Quantum formula sheet

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QUANTUM MECHANICS FORMULA SHEET

Contents

1. Wave Particle Duality

1.1 De Broglie Wavelength

1.2 Heisenberg’s Uncertainty Principle

1.3 Group velocity and Phase velocity

1.4 Experimental evidence of wave particle duality

1.4.1 Wave nature of particle (Davisson-German experiment)

1.4.2 Particle nature of wave (Compton and Photoelectric Effect)

2. Mathematical Tools for Quantum Mechanics

2.1 Dimension and Basis of a Vector Space

2.2 Operators

2.3 Postulates of Quantum Mechanics

2.4 Commutator

2.5 Eigen value problem in Quantum Mechanics

2.6 Time evaluation of the expectation of A

2.7 Uncertainty relation related to operator

2.8 Change in basis in quantum mechanics

2.9 Expectation value and uncertainty principle

3. Schrödinger wave equation and Potential problems

3.1 Schrödinger wave equation

3.2 Property of bound state

3.3 Current density

3.4 The free particle in one dimension

3.5 The Step Potential

3.7 Potential Barrier

3.7.1 Tunnel Effect





3.8 The Infinite Square Well Potential

3.7.1 Symmetric Potential

3.9 Finite Square Well Potential

3.10 One dimensional Harmonic Oscillator

4. Angular Momentum Problem

4.1 Angular Momentum

4.1.1 Eigen Values and Eigen Function

4.1.2 Ladder Operator

4.2 Spin Angular Momentum

4.2.1 Stern Gerlach experiment

4.2.2 Spin Algebra

4.2.3 Pauli Spin Matrices

4.3 Total Angular Momentum

5. Two Dimensional Problems in Quantum Mechanics

5.1 Free Particle

5.2 Square Well Potential

5.3 Harmonic oscillator

6. Three Dimensional Problems in Quantum Mechanics

6.1 Free Particle

6.2 Particle in Rectangular Box

6.2.1 Particle in Cubical Box

6.3 Harmonic Oscillator

6.3.1 An Anistropic Oscillator

6.3.2 The Isotropic Oscillator

6.4 Potential in Spherical Coordinate (Central Potential)

6.4.1 Hydrogen Atom Problem





7. Perturbation Theory

7.1 Time Independent Perturbation Theory

7.1.1 Non-degenerate Theory

7.1.2 Degenerate Theory

7.2 Time Dependent Perturbation Theory

8. Variational Method

9. The Wentzel-Kramer-Brillouin (WKB) method

9.1 The WKB Method

9.1.1 Quantization of the Energy Level of Bound state

9.1.2 Transmission probability from WKB

10. Identical Particles

10.1 Exchange Operator

10.2 Particle with Integral Spins

10.3 Particle with Half-integral Spins

11. Scattering in Quantum Mechanics

11.1 Born Approximation

11.2 Partial Wave Analysis

12. Relativistic Quantum Mechanics

12.1 Klein Gordon equation

12.2 Dirac Equation





1.Wave Particle Duality

1.1 De Broglie Wavelengths

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

mvh

ph where h is Plank’s constant

For relativistic case, the mass becomes

2

2

0

1cv

mm

where m0 is rest mass and v is

velocity of body.

1.2 Heisenberg’s Uncertainty Principle

“It is impossible to determine two canonical variables simultaneously for microscopic

particle”. If q and qp are two canonical variable then

2

qpq

where ∆q is the error in measurement of q and ∆pq is error in measurement of pq and h is

Planck’s constant ( / 2 )h .

Important uncertainty relations

2

xPx (x is position and xp is momentum in x direction )

2

tE ( E is energy and t is time).

2

L (L is angular momentum, θ is angle measured)





1.3 Group Velocity and Phase Velocity

According to de Broglie, matter waves are associated with every moving body. These

matter waves moves in a group of different waves having slightly different wavelengths.

The formation of group is due to superposition of individual wave.

Let If tx,1 and tx,2 are two waves of slightly different wavelength and frequency.

tdxdkkAtkxA sin,sin 21

21 tkxtddkA

sin

22cos2

The velocity of individual wave is known as Phase

velocity which is given ask

vp

. The velocity of

amplitude is given by group velocity vg i.e.dkdvg

The relationship between group and phase velocity is given by

ddv

vvdkdv

kvdkdv p

pgp

pg ;

Due to superposition of different wave of slightly different wavelength resultant wave

moves like a wave packet with velocity equal to group velocity.

1.4 Experimental evidence of wave particle duality

1.4.1 Wave nature of particle (Davisson-German experiment)

Electron strikes the crystals surface at an

angle . The detector, symmetrically located

from the source measure the number of

electrons scattered at angle θ where θ is the

angle between incident and scattered electron

beam.

The Maxima condition is given by phdnor

dn

where2

cos2

sin2

S D

gv

phv

t





1.4.2 Particle nature of wave (Compton and Photoelectric Effect)

Compton Effect

The Compton Effect is the result of scattering of a photon by an electron. Energy and

momentum are conserved in such an event and as a result the scattered photon has less

energy (longer wavelength) then the incident photon.

If λ is incoming wavelength and λ' is scattered

wavelength and is the angle made by

scattered wave to the incident wave then

cos1' cm

ho

where cm

ho

known as c which is Compton wavelength ( c = 2.426 x 10-12 m) and mo is

rest mass of electron.

Photoelectric effect

When a metal is irradiated with light, electron may get emitted. Kinetic energy k of

electron leaving when irradiated with a light of frequency o , where o is threshold

frequency. Kinetic energy is given by

max 0k h h

Stopping potential sV is potential required to stop electron which contain maximum

kinetic energy maxk .

0seV h h , which is known as Einstein equation

photonincident photon scattered

ElectronTarget Electron Scattered





2. Mathematical Tools for Quantum Mechanics

2.1 Dimension and Basis of a Vector Space

A set of N vectors N ,........, 21 is said to be linearly independent if and only if the

solution of equation

N

iiia

1

0 is 1 2 Na = a = ..... a =0

N dimensional vector space can be represent as

N

iiia

1

0 where i = 1, 2, 3 … are

linearly independent function or vector.

Scalar Product: Scalar product of two functions is represented as , , which is

defined as dx * . If the integral diverges scalar product is not defined.

Square Integrable: If the integration or scalar product 2, dx is finite then the

integration is known as square integrable.

Dirac Notation: Any state vector can be represented as which is termed as ket

and conjugate of i.e. * is represented by which is termed as bra.

The scalar product of and ψ in Dirac Notation is represented by (bra-ket). The

value of is given by integral rdtrtr 3* ,, in three dimensions.

Properties of kets, bras and brakets:

*

** aa

Orthogonality relation: If and are two ket and the value of bracket 0

then , is orthogonal.

Orthonormality relation: If and are two ket and the value of bracket 0

and 1 1 then and are orthonormal.

Schwarz inequality: 2





2.2 Operators

An operator A is mathematical rule that when applied to a ket transforms it into

another ket i.e.

A

Different type of operator

Identity operator I

Parity operator rr

For even parity rr , for odd parity rr

Momentum operator xP ix

Energy operator H it

Laplacian operator 2

2

2

2

2

22

zyx

Position operator rxrX

Linear operator

For 2211 aa if an operator A applied on results in 1 1 2 2a A a A

then operator A is said to be linear operator.





2.3 Postulates of Quantum Mechanics

Postulate 1: State of system

The state of any physical system is specified at each time t by a state vector t which

contains all the information about the system. The state vector is also referred as wave

function. The wave function must be: Single valued, Continuous, Differentiable, Square

integrable (i.e. wave function have to converse at infinity).

Postulate 2: To every physically measurable quantity called as observable dynamical

variable. For every observable there corresponds a linear Hermitian operator A . The

Eigen vector of A let say n form complete basis. Completeness relation is given

by1

n nn

I

Eigen value: The only possible result of measurement of a physical quantity na is one of

the Eigen values of the corresponding observable.

Postulate 3: (Probabilistic outcome): When the physical quantity A is measured on a

system in the normalized state the probability P(an) of obtaining the Eigen value an of

corresponding observable A is

2

1

ngin

in

aP a

where gn is degeneracy of state and

nu is the Normalised Eigen vector of A associated with Eigen value an.

Postulate 4: Immediately after measurement.

If the measurement of physical quantity A on the system in the state gives the result

an (an is Eigen value associated with Eigen vector na ), Then the state of the system

immediately after the measurement is the normalized projection

n

n

P

P where Pn is

projection operator defined by n n .





Projection operator P : An operator P is said to be a projector, or projection operator, if

it is Hermitian and equal to its own square i.e. PP ˆˆ 2

The projection operator is represented by n nn

Postulate 5: The time evolution of the state vector t is governed by Schrodinger

equation: ttHtdtdi , where H(t) is the observable associated with total

energy of system and popularly known as Hamiltonian of system. Some other operator

related to quantum mechanics:

2.4 Commutator

If A and B are two operator then their commutator is defined as A,B AB-BA

Properties of commutators

† † †

, , ; , , ,

, , , ; , ,

, , , , , 0 (Popularly known as Jacobi identity).

C C

C C B C

C C C

, 0f

If X is position and xP is conjugate momentum then

1,n nxX P nX i and 1, n n

x xX P nP i

If b is scalar and A is any operator then , 0b

If [A, B] = 0 then it is said that A and B commutes to each other ie AB BA .

If two Hermition operators A and B , commute ie , 0 and if A has non

degenerate Eigen value, then each Eigen vector of A is also an Eigen vector of B .

We can also construct the common orthonormal basis that will be joint Eigen state of

A and B .

The anti commutator is defined as ,





2.5 Eigen value problem in Quantum Mechanics

Eigen value problem in quantum mechanics is defined as

n n na

where an is Eigen value and n is Eigen vector.

In quantum mechanics operator associated with observable is Hermitian, so its Eigen

values are real and Eigen vector corresponding to different Eigen values are

orthogonal.

The Eigen state (Eigen vector) of Hamilton operator defines a complete set of

mutually orthonormal basis state. This basis will be unique if Eigen value are non

degenerate and not unique if there are degeneracy.

Completeness relation: the orthonormnal realtion and completeness relation is given by

In

nnmnmn

1

,

where I is unity operator.

2.6 Time evaluation of the expectation of A (Ehrenfest theorem)

1 AA,Hd Adt i t

where ,A H is commutation between operator A and

Hamiltonian H operator .Time evaluation of expectation of A gives rise to Ehrenfest

theorem .

