TransWikia.com

How to prove these relations for Pauli matrices?

Physics Asked by suna-neko on January 28, 2021

I am reading Schwartz’s QFT book and I am trying to verify (10.141) and (10.142). σ means Pauli matrix and $ϵ:=−iσ_2$.
How to prove these relations?
$$sigma^{mu}_{alphadot{alpha}}sigma^{nu}_{betadot{beta}}g_{munu}=2epsilon_{alphabeta}epsilon_{dotalphadotbeta} tag{10.141}$$
$$epsilon_{alphabeta}epsilon_{dotalphadotbeta}sigma^{mualphadot{alpha}}=sigma^{mu}_{betadot{beta}}.tag{10.142} $$

One Answer

(10.142):

For the second relation we will justify the need of covariant components of Weyl-spinors in a similar way as covariant vector components are needed in special relativity (SRT). Remind that in SRT contravariant vector components transform as ($g equiv (g)_{ik}$ as Minkowski metric tensor):

$$bar{u}^k = L^k_{,m} u^m ,(bar{u}=Lu) quad text{whereas covariant components transform like}quad bar{u}_i = L_i^{, m} u_m , (bar{u} = L^{-1 T} u)quadtext{because of}quad L_i^{,m} = g_{ik} L^k_{,j} g^{jm}$$

The latter relation can be written in matrix notation:

$L^{-1T}=gLg^{-1}quadquadquad (*)$

$L^i_{,k}$ are the components of the defining representation, whereas $L_i^{,m}$ are the components of the contragredient representation. Covariant components would not be necessary if $gequiv I_{4x4}$. But that's not the case.

A similar thing occurs to Weyl-spinors (undotted and dotted) ones. To that end we have to study the representation theory of Weyl-spinors.

These transformations are realised by 2x2 matrices $Ain SL(2,mathbf{C})$:

$Psi' = APsi quadquadtext{ or} quadquad Psi'^J = A^J_{,K} Psi^K$

As for the vector representation in SRT a contragredient representation exists $Arightarrow A^{-1T}$ in spinor theory. We are going to construct it. To that purpose we need another relationship of Pauli-matrices (check it by direct calculation):

$sigma_2 vec{sigma}sigma_2^{-1} = -vec{sigma}^T$

Then a matrix $Ain SL(2,mathbf{C})$ can be written in 2 ways ($vec{alpha}$ and $vec{u}$ are the rotation and velocity parameters of the corresponding Lorentz transformation):

$$A=expleft(-frac{i}{2}(vec{alpha}-i vec{u})vec{sigma}right)=sigma_2expleft(frac{i}{2}(vec{alpha}-ivec{u})vec{sigma}^Tright)sigma_2^{-1} = sigma_2 A^{-1T} sigma_2^{-1}$$

or as we will need in the following according to $sigma_2 =sigma_2^{-1}$:

$$A = sigma_2^{-1} A^{-1T} sigma_2quad text{respectively}quad A^{-1T} = sigma_2 A sigma_2^{-1} $$

Actually this relation shows that the defining and contragredient representations of $SL(2,mathbf{C})$ are equivalent. Moreover we can observe that $sigma_2$ respectively $epsilon$ now assumes the role of the metric tensor $g$ in (*).

We can now define a spinor $Phi = epsilon Psi$ and $Phi' = epsilon Psi'$. Then the transformation between $Phi'$ and $Phi$ is the following (note that $epsilon = isigma_2$):

$$Phi'=epsilon Psi' = epsilon APsi = epsilon Aepsilon^{-1} Phi =sigma_2 Asigma_2^{-1}Phi = A^{-1 T}Phi$$

which then can be written conveniently in covariant components:

$Phi'_J = A_J^{,K} Phi_K quadtext{with}quad A_J^{,K} A^J_{,L} = delta^{K}_L$

Therefore $Phi =epsilon Psi$ actually corresponds to $Phi_J =epsilon_{JK}Psi^K$ as $u_i = g_{ik} u^k$ in SRT. The same conclusion can be drawn for dotted Weyl-spinors: $Phi_dot{J}=epsilon_{dot{J}dot{K}} Psi^{dot{K}}$.

We could also have used $sigma_2$ as "metric tensor" in spinor theory. But in order to make this tensor real $epsilon= isigma_2$ is used (it's just convention).

Then the second relation (10.142)

$ sigma_{mubetadot{beta}}:= epsilon_{alphabeta}epsilon_{dot{alpha}dot{beta}}sigma^{mualphadot{alpha}}$

is just the definition of the covariant components of the $sigma^{mu}$ and its relation to its contravariant components.

(10.141):

The first relation can be just be verified by using the definitions of the symbols ($mu =(0,1,2,3))$: $sigma^mu = (I,vec{sigma})$ with $I$ as 2x2 identity matrix. The 3 Pauli-matrices fulfill the well-known relations $sigma_a sigma_b = delta_{ab} cdot I + iepsilon_{abc}sigma_c$ with $a,b,c = 1,2,3$. And $$epsilon := isigma_2= left( begin{array}{cc} 0 & 1 -1 & 0 end{array}right)$$.

Answered by Frederic Thomas on January 28, 2021

Add your own answers!

Ask a Question

Get help from others!

© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP