0
$\begingroup$

Considering even dimension.

From the definition of $\gamma^{(d+1)}$ (all products of gamma matrices) and its anti commutation, $\{ \gamma^\mu, \gamma^{(d+1)}\}=0$, if we choose $\gamma^{(d+1)}$ as diagonal matrix, $\gamma^\mu$ must be a form of off-block diagonal. One simple choice for this $\gamma^\mu$ is \begin{align} \gamma^\mu = \begin{pmatrix} 0 & \sigma^\mu \\ \bar{\sigma}^\mu & 0 \end{pmatrix} \end{align} Of course $\gamma^\mu$ satisfy Clifford algebra and this gives \begin{align} &\sigma^\mu \bar{\sigma}^\nu + \sigma^\nu \bar{\sigma}^\mu = 2 \eta^{\mu\nu} \label{1} \\ &\bar{\sigma}^\mu \sigma^\nu + \bar{\sigma}^\nu \sigma^\mu = 2 \eta^{\mu\nu} \label{2} \end{align} what I want to do is prove two algebra are equivalent.

I can only see the equvialence for $\mu=\nu$ case, which gives $\sigma^\mu = \eta^{\mu\mu} (\bar{\sigma}^{\mu})^{-1}$. but have no clue for $\mu\neq \nu$.

$\endgroup$

closed as off-topic by ACuriousMind, Norbert Schuch, Wolpertinger, DanielSank, Gert Sep 11 '16 at 0:55

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – ACuriousMind, Norbert Schuch, Wolpertinger, Gert
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ This post looks like it's going to be closed under the homework policy. Please edit the post so that you show effort and explain exactly what conceptual item is confusing you. $\endgroup$ – DanielSank Sep 10 '16 at 18:36
1
$\begingroup$

I assume that you want to prove that from the first line

$\sigma^{\mu} \bar{\sigma}^{\nu}+\sigma^{\nu} \bar{\sigma}^{\mu} = 2\eta^{\mu \nu}$

the second line (i.e. same thing with $\sigma$ and $\bar{\sigma}$ switched) follows. From the definition we have $\sigma^{\mu}= (1,\sigma^i)$ and $\bar{\sigma}^{\mu}= (1,-\sigma^i)$. When both $\mu$ and $\nu \neq 0$ we have

$\sigma^{i} \bar{\sigma}^{j}=-\sigma^{i} \sigma^{j}=\bar{\sigma}^{i} \sigma^{j}$

i.e you can freely move the bar from one $\sigma$ to another, since the product does not care where that minus sign comes from. Doing this to both terms in the first line gives you the second line.

When $\mu = \nu = 0$ one has $\sigma^0 = \bar{\sigma}^0 = 1$ and so the two lines are trivially equal.

When $\mu = 0$ but $\nu = i \neq 0$ both lines give you $\sigma^i=-\bar{\sigma}^i$ which is always satisfied by definition.

This is my first answer, so I hope it is useful. If I misinterpreted your question please let me know and I will fix the answer.

$\endgroup$
  • $\begingroup$ actually i do not assume the form $\sigma^\mu = (1, \sigma^i)$ and $\bar{\sigma}^\mu = (1, -\sigma^i)$. Apparently, the indicated form was given a particular solution of $\sigma$ in $3+1$ dimension. Here i want to cover general case, which do not have explicit form of $\sigma$. $\endgroup$ – phy_math Sep 8 '16 at 10:56
  • $\begingroup$ anyway, thanks. I just figure out how to cope with that problem. By some sandwich manipulation i can obtain the same equation from the first one to the second one. $\endgroup$ – phy_math Sep 8 '16 at 10:57
  • $\begingroup$ I see! Sorry for the misunderstanding. In 3+1 dimensions you can choose the $\sigma^i$ to be the Pauli matrices, but in my answer they do not need to be that, since I am not making use their particular algebra. But I used a very specific definition for the barred $\sigma$. So could you solve the problem in the case when $\sigma$ and $\bar{\sigma}$ are completely independent entities, or you had to impose some conditions? $\endgroup$ – DelCrosB Sep 8 '16 at 13:31
  • $\begingroup$ it turns out for $\mu=\nu$ case, the bar and unbarred sigma is kind of inverse relation to each other. Using this and consider for $\mu\neq \nu$ manipulating sandwich product i can obtain same second equation from the first one. $\endgroup$ – phy_math Sep 9 '16 at 4:37

Not the answer you're looking for? Browse other questions tagged or ask your own question.