1d R Pdt m

, ,d P V R tdt

where R is position, P is momentum and ,V R t is potential operator.

2.7 Uncertainty relation related to operator

If A and B are two operator related to observable A and B then

B,A21B A

where 22 AA A and

22 BB A .





2.8 Change in basis in quantum mechanics

If k are wave function is position representation and k are wave function in

momentum representation, one can change position to momentum basis via Fourier

transformation.

12

12

ikx

ikx

x k e dk

k x e dx

2.9 Expectation value and uncertainty principle

The expectation value A of A in direction of is given by

AA

or

A n na P where nP is probability to getting Eigen value an in state .





3. Schrödinger Wave Equation and Potential Problems

3.1 Schrödinger Wave Equation

Hamiltonian of the system is given by 2

2PH Vm

Time dependent Schrödinger wave equation is given by t

iH

Time independent Schrödinger wave equation is given by EH

where H is Hamiltonian of system.

It is given that total energy E and potential energy of system is V.

3.2 Property of bound state

Bound state

If E > V and there is two classical turning point in which particle is classically trapped

then system is called bound state.

Property of Bound state

The energy Eigen value is discrete and in one dimensional system it is non degenerate.

The wave function xn of one dimensional bound state system has n nodes if n = 0

corresponds to ground state and (n – 1) node if n = 1 corresponds to ground state.

Unbound states

If E > V and either there is one classical turning point or no turning point the energy

value is continuous. If there is one turning point the energy eigen value is non-

degenerate. If there is no turning point the energy eigen value is degenerate. The particle

is said to be unbounded.

If E < V then particle is said to be unbounded and wave function will decay at ± ∞. There

is finite probability to find particle in the region.





3.3 Current density

If wave function in one Dimension is x then current density is given by

xxim

J x

**

2

which satisfies the continually equation 0 J

t

Where * in general J v 2J v where v is velocity of particle?

If Ji, Jr, Jt are incident, reflected and transmitted current density then reflection coefficient

R and transmission coefficient T is given by

r

i

JRJ

and t

i

JTJ

3.4 The free particle in one dimension

Hψ = Eψ xEdxd

m

2

22

2

ikxikx AeeAx

Energy eigen value E mk

2

22 where 2

2

mEk the eigen values are doubly degenerate

3.5 The Step Potential

The potential step is defined as

000

xVx

xVo

Case I: E > Vo

00

22

11

2

1

xBeAexBeAe

xikxik

xikxik

Hence, a particle is coming from left so D = 0.

R = reflection coefficient = incident

reflected

JJ

=

2

1 2

1 2

k kRk k

xo

oV





T = transmitted coefficient = incident

dtransmitte

JJ

=

1 22

1 2

4k kTk k

where 2221

22

oVEmkmEk

Case II: E < Vo

212011

mEkxBeAe xikxikI

22

202

EVmkxce oxkII

1r

t

JRJ

and 0t

i

JTJ

For case even there is Transmission coefficient is zero there is finite probability to find

the particle in x > 0 .

3.7 Potential Barrier

Potential barrier is shown in figure.

Potential barrier is given by

axaxV

xxV

00

00

0

Case I: E > Vo

0

00

1

22

11

33

22

11

xEexx

axDeCexxxBeAexx

xik

xikxik

xikxik

Where 212

mEk 22

2

oVEmk

Transmission coefficient 1

2 1sin14

11

T

a

Energy

o x

oV





Reflection coefficient

1

2 1sin141

R Where 2

2,

o

o

mVaVE

Case II: E < Vo

3.7.1 Tunnel Effect

1 1

2 2

1

0

0

0

ik x ik xI

k x k xII

ik xIII

Ae Be x

Ce Be x aFe x

21 sin 14 1

R h

and 211 sin 1

4 1T h

Where o

o

VEmVa ,2

2

For E << Vo

222

116

EVma

oo

o

eVE

VET

Approximate transmission probability akeT 22 where 22

2

EVmk o

3.8 The Infinite Square Well Potential

The infinite square well potential is defined as as shown in the figure

axax

xxV 00

0

Since V(x) is infinite in the region 0x and x a so the wave function corresponding

to the particle will be zero.

The particle is confined only within region 0 ≤ x ≤ a.

Time independent wave Schrödinger wave equation is given by

xV

o a x

ao

oV





Edxd

m 2

22

2

sin cosA kx B kx

B = 0 since wave function must be vanished at boundary ie 0x so sinA kx

Energy eigen value for bound state can be find by ka n where 1,2,3...n

The Normalized wave function is for thn state is given by

2 sinnn xx

a a

Which is energy Eigen value correspondence to thn

2

222

2manEn

where n = 1, 2, 3 .....

othonormality relation is given by

0

sin sin i.e.2

0

12

a

mnm x n x adx

L Lm n

a m

If x is position operator x P Px is momentum operator and n x is wave function of

particle in nth state in one dimensional potential box then 2 sinnn xx

a a then

*

2 22 * 2

2 2

*

2 2 2 22 *

2 2

2

2 2

0

2

n n

n n

x n n

x n n

ax x x x

a ax x x xn

P x i xx

nP x xm x a





The uncertainty product is given by 2221

121

nnPxx

The wave function and the probability density function of particle of mass m in one

dimensional potential box is given by

3.8.1 Symmetric Potential

The infinite symmetric well potential is given by

220

22axa

axoraxxV

Schrondinger wave function is given by

)(02

cos2

sin

)(02

cos2

sin

0,2

2cossin

22;

