The Dirac Equation

•  Relativistic Quantum Mechanics for spin-1/2 Particles •  Klein-Gordon Equation •  Dirac g-matrices & Dirac Spinors •  Summing over Spin States •  Summary: Transformation Properties of Dirac Spinor Bilinears

For reference see Halzen&Martin pages 100-112

•  the quantum relativistic treatment of particles with spin ½ - aim is to learn the notation with which the ideas are expressed

•  concentrate on the basic concepts, symmetries and calculational techniques needed for the future studies and minimize repetition of standard text-book material (see, for example: Halzen&Martin)

The free-particle non-relativistic Schrödinger equation is ”derived” by starting with the energy-momentum relation:

Relativistic formulation...

)///( 2

:yields This

ˆ and /ˆ:onssubstituti canonical with the

ˆ21ˆ

222222222

2

zyxmt

i

iptiH

pm

H

∂∂+∂∂+∂∂=∇Ψ∇−=∂

Ψ∂

∇−=∂∂=

Ψ=Ψ

!!

"!

"!

•  the relativistic generalization of the Schrödinger equation is:

•  the Klein-Gordon equation that was first rejected because the quantity that

would be to interpreted as a probability density, can be negative (it is second order in time).

{Note: In 1934 Wolfgang Pauli and Vickie Weisskopf showed that the ”probability density” could be reinterpreted as a charge density and everything would be OK - ”probability” is not conserved, because –relativistically – one can create pairs of particles. This equation correctly describes relativistic spin-0 particles.} •  Paul Dirac in 1927 looked for an equation which would be of first order in time.

Since the equation is first order in time, it must be first order in space in order to be covariant.

Klein-Gordon Equation

Ψ+Ψ∇−=∂

Ψ∂

Ψ+Ψ=

422222

2

42222

t-

toleads which ,ˆˆ

cmc

cmpcH

!!

0componentsother and ,1,1: tensormetric theis g where 33221100 =−========⋅ ggggBAgBAgBABABA µννµµννµ

µνµµµ

µRemember: where summation over repeated indices is implied. Upper (lower) index vectors are called contravariant (covariant) vectors. The rule for forming Lorentz invariants is to make the upper indices balance the lower indices. If an equation is Lorentz covariant, we must ensure that all unrepeated indices (upper and lower separately) balance on either side of the equation, and that all repeated indices appear once as an upper and once as a lower index.

Therefore,

Inserting these in,

Thus we need: This is clearly impossible if the α’s and β are numbers. However, it could be possible with matrices. It turns out that the lowest dimension matrices that will work are 4x4 matrices. This means that Ψ will be a four-component vector.

Dirac Equation

ip.relationsh p-E hesatisfy t order toin )ˆ(ˆsatisfy tohave will thisand )ˆ(ˆ

42222

2

Ψ+=Ψ

Ψ+⋅=Ψ

cmpcH

mcpcH!

!!βα

Ψ+⋅=Ψ 222 )ˆ(ˆ mcpcH βα!!

[ ]

Ψ+=

Ψ+++++=∑≠

)ˆ(

ˆ)(ˆˆ)(ˆ

4222

,

4223222

cmpc

cmpmcppcpcjiji

iiijiijjiiii

!

!!!!ββαβααααααα

0,0,1223

22

21 =+=+==== iiijji βαβαααααβααα

•  first transform the equation into a more covariant looking form:

by multiplying on left by β/c:

. Note that γµ pµ is not an invariant since the γµ are just a bunch of matrices that do not change with Lorentz transformations – need to make them Lorentz invariant. The (anti-)commutation relations for the γ-matrices are:

^

^

Dirac Equation...

Ψ+∂

Ψ∂−=

∂

Ψ∂⇒Ψ+⋅=Ψ ∑ 22

ti )ˆ(ˆ mc

xcimcpcHi i

i βαβα !!""

0)ˆ(

i.e. , and where

0)( :as can write which we,

=Ψ−

==

=Ψ−∂Ψ=∂

Ψ∂+

∂

Ψ∂ ∑

mcp

mcimcx

itc

i

oi

i ii

i

µµ

µµ

γ

βγβαγ

γβαβ

!

"""

Dirac equation

νµγγγγγγγγ µννµ ≠=+−==== for 0 ;1)()()( ;1)( 2322212o

This can all be written as

One set of matrices which will work is:

These are not unique and sometimes other ones are used (see e.g. Perkins). A simple way to write these down is: where I is the 2x2 identity matrice and σi’s are the Pauli matrices.

