Course-2, PHY502, Block-VI Relativistic Quantum Mechanics

Course-2, PHY502, Block-VI

Relativistic Quantum Mechanics

Prof O. P. S. Negi

Vice Chancellor

Uttarakhand Open University

Haldwani (Nainital)

Uttarakhand

Physical Quantity Operators

symbol actual operation

Momentum

Total Energy

Coordinate

Potential Energy

Quantum Mechanical Operators

)()(ˆ)(

ˆ

ˆ

ˆ

xUxUxU

xxx

tiEE

xipp xx

Motivation?

Schrödinger Equation

• Schrödinger developed the wave equation which can be solved to find the wave-function by translating the equation for energy of classical physics into the language of waves

• For fixed energy, we obtain the time-independent Schrödinger equation, which describes stationary states

• the energy of such states does not change with time– ψn(x) is an eigen-function or eigen-state– U is a potential function representing the particle interaction with the

environment

x

tixxU

x

x

m

)(2 2

22

xExxU

x

x

m

)(

2 2

22

ExUm

p )(

2

2

• Schrödinger, building on the matrix formulation of Heisenberg, realized the

state of a particle could be described as a complex-valued wave function.

• Any wave function could be expressed as linear combinations of basis

eigen functions, whose evolution through time was given by the

Schrodinger’s Equation

Special Relativity: Postulates

The two fundamental postulates of special

relativity are :1. The principle of relativity: The laws of physics take the

same form in all inertial reference frames.

2. The constancy of the speed of light: The speed of light

in vacuum is same in all inertial reference frames.

• Lorentz

Transformations

2

2

1

1

c

v

Special Theory of Relativity: Four vector• We live in a four-dimensional world of space-time continuum.

• Einstein introduced the concept of four vectors such that the scalar product of any two four-vectors is invariant under Lorentz transformations.• It is similar to the concept that the scalar product of any two three-vectors in the three dimensional space is invariant under rotation of coordinate system.• We list below some of the four-vectors.

• Space-time four vector (four-space)• Four differential operator

(Space-time gradient)• Energy-momentum four vector

(Four-momentum)• Scalar product of four vectors (length) listed above and given below remain invariant under Lorentz Transformations.

• Line element

• D’ Alembertian operator

• Conservation of energy momentum

),,,(

),,,(

)4,3,2,1(),,,(

c

iEpppp

tci

zyxx

ictzyxx

zyx

x

2

02

2222

2

2

22

2

2

2

2

22

222222

1

cmc

Eppppp

tczyx

dtcdzdydxdxdxds

zyx

Relativistic Limits

• The Schrödinger equation breaks down at relativistic limits.

• This is because it is not Lorentz covariant.

• Time and space do not enter the equation symmetrically.

Question-What is the remedy?

Friday, November 5, 2010

x

tixxU

x

x

m

)(2 2

22

Drawbacks of Schrödinger Equation

• It does not confirm with the principle of relativity• It is not invariant under Lorentz transformations • It is one particle non relativistic equation• Space and time are not on equal footings• Homogeneity of space and time is lost• Spin and magnetic moments are silent features• Exchange of particles is difficult.• Question-What is the remedy?

Natural Units1c

c = 1 → [ L ] = [ T ]

= 1 → [ E ] = [ T ]1

Choosing [ E ] = MeV, we have

• [ M ] = [ P ] = [ E ] = MeV

• [ L ] = [ T ] = [ E ] 1 = (MeV) 1

c = 2.997 108 m/s

= 6.582 1022 MeV s

1 MeV = 1.602 1013 J

T( s ) = 6.582 1022 T( MeV 1 )

L( m ) = 1.937 1013 L( MeV 1 )

M( kg ) = 1.783 1030 M( MeV )

Reconciliation of Relativity with Quantum mechanics

Relativistic Quantum Mechanics

Role of c and ћ are respectively Known as the scales used for Relativity and quantum mechanics

Domains of Physics

Speed

Size

RelativisticQuantumMechanics

QuantumMechanics

Relativity

ClassicalMechanics

NewtonianCosmology

RelativisticCosmology

c

Nucleus(10-14 m)

Atom(10-10 m)

Galaxy(1020 m)

c/10

Klein-Gordon Equation: Relativistic Generalization of S.E.• Using Einstein’s relation for a free particle:

• Einstein’s relation is rewritten in a covariant form:

• one obtains:

• K. G. Equation:

• Now let’s introduce covariant derivative:

• In this notation, Gordon-Klein equation becomes:

)ˆˆ;ˆ(;42

0

222

ipp

tiEEcmcpE

2

0 )( cmpp

),(),()(

)ˆ()ˆ())((

),(),(ˆˆ),(ˆ(ˆ

2

22

0

2

2

12

2

0

2

2

0

2

2 txcm

txt

miict

it

i

trmtrppctrEE

c

),,,(ticzyxx

),(

4xii

xip

))(

0),(

0

2

lengthwaveComptoninverseiscm

Here

tx

Problems of indefinite Probability withKlein-Gordon Equation:

1. Unlike the Schrödinger equation, it is the second order in time and therefore the solutions are specified on botheverywhere

• Probability and current source densities are

2. Probability is not positive definiteand can not be interpreted as position probability

• But satisfy Continuity equation guarantees that “No probability is lost”

tand

*)(*2

),(

**

2),(

0

2

0

imtrJ

ttcm

itr

0),(),(

trJ

t

tr

Problems of Negative energies with with Klein Gordon Equation

Free particle K.G. Equation

Plane wave solution

Energy eigen values are

• We have additional ‘negative energy’ solutions and energy spectrum is not bound from below.

• Then arbitrary large energy can be extracted from the system if external perturbation leads to transition between positive and negative energy states.

• K.G. Equation may be interpreted as Field equation as the relativistic wave equation for spin less particle in the frame work of may particle theories.

)( 42

0

222

2

22 cmc

t

)(

;

),(

2242

0

)(

cpcmE

kpE

eAtrEtrpi

Non Relativistic limit of K.G.Equation

• To study the nonrelativistic limit, we use the ansatz

• In nonrelativistic limit

We get

• Which on inserting to

K.G. Equation leads to the

Free Schrödinger equation for spin less particle.

• So, the wave equation does not depend

upon whether the particle is relativistic or non-relativistic, the Klein Gordon equation describes spin- zero particles. This is also known as the Schrodinger form of the free Klein-Gordon equation,

)( 20),(),(

tcmi

etrtr

2

0

2

0 '' cmEwherecmEE

)(

2

42

0

2

0

)(2

0

2

2

)(2

0)(

2

0

20

20

20

20

2

)(

)(

tcmi

tcmi

tcmi

tcmi

ecm

t

cmi

ecmi

ttt

ecmi

ecmi

tt

2

2

2

mti

Issues with Klein-Gordon

• Klein-Gordon Eqn. is Lorentz covariant .

• It fails some of the requirements of the QM

postulates.

• It is second order differential equation with respect to the time unlike Schrödinger's equation of quantum mechanics.

Being second order in time, determining a particular solution required information about both ψ as well as ∂ψ/∂t. But QM says wave-function must be a complete description.

Also, Klein-Gordon admits solutions where norm and hence, probability is not conserved.

So, Klein-Gordon is necessary but not sufficient.

• Klein-Gordon Equation was obtained in 1927 by Oskar Klein and Walter Gordon. • But certain problems arise which stand in its way of being a complete description of the dynamics of a relativistic electron.

Klein-Gordon Equation: Problems• It was actually found by Schrödinger before his non-relativistic equation.• K.G. equation is the relativistic generalization of Schrödinger’s equation and thus

it may be written in covariant way.• Unlike Schrödinger’s equation it is second order differential equation in time.• However, didn’t work for the hydrogen atom, and hence was abandoned

– This is because it’s an equation for spin-0, not spin ½ particles• Also incompatible with statistical interpretation of |Y(r)|2 as the probability of

finding particle at point r

• However, the Klein-Gordon equation does not lead to a positive definite probability density and admits positive and negative energy solutions – these features led to it being abandoned as a viable candidate for a relativistic quantum mechanical theory.

• Consequently, the arbitrary large energies can not be extracted from the system if external perturbation leads to transition between positive and negative energies.

So, Klein-Gordon is necessary but not sufficient and hence can not be accepted as the equation of motion of a particle with spin angular momenta.

• Dirac’s strategy was to find first order equation still compatible with the relativistic energy/momentum relation.

Dirac’s Insight

We need an equation that was both first-order in time as well as space.

Paul Dirac realized that this might be possible if we interpret the square root in the Einstein formula differently.

Significance:• Among other virtues, Dirac's equationaccounted for spin, discovered experimentally in 1925 but without a theoretical basis at the time. • Dirac's theory also predicted an antiparticle for every particle, which has since been found to

be correct

•Dirac placed emphasis on two constraints:1. Relativistic equation must be first order in time derivative.2. Elements of wave- function must obey Klein-Gordon equation.

Paul Dirac (1902-84)

• First to try to combine quantum mechanics with special relativity

• Obtained the relativistic version of Schrödinger’s equation in 1928

• Known as the Dirac equation: The - yet unknown -coefficients α, and β cannot be simple numbers.

2

0

2

0

3

1

2

0

2

0

)(

)(

cmzyx

cicmpcH

where

cmzyx

cit

i

HcmpppcEt

i

zyx

j

jj

zyx

zzyyxx

Dirac equation……….