2 2

22

iikaBkaA

ikaBkaAso

xaxat

mEkwherekxBkxA

axaEdxd

m

Hence parity ( ) commute with Hamiltonian ( )H then parity must conserve

So wave function have to be either symmetric or anti symmetric

2a

2aa

xV

mEn

21

22

1

12 42 EEn

13 93 EEn

ax

a

sin21

ax

a

2sin2

2

ax

a

3sin2

3

21

22

23





For even parity

cosB kx and Bound state energy is given by 02

cos ka

,....5,3,102

na

nkka

Wave function for even parity is given as a

xna

cos2

For odd parity

Ψ (x) = A sin kx and Bound state energy is given by

sin 02ka

, 02ka nk

a

2,4,6,........n

For odd parity Wave function is given as a

xna

sin2

The energy eigen value is given by ,......3,2,12 2

222

nma

nEn

First three wave function is given by

a

xa

x

ax

ax

ax

ax

3cos2

2sin2

cos2

3

2

1

where a2 is normalization constant.

2

22

2ma

2a

2a

2

22

29

ma

2

22

24

ma

2a

2a

2a

2a

x3

x2

x1





3.9 Finite Square Well Potential

For the Bound state E < Vo

Again parity will commute to Hamiltonian

So wave function is either symmetric or

Anti symmetric

For even parity

2

22cos

2

axAex

axakxCx

axAex

xIII

II

xI

wave function must be continuous and differentiable at boundaries so using boundary

condition at 2a

one will get 2

tan kak

tan where 2a

2ka

For odd parity

2

22sin

2

2

1

axAex

axakxDx

axAex

xIII

x

using boundary condition one can get

cot where 2

2

EVm o 22

mEk

and 2

222

2amVn o which is equation of circle.

The Bound state energy will be found by solving equation

tann for even

2a

2a x

oV

xV





cotn for odd

222

2

amVn o

one can solve it by graphical method.

The intersection point of = tan (solid curve) and 2

2 222

omV a

(circle) give

Eigen value for even state and intersection point at = - cot (dotted curve) and

22 2

22omV a

(circle) give Eigen value for odd state.

The table below shows the number of bound state for various range of 20V a where Voa2

is strength of potential.

Voa2 Even eigen function Odd eigen function No. of Bound

states

m2

22 1 0 1

maV

m 24

2

222

0

22

1 1 2

maV

m 29

2

222

0

22

2 1 3

maV

m 216

29 22

20

22

2 2 4

n

tann

o2

23 2





The three bound Eigen function for the square well

3.10 One dimensional Harmonic Oscillator

One dimensional Harmonic oscillator is given by

xxmxV 22

21

The schrodinger equation is given by

Exmdxd

m

222

22

21

2

It is given that xm

and Ek 2

The wave function of Harmonic oscillator is given by.

22/14/1 2

!21

eH

nm

nnn

and energy eigen value is given by

nE = (n+1/2) ; n = 0, 1, 2, 3, ....

The wave function of Harmonic oscillator is shown

0H ( ) = 1 , 1H ( ) = 2 , 22H ( ) = 4 -2

It n and m wave function of Harmonic oscillator then

mnnm dxxx

x2a

2a

x2

x2a

2a

x1

x

x3

1n

20E

xV

0n

2n

23

1E

25

2

E





Number operator

The Hamiltonian of Harmonic operator is given by 222

21

2Xm

mPH x

Consider dimensioned operator as HPX ˆ,ˆ,ˆ where

Xm

X ˆ

ˆx xP m P HH ˆ so 22 ˆˆ

21ˆ XPH x .

Consider lowering operator xiPXa ˆ2

1 and raising † 1 ˆ2 xa X iP

2/1ˆ NH

where †N a a and 1/ 2H N N is known as number operator

n is eigen function of N with eigen value n.

N n n n and 12

H n N n

nnnH

21 where n = 0, 1, 2, 3,………

Commutation of a and †a : †[a, a ] = 1 , †[a , a] = -1 and [N, a] = -a , † †[N a ] = a

Action of a and †a operator on n

11

001

nnna

abutnnna

Expectation value of 22 ,,, XX PPXX in stationary states

mnPnm

X

PX

miaaPaa

mX

X

X

X

21,12

2

0,02

,2

22

2;0,

21

XX PXnfornPX





4. Angular Momentum Problem

Angular momentum in Quantum mechanics is given kLjLiLL zyxˆˆˆ

Where xyzzxYyzX YPXPLXPZPLZPYPL ;;

andx

iPX

, y

iPY

, z

ipZ

Commutation relation

,x y zL L i L , ,y z xL L i L , ,z x yL L i L

2 2 2 2X Y ZL L L L and 2 , 0XL L , 2 , 0YL L , 2 , 0ZL L

4.1.1 Eigen Values and Eigen Function

iL

iL

SiniL

Z

Y

X

cotsincos

cotcos

Eigen function of ZL is

ime

21 .

and Eigen value of ZL m where m = 0, ± 1, ± 2...

