130_notes.dvi

(Frankie) #1
S−boost^1 = cosh

χ
2

−iγiγ 4 sinh

χ
2
S†rot= cos
θ
2

+γjγisin
θ
2

= cos
θ
2

−γiγjsin
θ
2
Sboost† = cosh

χ
2

−iγ 4 γisinh

χ
2

= cosh

χ
2

+iγiγ 4 sinh

χ
2
γ 4 Srot† γ 4 = cos

θ
2

−γiγjsin

θ
2

=Srot−^1

γ 4 Sboost† γ 4 = cosh

χ
2
−iγiγ 4 sinh

χ
2
=S−boost^1

γ 4 S†γ 4 =S−^1
ψ ̄′= (Sψ)†γ 4 =ψ†γ 4 γ 4 S†γ 4 =ψ†γ 4 S−^1 =ψS ̄ −^1

This also holds forSP.


SP=γ 4
SP†=γ 4
S−P^1 =γ 4
γ 4 SP†γ 4 =γ 4 γ 4 γ 4 =γ 4 =S−P^1

From this we can quickly get thatψψ ̄ is invariant under Lorentz transformations and hence is a
scalar.
ψ ̄′ψ′=ψS ̄ −^1 Sψ=ψψ ̄


Repeating the argument forψγ ̄μψwe have


ψ ̄′γμψ′=ψS ̄ −^1 γμSψ=aμνψγ ̄νψ

according to our derivation of the transformationsS. Under the parity transformation


ψ ̄′γμψ′=ψS ̄ −^1 γμSψ=ψγ ̄ 4 γμγ 4 ψ

the spacial components of the vector change sign and the fourthcomponent doesn’t. It transforms
like aLorentz vectorunder parity.


Similarly, forμ 6 =ν,
ψσ ̄μνψ≡ψiγ ̄ μγνψ


forms arank 2 (antisymmetric) tensor.


We now have 1+4+6 components for the scalar, vector and rank 2 antisymmetric tensor. To get an
axial vector and a pseudoscalar, wedefine the product of all gamma matrices.


γ 5 =γ 1 γ 2 γ 3 γ 4

which obviouslyanticommuteswith all the gamma matrices.


{γμ,γ 5 }= 0
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