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Anticommutation Relations and the Exclusion Principle

The Dirac Field

Physics 217 2013, Quantum Field Theory

Michael DineDepartment of Physics

University of California, Santa Cruz

October 2013

Physics 217 2013, Quantum Field Theory The Dirac Field


Lorentz Transformation Properties of the DiracField

First, rotations. In ordinary quantum mechanics,

ψ†σiψ (1)

is a vector under rotations. How does this work?

Under infinitesimal rotations,

ψ → (1 + iωiSi)ψ (2)

Soψ†σiψ → ψ†σiψ + iωjψ†[σi , σj ]ψ (3)

= (δik − εijkωj)ψ†σkψ



This is the transformation law for a vector. We seek ananalogous construction with γµ.

First we rewrite the transformation law under rotations in a waythat is closer to the form in which we have written infinitesimalLorentz transformations. Replace ωi by

ωij = εijkωk (4)

(compare this with rewriting the magnetic field, ~B, as anantisymmetric tensor; also useful in considering Lorentztransformation properties, Fij ).

Similarly, Jij = εijkJk for the angular momentum operators(generators of rotations).



In terms of such tensors, the ordinary orbital angularmomentum operator is simply

Lij = −i(xi∂j − xj∂i) (5)

and similarly for other angular momentum operators. Theangular momentum commutation relations are:

[Jij , Jkl ] = i(δikJjl − δilJjk − δjkJil + δjlJik ) (6)

One can check these commutation relations for the Lij ’s, forexample. One can think of these as the defining relations of therotation group. E.g.

eiαij Jij ≈ (1 + iαijLij + iαijSij). (7)



Group Theory of the Lorentz Group

Similarly, we can start with the transformation law for a scalarunder Lorentz transformations:

φ′(x) = φ(Λ−1x) (8)

where Λµν ≈ 1 + ωµν . In other words:

δφ(x) = −ωµνxµ∂νφ(x) (9)

So we can define:

Lµν = i(xµ∂ν − xν∂µ) (10)

as the analog of the generators of orbital rotations; indeed, Lij

are just the angular momentum operators. We can evaluatetheir commutators, and abstract the basic commutationrelations for the generators of Lorentz transformations:

[Jµν , Jρσ]− i(gνρJµσ − gµρJνσ − gνσJµρ + gµσJνρ). (11)



Representations of the Lorentz Group

We have the analog of orbital angular momentum. For ordinaryrotations, the spin one generator is (G)i

jk = −iεijk , or in ourantisymmetric index notation:

(Gij)kl = i(δikδ

jl − δ

ikδ

j`) (12)

The analog for the Lorentz group, corresponding to thetransformation law for vectors, Vµ, is

Gµναβ = i(δµαδνβ − δ

µβδ

να) (13)

You can check that this:1 gives the correct infinitesimal transformation for a vector,

xµ

2 obeys the Lorentz group commutation relations.



Lorentz generators in the spinor representation

Starting with:{γµ, γν} = 2gµν (14)

we construct the matrices, analogous to the spin-1/2 matrices:

Sµν =i4

[γµ, γν ] (15)

(for µ, ν = i , j it is easy to check that these are the spinmatrices).

These are readily seen to obey the Lie algebra of the Lorentzgroup.



Now, however, it is not ψ†γµψ which transforms as a vector, butψγµψ, where

ψ = ψ†γ0. (16)

To check this, we need certain properties of the γµ matrices,easily seen to be true in our representations:

1 (γ0)† = γ0 (γ i)† = −γ i

2 From which it follows that: (γµ)† = γ0γµγ0.



Let’s start, in fact, by checking that ψψ is a scalar.

ψψ → ψ†(1− iωµν−i4γ0[γν , γµ]γ0)γ0(1 + iωρσ

i4

[γρ, γσ])ψ (17)

= ψψ.

Now let’s do the vector:

ψγµψ → ψ†(1− iωµν−i4γ0[γν , γµ])γ0γ0γµ(1 + iωρσ

i4

[γρ, γσ])ψ

(18)Using the commutation relations:

[γµ,Sρσ] = Gρσµν γν (19)

the right hand side of eqn. 18 becomes

ψγµψ − i2ωρσ(Gρσ)µν ψγ

νψ. (20)



Introducing also the matrix

γ5 = iγ0γ1γ2γ3 (21)

which anti commutes with all of the other γ’s, we can constructthe following bilinears in the fermion field which transform asirreducible tensors:

1 Scalar: ψψ2 Pseudoscalar: ψγ5ψ

3 Vector: ψγµψ4 Pseudo vector: ψγµγ5ψ

5 Second rank tensor: ψσµνψ.(You will get to familiarize yourselves with these objects forhomework; the "pseudo" character will be discussed shortly.)



With these results, it is easy to construct a relatiavitsicallyinvariant lagrangian:

L = iψ∂µγµψ −mψψ ≡ iψ 6∂ψ −mψψ. (22)

Euler-Lagrange equations (varying with respect to ψ, ψindependently:

i 6∂ψ −mψ = 0 (23)

the Dirac equation.We want to interpret now as quantum fields.