L2 operator is given by

2

2

222

sin1sin

sin1

L

Eigen value of 2L is 2( 1)l l where l = 0, 1, 2, ... l

Eigen function of 2L is mlP where m

lP is associated Legendre function

L2 commute with Lz so both can have common set of Eigen function.

, ( )m m iml lY P e is common set of Eigen function which is known as spherical

harmonics .

The normalized spherical harmonics are given by





2 1 !

( , ) 1 cos4 !

mm m iml l

l l mY P e

l m

l m l

41,0

0 Y

ieY sin83,1

1 , 01

3, cos4

Y

, 11

3, sin8

iY e

,1, 22 ml

ml YllYL l = 0, 1, 2, ……. And

,, ml

mlZ YmYL m = -l ,(-l +1) ..0,….. (l – 1)( l ) there is 2 1l

Degeneracy of 2L is 2 1l .

Orthogonality Relation 2

' '0 0

, , sinm ml l ll mmd Y Y d

4.1.2 Ladder Operator

Let X YL L iL and X YL L iL

Let us assume ml, is ket associated with 2L and ZL operator.

22 , 1 ,L l m l l l m 0,1, 2,.......l

, , , ...0,...zL l m m l m m l l

Action of L+ and L- on ml, basis

, 1 1 , 1L l m l l m m l m

, 1 1 , 1L l m l l m m l m

Expectation value of XL and YL in direction of ml,

0xL , 0yL 2

2 2 212X YL L l l m

2

212X YL L l l m





4.2 Spin Angular Momentum

4.2.1 Stern Gerlach experiment

When silver beam is passed to the inhomogeneous Magnetic field, two sharp trace found

on the screen which provides the experimental evidence of spin.

4.2.2 Spin Algebra

ˆˆ ˆX Y ZS S i S j S k , 2 2 2 2

X Y ZS S S S

2 , 0XS S , 2 , 0YS S 2 , 0ZS S and

,X Y ZS S iS ,Y Z XS S iS ,Z X YS S iS

2 2, 1 ,s sS s m s s s m , ,z s sS s m m s m where ss m s

X YS S iS and X YS S iS

, 1 1 , 1s s s sS s m s s m m s m

4.2.3 Pauli Spin Matrices

For Spin 12

Pauli matrix 12

s , 1 1,2 2sm

Pauli Matrix is defined as

e.anticommutMatrix Spin Pauli;0

1001

00

0110

222

kjI

ii

jkkj

zyx

zyx

1 if is an even permutation of , ,1 if is an odd permutation of , ,

0 if any two indices among , , are equal jkl

jkl x y zjkl x y z

j k l

zzyyxx SSS 2

,2

,2

, 1 1 , 1s s s sS s m s s m m s m





1001

200

20110

2

zyx Si

iSS

01

000010

SS

1001

43 2

2 S

For spin ½ the quantum number m takes only two values 21

sm and 21

. So that two

states are

21,

21and

21,

21, smS

21,

21

43

21,

21 2

2 S ,

21,

21

43

21,

21 2

2 S

21,

21

21

21,

21

zS , 21,

21

21

21,

21

zS

021,

21

S , 21,

21

21,

21

S

21,

21

21,

21

S , 021,

21

S

4.3 Total Angular Momentum

Total angular momentum J = L + S , kJjJiJJ zyxˆˆˆ

jmj, is the Eigen ket at J2 and Jz and x yJ J iJ x yJ J iJ

jj mjjjmjJ ,1, 22 , jjjz mjmmjJ ,,

1,11, jjjj mjmmjjmjJ

1,11, jjjj mjmmjjmjJ

J L S l s j l s

jmjSLJ jzzz and l s jm m m





5. Two Dimensional Problems in Quantum Mechanics

5.1 Free Particle

Hψ = Eψ

Eyxm

2

2

2

22

2

x and y are independent variable. Thus

1,21

2

yx ik yik xn

i k r

x y e e

e

Energy Eigen value 22

222

22k

mkk

m yx

As total orientation of k

which preserve its magnitude is infinite. So energy of free

particle is infinitely degenerate.

5.2 Square Well Potential

0V x x a and 0 y a

otherewise

2 2 2

2 22H E

m x y

The solution of Schrödinger wave equation is given by Wave function

, 24 sin sin

x y

yxn n

n yn xa aa

Correspondence to energy eigen value

2

2

2

222

, 2 an

an

mE yx

nynx 1,2,3...xn and 1,2,3...yn





Energy of state (nx, ny) Degeneracy

Ground state 2

22

22

ma

(1, 1)

Non degenerate

First state 2

22

25

ma

(1, 2), (2,1)

2

Second state 2

22

28

ma

(2, 2)

Non-degenerate

5.3 Harmonic oscillator

Two dimensional isotropic Harmonic oscillators is given by

22

2 2 212 2 2

yx ppH m x y

m m

1x yn n x yE n n where 0,1,2,3...xn 0,1,2,3...yn

1nE n

degeneracy of the nth state is given by (n + 1) where n = nx + ny.





6. Three Dimensional Problems in Quantum Mechanics

6.1 Free Particle

EH

Ezyxm

2

2

2

2

2

22

2

Hence x, y, z are independent variable. Using separation of variable one can find the

zikyikxikk

zyn eeezyx 2/321,,

3/ 2 .2 ik re

Energy Eigen value 22

2222

22k

mkkk

m zyx

As total orientation of k which preserve its magnitude is infinite, the energy of free

particle is infinitely degenerate.

