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  Indices of a Pauli matrix transformed in the Lorentz representation

+ 3 like - 0 dislike

When Peskin and Schroeder want to prove a Fierz identity on page 51, they make use of the identity $$(\sigma^{\mu})_{\alpha \beta} (\sigma_{\mu})_{\gamma\delta} = 2 \epsilon_{\alpha \gamma} \epsilon_{\beta \delta}.$$ where $\sigma^{\mu} \equiv (1,\mathbf{\sigma}).$ They state "One can understand this relation by noting that the indices $\alpha, \gamma$ transform in the Lorentz representation of $\psi_L$, while $\beta,\delta$ transform in the separate representation of $\psi_R$, and the whole quantity must be a Lorentz invariant." What do they want to say?

This post imported from StackExchange Physics at 2015-04-17 07:39 (UTC), posted by SE-user L. Su
asked Apr 16, 2015 in Theoretical Physics by DarKnightS (125 points) [ no revision ]
reshown Apr 17, 2015 by DarKnightS
@Jia Yiyang You had the answer. The question may not be "high-level" enough to go to Overflow.

This post imported from StackExchange Physics at 2015-04-17 07:39 (UTC), posted by SE-user L. Su

Hi @L.Su,

it is ok to ask such graduate-level technical questions on PO, they are welcome.

@Dilaton Haha. Thank you. I certainly have no idea how to define "graduate-level."

Let's not water PO down.

3 Answers

+ 3 like - 0 dislike

Ok, here's what I figured out after you asked me the question:

Recall how (the spinor representation of) Lorentz transformations act on gamma matrices:

\[ S^{-1}(\Lambda)\gamma^{\mu}S(\Lambda)=\Lambda^\mu_{\ \ \nu}\gamma^\nu\cdots(1),\]

where according to Peskin and Schroeder, 

\[S(\Lambda)=\begin{bmatrix} S_L(\Lambda) & 0\\0& S_R(\Lambda) \end{bmatrix}\cdots(2),\]

where $S_L$ and $S_R$ are transformations that act on  left-handed spinor $\psi_L$ and right-handed spinor $\psi_R$(see P&S's equation (3.37)). And 

\[\gamma^\mu=\begin{bmatrix} 0 & \sigma^\mu\\ \bar{\sigma}^\mu& 0 \end{bmatrix}\cdots(3).\]

Plug (2) and (3) into (1) you immediately see 

\[S_L^{-1}\sigma^\mu S_R=\Lambda^\mu_{\ \ \nu}\sigma^\nu\cdots(4).\]

Note $S_L$ acts on the row index while $S_R$ acts on the column index, and this is the meaning of 

.... that the indices $\alpha, \gamma$ transform in the Lorentz representation of $\psi_L$, while $\beta, \delta$ transform in the separate representation of $\psi_R$...

Clearly this implies that $\sigma^\mu\otimes\sigma_\mu$ is invariant under the transformation of the LHS of (4). In terms of matrix entries, if we define




We need to solve for $I_{\alpha\gamma\beta\delta}$. Now clearly $\epsilon_{\alpha \gamma} \epsilon_{\beta \delta}$ is a solution, because of the identity $\epsilon_{ij}A_{li}A_{kj}=\det(A)\epsilon_{lk}$, and our $S_L, S_R$ both have determinant 1. The proportionality constant 2 can be obtained by comparing appropriate entries on both sides of the equation $(\sigma^\mu)_{\alpha\beta}(\sigma_\mu)_{\gamma\delta}=\text{const}\times\epsilon_{\alpha \gamma} \epsilon_{\beta \delta}$.

Now the only gap remained is the uniqueness of the solution. To prove uniqueness it is convenient to re-write (6) as a matrix equation:\[I(S_R\otimes S_R)=(S_L \otimes S_L)I \cdots(7),\]

where $I$ and $S\otimes S$ are $4\times 4$ matrices, in particular, the row index for $I$ is the pair $\alpha\gamma$ and column index is the pair $\beta\delta$. We are going to apply Schur's lemma to (7).

Recall in the standard representation theory analysis of Lorentz group, $S_L$ is in the $(\frac{1}{2}, 0)$ representation and $S_R$ is in the $(0,\frac{1}{2})$ representation, hence $S_L \otimes S_L\approx (1,0)\oplus (0,0)$ and $S_R \otimes S_R\approx (0,1)\oplus (0,0)$. Note they only share the 1-dimensional representation $(0, 0)$. Then Schur's lemma basically says in suitable basis, matrix $I$ is block diagonal with a 3 by 3 block and a 1 by 1 block, and the 3 by 3 block is a zero matrix, and the 1 by 1 block of course is unique up to scaling. Then return to the original basis you start with, we conclude the matrix $I$ must be unique up to scaling.


A small caveat: the choice of basis (to have block diagonalization)is only unique up to an arbitrary linear combination within each invariant subspaces, and this freedom is implemented by multiplying a 3+1 block diagonal matrix to your original similarity transfomation, and you can easily show this does not affect the uniqueness.   

answered Apr 17, 2015 by Jia Yiyang (2,635 points) [ revision history ]
edited Apr 17, 2015 by Jia Yiyang

Very detailed!

+ 2 like - 0 dislike

They mean that - apart from being able to verify the equation by brute force evaluation of both sides - one can see that it must be true based on symmetry consideration.

One has a Lorentz scalar if one multiplies the left hand side by spinors $u_L^\alpha$, $v_R^\beta$, $w_L^\gamma$, and $z_R^\delta$ (whose chirality is given by the index $L$ or $R$) and sums over repeated indices. Thus the result must be a scalar formed out of these spinors, and linear in each of them. This gives a linear combination of the possibilities $(u_L\epsilon w_L)(v_R\epsilon z_R)$, $(u_Lv_R)(w_L z_R)$, and $(u_Lz_R)(w_L v_R)$. Only the first one has the right tensor product structure to work. Thus the formula holds up to a constant factor, which is obtained by evaluating the left hand side for the particular choice $\alpha=\beta=1,\gamma=\delta=2$, say.

answered Apr 17, 2015 by Arnold Neumaier (13,189 points) [ revision history ]
edited Apr 17, 2015 by Arnold Neumaier
+ 1 like - 2 dislike
  1. The Pauli matrices anticommute, so the product of two of them has to be antisymmetric in its indices.

  2. The only antisymmetric 2-tensor in two dimensions is the Levi-Civita symbol $\epsilon$.

Hence, you can "guess" the structure of the product in question up to a constant without calculating anything explicitly.

This post imported from StackExchange Physics at 2015-04-17 07:39 (UTC), posted by SE-user ACuriousMind
answered Apr 16, 2015 by ACuriousMind (820 points) [ no revision ]
How do you attach the indices? I would like to know you understanding of the statement as well.

This post imported from StackExchange Physics at 2015-04-17 07:39 (UTC), posted by SE-user L. Su
-1. this is a tensor product, you cannot apply "anticommute"argument this way, besides even if you can, a pauli matrix does not anticommute with itself.

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