The canonical momentum is curious:

Π =∂L∂ψ

= iψ† (24)

With this we can construct the Hamiltonian; it is reminiscent ofDirac’s original expression:

H =

∫d3xH =

∫d3xψ†(x)(−iγ0~γ · ~∇+ βm)ψ(x). (25)

We will take a step which we will see is necessary for asensible interpretation of the theory: we require

{ψ(~x , t),Π(~x ′, t)} = iδ(~x − ~x ′). (26)

In order to develop a momentum space expansion of thefermion fields, as we did for scalars, we first, we need morecontrol of the solutions of the free field equations in momentumspace (for the scalar, these were trivial).



Momentum Space Spinors

In your text, various relations for spinors, includingorthogonality relations and spin sums, are worked out bylooking at explicit solutions. We can short circuit thesecalculations in a variety of ways. Here is one:

Two things slightly different than your text:1 Dirac matrices: It will be helpful to have an explicit

representation of the Dirac matrices, or more specifically ofDirac’s matrices, somewhat different than the one in yourtext:

γ0 =

(1 00 −1

)~γ =

(0 ~σ−~σ 0

)(27)



Quick calculation of spin sums, Normalizations,etc.

We first consider the positive energy apinors, p0 =√~p2 + m2.

If χ is a constant spinor,

u(p) = N( 6p + m)χ

solves the Dirac equation. Now take χ to be a solution of theDirac equation with ~p = 0. We can work, for this discussion, inany basis, so let’s choose our original basis, where the ~p = 0spinors are particularly simple, and take the twolinearly-independent spinors to be

χ1 =

1000

;χ2 =

0100



Let’s first get the normalizations straight. We will require:

u(p)u(p) = 2m

(note this is Lorentz invariant). With our solution the left handside is

N2χ†(6p† + m)γo(6p + m)χ

= N2χ†γoγo(6p† + m)γo(6p + m)χ

= N2χ†γo(6p + m)(6p + m)χ

= N2χ†2m(6p + m)χ

From the explicit form of the Dirac matrices and the χ’s,χ† 6pχ = E . So

N2 =1

(E + m).



With this we can do the spin sums. First note that for the χ’s, lookingat their explicit form:∑

s

χχ† =

(1 00 0

)=

12

(1 + γo).

So now ∑s

u(p, s)u(p, s) =∑

s

u(p, s)u†(p, s)γo

=12

N2( 6p + m)(1 + γo)( 6p† + m)γo

=12

N2( 6p + m)(1 + γo)γoγo( 6p† + m)γo

=12

N2(6p + m)(1 + γo)( 6p + m)

=12

N22(m + po)(6p + m)

= (6p + m).

(in the next to last step, just multiply out the terms).Physics 217 2013, Quantum Field Theory The Dirac Field


Finally, we can compute the inner products:

u†(p, s)u(p, s′) = N2χ†( 6p† + m)(6p + m)χ

= N2χ†γo( 6p + m)γo(6p + m)χ

= N2χ†γo(p2 −m2 + 2po(6p + m)γoχ

= 2Eδs,s′



Exercise:Work out the corresponding relations for the negative energyspinors, v(p, s), including the spin sums, normalization, andorthogonality relations.∑

s

uα(p, s)uβ(p, s) = ( 6p+m)αβ∑

s

vα(p, s)vβ(p, s) = ( 6p−m)αβ

u†α(p, s)uα(p, s′) = 2Epδss′ = v†α(p, s)vα(p, s′)

uα(p, s)uα(p, s′) = 2mδss′ = vα(p, s)vα(p, s′)



Now we want to write momentum space expansions of thespinor fields, analogous to those for scalar fields. Suppose wehave operators a, a† which obey the anticommutation relations:

{a,a†} = 1; {a,a} = 0; {a†,a†} = 0

Then construct the “number operator"

N = a†a

and the state |0〉 by the condition

a|0〉 = 0

(a is a destruction operator). Then

Na†|0〉 = a†aa†|0〉.



Using the anticommutation relations to move a to the right,

= −a†a†a|0〉+ a†|0〉.

In other wordsNa†|0〉 = |0〉

so a† creates a one particle state. But since (a†)2 = 0 (from theanticommutation relations) there is no two particle state.



So anticommutation relations of this kind build in the exclusionprinciple; only the zero and one-particle states are allowed.Consider, first, finite volume, as we did for the scalar field. Thefield operator Ψ(~x , t) should satisfy the Dirac equation. So wewrite

ψα =∑

s

∑p

1√2Ep

(a(p, s)uα(p, s)e−ip·x + b†(p, s)vα(p, s)eip·x

).

(28)

ψ†α =∑

s

∑p

1√2Ep

(a†(p, s)u†α(p, s)e−ip·x + b(p, s)v†α(p, s)eip·x

).