Dirac Equation...

{ } { } { } BAABBAg +== , :commutator-anti theis where,2, µννµ γγ

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

−=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

−

=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

−=

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

−=

001000011000

0100

000000000

000

0001001001001000

1000010000100001

321 γγγγ

iii

i

o

⎥⎦

⎤⎢⎣

⎡

−=⎥

⎦

⎤⎢⎣

⎡

−=

00

0

0

i

iio

II

σ

σγγ

Let us look at the solutions for: The solutions are:

From the normal exp(-iEt/!) factor, identify ΨA as spin-up and spin-down states of a spin-½ particle. -But what is ΨB with E < 0? Instead of E < 0, can take t < 0, and say that ΨB represents a particle going backwards in time. ΨB thus represents an antiparticle.

Dirac Equation...

and where0

0

and t

i : then,00

4

3B

2

1A

2

2o

⎥⎦

⎤⎢⎣

⎡

Ψ

Ψ=Ψ⎥

⎦

⎤⎢⎣

⎡

Ψ

Ψ=Ψ⎥

⎦

⎤⎢⎣

⎡

Ψ

Ψ=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

∂

Ψ∂∂

Ψ∂

⎥⎦

⎤⎢⎣

⎡

−

Ψ=∂

Ψ∂=

∂

Ψ∂=

∂

Ψ∂=

∂

Ψ∂⇒=

B

A

B

A

mc

t

tI

Ii

mczyx

p

!

!"

γ

)0(exp)(

)0(exp)(

2

2

BB

AA

timct

timct

Ψ⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧+=Ψ

Ψ⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧−=Ψ

!

!

•  the necessary algebra can be found in the text books, will not go through all the details here, solve for the p≠ 0 plane wave states. •  define u(pµ) such that

Dirac Equation...

)(exp)( µµµµ pupxix

⎭⎬⎫

⎩⎨⎧−=Ψ!

)( )(01

)(

01

2242

2

2)2(

2

2)1( cpcmE

mcEcpmcEippcNu

mcEippcmcEcpNu

z

yx

yx

z +=

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

+

−+

−=

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

+

++

=

rpEt!!⋅−

This transforms us to the momentum space

The spin operator is:

For example:

Dirac Equation...

)(

10

)(

01

)( 22422

2

)4(2

2

)3( cpcmEmcEcpmcEippc

NumcEippcmcEcp

u z

yx

yx

z

+−=

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−−

−

=

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−

+−

=

⎥⎦

⎤⎢⎣

⎡

σ

σ!

!"0

02

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−

−=

1000010000100001

2!

zσ

Thus, the u’s are not eigenstates of σz unless the spin direction is aligned with the momentum. Then u(1) and u(3) are spin-up and u(2) and u(4) are spin-down. To write u(3) and u(4) in terms of the physical energy, E and have to be taken to –E and . It is conventional to define new Dirac spinors, v(1) and v(2):

Dirac Equation...

)(

10

)(

),(),( 22422

2

)4()1( cpcmEmcEcpmcE

ippc

NpEupEv z

yx

+=

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

+

−+

−

=−−=!!

p!

p!

−

The u Dirac spinors satisfy the (normal) Dirac equation:

while the v Dirac spinors satisfy the equation with pµ → –pµ:

Dirac Equation...

)(

01

)(),(),( 2242

2

2

)3()2( cpcmEmcEippcmcEcp

NpEupEv yx

z

+=

⎥⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢⎢

⎣

⎡

+

++

=−−−=!!

0)( =− umcpµµγ

0)( =+ vmcpµµγ

•  next, sum over spin states, and use a completeness relation •  for a single spin state, develop a simple trick to get it without going back to the

individual states. Dirac spinors, although they have four components, are not four-vectors - to determine how a Dirac spinor transforms, need to consider Ψ as seen in two inertial frames, then

There must exist a 4x4 matrix S, such that: •  for a Lorentz transformation along the x-axis (show this!)

Dirac Equation...

νν

µµµ

µµ

µµµ

γ

γ

XXxmx

xi

xmx

xi

Λ==Ψ−∂

Ψ∂

=Ψ−∂

Ψ∂

' where, 0)'(''

)'('

and 0)()(

!

!