6.2 Particle in Rectangular Box

Spinless particle of mass m confined in a rectangular box of sides Lx, Ly, Lz

, , 0 ,xV x y z x L 0 ,yy L 0 ,zz L

= other wise .

The Schrodinger wave equation for three dimensional box is given by

Ezyxm

H

2

2

2

2

2

22

2

Solution of the Schrödinger is given by Eigen function x y zn n n and energy eigen value

isx y zn n nE is given by

z

z

y

y

x

x

zyxnnn L

znL

ynL

xnLLLzyx

sinsinsin8

2

2

2

2

2

222

2 z

z

y

y

x

xnnn L

nLn

Ln

mE

zyx





6.2.1 Particle in Cubical Box

For the simple case of cubic box of side a, the

i.e. Lx = Ly = Lz = a

azn

ayn

axn

azyx

nnn zyx

sinsinsin8

3

2222

22

2 zyxnnn nnnma

Ezyx

1,2,3...xn 1,2,3...yn 1, 2,3...zn

Energy of state (nx, ny, nz) Degeneracy

E of ground state 2

22

23

ma

(1, 1, 1)

Non degenerate

E of first excited state 2

22

26

ma

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

3

E of 2nd excited state 2

22

29

ma

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

3

6.3 Harmonic Oscillator

6.3.1 An Anistropic Oscillator

222222

21

21

21,, ZmYmXmZYXV ZYX

zzyyxxnnn nnnEzyx

21

21

21 where

0,1,2,3...xn 0,1, 2,3...yn 0,1,2,3...xn

6.3.2 The Isotropic Oscillator

x y z

32x y zn n n x y zE n n n

where x y zn n n n 0,1,2,3...n

Degeneracy is given by = 2121

nn





6.4 Potential in Spherical Coordinate (Central Potential)

Hamiltonion in spherical polar co-ordinate

2 2

22 2 2 2 21 1 1sin

2 sin sinr V r E

m r rr r r

2 2

22 21

2 2Lr V r

m r rr mr

L2 is operator for orbital angular momentum square.

So ,mlY are the common Eigen state of and L2 because [H, L2] = 0 in central force

problem and 2 2, 1 ( , )m ml lL Y l l Y

So ,,r can be separated as ,mlf r Y

2 22

2 212 0

2 2

d f r l lf f r V r E f rm r rdr mr

To solve these equations rrurf

So one can get

02

12 2

2

2

22

uE

mrllrV

drud

m

Where 2

2

21

mrll is centrifugal potential and

2

2

21

mrllrV

is effective potential

The energy Eigen function in case of central potential is written as

, , , ,m ml l

u rr f r Y Y

r

The normalization condition is

2, , 1r d

2 2

22

0 0 0, sin 1m

lu r

r dr Y d dr

2

0

1u r dr

or 2 2

0

1R r r dr





6.4.1 Hydrogen Atom Problem

Hydrogen atom is two body central force problem with central potential is given by

2

04eV r

r

Time-independent Schrondinger equation on centre of mass reference frame is given by

2 2

2 2 , ,2 2R r V r R r E R rm

where R is position of c.m and r is distance between proton and electron.

R,r R r The Schronginger equation is given by

h2

2M1

R R2 R

h2

21

r R2 r V r

E

Separating R and r part

RERM R 2

2

2

rrVrr

22

2

Total energy R rE = E + E

ER is Energy of centre of mass and Er is Energy of reduce mass µ

3/ 2

12

i k RR e

2 2

2RkEM

i.eCentre of mass moves with constant momentum so it is free particle.

Solution of radial part

h2

2d2u r

dr2 l l 1 h2

2r2 e2

4 0 r

u r E r

For Hydrogen atom em





The energy eigen value is given by nE =

4

2 2 202 4em e

n

eV

n2

6.13 1,2,3...n

And radius of thn orbit is given by 2 2

02

4n

nrm e

1,2,3...n where me is mass of

electron and e is electronic charge.

n is known as participle quantum number which varies as 1, 2,...n l n For the

given value of n the orbital quantum number n can have value between 0 and 1n (i.e. l

= 0, 1, 2, 3, ….. n – 1) and for given value of l the Azimuthal quantum number m varies

from – l to l known as magnetic quantum mechanics .

Degeneracy of Hydrogen atom without spin = 2n and if spin is included the degeneracy

is given by 1

2

02 2 1 2

n

nl

g l n

For hydrogen like atom

2

26.13n

zEn where z is atomic number of Hydrogen like atom.

Normalized wave function for Hydrogen atom i.e. Rnl (r) where n is principle quantum

number and l is orbital quantum.

n l E(eV) Rnl

1

0

2-13.6z

0

3/ 2/

02 zr az e

a

2

0

2- 3.4 z

0

3/ 2/ 2

0 02

2zr az zr e

a a

2

0

2- 3.4 z

0

3/ 2/ 20

0 0

124

zr az zr ea a





The radial wave function for hydrogen atom is Laguerre polynomials and angular part of

the wave function is associated Legendre polynomials .