• ThenΨ cannot be a simple scalar, but has to be a column vector

• So that the Dirac equation reduces to be

• Thus the Schrodinger-like equation represents a system of N coupled first-order differential equations of the spinor componentsΨ for i = 1,2,... ,N

),(

),(

),(

),(

3

2

1

tr

tr

tr

tr

N

N N

zy

N

x Hcmzyx

cit

i1 1

2

0

1

)()(

How to determine α and β in Dirac equation………. • To continue, we demand the following natural properties: 1. The correct energy-momentum relation for a relativistic free

particle

2. the continuity equation for the density and 3. the Lorentz covariance (i. e. Lorentz form-invariance) for various

form of Dirac equation

42

0

22222 )( cmpppcE zyx

0),(),(

trJ

t

tr

42

0

23

1

3

0

23

1,

22

2

22

42

0

222

42

0

222

2

22

)(

2

cmx

cmi

xxc

t

torisegivesiterationonwhich

cmcpEor

cmct

j

jj

j

kjkj

jkkj

• To fulfil the requirement that

every single component of the spinor Ψ has tosatisfy the Klein-Gordon equation i.e.

Dirac’s InsightComparing coefficients ,

we have:

• These anti-commutation

relations define an

algebra for the matrices

• Hermiticity of the Dirac Hamiltonian leads

Since, ; the eigenvalues can only have the values ±1.

• This has no solutions in scalars, but it does have if

we allow α and β to be square matrices.

12

0

120

1);3,2,1(1 22

generalinor

kj

j

jj

jkjkkj

jkkj

j

andjj

1);3,2,1(1 22

jj

What should be the minimum rank for α and β in Dirac equation ?

• The anti-commutation relations

• leads the trace(the sum of the diagonal elements of the matrix) of each α and β has to be zero i.e.

• Because the eigenvalues of α and β are equal to ±1, each matrix α and β has to possess as many positive as negative eigen values, and therefore has to be of even dimension. The smallest even dimension, N= 2, cannot be right, because only three anti-commuting matrices exist, namely the three Pauli matrices . Therefore, the smallest dimension for which the requirements must be fulfilled is N =4.

jjjj and

02 jjjjj trtrtrtrtr

• One of possible explicit representation of the Dirac matrices, is

which may be written in its explicit form in terms of 4 x 4 matrices as

• This representation is also referred as the Dirac Pauli representation• In contrast to KG Eqn. DE is of first order in all coordinates .• Space and time are on the same footings unlike the Schroedinger wave eqn.

Dirac-Pauli Representation of α and β

00

000

10

011

10

01;

0

0;

01

10

10

01

0

0

22321

22

22

2

2

andwithi

iare

matricesPaulithewhereandj

j

j

1000

0100

0010

0001

0010

0001

1000

0100

000

000

000

000

0001

0010

0100

1000

3

21

i

i

i

i

Unitary transformations for Dirac Equation

• Dirac Pauli representation of α and β has been chosen to satisfy the properties of these matrices

• The representation can be changed to another representation by means of unitary transformations to yield the properties of α and β

Which leads to

11 ';' SSSS kk

0'''

02

0

2''''

12

12

11111

11

111111

111

jj

jj

jj

jkjkkj

jkjkkj

jkjkkj

SSSSSSSSSS

SSSS

and

SSSSSSSSSSSS

SSSSSS

Dirac equation: Probabilistic interpretation

),,,(† 4321

3

1

††2

0

†††

k

k

k

cmx

cit

i

††

k ,k

†2

0

3

1

††

3

1

††2

0

†††

cmx

cit

i

cmx

cit

i

kk

k

k

k

k

2

0. cmpcH

3

1

2

0

k k

k cmx

cit

i

Dirac equation: Probabilistic interpretation• subtraction with the use of hermiticity provides

• Positive definite probability /charge source density is

• Current source density is

• So the conservation law is described as

•Thus we can accept the interpretation of ρ as a probability density [in contrast to the density for the Klein-Gordon equation which was not positive definite].•Accordingly, we call J the probability current density.

0

0)()(

)()(

)(

3

1

††

3

1

††

3

1

3

1

††

††

Jdivt

cxt

xci

ti

xxci

tti

k

k

k

k

k

k

k k

k

kk

k

4

1

*†

†

†

cJor

cJ kk

03†3 xdJdivxd

t

Probability Conservation

• That the Dirac equation is first-order in time and space is straightforward.

• The equation permits, when the matrices αk and β are Hermitian, a continuity equation.

• The density is Y†Y, which consists of a positive definite entry that can be interpreted as probability density.

• Probability can hence remain conserved.

• Dirac equation is thus a happy marriage of QM and STR.

Arpan Saha, Sophomore, IITB, Engineering Physics with Nanoscience

The Dirac Equation : Covariant formulation• Before deriving the consequences of Dirac eqn. we first cast into

more symmetric form. Let us introduce the new operators

• On multiplying from left by β to the Dirac equation and solving we get

• Then we get the completely symmetric covariant form of Dirac eqn.