)()'(' xSx Ψ=Ψ

2

21 1 and )1(

21 where,

cvaaIaS o −=±±=+= ±−+ γγγγ

We want to create Lorentz covariants out of bilinear combinations of Ψ’s. For example, we might guess that Ψ†Ψ =

is an invariant. Unfortunately, it is not since: and . (see the exercises!). However, (see the exercises again!) Therefore, if we define an adjoint Dirac spinor: Ψ ≡ Ψ†γo:

_

Dirac Equation...

24

23

22

21

4

3

2

1

4321 *)*,*,*,( Ψ+Ψ+Ψ+Ψ=

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

Ψ

Ψ

Ψ

Ψ

ΨΨΨΨ

ΨΨ=ΨΨ=ΨΨ ++++ SSSS )()(''1≠+SS

ooSS γγ =+

ΨΨ=ΨΨ=ΨΨ=ΨΨ=ΨΨ

ΨΨΨ≡Ψ++++

+

ooo

o

SS γγγ

γ

''''

:scalar Lorentz a is then ,

The next question: is ΨΨ a scalar or a pseudoscalar? I.e., how does it transform under parity? It is easy to see that γo is the parity operator, since, in the Dirac equation:

Under parity transformation, Ψ’ = P Ψ

and this equation is returned to the previous form by taking P = γo and multiplying on the left by γo. Then

This is a scalar.

_ Dirac Equation...

0

0)(

321 =Ψ⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−∂

∂+

∂

∂+

∂

∂+

∂

∂

=Ψ−∂

mcz

iy

ix

itc

i

mci

oγγγ

γ

γ µµ

!!!!

!

0'321 =Ψ⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

−∂

∂−

∂

∂−

∂

∂−

∂

∂ Pmczyxtc

io

γγγγ

!

ΨΨ=ΨΨ=ΨΨ=ΨΨ=ΨΨ +++ oooo γγγγ'')'(

•  now define: The γ5 has the property that it anti-commutes with every γ matrix, i.e.

It is easy to show that Ψ γ5 Ψ is a pseudoscalar. Since there are 16 elements to Ψ1

†Ψ2, their linear combinatios form 16 bilinear covariants:

Any other combination of γ matrices can be written as a combination of these.

_

Dirac Equation...

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=≡

0010000110000100

3215 γγγγγ oi

{ } 0,1,2,3 0, 5 == µγγ µ

)(2

µννµµν γγγγσ −=i

ΨΨΨΨ 5γ

ΨΨ µγ

ΨΨ 5γγ µ

ΨΨ µνσ

1 element scalar P=+1 1 element pseudoscalar P=-1

4 elements vector P=-1

4 elements axial vector

6 elements antisymmetric tensor P=+1

•  the completeness relations have an important role to play:

•  note that is ”backwards”; it is not a scalar, but a 4x4 matrix. •  these relationships will allow to sum over spins. By now, the fermions have been discussed; how about the photons? They are described by the Maxwell’s equations; these equations have to be in a covariant form.

Dirac Equation...

mcpvv

mcpuu

i

ii

i

ii

−=

+=

∑

∑

=

=

µµ

µµ

γ

γ

2,1

2,1

uu

The Maxwell’s equations are:

In Eq. (3):

The use of is equivalent to Eqs. (2) and (3). Take , then it is straightforward to verify that the remaining two Maxwell’s equations are equivalent to:

Maxwell Equations

jct

EB

tB

cEE

!!

!!!!

!!!!!

π

πρ

4c1-B (4) 0 )3(

01 (2) 4 )1(

=∂

∂×∇=⋅∇

=∂

∂+×∇=⋅∇

VtA

cE

tA

cEAB ∇−=

∂

∂+=

∂

∂+×∇×∇=

!!

!!

!!!!! 1 ,0)1( :(2) from ,

),( ,4)( JcJJc

AA!

ρπ µνµ

µννµ

µ ==∂∂−∂∂

AV!

and

),( AVA!

=µ

By defining: , , the Maxwell’s equations can be written in their most elegant form:

Aµ is not unique since Aµ’ = Aµ + ∂ µλ, where λ is an arbitrary function of space and time, will give the same result: This is known as a gauge transformation. The freedom to make gauge transformations characterize the theories of all the three forces. We can always use our gauge freedom to choose ∂µ Aµ = 0. This is called the Lorentz condition. Therefore,

Note that for free space (Jν = 0), this is just the Klein-Gordon equation for a massless particle:

Maxwell Equations µννµµν AAF ∂−∂=

νµνµ

π Jc

F 4=∂

0)()()( =∂∂∂−∂∂∂+∂∂−∂∂=∂+∂∂−∂+∂∂ λλλ µµ

ννµµ

µµ

ννµµ

µµµ

νννµµ AAAA

ννµµ J

cpA 4

=∂∂

00 22 =−⇒= pEpp!µ

µ

There is still more gauge freedom in Aµ, since any function λ can be added to Aµ

if: For example, pick , so that the Lorentz condition is This is known as the Coulomb gauge. It clearly breaks the Lorentz invariance, but is the simplest course of action. Just as the u Dirac spinor was defined for Ψ , now define εµ by

this takes to the momentum space; εµ is called the polarization vector.

by the Coulomb gauge. Thus there are only two independent components in εµ, corresponding to the two states of transverse polarization. Normally a spin-1 particle has three spin states µ= +1, 0, -1, but a massless spin-1 particle has only two, µ= +1 and m= -1. The completeness relation is:

Polarization vector

0=∂∂ λµµ

0=oA 0=⋅∇ A!!

)()exp()( pxpiaxA!!!

"µµµ ε⋅−=

,00 and condition, Lorentz by the 0 =⋅⇒== εεε µµ!"

pp o

jiijjsis

s pp ˆˆ)*()(2,1

−=∑=

δεε

Transformation Properties Dirac spinors do not transform as four-vectors when moving from one reference frame to another. - They are not four-vectors. Need to construct a quantity that is invariant for a Lorentz invariant theory. The linear transformation rule is: Apply to a scalar constructed from a Dirac spinor: Is this invariant? because:

21

110 1/1 and )1(21 with ,

:matrix 44 theis S where,'

βγγσ

σγγ −=±±=⎟⎟

⎠

⎞⎜⎜⎝

⎛=+=

×Ψ=Ψ→Ψ

±+−

−+−+ a

aaaa

aaS

S

⎟⎟⎠

⎞⎜⎜⎝

⎛

−

−==

ΨΨ≠ΨΨ=ΨΨ=ΨΨ

Ψ+Ψ+Ψ+Ψ=ΨΨ

+

+++++

+

11

)(')'()'(

1

12

24

23

22

21

βσ

βσγSSS

SS

Transformation Properties... Try a different spinor instead of the complex transpose: Note: there is minus signs for the 3rd and 4th components. Next, do the transformation: What happens to under Parity? The Parity transformation is Remember that the Parity operator changes the sign of the 3 space components of a vector but not the time component. Now do the P transformation: Therefore is invariant under P and is true scalar (parity=+1).

)ΨΨΨ(ΨΨΨ **** 43210 −−=≡ +γ

ΨΨ==== ++++ ΨΨSΨSΨΨ'(ΨΨ)'Ψ( 000)' γγγ

ΨΨ

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

−

−==

1000010000100001

0γP

0 where γ==→ PPΨΨ'Ψ

ΨΨΨγΨΨγγ)(γΨPΨγPΨΨΨΨ)'Ψ( ===== ++++++ 000000 ')'( γ

ΨΨ

Transformation Properties... Try another Scalar (Pseudoscalar): We can construct a different scalar What happens under Parity? We have used the fact that and therefore is a pseudoscalar with parity = -1.

ΨΨ 5γ

ΨΨ−=ΨΨ−=

ΨΨ=ΨΨ==+

+++++

05

050050505 )(')'()'(

γγ

γγγγγγγγγ PPΨΨΨΨ

100 =γγ 0550 γγγγ −=

ΨΨ 5γ

1 1 2 2 1 22 2 1 1 2 1

1 1 2 2 1 25 2 5 2 5 1 5 1 5 2 5 1

1 1 2 2 1 22 2 1 1 2 1

1 1 2 2 1 25 2 5 2 5 1 5 1 5 2 5 1

1 1 22 2 1

:

:

:

:

:

P C CP T CPT

S

P

V

A

T

µ µ µµ µ µ

µ µ µµ µ µ

µνµν µν

ψ ψ ψ ψ ψ ψ ψ ψ ψ ψ ψ ψ

ψ γ ψ ψ γ ψ ψ γ ψ ψ γ ψ ψ γ ψ ψ γ ψ

ψ γ ψ ψ γ ψ ψ γ ψ ψ γ ψ ψ γ ψ ψ γ ψ

ψ γ γ ψ ψ γ γ ψ ψ γ γ ψ ψ γ γ ψ ψ γ γ ψ ψ γ γ ψ

ψ σ ψ ψ σ ψ ψ σ ψ

→

→ − − −

→ − − − −

→ − − − −

→ − 1 2 22 1 1µν µν

µνψ σ ψ ψ σ ψ ψ σ ψ− −

Transformation Properties..