For the nth state there is n – l – 1 node

If Rnl is represented by ln, then

20

1 3 12

nl r nl n l l a

2 2 2 20

1 5 1 3 12

nl r nl n n l l a

12

0

1nl r nln a

2

3 20

22 1

nl r nln l a

rR10

0/ arr

rR20

0/arr

21R

0/ar





7. Perturbation Theory

7.1 Time Independent Perturbation Theory

7.1.1 Non-degenerate Theory

For approximation methods of stationary states

o PH H H

Where H Hamiltonian can be divided into two parts in that oH can be solved exactly

known as unperturbed Hamiltonian and pH is perturbation in the system eigen value

of oH is non degenerate

It is known nonno EH and 1 pH W where 1

Now o n n nH W E where n is eigen function corresponds to eigen value nE

for the Hamiltonian H

Using Taylor expansion

........2210 nnnn EEEE and .........221 nnnn

First order Energy correction 1nE is given by 1

n n nE W

And energy correction up to order in is given by 1n n n nE E W

First order Eigen function correction 10 0

m nn m

m n n m

WE E

And wave function up to order correction in 0 0m n

n n mm n n m

WE E

Second order energy correction

2

20 0

m nn

m n n m

WE

E E

Energy correction up to order of 2

2

20 0 0

m nn n n

m n n m

WE E W

E E





7.1.2 Degenerate Theory

( )n o p n n nH H H E

0nE is f fold degenerate fEH nnno ,.....3,2,10

To determine the Eigen values to first-order and Eigen state to zeroth order for an f-fold

degenerate level one can proceed as follows

First for each f-fold degenerate level, determine f x f matrix of the perturbation pH

ff

f

f

ppfpf

ppp

ppp

p

HHH

HHH

HHH

H

...........

.........

.........

ˆ

21

22221

11211

where p n p nH H

then diagonalised pH and find Eigen value and Eigen vector of diagonalized pH which

are nE and n respectively.

10 nnn EEE and

f

nn q1

7.2 Time Dependent Perturbation Theory

The transition probability corresponding to a transition from an initial unperturbed

state i to perturbed f is obtained as

1

'

0

'fiiw tif f i

iP t V t e dt

Where f ifi

E Ew

and

f o f i o ifi

H Hw





8. Variational Method

Variational method is based on energy optimization and parameter variation on the basis

of choosing trial wave function.

1. On the basis of physical intuition guess the trial wave function. Say

noo ,.....,, 321 where 321 ,, are parameter.

2. Find nn

nnn

HE

,.....,,,.....,,

,.....,,,.....,,,.....,,

32103210

321032103210

3. Find 0,.....,, 321

ni

oE

Find the value of ,.....,, 321 so that it minimize E0.

4. Substitute the value of ,.....,, 321 in ,.....,, 3210E one get minimum value

of E0 for given trial wave function.

5. One can find the upper level of 1 on the basis that it must be orthogonal to 0 i.e.

1 0 0

Once 1 can be selected the 2, 3, 4 step can be repeated to find energy the first Eigen

state.





9. The Wentzel-Kramer-Brillouin (WKB) method

WKB method is approximation method popularly derived from semi classical theory

For the case 1ddx where ( )

2 ( ( )x

m E V x

If potential is given as V(x) then and there is three region

WKB wave function in the first region i.e 1x x

11 1exp ' '

x

Ix

c P x dxP x

WKB wave function in region II: i.e 1 2x x x

' "2 2exp ' exp ' 'II

x

c ci iP x dx P x dxP x P x

WKB wave function in region III: i.e 2x x

2

3 1exp ' 'x

IIIx

c P x dxP x

9.1.1 Quantization of the Energy Level of Bound state

Case I: When both the boundary is smooth

1( ) ( )2

p x dx n where 0,1,2...n

2

1

12 22

x

nx

m E V x dx n where 0, 1, 2,......n and 1x and 2x are turning

point

Case II: When one the boundary is smooth and other is rigid

2

1

3( ) ( )4

x

x


2

1

324

x

nx


points

xV

I II III





Case III: When both boundary of potential is rigid

2

1

( ) ( 1)x

x


2

1

2 1x

nx


points .

9.1.2 Transmission probability from WKB

T is defined as transmission probability through potential barrier V is given by

exp 2T where 2

1

1 2x

nx

m E V x dx and 1x and 2x are turning points.





10. Identical Particles

Identical particle in classical mechanics are distinguishable but identical particle in

quantum mechanics are Indistinguishable.

Total wave function of particles are either totally symmetric or totally anti-symmetric.

11.1 Exchange Operator

Exchange operator ijP as an operator as that when acting on an N-particle wave

function 1 2 3( .... ... ... )i j N interchanges i and j.

i.e 1 2 3 1 2 3( .... ... ... ) ( .... ... ... )ij i j N j i NP

sign is for symmetric wave function s and sign for anti symmetric wave

function a .

11.2 Particle with Integral Spins

Particle with integral spins or boson has symmetric states.

1 2 1 2 2 11 , ,2s

For three identical particle:

123213312

132231321321 ,,,,,,

,,,,,,6

1

s

For boson total wave function(space and spin) is symmetric i.e if space part is symmetric

spin part will also symmetric and if space part is ant symmetric space part will also also

anti symmetric.

11.3 Particle with Half-integral Spins

Particle with half-odd-integral spins or fermions have anti-symmetric.