),(;;)3,2,1( 4 ictxxki kk

4

1

03

14

4

3

1

2

0

3

1

2

0

2

3

1

2

0

0

)(

)(

)(

x

cm

xx

c

cm

xi

tic

cmx

ict

i

cmx

cit

i

k k

k

k k

k

k k

k

k k

k

)(0)(

x

0),(0

tx

cm

x

The Dirac Equation : Covariant formulation

• Dirac equation in covariant form can also be written in slash notations as

• Dirac equation may also be written in covariant form as observable equation

• Derivation

)(0)(

0)(0)( 00 cimporcimp

0)(

0)(

0)(

)(

)(

)(

0

0

4

1

044

3

1

44

2

0

3

1

2

0

3

1

2

2

0

cimp

cimp

cimpp

pccimpc

Eicmipci

Ecmpc

k

kk

k

kk

k

kk

Dirac Algebra• Dirac matrices satisfy the property

• So we may define the entity

• It also ant commutes with other gammas

• Next we may consider the following sixteen identities

• Denoting the general element of this array by ГA (A=1,2,3…….16), it can easily be verified that

)4,3,2,1,(2

43215

555 2

4321

432413421321

434241211332

4321

;,,,

;,,,,,

;,

);1(;1

iiii

iiiiii

dTrTr

Aif

iaa

A

ABAB

CBA

A

1;0

)1(

,1

12

Dirac Algebra

• The sixteen quantities ГA (A=1,2,3…….16), are linearly independent i.e.

• ГA form the Dirac algebra

• This algebra has a unit element but is not commutative

• If ψ denotes a n-dimensional “vector” in the representation space, then any operator X acting on ψ can be expressed as a linear combination of the sixteen ГA i.e.

• If , for an algebra det[ԌAB ] ≠0 where ԌAB ≡ Tr ГA ГB , then the algebra is called semi simple algebra.

• Here Tr ГA ГB = δAB , Tr I= d δAB

• Hence det[ԌAB ] =d16 ≠0 , so that the Dirac Algebra is semi simple.

16

1

1621 0......0A

AA aaaunlessa

)(; 1

16

1

AdA

a

AA XTrxxX

Dirac Algebra

• Frobenius theorem states that the number ‘r’ of possible

irreducible representation of a (finite) semi simple algebra

(which possesses an unit element) is equal to the number

of elements in the algebra that commute with all other

elements of the algebra

where N is the total number of elements of the algebra and dn

is the dimensionality of nth irreducible representation.

• For Dirac algebra the only commutating element is unit

matrix.

• So, up to the equivalence ,there exists only one irreducible

representation. Further N=16 so that 16=(d1 )2 gives d1 =4.

• Hence up to the equivalence , the Dirac algebra has only

one irreducible representation and this is four dimensional.

22

2

2

1 ......... rdddN

Derivation of Spin from Dirac equation• Dirac equation can be shown to describe a particle with spin ½.• For this we start from the commutation relation between the Dirac

Hamiltonian and the orbital momentum of the particle

• The commutators are non vanishing and neither total angular momentum nor its x-,y- and z- components are constants of motion in relativistic quantum mechanics

• Hence in contrast to what we expect from a free particle, the orbital angular momentum L is not a constant of motion

0)(,

0)()(,

0)()(,

0)()(,

)(),(, 2

0

pciLH

generalinor

pcippciLH

pcippciLH

Similarly

pcippciLH

prcmpcLH

zzxxzy

xyzzyx

zxyyxz

zz

Derivation of Spin from Dirac equation…..• Law of conservation of angular momentum fails and to save the law

of conservation of angular momentum, we must amend the orbital angular momentum by an additional intrinsic angular momentum.

• So, we must assume that the particle carries an intrinsic angular momentum S besides its orbital angular momentum such that total angular momentum J=L+S is conserved so that [H,J]=0.

• Here S must be determined in such a way that

1. It satisfies the commutation rules

2. It commutes with L (so that the familiar rules of angular momentum should be applicable to obtain J from L and S.

3. The total angular momentum J=L+S commutes with the Dirac Hamiltonian H that is it is constant of motion.

• It can easily be verified that we may define S from four dimensional representation of Pauli matrices

ljklkj SiSS

,

jj

j

j

j Sand

2

1

0

0

Derivation of Spin from Dirac equation……..• Contrary to L, the intrinsic angular momentum S does not depend

on the state of motion (i.e. on space time coordinates)

• The third component of S i.e. Sz is diagonal

• with the eigen values sz =±½ ћ each being two fold degenerate.

