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How to prove $(\gamma^\mu)^\dagger=\gamma^0\gamma^\mu\gamma^0$?

+ 7 like - 0 dislike
386 views

Studying the basics of spin-$\frac{1}{2}$ QFT, I encountered the gamma matrices. One important property is $(\gamma^5)^\dagger=\gamma^5$, the hermicity of $\gamma^5$. After some searching, I stumbled upon this interesting Phys.SE answer to an earlier question on this forum. Specifically, I am interested in the formula \begin{equation} (\gamma^\mu)^\dagger=\gamma^0\gamma^\mu\gamma^0 \end{equation} which is mentioned but not proven. After consulting a faculty member of my university, I pieced together that the proof must rely somehow on the fact that the $(\gamma^\mu)^\dagger$ also obey the Clifford algebra: $$\{(\gamma^\mu)^\dagger,(\gamma^\nu)^\dagger\}=-2\eta^{\mu\nu}$$ $$\{\gamma^\mu,\gamma^\nu\}=-2\eta^{\mu\nu}$$ (for clarity, I am using $- + + +$ signature for the Minkowski metric). This should imply that there is some similarity transformation relating the two, but I am not well-versed in group theory. I guess that it should somehow turn out that the matrix that acts to transform the two representations of this algebra into each other is $\gamma^0$, which is equal to its inverse $\gamma^0=(\gamma^0)^{-1}$, as can be seen immediately from taking $\mu=\nu=0$ in the Clifford algebra. Then, the similarity transform is in the right form:

$$ (\gamma^\mu)^\dagger=S\gamma^\mu S^{-1}=\gamma^0\gamma^\mu\gamma^0 $$

I have the feeling I've got most of the necessary ingredients. However, I can't seem to be able to make this argument explicit and clear (due to my lack of proper knowledge of group theory). Could someone help me out? It would be much appreciated.

EDIT: I am looking for an answer that does not rely on using a particular representation of the gamma matrices.


This post imported from StackExchange Physics at 2014-03-06 21:57 (UCT), posted by SE-user Danu

asked Dec 9, 2013 in Mathematics by Danu (145 points) [ revision history ]
recategorized Apr 11, 2014 by dimension10
Most voted comments show all comments
No, that would not cut it for me. I know that it can apparently be proven that all representations of the Clifford algebra are related by unitary similarity transformations, and that these preserve hermiticity, giving you the full result if you can prove it for one representation. However, it is clearly much more elegant not to resort to picking a particular representation, and I know that it should be possible this way, too!

This post imported from StackExchange Physics at 2014-03-06 21:57 (UCT), posted by SE-user Danu
Use the fact that Gamma matrices are all unitary i.e. $(\gamma^{\mu})^{\dagger}=(\gamma^{\mu})^{-1}$ (they can be chosen to be so since they form a representation of a finite group). Now using this fact along with the commutation relations we find that $(\gamma^0)^{\dagger}=\gamma ^0$ and $(\gamma ^i)^{\dagger}=-\gamma^i$. Again using the commutation relations we have $\gamma^0 \gamma^i \gamma^0 = -\gamma^i =(\gamma^i)^{\dagger}$. Also $\gamma^0 \gamma^0 \gamma^0=\gamma^0 =(\gamma^0)^{\dagger}$

This post imported from StackExchange Physics at 2014-03-06 21:57 (UCT), posted by SE-user user10001
Could you point me to a reference where I could learn why it is true that a representation of a finite group can be chosen to consist of unitary elements?

This post imported from StackExchange Physics at 2014-03-06 21:57 (UCT), posted by SE-user Danu
The proof that any finite dimensional representation of a finite group can be chosen to be unitary is not diffcult. Let $V$ be a finite dimensional complex representation of a finite group $G$. Let ( , ) be any Hermitian product on $V$. Define a new Hermitian product as $(x , y)' = \displaystyle\sum_{g\in G} (gx,gy)$. To see that the new product is unitary note that for any $h\in G$, $(hx,y)'=\displaystyle\sum_{g\in G} (ghx,gy) =\displaystyle\sum_{g\in G} (ghx,gh h^{-1}y)=(x,h^{-1}y)'$.

