P H Y 4 8 9 / 1 4 8 9

T H E D I R A C E Q U AT I O N

S O FA R :• Quick introduction to:

• special relativity, relativistic kinematics

• quantum mechanics, golden rule, etc.

• Phase space:

• how to set up and integrate over phase space to determine integrated and differential rates/cross sections

• Now we move to the particles themselves

• start with the Dirac equation

• describes “spin 1/2” particles

• quarks, leptons, neutrinos

R E L AT I V I S T I C WAV E F U N C T I O N

• In non-relativistic quantum mechanics, we have the Schrödinger Equation:

• Inspired by this, Klein and Gordon (and actually Schrödinger) tried:

H� = i� ⇥

⇥t� p⇥ �i�⇤H =

p2

2m

� �2

2m⇥2� = i� ⇥

⇥t�

“Manifestly Lorentz Invariant”

E2 = p2 +m2 ) (�r2 +m2) = � @2

@t2

✓� @2

@t2 +r2

◆ = m2

@µ ) (@0, @1, @2, @3) =

✓@

@t

,

@

@x

,

@

@y

,

@

@z

◆(�@µ@µ +m2) = 0

I S S U E S W I T H K L E I N - G O R D O N• Within the context of quantum mechanics, this had some issues:

• As it turns out, this allows negative probability densities:

• Dirac traced this to the fact that we had second-order time derivative

• “factor” the E/p relation to get linear relations and obtained:

• and found that:

• Dirac found that these relationships could be held by matrices, and that the corresponding wave function must be a “vector”.

↵ = �

{�µ, �⌫} = �µ�⌫ + �⌫�µ = 2gµ⌫

| |2 = 0

pµpµ �m2c2 = 0 ) (↵p +m)(��p� �m)

�µpµ �m = 0 ) (i�µ@µ �m) = 0

T H E D I R A C E Q U AT I O N I N I T S M A N Y F O R M S

⇥a � aµ�µ

⇤a ⇥ aµ�µ = a0�0 � a1�

1 � a2�2 � a3�

3

(i�µ@µ �m) = 0

(i/@ �m) = 0

@µ ) (@0, @1, @2, @3) =

✓@

@t

,

@

@x

,

@

@y

,

@

@z

◆

i

✓�

0 @

@t

� �

1 @

@x

� �

2 @

@y

� �

3 @

@z

◆�m

� = 0

⇥i(�0@0 � �1@1 � �2@2 � �3@3)�m

⇤ = 0

D I R A C Q U O T E S :• Q: What do you like best about America?

• A: Potatoes.

• Q: What is you favourite sport?

• A: Chinese chess.

• Q: Do you go to the movies?

• A: Yes.

• Q: When?

• A: 1920, perhaps also 1930.

• Q: How did you find the Dirac Equation?

• A: I found it beautiful.

• Q: Professor Dirac, I did not understand how you derived the formula on the top left

• A: That is not a question. It is a statement. Next question, please.

• I was taught at school never to start a sentence without knowing the end of it.

• “Dirac’s spoken vocabulary consists of “yes”, “no”, and “I don’t know.”

“ G A M M A ” M AT R I C E S :

• Note that this is a particular representation of the matrices

• Any set of matrices satisfying the anti-commutation relations works

• There are an infinite number of possibilities: this particular one (Björken-Drell) is just one example

�µ = (�0, �1, �2, �3)

�0 =

�

⇧⇧⇤

1 0 0 00 1 0 00 0 �1 00 0 0 �1

⇥

⌃⌃⌅

�2 =

�

⇧⇧⇤

0 0 0 �i0 0 i 00 i 0 0�i 0 0 0

⇥

⌃⌃⌅

�3 =

�

⇧⇧⇤

0 0 1 00 0 0 �1�1 0 0 00 1 0 0

⇥

⌃⌃⌅

�1 =

�

⇧⇧⇤

0 0 0 10 0 1 00 �1 0 0�1 0 0 0

⇥

⌃⌃⌅

=�

1 00 �1

⇥

=�

0 �1

��1 0

⇥

=�

0 �2

��2 0

⇥

=�

0 �3

��3 0

⇥

⌅� = (�1, �2, �3) =⇤�

0 11 0

⇥,

�0 �ii 0

⇥,

�1 00 �1

⇥⌅

{�µ, �⇥} = �µ�⇥ + �⇥�µ = 2gµ⇥

I N F U L L G L O RY

(� · p)2 = p · p(� · a)(� · b) = a · b + i� · (a⇥ b)From problem 4.20c

�A

�B

• This is just the KG equation four times

• Wavefunctions that satisfy the Dirac equation also satisfy KG

2

6664i

0

BBB@

@

@t

0 � @

@z

� @

@x

+ i @

@y

0 @

@t

� @

@x

� i @

@y

@

@z

@

@z

@

@x

� i @

@y

@

@t

0@

@x

+ i @

@y

� @

@z

0 � @

@t

1

CCCA�

0

BB@

m 0 0 00 m 0 00 0 m 00 0 0 m

1

CCA

3

7775

0

BB@

1

2

3

4

1

CCA =

0

BB@

0000

1

CCA

pµ , �@µ

✓p20 �m2 � p2 0

0 p20 �m2 � p2

◆✓ A

B

◆=

✓00

◆

✓p0 �m �p · �p · � �p0 �m

◆✓ A

B

◆=

✓00

◆

✓p0 +m �p · �p · � �p0 +m

◆✓p0 �m �p · �p · � �p0 �m

◆=

✓p20 �m2 � (p · �)2 0

0 p20 �m2 � (p · �)2◆

S O L U T I O N S T O T H E D I R A C E Q U AT I O N• Particle at rest

• General plane wave solution

• A particle at rest has only time dependence.

• The equation breaks up into two independent parts:

(x) ⇠ e

�ik·x

(i�0@

@t�mc) = 0

✓1 00 �1

◆✓@@t A@@t B

◆= �im

✓@@t A@@t B

◆

@

@t A = �im A � @

@t B = �im B

A = e�imt A(0) B = e+imt B(0)

D I R A C ' S D I L E M M A• ψB appears to have negative energy

• Why don’t all particles fall down into these states (and down to -∞)?

• Dirac’s excuse: all states in the universe up to a certain level (say E=0) are filled.

• Pauli exclusion prevents collapse of states down to E = -∞

• We can “excite” particles out

of the sea into free states

This leaves a “hole” that looks like

a particle with opposite properties

(positive charge, opposite spin, etc.)

kF =0ee

e

hole

Dirac originally proposed that this might be the proton

A = e�imt A(0) B = e+imt B(0)

E X C U S E T O T R I U M F• 1932: Anderson finds “positrons”

in cosmic rays

• Exactly like electrons but

positively charged:

Fits what Dirac was looking for

Dirac predicts the existence of anti-matter and it is found

S O L U T I O N S T O T H E D I R A C E Q U AT I O N• Note that all particles have the same mass

�1(t) = e�imc2t/�

�

⇧⇧⇤

1000

⇥

⌃⌃⌅

positive energy solutions (particle)

“negative” energy solutions (anti-particle)

“spin up” “spin down”

“spin down” “spin up”

2(t) = e�imt

0

BB@

0100

1

CCA

3(t) = e+imt

0

BB@

0010

1

CCA 4(t) = e+imt

0

BB@

0010

1

CCA

P E D A G O C I A L S O R E P O I N T• Discussion we had thus far about difficulties with relativistic

equations, (negative probabilities, negative energies) is historic

• The framework for dealing with quantum mechanics and special relativity (i.e. quantum field theory) had not been developed

• The old tools of NR quantum mechanics had reached their limit and new ones were necessary.

• In particular, the idea of a “wavefunction” had to be revisited

• Until this was done, there were many difficulties!

• Once QFT was developed, all of these problems go away.

• Both KG and Dirac Equations are valid in QFT

• No negative probabilities, no negative energies

P L A N E WAV E S O L U T I O N S T O T H E D I R A C E Q U AT I O N

• Consider a solution of the form:

• and place it in the Dirac equation:

Column vector of 4 elements with space-time dependence

space-time dependence

column vector

�(x) = e�ik·xu(k)

(i�µ@µ �m) = 0

(�µkµ �m)u(k) = 0(�µkµ

�m)e�ik·xu(k) = 0

M O R E E X P L I C I T LY:• By definition:

• So that the Dirac equation reads:

�µkµ = �0k0 � �1k1 � �2k2 � �3k3�

k0 �k · �k · � �k0

⇥

Note 2x2 notation

u(k)��

uA

uB

⇥

✓k0 �mc �k · �k · � �k0 �mc

◆✓uA

uB

◆

(k0 �m)uA � (k · �)uB = 0 uA =k · �

k0 �muB

(k · �)uA � (k0 +m)uB = 0 uB =k · �

k0 +muA

(�µkµ �m)u(k) = 0

D E T E R M I N G u• this means we can identify k ↔ ±p

• We can now construct the column vector u:

Use positive solutions

Used negative k solutionspositrons

electrons

p · � =�

pz px � ipy

px + ipy �pz

⇥

uA =k · �

k0 �muB

k · �k0 +m

k · �k0 �m

= 1

u1 = N

0

BB@

10

pz

/(E +m)(p

x

+ ipy

)/(E +m)

1

CCA u2 = N

0

BB@

01

(px

� ipy

)/(E +m)�p

z

/(E +m)

1

CCA

uB =k · �

k0 +muA

pµ = ±kµ

u3 = N

0

BB@

pz

/(E +m)(p

x

+ ipy

)/(E +m)10

1

CCA u4 = N

0

BB@

(px

� ipy

)/(E +m)�p

z

/(E +m)01

1

CCA

Used positive k solutions

N O R M A L I Z AT I O N :• Choose “normalization” of the wavefunctions

• Note that multiples of the solutions are still solution

• normalization convention simply fixes this arbitrary choice:

• for u1

u =

�

⇧⇧⇤

u1

u2

u3

u4

⇥

⌃⌃⌅ � u† = (u�1, u�2, u�3 u�4)u† � (uT )�u†u = 2E

u†1u1 = N2

1 +

p2

(E +m)2

�= 2E N =

pE +m

L O R E N T Z C O VA R I A N C E• The Dirac equation “works” in all reference frames.

• What exactly does this mean?

• i, m, γ are constants that don’t change with reference frames.

• ∂µ and ψ will change with reference frames, however.• ∂µ is a derivative that will be taken with respect to the space-time

coordinates in the new reference frame. We’ll call this ∂’µ• how does ψ change?

• ψ’ = Sψ where ψ’ is the spinor in the new reference frame

• Putting this together, we have the following transformation of the equation when evaluating it in a new reference frame

What properties does S need to make this work?

“Lorentz Covariant”(i�µ@µ �m) = 0

(i�µ@µ �m) = 0 ) (i�µ@0µ �m) 0 = 0

i�µ@0µ(S )�m(S ) = 0

P R O P E RT I E S O F S• In general, we know how to relate space time coordinates in one

reference with another (i.e. Lorentz transformation), we can do the same for the derivatives

• Using the chain rule, we get:

• where we view x as a function x’ (i.e. the original coordinates as a function of the transformed or primed coordinates).

• Note the summation over ν

• if the primed coordinates moving along the x axis with velocity β:

��µ �

�

�xµ� =�x⇥

�xµ��

�x⇥

x0 = ⇥(x0� + �x1�)x1 = ⇥(x1� + �x0�)x2 = x2�

x3 = x3�

(⇤ = 0, µ = 0)� ⇧x0

⇧x0� = �

(⌅ = 0, µ = 1)� ⌃x0

⌃x1� = ⇥�

etc.

T R A N S F O R M I N G T H E D I R A C E Q U AT I O N

S is constant in space time, so we can move it to the left of the derivatives

Now slap S-1 from the left

Since these equations must be the same, S must satisfy

(i�µ@µ �m) = 0

�⇥ = S�1�µS⇥x⇥

⇥xµ⇥

i�µ@0µ(S )�m(S ) = 0

i�µ@µ �m = 0 ) i�µ@0µ 0 �m 0 = 0

i�

µ @x⌫

@x

µ0 @⌫(S )�m(S ) = 0

i�

µS

@x

⌫

@x

µ0 @⌫ �mS = 0

S

�1 ! i�

µS

@x

⌫

@x

µ0 @⌫ �mS = 0

PA R I T Y O P E R AT O R• For the parity operator, invert the spatial coordinates

while keeping the time coordinate unchanged:

• We then have

P =

�

⇧⇧⇤

1 0 0 00 �1 0 00 0 �1 00 0 0 �1

⇥

⌃⌃⌅

�x0

�x0� = 1

�x1

�x1� = �1

�x2

�x2� = �1

�x3

�x3� = �1

�0 = S�1�0S

�1 = �S�1�1S

�2 = �S�1�2S

�3 = �S�1�3S

�0 =�

1 00 �1

⇥

(�0)2 = 1

{�µ, �⇥} = 2gµ⇥ ⇥ �0�i = ��i�0

�0 = �0�0�0 = �0�⇥ = S�1�µS⇥x⇥

⇥xµ⇥

Recalling We find that γ0 satisfies our needs

�i = ��0�i�0 = �0�0�i = �i

SP = �0

N E X T T I M E• Read 4.6-4.9 and Chapter 5

• Lots of notation, lots of stuff going on . . . .

• please stop by if you have questions!