Thursday, March 10, 2016

special relativity - Transformation of the spinor indices of Hermitian 2times2 matrices under the Lorentz group


The left-handed Weyl operator is defined by the 2×2 matrix


pμˉσμ˙βα=(p0+p3p1ip2p1+ip2p0p3),


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:



pμp˙βαpμˉσμ˙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(Λ1LpΛ1L)˙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˙ββpμqμ? What is the Lorentz transformation for pμ due to the transformation ΛL for p˙αα?



Answer



Here are some navigation points:





  1. 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.




  2. 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˙ββδ˙αβδ˙βα. In the second equality of eq. (4) we have proven a Fierz identity/completeness relation for the Pauli-matrices.




  3. 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.





  4. Lie group elements gSL(2,C) act on Hermitian 2×2 matrices as x = gxg,xα˙α = gαβ xβ˙β(g)˙β˙α.




  5. 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 xμ = Λμν xν,Λμν = 12tr(σμgˉσνg),Λ  O(1,3). In fact, one may show that the Lorentz transformation ΛSO+(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

classical mechanics - Moment of a force about a given axis (Torque) - Scalar or vectorial?

I am studying Statics and saw that: The moment of a force about a given axis (or Torque) is defined by the equation: $M_X = (\vec r \times \...