The left-handed Weyl operator is defined by the 2×2 matrix
pμˉσμ˙βα=(p0+p3p1−ip2p1+ip2p0−p3),
where ˉσμ=(1,−→σ) are sigma matrices.
One can use the sigma matrices to go back and forth between four-vectors and 2×2 matrices:
pμ⟺p˙βα≡pμˉσμ˙βα.
Given two four-vectors p and q written as 2×2 matrices,
ϵ˙α˙βϵαβp˙ααq˙ββ=2pμqμ.
Given a complex 2×2 matrix ΛL with unit determinant, it can be shown that the transformation p˙βα→(Λ−1†LpΛ−1L)˙βα preserves the product ϵ˙α˙βϵαβp˙ααq˙ββ.
How does it then follow that ΛL is a Lorentz transformation? Do we have to use the fact that ϵ˙α˙βϵαβp˙ααq˙ββ∼pμqμ? What is the Lorentz transformation for pμ due to the transformation ΛL for p˙αα?
Answer
Here are some navigation points:
4-vectors xμ are realized as Hermitian 2×2 matrices x = (x1˙1x1˙2x2˙1x2˙2) = xμˉσμ,ˉσμ = (1,→σ) = σμ, with components1 xα˙α = xμ(ˉσμ)α˙α,xμ = 12tr(σμx) = 12(σμ)˙ααxα˙α, cf. this Phys.SE post.
The quadratic form of the Minkowski metric (+,−,−,−) becomes the determinant ||x||2 = xμημνxν = xα˙αη˙αα,˙ββxβ˙β = det(x), 2η˙αα,˙ββ = 12(σμ)α˙αημν(σν)β˙β = ϵ˙α˙βϵαβ = δ˙ααδ˙ββ−δ˙αβδ˙βα. In the second equality of eq. (4) we have proven a Fierz identity/completeness relation for the Pauli-matrices.
The Minkowski inner product is determined from the quadratic form (3) via polarization ⟨x,y⟩ = xμημνyν = xα˙αη˙αα,˙ββyβ˙β = −12tr(xϵytϵ) = 14∑±||x±y||2.
Lie group elements g∈SL(2,C) act on Hermitian 2×2 matrices as x′ = gxg†,x′α˙α = gαβ xβ˙β(g†)˙β˙α.
Since the determinant is clearly preserved in eq. (6), ||x′||2 (2)= det(x′) (4)= det(gxg†) = det(x) (2)= ||x||2, the transformation (6) corresponds to a Lorentz transformation x′μ = Λμν xν,Λμν = 12tr(σμgˉσνg†),Λ ∈ O(1,3). In fact, one may show that the Lorentz transformation Λ∈SO+(1,3) belongs to the restricted Lorentz group.
--
1 Be aware that different authors use different conventions. Here unbarred (barred) sigma matrices have dotted (undotted) row index and undotted (dotted) column index, respectively, which is usually the other way around, cf. e.g. A. Zee, QFT in a Nutshell, and M. D. Schwartz, QFT & SM.
No comments:
Post a Comment