(29)



Taking{a(~p, s),a†(~p′, s′)} = δss′δ~p,~p′

satisfies the (anti) commutation relations above.For the Hamiltonian we obtain (as we will see in a moment),

H =∑~p,s

E(p)(a†(p, s)a(p, s) + b†(p, s)b(p, s)) +∞

only because we assumed anti commutation relations. So wesee that we can have states with zero or one electron and zeroor one positron for each momentum and spin.



Momentum Space Expansion of the Dirac Field,Infinite volume

Infinite volume:

ψα =∑

s

∫d3p

(2π)3√

2Ep(a(p, s)uα(p, s)e−ip·x (30)

+b†(p, s)vα(p, s)eip·x ).

ψ†α =∑

s

∫d3p

(2π)3√

2Ep(a†(p, s)u†α(p, s)e−ip·x (31)

+b(p, s)v†α(p, s)eip·x ).



Now consider the Hamiltonian:

H =

∫d3xψ

(−i~γ · ~∇+ m

)ψ. (32)

Plug in the expansions of the fields, and use:

(~γ · ~p + m)u = p0γ0u (−~γ · ~p + m)v = −p0γ0v (33)

andu†(p, s)u(p, s′) = v†(p, s)v(p, s′) = 2p0δs,s′ (34)

to find:

H =

∫d3p

(2π)3

∑s

Ep

[a†(p, s)a(p, s)− b(p, s)b†(p, s)

]. (35)



Now if we quantize as for the scalar field with commutators, weobtain a negative contribution from the negative energy states,and the energy is unbounded below. If we quantize with anticommutators, we have:

H =

∫d3p

(2π)3

∑s

Ep

[a†(p, s)a(p, s) + b†(p, s)b(p, s)− 1

].

(36)So we now have a sensible expression in terms of numberoperators, with an infinite contribution which we can think of asrepresenting the energy of the Dirac sea.



The Dirac Propagator

Just as Dirac fields obey anti commutation relations, thetime-ordered product for fermions is designed with an extraminus sign, for example:

T (ψ(x1)ψ(x2)) = θ(x01 − x0

2 )ψ(x1)ψ(x2)− θ(x02 − x0

1 )ψ(x2)ψ(x1).(37)

The basic fermion propagator is:

SF (x1 − x2) = T 〈0|ψ(x)ψ(y)|0〉. (38)



Let’s take a particular time ordering, x01 > x0

2 ,

SF =∑s,s′

∫d4x

d3pd3p′

(2π)6√

E(p)E(p′)e−ip·(x1−x2)u(p, s)u(p′, s′).

(39)The x integral gives δ(p − p′); then we have, from the sum overspins, 6p + m. Indeed, up to the factor of 6p + m, this is what wehad for the same time ordering for the scalar propagator.

For the other time ordering, we obtain, again, the same resultas for the scalar, except with a factor − 6p + m, from the spinsum. Changing p → −p gives

SF (p) =6p + m

p2 −m2 + iε. (40)



The Discrete Symmetries P and CParity: The Dirac lagrangian is unchanged if we make thereplacement:

ψ(~x , t)→ γoψ(−~x , t) (41)

Let’s see what effect this has on the creation and annihilationoperators, a, b, etc.

ψp(~x , t) = γoψ(−~x , t) (42)

=

∫d3p

(2π)3√

Ep(a(~p, s)γou(~p, s)e−ipoxo−i~p·~x +b†(~p, s)γov(~p, s)eipoxo+i~p·~x ).

We can easily determine what γo does to u and v using our explicitexpressions (ignoring the normalization factor, which is unimportantfor this discussion):

γo(6p + m)χ = (poγo + ~p · ~γ + m)γoχ (43)

= u(−~p, s)

γov(~p, s) = −v(−~p, s) (44)



So making the change of variables ~p → −~p in our expressionfor ψp, gives

ψp(~x , t) = γoψ(−~x , t) (45)

=

∫d3p

(2π)3√

Ep(a(−~p, s)u(~p, s)e−ip·x + b†(−~p, s)v(~p, s)eip·x )



Charge Conjugation: Now we can do the same thing for C.Here:

ψc(x) = γ2ψ∗(x) (46)

=

∫d3p

(2π)3√

Ep(a†(~p, s)γ2u∗(~p, s)eip·x +b(~p, s)γ2v∗(~p, s)e−ip·x ).

Now we consider the action of γ2 on u∗, v∗:

γ2u∗ = γ2( 6p∗+m)χ∗ = γ2(poγo−p1γ1+p2γ2−p3γ3+m)χ (47)

= (− 6p + m)γ2χ.



Here we have not been ashamed to use the explicit propertiesof the γ matrices; γ1 and γ3 are real, while γ2 is imaginary; thefirst two anticommute with γ2 while the third commutes. Nowwe use the explicit form of γ2 to see that it takes the positiveenergy χ to the negative energy χ, with the opposite spin. So,indeed, we have that

ac(p, s) = b(p,−s) bc(p, s) = a(p,−s) (48)

i.e. it reverses particles and antiparticles and flips the spin.

Exercise: Determine the action of P and C on particle andantiparitcle states of definite momentum.


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