In the Weyl representation of the Dirac $\gamma$-matrices, the spinor transformations $S=e^{\frac{1}{2} \omega_{\alpha \beta} \Sigma^{\alpha \beta}} \in \,\mathrm{G}_{\mathrm{L}} \leq \mathrm{GL}(4, \mathbb{C})$ become block diagonal. This can be seen at the level of the generators. The boost generators become \begin{align} \Sigma^{0 i}=\frac{1}{4}\left[\gamma^{0}, \gamma^{i}\right]=\left(\begin{array}{cc}{-\frac{\sigma^{i}}{2}} & {0} \\ {0} & {\frac{\sigma^{i}}{2}}\end{array}\right)=: K^{i}, \end{align} and the generators of rotations are \begin{align} \Sigma^{i j}=\frac{1}{4}\left[\gamma^{i}, \gamma^{j}\right]=-i \epsilon^{i j k}\left(\begin{array}{cc}{\frac{\sigma^{k}}{2}} & {0} \\ {0} & {\frac{\sigma^{k}}{2}}\end{array}\right)=: i \epsilon^{i j k} \Sigma^{k}. \end{align} We write \begin{align} \psi=\left(\begin{array}{l}{\psi_{L}} \\ {\psi_{R}}\end{array}\right) \end{align} for the Dirac spinors, and further define \begin{align} \eta_{k} :=-\omega_{0 k} \qquad \text { and } \qquad \omega_{i j}=:-\epsilon_{i j k} \theta_{k} \end{align} such that the argument of the exponent is \begin{align} \frac{1}{2} \omega_{\alpha \beta} \Sigma^{\alpha \beta}&=\omega_{0 i} \Sigma^{0 i}+\frac{1}{2} \omega_{i j} \Sigma^{i j}\\ &=-\left(i \theta_{k} \Sigma^{k}+\eta_{k} K^{k}\right)\\ &=i \theta_{k}\left(\begin{array}{cc}{\sigma^{k}} & {0} \\ {0} & {\sigma^{k}}\end{array}\right)+\eta_{k} \frac{1}{2}\left(\begin{array}{cc}{\sigma^{k}} & {0} \\ {0} & {-\sigma^{k}}\end{array}\right). \end{align} Recalling the transformation rule for spinor fields, $\psi^{\prime}\left(x^{\prime}\right) :=S \psi\left((\Lambda(S))^{-1} x^{\prime}\right)$, we see that under a Lorentz transformation \begin{align} \left(\begin{array}{c}{\psi_{L}} \\ {\psi_{R}}\end{array}\right)(x) &\mapsto\left(\begin{array}{c}{\psi_{L}^{\prime}} \\ {\psi_{R}^{\prime}}\end{array}\right)\left(x^{\prime}\right)\\\tag{*} &=\left(\begin{array}{cc}{e^{(i \boldsymbol{\theta}+\boldsymbol{\eta}) \cdot \frac{\boldsymbol\sigma}{2}}} & {0} \\ {0} & {e^{(i \boldsymbol{\theta}-\boldsymbol{\eta}) \cdot \frac{\boldsymbol\sigma}{2}}}\end{array}\right)\left(\begin{array}{c}{\psi_{L}} \\ {\psi_{R}}\end{array}\right)\left(\Lambda^{-1} x^{\prime}\right)\\ &=\left(\begin{array}{c}{e^{(i \boldsymbol{\theta}+\boldsymbol{\eta}) \cdot \frac{\boldsymbol\sigma}{2}}\psi_{L}} \\ {e^{(i \boldsymbol{\theta}-\boldsymbol{\eta}) \cdot \frac{\boldsymbol\sigma}{2}}\psi_{R}}\end{array}\right)\left(\Lambda^{-1} x^{\prime}\right). \end{align}

My understanding is that the matrix in (*) is a reducible representation of $\mathrm{SL}(2, \mathbb{C})$; the two irreducible representations are sitting on the diagonal of that matrix. My problem with this is that its hardly clear to me that the determinant of any of these representations is unity, as is required to be a representation of $\mathrm{SL}(2, \mathbb{C})$.

The determinant of the full matrix is $e^{i\boldsymbol \theta \cdot \boldsymbol \sigma}$. I am stuck on showing that the argument of the exponential should vanish.

Next, using that $\operatorname{det}\left(e^{X}\right)=e^{\operatorname{tr}(X)}$, we have that the upper block of the matrix has determinant $\operatorname{det}e^{(i \boldsymbol{\theta}+\boldsymbol{\eta}) \cdot \frac{\boldsymbol{\sigma}}{2}} = e^{\operatorname{tr}(i \boldsymbol{\theta}+\boldsymbol{\eta}) \cdot \frac{\boldsymbol{\sigma}}{2}}$, and I'm unsure how to go about evaluation of this expression. The lower block in the matrix gives me similar troubles.

  • 2
    $\begingroup$ What you have to use is that $tr(\sigma_j)=0$ and that the trace is linear...the trace is not correctly inserted in the right-hand side of your last identity. $\endgroup$ – Valter Moretti Jun 26 at 21:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.