Transformation properties of Dirac spinor bilinears:

†

2 00

†

†

( , ) ( , ) ( , ): ( , ) ( , ) ( , ): ( , ) ( , ) ( , ): ( , ) ( , ) ( , ): ( , ) ( , ) ( , )

T

P Cx t x t x t

Scalar Field x t x t x tDirac Field x t x t i x tVector Field V x t V x t V x tAxial Field A x t A x t A x t

µµ µ

µµ µ

φ φ φ

ψ γ ψ γ γ ψ

→ −

→ −

→ −

→ − −

→ − −

r r rr r rr r rr r rr r r

00

Feynman Metric: , k

kQ Q Q Q= = −

(Ignoring arbitrary phases)

c→c* c→c*

Transformation Properties.. Non-Relativistic limit: In nuclear β-decay the matrix elements can be simplified. The Dirac spinor of a proton or neutron transforms like. Therefore: Scalar, Vector: S,V → Tensor, Axial: T,A → Pseudoscalar: P → This leads to: The Scalar and Vector transitions are called Fermi transitions. The spin of the nucleus remains unchanged (ΔJ=0). The Tensor and Axial transitions are called Gamow-Teller transitions. The spin of the nucleus may change (ΔJ=0,±1) For example, β-decays: spin goes Fermi only. Spin goes Gamow-Teller only. Spin goes Gamow-Teller only. Spin goes: Both can happen.

limit) icrelativist-(non 0 ⎟⎟⎠

⎞⎜⎜⎝

⎛→

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

+

⋅=Ψ

ΨmEpΨ

Ψ!!

σ

np ΨΨ +

np ΨΨ σ+

0

eeNO ν++→ +1414 .0 and 00 =Δ→ ++ JeeLiHe ν++→ −66 .1 and 10 +=Δ→ ++ J eeNiCo ν++→ −6060

.1 and 45 −=Δ→ ++ Jeepn ν++→ −

.1 and 21

21

+=Δ→++

J

Spin, Helicity and Chirality The spin matrix S of a Dirac spinor is: The operator to project out positive and negative helicity states from an arbitrary spinor is: If we have an arbitrary spinor u: But is not Lorentz invariant. If we take the massless limit: are the projection operators on states of positive and negative chirality.

pSSpin ˆHelicity 0

0 with

2 ⋅Σ=Λ⎟⎟

⎠

⎞⎜⎜⎝

⎛=ΣΣ=

σ

σ!

2

ˆ1ˆ Λ=±∓P

)()(

)()(

ˆ and ˆthen

where,

−−

++

±±−+

==

±=Λ+=

uuPuuP

uuuuu

p̂⋅Σ=Λ

uuP

pmEpp

)1(21)ˆ1(

21ˆ Therefore

ˆˆ

5

50

5

γ

γβ

αγ

±≅Λ±=

≅−

=⋅=⋅Σ

±

)1(21 5γ±

(1)

Spin, Helicity and Chirality...

On massless particles projects out the helicity. For example: if u(p) has helicity =+1 if u(p) has helicity =-1 On massive particles, only in the relativistic limit does Equation 1 (see previous transparency) hold. Therefore, the statement that electrons always have negative helicity is only an approximation. In the case of pion decay, the electron has a small amount of positive helicity but negative chirality. When we talk about left-handed spinors in the WI theory, this does not mean helicity =-1 except in the case where the particle mass is zero. They really mean chirality.

)1(21 5γ±

⎩⎨⎧

=±)(

0)()1(

21 5

pupuγ

Summary:Matrices Let A=Aij be an nxm matrix i=1,...,m; j=1,...,n:

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

=

mnmm

n

AAA

AAAAA

A

.....................

...

21

2221

11211

Then we have:

ijAA **=Complex Conjugate: Transpose: Hermitian Conjugate: Hermitian Matrix: Unitary Matrix: Unitary transformation: Unitary transformation: For every hermitian matrix, there exists a unitary matrix that will create a real diagonal matrix:

ijijT AA =

*)(or ** Tji

Tijji AAAAA === ++

jiij HHHH *or == +

+− =UU 1

+− ==→ UAUUAUAA 1'

DHUHUH == +'