For the first point, the transformed spinor bilinear is ψ † R Λ † R σ µ Λ R ψ R . Let us expand the quantity between the spinors infinitesimally:

Quantum Field Theory

Set 13: solutions

Exercise 1

For the first point, the transformed spinor bilinear is ψ ^† _R Λ ^† _R σ ^µ Λ R ψ R . Let us expand the quantity between the spinors infinitesimally:

Λ ^† _R σ ^µ Λ R = e

^{(~ η+i~ θ)~ σ} σ ^µ e

^{(~ η−i~ θ)~ σ} ' (1 + 1

2 (~ η − i~ θ)~ σ)σ ^µ (1 + 1

2 (~ η + i~ θ)~ σ) = σ ^µ + 1

2 η ⁱ {σ ⁱ , σ ^µ } + i

2 θ ⁱ [σ ⁱ , σ ^µ ] Let us distinguish the cases µ = 0, µ = i. Knowing the equations [σ ⁱ , σ ⁰ ] = 0, [σ ⁱ , σ ^j ] = 2i ^ijk σ ^k , {σ ⁱ , σ ⁰ } = 2σ ⁱ , {σ ⁱ , σ ^j } = 2δ ^ij σ ⁰ , we find:

Λ ^† _R σ ⁰ Λ R ' σ ⁰ + η ⁱ σ ⁱ Λ ^† _R σ ⁱ Λ _R ' σ ⁱ + η ⁱ σ ⁰ − θ ^{j jik} σ ^k

We can guess that this represents the infinitesimal Lorentz transformation of a vector, with the boost part parametrized by η ⁱ and the rotation part by θ ⁱ . Indeed, the vector (spin 1) representation of the Lorentz trans- formation is given by Λ(η, θ) ^µ _ν = e ^ω

' δ ^µ _ν + ω ^µ _ν , with the constraint that ω ^αν = η ^να ω ^µ _α is antisymmetric:

ω ^µν = −ω ^νµ , and the identification θ ⁱ = ¹ ₂ ^ijk ω ^jk , η ⁱ = ω ⁱ⁰ .

Λ ^† _R σ ⁰ Λ R ' σ ⁰ + ω ⁱ⁰ σ ⁱ = σ ⁰ + ω ⁰ _i σ ⁱ Λ ^† _R σ ⁱ Λ _R ' σ ⁱ + ω ⁱ⁰ σ ⁰ − 1

2 ^jlm ω ^{lm jik} σ ^k = σ ⁱ + ω ⁱ ₀ σ ⁰ + ω ⁱ _k σ ^k

where we used the obvious identities ω ^µi = −ω ^µ _i , ω ^µ0 = ω ^µ ₀ . Thus, we have proved the infinitesimal form of Λ ^† _R σ ^µ Λ R = Λ ^µ _ν σ ^ν .

For the second point, the calculation is very simple:

⁻¹ Λ L = ⁻¹ e ⁻

^{(~ η+i~ θ)~ σ} = e ⁻

^{(~ η+i~ θ)}

^{~ σ} = e ⁻

^{(~ η+i~ θ)(−~ σ}

⁾ =

e

^{(~ η−i~ θ)~ σ} ∗

= Λ R ∗

by using the properties of , this implies also ⁻¹ Λ R = Λ ^∗ _L , ⁻¹ Λ ^† _R = Λ ^T _L .

For the third point, we can take the first result and do a similarity transformation through :

⁻¹ Λ ^† _R σ ^µ Λ R = ⁻¹ Λ ^µ _ν σ ^ν Since ⁻¹ σ ^µ = ¯ σ ^∗µ , the previous equation is equivalent to:

Λ ^T _L ¯ σ ^∗µ Λ ^∗ _L = Λ ^µ _ν σ ¯ ^∗ν

The complex conjugate of this relation is Λ ^† _L ¯ σ ^µ Λ = Λ ^µ _ν σ ¯ ^ν . This shows that the spinor bilinear ψ ^† _L ¯ σ ^µ ψ L transforms

as a vector.

Exercise 2

Given the results in the previous exercise, it is now easy to show that Λ ⁻¹ _D γ ^µ Λ _D = Λ ^µ _ν γ ^ν . Indeed, given that Λ ^† _R = Λ ⁻¹ _L and Λ ^† _L = Λ ⁻¹ _R , then

Λ ⁻¹ _D =

Λ ⁻¹ _L 0 0 Λ ⁻¹ _R

=

Λ ^† _R 0 0 Λ ^† _L

,

and thus

Λ ⁻¹ _D γ ^µ Λ _D =

Λ ^† _R 0 0 Λ ^† _L

0 σ ^µ

¯ σ ^µ 0

Λ L 0 0 Λ R

=

0 Λ ^† _R σ ^µ Λ R

Λ ^† _L σ ¯ ^µ Λ L 0

= Λ ^µ _ν γ ^ν . To summarize:

Λ ^† _R σ ^µ Λ R = Λ ^µ _ν σ ^µ , Λ ^† _L ¯ σ ^µ Λ L = Λ ^µ _ν σ ¯ ^µ , Λ ⁻¹ _D γ ^µ Λ D = Λ ^µ _ν γ ^ν .

Recalling the definition of σ ^µ = (1, σ ⁱ ) and ¯ σ ^µ = (1, −σ ⁱ ), one easily verifies:

σ ^µ ¯ σ ^ν + σ ^ν σ ¯ ^µ =







µ = 0, ν = 0 σ ⁰ σ ¯ ⁰ + σ ⁰ σ ¯ ⁰ = 2, µ = 0, ν = i σ ⁰ σ ¯ ⁱ + σ ⁱ σ ¯ ⁰ = −σ ⁱ + σ ⁱ = 0,

µ = i, ν = j σ ⁱ σ ¯ ^j + σ ^j σ ¯ ⁱ = −σ ⁱ σ ^j − σ ^j σ ⁱ = −{σ ⁱ , σ ^j } = −2δ ^ij . The same holds for the identity ¯ σ ^µ σ ^ν + ¯ σ ^ν σ ^µ = 2η ^µν . Therefore, given the definition of the Dirac matrices,

γ ^µ =

0 σ ^µ

¯ σ ^µ 0

one can deduce the anticommutation relation {γ ^µ , γ ^ν } = 2η ^µν : 0 σ ^µ

¯ σ ^µ 0

0 σ ^ν

¯ σ ^ν 0

+ (µ ↔ ν) =

σ ^µ ¯ σ ^ν 0 0 σ ¯ ^µ σ ^ν

+ (µ ↔ ν ) = 2η ^µν 1 ₄ .

2