 which represent the first and the last two components of respectively.
2
i.e.
0
1
2
1 0
1 0 1
c
p 0 1 mc V (r ) 0 1 E
2
0
2
i.e.
c p 2
mc 2 1
2
mc 2
c p 1
V 1 E 1
V E
2
2
This equation is equivalent to following two equations
c p 1 mc2 1 V 1 E 1
and
c p 1 mc2 2 V 2 E 2
( E V mc2 ) 1 c p 2 0
( E V mc2 ) 21 c p 1 0
(5.65)
Assuming the 1 and 2 together constitute a non-relativistic energy eigen –function, which
means that 136
E E mc
2
is regarded as a number rather than an operator, the non-relativistic energy E and V are
assumed to be smaller in comparison with mc 2 .
The wave equations (5.65) then become
E V (r ) 1 c p 2 0 ...(a)
E 2mc V (r )
2
2 c p 1 0 ...(b)
(5.66)
From (5.66b), we have
2
c p
ψ1
E 2mc2 V (r )
Substituting this value of 2 in (5.66b), we get
E V (r)1 c2 ( p) [E 2mc2 V (r)] 1 p 1
=
p E V (r ) 1
1
p 1
2m
2mc2
p E V
1
2m
2mc2
=
=
p 1
1
( E V )
( p) (( p) 1 ( p)
p 1
2m
4m 2 c 2
Using the identity
( B) ( C) B C i B x C
we have ( p) ( p) p p i p x p p
2
(5.67A)
For any any function u,
p, Vu (p V Vp) u p V u V p u
= (Vu ) V u
i
i
(5.67) 137
= Vu u V V u
i
=
i
u V V u
i
i
i.e.
(pV Vp)u (i V) u
or
(pV Vp)u i V
pV V p i V
( p) V V ( p) i V
( p)
(5.68)
( E V )
E
1
p
( p) ( p) ( p) ( p)
2 2
4m c
4m 2 c 2
4m 2 c 2
=
E'
1
( p) ( p) V ( p) ( p) - (i V )( p) )
4m 2 c 2
4m 2 c 2
using eqn. (5.68)
= E2
4m c
E V
2 2
4m c
=
2
p2 -
1
i
Vp 2
( V ) ( p) using eqn.(5.67a)
2 2
4m c
4m 2 c 2
i
2
p 2 2 V p V x p
4m c
(5.69)
Using (5.67A) and (5.69), equation (5.67) gives
( E V ) 1
1 2
p 1
2m
E V 2
i
p
( V p i V x p 1 .
2 2
2 2
4m c
4m c
E 'V p 2
E ' 1 1
V
2mc 2 2m
2
V 1
V x p 1
1
2 2
4m c
4m 2 c 2
Now if V is spherically symmetric, we have
and
V
1 dV
r
r dr
(V )
dV
dr r
(5.70) 138
1 dV
1 dV
V x p
rx p
(r x p)
r dr
r dr
2
and noting that E V p , equation (5.70) gives
2m
( p 2 / 2m)
E 1 1
2mc2
p2
2 dV 1
1 dV
V 1 2 2
r x p 1
2 2
4m c dr r
4m c
r dr
2m
Using 1 σ S and r x p L, we have
2
p2
p4
2
dV
1 1 dV
E 1
V
S .L 1 (5.71)
3 2
2 2
2 2
8m c
4m c dr r 2m c r dr
2m
The first and second terms on right hand side of above equation give the non-relativistic
Schroedinger equation. The third term is the classical relativistic mass correction term which
can be obtained by the expression.
E E mc2 ( p 2c 2 m2c 4 ) 1 / 2 mc2
1/ 2
p 2c 2
= mc 1 2 4
mc
2
2
p
= mc 1
2m 2 c
2
=
p2
p4
2m 8m3c 2
1/ 2
p2
mc mc 1 2 2
mc
2
2
p
2
4
4
8m c
4
mc2
......... mc
2
(5.72)
The fourth term is a similar relativistic correction to the potential energy, which does not
have a classical analogue and the last term is the spin orbit coupling energy which appears as
an automatic consequence of the Dirac equation.
Thus the spin-orbit coupling energy is
Us 0
1 1 dV
S .L,
2m2c 2 r dr
(5.73)
Related Documents
- Sample Student Responses and Scoring Commentary
- Permutations and Combinations Worksheet Answer Key
- MAT 172 SP24 Unit 3 Test Key A
- Lecture 1 - Derivative of a Polynomial Functions
- Guided Notes 5.2: The Definite Integral
- Triangle Exterior Angle Sum Theorem
- More Graphing Exponentials Practice Key
- Limits of Trigonometric Functions
The Schrodinger Equation and Stationary States
of 142
Recommended Documents
New Documents from this Course
Report
Tell us what’s wrong with it:
Thanks, got it!
We will moderate it soon!
Report
Tell us what’s wrong with it:
Free up your schedule!
Our EduBirdie Experts Are Here for You 24/7! Just fill out a form and let us know how we can assist you.
Take 5 seconds to unlock
Enter your email below and get instant access to your document