• Therefore the first and third rows and columns correspond to spin up (sz =+1/2 ћ) while the second and fourth columns leads to spin down (sz =-1/2 ћ) and

• It shows that total intrinsic angular momentum S2 has the eigen values s(s+1) ћ² where s=1/2

1000

0100

0010

0001

2

1

zS

22

432222 )1(

1000

0100

0010

0001

ssSSSS zyx

Derivation of Spin from Dirac equation………..• Thus, we are led to consider as S the spin operator

• Where

• Hence the Dirac equation admits spin and thus describes the particles of spin ½.

• The spin has been automatically incorporated into the wave equation because of the unavoidable increase in the number of components of Dirac spinor ψ

j

j

jjj withS

0

0

2

1

ljklkj

yxxzzy

i

SiSSorSiSS

iS

kji

iii

,

)(

222

;;

),,()(

41

21

213132321

3212

0)()(,,,

0)(,

0)()(,

0)()(,

)()(,

)(),(, 2

0

pcipciSHLHJH

pciSH

pcippciSH

pcippciSH

pcippciSH

prcmpcLH

zzxxzy

xyzzyx

zxyyxz

zz

Consequently, total angular momentum commutes with Dirac Hamiltonian and is thus verifies the law of conservation of angular momentum

Free Motion of a Dirac Particle• Let us examine the solution of the free Dirac equation (that is, the

Dirac equation without potentials)

• Its stationary states are found with the ansatz

• Which transforms D.E. into

• Again the quantity E describes the time evolution of the stationary state ψ (r).

• For many applications it is useful to split up the four-component spinor into two two-component spinors ф and χ i.e.

)()(),(

Eti

ertr

)()(]).([ 2

0 rErcmci

)(

)()(;

)(

)()(

)(

)(

)(

)(

)(

)(

)(

4

3

2

1

4

3

2

1

r

rr

r

rr

r

r

r

r

r

r

r

Free particle solutions of Dirac equation……….

• Note that, since H is only a function of P then [P, H ]=0 so that the eigenvalues pj of P can be used to characterize the states.

• In particular, we look for free-particle (plane-wave) solutions of the form

• Where u(p) is a four-component vector which satisfies

• and

• Therefore, writing the equation in matrix form, we find

• which yields two equations

).()()(

rpi

epur

)()().( 2

0 pEupucmpc

)(

)()(

p

ppu

0)(

)(

0

0

)(

)(2

0

2

0

2

0

2

0

cmEpc

pccmE

E

E

cmpc

pccm

222

0

2

0

222

0

2

0

2

0

2

0

2

0

2

0

)())((

)())((

)(

)(0)()(

)(

)(0)()(

pccmEcmEor

pccmEcmE

eithergiveswhich

cmE

pccmEpc

cmE

pcpccmE

Free particle solutions of Dirac equation……….• Thus a free solution of Dirac equation can be written with the

spinor amplitude expressed in either of the two alternative forms

• Where N is normalization constant.

• However

• Here we have used the identity

• Hence, we get

• It means the relativistic energy eigen values in terms of relativistic mass energy relationship are obtained either for φ or χ or for both.

• Thus we get the energy eigen values

• We see that the eigenvalues can be positive or negative and E is the energy of a particle described by the spinors

2

02

0

)()( cmE

pc

Npuor

cmE

pcNpu

2222 )())(()( pppippppcpc

)())(( BAiBABA

)()( 42

0

22242

0

222 cmpcEorcmpcE

)( 42

0

22 cmpcE

2

02

0

)()( cmE

pc

Npuor

cmE

pcNpu

Free particle solutions of Dirac equation……….

• Here, we also obtain a negative energy as a possible value like the KG equation.

• These negative energy eigen values are not accepted in classical and non relativistic mechanics since we do not observe a free particle with negative energy and that’s why we have rejected them.

• Thus, the Dirac equation admits positive as well as negative energy solutions and both are compatible with Dirac spinors.

• So it is customary to write the positive energy solutions (E=+ε)using first form for φ i.e.

• And the negative energy solutions (E=-ε)using second form for χ

i.e.

)()(2

0

)(

E

cm

pcNpu

)()( 2

0

)(

Ecm

pc

Npu

Free particle solutions of Dirac equation……….• This convention has the advantages

– Since cp< ε the quantity c σ.p/(ε +m0 c2 ) is always smaller than one and tends to zero in non relativistic limit.

– Therefore the lower components of the spinor amplitude u(+) are negligible in non relativistic limit.

– The same happens to the upper components of u(-) which is negligible in NR limit

– In the special case in which φ and χ are each equal to

• We have

• Which expands to

1

0

0

1 or

2

0

)(

2

0

)(

)(

)(;)()( cm

pc

Npuand

cm

pcNpu

),(

1

0

)(

)();,(

0

1

)()(

),()(

1

0

)();,(

)(

0

1

)(

2

0

2

0

)4(

2

0

2

0

)3(

2

0

2

0

)2(

2

0

2

0

)1(

Ecm

cp

cm

ipp

puEcm

ipp

cm

cp

pu

E

cm

cp

cm

ipppuE

cm

ipp

cm

cppu

z

yx

yx

z

z

yx

yx

z

Free particle solutions of Dirac equation……….

