I was looking at how to derive the (1/2, 0) representation of the Lorentz group when acting on fields. Specifically, I'm interested in understanding the logic behind replacing the "symbols" $A,B$ with particular matrices. Here $A, B$ are the generators of the matrices $D$ which transform fields via

$$\phi'(x)_{\sigma}=\sum D_{\sigma \sigma'}\phi_{\sigma'}(\Lambda x)$$

so they are elements of the (complexified) Lie Algebra.

From what I understand, the information I have about these matrices are:

  1. The commutation relations, which were derived because $D$ is a repr of the lorentz group:

$$[A_i, A_j]=i\epsilon _{ijk}A_k$$ $$[B_i, B_j]=i\epsilon _{ijk}B_k$$ $$[A_i,B_j]=0$$

  1. The eigenvalues of $A^2$ and $B^2$: We have chosen to look at the case $(1/2, 0)$.

From this (I checked) it follows that an irreducible representation of $A_i$ has 3 matrices of the form

$$\frac{1}{2}\begin{bmatrix} \alpha & \bar{\beta}\\ \beta & -\alpha \end{bmatrix}$$

With $\alpha$ real, $\beta$ complex, and $|\alpha|^2 + |\beta|^2=1$.

This is (I believe) as far as I can get without making further assumptions.

In my class, we didn't think about any of this and just wrote down the pauli matrices as the generators. But it seems any similarity transform on the pauli matrices will also do. If I were to replace $\sigma_x$ in $e^{i\sigma_x}$ with $S\sigma_x S^{-1}$, I wouldn't get the same result. So how do I know that the pauli matrices are in fact the correct matrices to generate transformations for Majorana spinors?

Edit: I think I was unclear in my last sentence. I want to know how I can derive that the generator whose coefficient is $i \theta_1 + \eta_1$ is in fact $\sigma_x$ - or how I can be sure that it doesn't matter which pauli matrix I place there. $\sigma_x$ is the most straightforward choice to write down for this example, but still I haven't seen any clear reasoning for exactly why it must be $\sigma_x$.

  • 1
    $\begingroup$ "the correct matrices"? The Pauli matrices are a popular choice of basis. Nothing prevents you from $\sigma_x,\sigma_y,\sigma_z ~\mapsto ~ \sigma_y,\sigma_z,\sigma_x$, for instance. Choices of basis are not correct or incorrect, no? $\endgroup$ – Cosmas Zachos Feb 18 '19 at 22:35
  • $\begingroup$ Matching up the coefficients of the generators gives that, for example, in the case of a pure rotation in the y-z plane with no boost, the corresponding lie group element is $exp(-i \theta_1 A_1)$. From there it's not clear to me how replacing $A_1$ with $S A_1 S^{-1}$ is just a change of basis. $\endgroup$ – doublefelix Feb 18 '19 at 22:48
  • $\begingroup$ In my example, which can result out of a similarity transformation, did you cyclically permute all space like indices, everywhere? $\endgroup$ – Cosmas Zachos Feb 18 '19 at 22:54
  • $\begingroup$ But it's a priori not clear to me that I should have started with $\theta_1 \sigma_x$ to begin with. Of course that's the most obvious thing to write down, but I'm looking for a logical implication. I agree that if I start with that and permute both indices I will get $\theta_2 \sigma_y$ which is reasonable, and that can easily be extended to $\vec{\sigma} \cdot \hat{n}$. $\endgroup$ – doublefelix Feb 18 '19 at 23:10
  • $\begingroup$ Notice that these are not the generators of the Lorentz group. They are the generators of $\text{SU}(2)\times\SU(2)$, which is, in contrast to the Lorentz group, compact. What is true, however, is that the complex extensions of these Lie algebras coincide. And, as already mentioned, physics is independent of the choice of basis. $\endgroup$ – user178876 Feb 19 '19 at 1:43

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