This post imported from StackExchange Physics at 2014-03-06 21:57 (UCT), posted by SE-user user10001
@joshphysics - There is a finite multiplicative group contained in the Clifford algebra generated by gamma matrices. E.g. in two dimensions and with metric of signature (1,1), this group consist of elements $1$, $-1$, $\gamma^0$, $-\gamma^0$, $\gamma^1$ ,$-\gamma^1$, $\gamma^0 \gamma^1$ and $-\gamma^0 \gamma^1=\gamma^1 \gamma^0$.

This post imported from StackExchange Physics at 2014-03-06 21:57 (UCT), posted by SE-user user10001
@user10001 Ah interesting! Thanks for the clarification.

This post imported from StackExchange Physics at 2014-03-06 21:57 (UCT), posted by SE-user joshphysics
@user10001 The gamma matrices generate a representation of a Clifford algebra, not a finite group, so I'm unconvinced.

This post imported from StackExchange Physics at 2014-03-06 21:57 (UCT), posted by SE-user joshphysics

2 Answers

+ 3 like - 0 dislike

How about just testing the two different cases?

I.e. if $\mu\not=0$ then the LHS becomes

\begin{equation} (\gamma^\mu)^\dagger= (\gamma^i)^\dagger= -\gamma^i \end{equation} while the RHS becomes

\begin{equation} (\gamma^\mu)^\dagger=\gamma^0\gamma^i\gamma^0 = -\gamma^0\gamma^0\gamma^i=-\gamma^i~~~~~~~~ (\text{OK}). \end{equation}

For $\mu=0$, the case is trivial.

See also the wiki page below: http://en.wikipedia.org/wiki/Gamma_matrices#Normalization

This post imported from StackExchange Physics at 2014-03-06 21:57 (UCT), posted by SE-user Love Learning
answered Dec 9, 2013 by Love Learning (160 points) [ no revision ]
I don't think it's easy to prove that $(\gamma^i)^\dagger=-\gamma^i$ without going to a particular representation. Can you show this? The reason I got stuck trying to show the hermicity of $\gamma^5$ myself in the first place is because I could not. The discussion in the linked question also seems to indicate it's not possible.

This post imported from StackExchange Physics at 2014-03-06 21:57 (UCT), posted by SE-user Danu
+ 1 like - 0 dislike

A partial answer, is that supposing the gamma matrices, block-diagonal , as $\begin{pmatrix}A&\\&\epsilon A\end{pmatrix}, \begin{pmatrix}&A\\\epsilon A&\end{pmatrix}$, where $A$ is hermitian or anti-hermitian, and $\epsilon =\pm1$, give constraints on $A$ and $\epsilon$ due to $(\gamma^0)^2= \mathbb Id_4, (\gamma^i)^2= - \mathbb Id_4$.

For instance, if $\gamma_0 = \begin{pmatrix}&A\\ \epsilon A&\end{pmatrix}$, then $(\gamma_0)^2 = \begin{pmatrix} \epsilon A^2&\\ &\epsilon A^2\end{pmatrix}$.

So, if $A$ is hermitian, we may choose $A$ such $A^2 = AA^\dagger = A^\dagger A = \mathbb Id_2$, and $\epsilon = 1$

If $A$ is anti-hermitian, we may choose $A$ such $A^2 = - AA^\dagger = - A^\dagger A= -\mathbb Id_2$, and $\epsilon=-1$

In the two cases, it is easy to see that $\gamma^0$ is hermitian.

So, with the above hypothesis about the gamma matrices, it is easy to see that $\gamma^0$ is hermitian and the $\gamma^i$ are anti-hermitian.

Now with the anti-commutation relations $\gamma^0 \gamma^i + \gamma^i \gamma^0 =0$, you have $\gamma^i= - \gamma^0 \gamma^i \gamma^0$ (remembering that $(\gamma^0)^2= \mathbb Id_4$), so you have $(\gamma^i)^\dagger= - \gamma^i = \gamma^0 \gamma^i \gamma^0$, and you have obviously $(\gamma^0)^\dagger= \gamma^0 = \gamma^0 \gamma^0 \gamma^0$

This post imported from StackExchange Physics at 2014-03-06 21:57 (UCT), posted by SE-user Trimok
answered Dec 9, 2013 by Trimok (950 points) [ no revision ]
Nice train of thought! However, I'd like to refrain from picking any representation whatsoever. The answer as outlined by user10001 in the comments to the original question does what I was looking for.

This post imported from StackExchange Physics at 2014-03-06 21:57 (UCT), posted by SE-user Danu

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