• Here we have used

• Obviously in each case

• Although the expression of φ and χ as either χ + or χ – imply that all the four spinors correspond to spin up or spin down relative to the z- axis. These spinors are not eigen function of Σ z which does not commute with Dirac Hamiltonian unless its unit vector is parallel to the momentum.

• So the direction of propagation must be considered as Z-axis which is placed parallel to momentum axis where p x = p y =0 in all of these four spinors. So we must consider

)4,3,2,1()exp()()(

rp

iu

pppp zz ;

zyx

yxz

z

z

y

y

x

x

zzyyxx

pipp

ippp

p

p

ip

ip

p

p

pppp

0

0

0

0

0

0

Free particle solutions of Dirac equation……• As such we get the spinor amplitudes

• In order to obtain the normalization constant N, we use

• Which yields the constant of normalization as

),(

1

0

0

)();,(

0

1

0)(

),(0

1

0

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0

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cm

cp

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cm

cp

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cp

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cp

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cpNpu

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cpN

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cmuu

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)(

)(1

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cm

cm

cm

pc

cmN

Free particle solutions of Dirac equation……

• However, the complete spinor ψ(r) is normalized as

• So, the complete free particle Dirac Spinor is written as

• The most general free particle solution of Dirac ‘s equation is

• It shows the obvious need of both positive and negative energy solutions since it is necessary to have four basic independent spinors to express an arbitrary spinor in terms of them.

1)(1 44332211

† dd

)()](.[exp),( )(2

0)(

EEtrpi

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)}(.[exp)()()(.[exp)()({

)(.[exp)()(),(

)2(

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2

1

)(2

0

4

1

)(2

0

trpi

pupctrpi

pupcV

cm

Etrpi

pupcV

cmtr

p

p

Dirac equation :-Particle Velocity…• Recalling the equation of motion for velocity

• If x is explicitly independent of time, the equation of motion for velocity reduces to

• Or in general v=cα or α = v /c which shows that α plays the role of the velocity of particle.

• Since the eigen values of α are either +1 or -1, it is to be concluded that the possible values of the velocity of a particle are ± c, in contradiction to the experimental fact.

• So the fact is that the velocity we measure is the expectation value of v=cα under the physical condition of a particle i.e. well defined energy and momentum. For example if the Z-axis is parallel to direction of propagation, the observed velocity is

• The phase and group velocity then accommodate the positive and negative energies

Hxit

x

dt

xdvx

,1

x

j

jjx ccmpcxi

Hxidt

xdv

3

1

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0 )(,1

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†

3

)(2

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pcvand

pv gph

Dirac’s interpretation of the negative energy solution

• In fact, it is impossible to ignore the negative energy solutions .

• The Dirac equation gives a non-zero probability of transition between a positive energy state and a negative energy state, so there must be some reason to explain why we do not observe such transitions.

• Moreover, we require a complete set of wave functions in order to be able to represent an arbitrary wave function as a (Fourier) expansion.

• If we exclude the negative energy states, we no longer have a complete set of wave functions.

• So, if negative energy states do exist, one had to explain why electrons did not fall down to lower energy levels.

• To handle this problem, Dirac appealed to the fact that electrons are fermions and no two electrons could take the same quantum state (the Pauli Exclusion Principle).

•

Interpretation of Negative Energy Solutions• Therefore, Dirac argued, there would be no problem if one could

imagine that all the negative energy states were already filled with electrons, forming a so-called Dirac sea of electrons with negative energies.

• Therefore, the exclusion principle forbids transitions from positive energy states to the negative energy states.

• Consequently, electrons with positive energy would not be able to fall down to the lower level energy levels.

• The existence of the “Dirac sea" of an infinite number of filled states may be philosophically unappealing, but it is at least mathematically consistent.

• Since these negative energy levels were inaccessible to the few “real” electrons, the existence of such sea would not be observable.

• However, Dirac himself soon noticed that there would be at least one observable manifestation of such sea.

Interpretation of Negative Energy Solutions……..

• Let us assume that real electrons are described only by positive energy states , these are the states with E = +E.

• All states of negative energy are occupied by electrons, one electron in each state of negative energy.

• This is illustrated in Fig. where the negative

energy states with E < m 0 c2 are occupied by

electrons and form the "Dirac sea".

• Negative energy electrons represent the

vacuum state and are unobservable; whereas

real,( i.e. observable) electrons in general

exist only in states of positive energy .