For two identical particle:

122121 ,,2

1 a

For three identical particle:





1 2 3 1 3 2 2 3 1

1 2 32 1 3 3 1 2 3 2 1

, , , , , ,1, , , , , ,6a

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

symmetric spin part will anti symmetric and if space part is ant symmetric space part will

also symmetric.

11. Scattering in Quantum Mechanics

Incident wave is given by .oik rinc r Ae

. If particle is scattered with angle θ which is

angle between incident and scattered wave vector ok

and k

Scattered wave is given by

.

,ik r

scer Af

r

, where ,f is called scattering amplitude wave function.

is superposition of incident and scattered wave

refeA

rikrik

oo ,

differential scattering cross section is given by 2,o

d k fd k

where is solid angle

For elastic collision 2,d fd

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

k

ok

2 22 ' 3

2, ' ' '4

ikrd f e V r r d rd





11.1 Born Approximation

Born approximation is valid for weak potential V(r)

20

2, ' ' sin ' 'f r V r qr drq

Where 2 sin2oq k k k

for and for ok k

22

4 20

4 ' ' sin ' 'd r V r qr drd q

11.2 Partial Wave Analysis

Partial wave analysis for elastic scattering

For spherically symmetric potential one can assume that the incident plane wave is in z-

direction hence exp cosinc r ikr . So it can be expressed in term of a superposition of

angular momentum Eigen state, each with definite angular momentum number l

cos

02 cosik r ikr l

l ll

e e i l l J kr p

, where Jl is Bessel’s polynomial function and Pl

is Legendre polynomial.

0

, 2 1 cosikr

ll l

l

er i l J kr P fr

1 2 1 sin coslil lf l e P

k

Total cross section is given by

22

0 0

4 2 1 sinl ll l

lk

Where σl is called the partial cross section corresponding to the scattering of particles in

various angular momentum states and l is phase shift .





12. Relativistic Quantum Mechanics

12.1 Klein Gordon equation

The non-relativistic equation for the energy of a free particle is

2

2p Em and

2

2p Em

2

2p im t

where

p i is the momentum operator ( being the del operator).

The Schrödinger equation suffers from not being relativistic ally covariant, meaning it

does not take into account Einstein's special relativity .It is natural to try to use the

identity from special relativity describing the energy: 2 2 2 4p c m c E Then, just

inserting the quantum mechanical operators for momentum and energy yields the

equation 2 2 2 2 4c m c it

This, however, is a cumbersome expression to work with because the differential operator

cannot be evaluated while under the square root sign.

which simplifies to2

2 2 2 2 42c m c

t

Rearranging terms yields 2 2 2

22 2 2

1 m c Ec t

Since all reference to imaginary numbers has been eliminated from this equation, it can

be applied to fields that are real valued as well as those that have complex values.

Using the inverse of the Murkowski metric we ge 2 2

2 0m c

where

2( ) 0





In covariant notation. This is often abbreviated as 2( ) 0 where mc

and

22

2 21c t

This operator is called the d’Alembert operator . Today this form is interpreted as the

relativistic field for a scalar (i.e. spin -0) particle. Furthermore, any solution to the Dirac

equation (for a spin-one-half particle) is automatically a solution to the Klein–Gordon

equation, though not all solutions of the Klein–Gordon equation are solutions of the Dirac

equation.

The Klein–Gordon equation for a free particle and dispersion relation

Klein –Gordon relation for free particle is given by 2

22 2

1 Ec t

with the same solution as in the non-relativistic case:

dispersion relation from free wave equation ( , ) exp ( . )r t i k r t which can be

obtained by putting the value of in 2

22 2

1 Ec t

equation we will get

dispersion relation which is given by 2 2 2

22 2

m ckk

.

12.2 Dirac Equation

searches for an alternative relativistic equation starting from the generic form describing

evolution of wave function:

Ht

i ˆ

If one keeps first order derivative of time, then to preserve Lorentz invariance, the space

coordinate derivatives must be of the first order as well. Having all energy-related

operators (E, p, m) of the same first order:

t

iE

ˆ and z

ipy

ipx

ip zyx

ˆ,ˆ,ˆ





mpppE zyx ˆˆˆˆ321

By acting with left-and right-hand operators twice, we get

21 2 3 1 2 3

ˆ ˆ ˆ ˆ ˆ ˆ ˆx y z x y zE p p p m p p p m

which must be compatible with the Klein-Gordon equation

22222 ˆˆˆˆ mpppE zyx

This implies that

,0 ijji for ji

0 ii

12 i

12

Therefore, parameters α and β cannot be numbers. However, it may and does work if they

are matrices, the lowest order being 4×4. Therefore, ψ must be 4-component vectors.

Popular representations are

00

i

ii

and

10

01

where i are 2 2 Pauli matrices:

1001

00

0110

321 i

i

The equation is usually written using γµ-matrices, where i i for

The equation is usually written using γµ-matrices, where i i for 1,2,3i and

0 (just multiply the above equation with matrix β and move all terms on one side of

the equation):

0

mx

i

where

0

0

i

ii

and

10

010

Find solution for particles at rest, i.e. p=0:





0

4

3

2

1

0

mt

i

B

A

B

A mt

i10

01

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

electrons:

01imt

A e and

10imt

A e

and two symmetrical negative-energy solutions

01imt

B e and

10imt

B e

Education

Quantum formula sheet