• If to the vacuum a positive energy electron

is added, the electron can suffer transitions only to the states

of energy between + m 0 c2 to + ∞, but transition into the

negative energy states are forbidden due to Pauli Exclusion Principle.

Dirac’s electron holes theory

• However, it is possible to communicate an energy E ≥ 2 m0 c2 to the

‘vacuum’ in the form of a photon (γ ray) of energy.

• So, the energy gap separating the negative and positive energy solutions is exactly 2 m0 c

2 leads to the possibility to ‘raise’ one negative energy electron from the vacuum to the positive energy state.

• As a result a “HOLE” is produced in the vacuum (i.e. the sea of negative energy electrons) and a positive energy electron is created which becomes observable.

• Such a ‘hole’ behaves, for all purposes, exactly as a particle with same mass as that of the electron but having an opposite sign of electric charge.

• Hence it may be identified with “positron” which has positive energy , because the “hole” which represents it means the missing of a certain amount of negative energy .

Dirac’s electron holes theory

• Thus the overall effect of the process will be the transformation of (radiative ) energy into the creation of electron-hole pair.

• If -p, -ε, and –e are respectively the momentum, energy and charge of the electron in its original negative energy state, the hole in the vacuum acts as a particle of momentum +p, energy + ε, and charge +e, which is called a ‘positron’.

• The process is described as

which is called ‘Pair Production ’since two observable particles are

simultaneously created.

• The reverse process would be a transition of a positive energy electron into a vacant negative energy state or hole which can further be expressed as

• Since as a result, both the electron and hole become unobservable the process is called ‘electron-positron annihilation’ so that two photon are produced to conserve both energy and momentum.

• Positrons are observed by C. D. Anderson in 1933 in cosmic rays.

)0()0()0()0( EeEeorEeEe

2)0()0(2)0()0( EeEeorEeEe

Concluding Remarks

• While the formulation in terms of ‘holes’ might seem to indicate some asymmetry, particles and their antiparticles are absolutely symmetrical.

• An operation called charge conjugation takes one from a particle’s wave function to that of its antiparticle in the same potential.

• In fact, as Feynman showed we might even regard antiparticles as particles traveling backwards through time.

Arpan Saha, Sophomore, IITB, Engineering Physics with Nanoscience ,

Friday, November 5, 2010

The equation of continuity

• To derive an equation of continuity, we first introduce the so called “Adjoint Dirac Spinor and adjoint Dirac equation”.

• Adjoint spinor is denoted as

• So the adjoint of DE is

• Derivation

),,,(),,,( 432143214

†

0)(

0)(

0)(

0)(0)()(

0)()(

0)()(

)(0)()(

0)()(

0)()()()(

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44

44

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†

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4

†

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jj

jj

jj

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jj

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conjugatehermitianitsTaking

The equation of continuity in covariant notations

• Multiplying D. Eqn. from the left by ψ and its adjoint equation from right by ψ and then adding, we get

• Introducing the four current density

• We get the covariant form of equation of continuity as

0)(

icj

)&;;(

0

†

44

icicjicj

where

j

kk

Dirac particle in electromagnetic field Here we introduce the case of an electromagnetic field whose

effects are defined by the four potential {Aμ} ≡(Ā, iФ) where the vector potential Ā (r,t) and the scalar potential Ф(r,t) modify the canonical momentum and energy with electric charge e as

So the Dirac equation

changes to

Thus the Dirac equation for a particle interacting with EM field is

Which can also be

written as

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epp ;

])([

])([

])([)(

2

0

2

0

2

0

ecmAc

epcH

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cmAc

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em

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k

kkk ecmAept

i

)(0)(0)(

)(0)(0)( 00

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ieDwhereDorD

OR

Ac

epwherecimorcim

Dirac particle in electromagnetic field……

• So, the Dirac equation in an electromagnetic field is described as

• Where is known as the covariant derivative. Here, one can also obtain the DE in EM field on replacing the partial derivative by covariant derivative.

• So, the adjoint Dirac equation becomes

• Which are the equations for anti electron (positron)and may also be obtained on replacing the spinor by adjoint spinor and charge –e by +e showing that if ψ is used for particle its adjoint is associated with anti particle.

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00)(

00

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)( Ac

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cimcimAc

epor

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Gauge invariance of Dirac equation in electromagnetic field

• Dirac equation in electromagnetic field

is invariant under the following local gauge transformations

• It shows that

• It supports the invariance quantum electrodynamics under local U(1) gauge symmetry of electromagnetic interactions.

• The symmetry group U(1) is the group of all 1X1 unitary matrices.

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00)(

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Non Relativistic limit of Dirac equation……• Consider a free electron at rest p=0 and E=m0c2

where m0 is the rest mass of the Dirac free particleand the Dirac equation reduces to

• Then the Four of the eigen spinors satisfying Dirac equation are

• Which are known for the free particle spinor amplitudes of a Dirac particle at rest . Arpan Saha, Sophomore, IITB, Engineering

Physics with Nanoscience ,Friday, November 5, 2010

2

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1

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Non Relativistic limit of Dirac equation……We may thus express the four component Dirac spinor as

• In non relativistic limit, we have for positive energy solutions

• Then shows that if the particle velocity is small compared to the velocity of light then two of the four component wave function are small compared to other two.

• For E=+ε, φ represents the large component and χ denotes the small component while for E=-ε, φ denotes small component and χ is used for large component. In case the state of the particle does not have well defined value of momentum, we have the relation

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2

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pc

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).(

cm

ci

Non Relativistic limit of Dirac equation……• For Dirac particle in electromagnetic field, we have

• For non relativistic limit in weak field approximation ,we have

• Which reveals for positive energy that χ represents the small component of the Dirac spinor ψ compared to large component φ .

• So, inserting we get

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)}.({)(

;

2

0

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Ac

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Ac

epccmeE

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0

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)}.({

2

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)()}.({

)(

cm

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).(

cm

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em

em

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0

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).(

2

)}.({

Non Relativistic limit of Dirac equation……• Here we have

• Which is recognized as Pauli equation where the second term on right hand side is potential energy associated with a magnetic dipole momentum

• Furthermore, since the charge of electron is negative is antiparallel to the spin. Pauli equation not only establishes a meaningful non relativistic analogue of Dirac equation but also shows that the Dirac particles (at least with positive energy ones) are electron.

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e

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Non Relativistic limit of Dirac equation……• Pauli equation

• Can also be obtained for first order approximation. For pure weak uniform magnetic field (i.e. in the absence of electric field), we have

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e

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Ac

ep

).(22

)(

00

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)(

.).()(

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;2

1

00

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00

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222

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2

22

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e

m

p

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Ac

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c

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rArpc

epA

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c

eppA

c

ep

ArA

Non Relativistic limit of Dirac equation……• The coefficient of the interaction of the spin with magnetic field

gives the magnetic moment of the electron corresponding to a ‘g’ factor with value ‘g=2’.

• The magnetic moment is then be called as spin magnetic moment as it occurs only for the particle with spin.

• Thus the Hamiltonian of Dirac equation in non relativistic limit contains a term which takes the intrinsic properties of the electron into account showing that the Dirac particle with positive energy is an electron.

• The consequence of the theory is in the excellent agreement with experiments for electron.

• As such the Dirac equation is a starting point to construct a relativistic theory of a particle like electron.

• The ratio of the spin magnetic moment to the spin angular momentum is equal to e/2m0 c which is twice the value for orbital angular momentum.

cm

e

02

Spin and magnetic moment from Dirac equation

• We may establish the second order differential equation by iterating the Dirac equation as

• So, we get

• Restricting ourselves for pure magnetic field (i.e. E=0↔Φ=0), we get

);,(

0)(2

1

0)(2

1

0)()(

2

2

AAFFc

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Fc

ieD

DDDDD

Ac

ieA

c

ie

At

A

c

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iii

i

i

i

F

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zyx

zxy

yxz

xyz

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0

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j

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c

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;;

0).(

44

22

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1

2

).().()(2

1

0).().(22

c

e

c

ieF

c

ie

c

e

c

ieD

Spin and magnetic moment from Dirac equation ….• Multiplying this equation by -ħ²/2m0 and substituting iħ∂ψ∕∂t=E’ ψ, we obtain

• For non relativistic limit

• We then get

• This equation can be easily compared with the corresponding Schrödinger energy equation and shows that the particle in the magnetic field has a magnetic energy

• Consequently , we have which implies that the particle has a magnetic dipole moment . Since the eigen values of Σ are ±1, this means that this intrinsic (i.e. independent of the motion) magnetic dipole moment is capable of taking on the values ±1 times Bohr magnetron.

• Since the charge ‘e’ of the electron is negative , the direction of its magnetic moment is antiparallel to the spin orientation. All these findings are in agreement with the experimental observations of spin ½ particles.

0).(22

1'

2

1)(

2 0

2

0

2

2

0

2

0

2

cm

ecmE

cmA

c

ie

mjj

2

0

42

0

22

0

2

0 '2'';'' cmcmEcmcmE

));(').(22 00

2

j

jx

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ep

cm

e

m

)2

(;..2 00

cm

e

cm

eVm

'2 0

2

mV

m

cm

e

02

Solvay Conference, 1927

Schrodinger

Heisenberg

Pauli

Debye

Bragg

Dirac

Compton

EinsteinLorentz

CuriePlanck

de Broglie

Born

Wilson

Bohr

“Anyone who is not

shocked by quantum theory has not

understood it.”― Niels Bohr

Concluding Remarks