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

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 $$(\gamma^\mu)^\dagger=\gamma^0\gamma^\mu\gamma^0$$ 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.

• How about just prove it using some kind of representation for the gamma-matrices? Would that be good enough for you? – Your Majesty Dec 9 '13 at 14:23
• 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! – Danu Dec 9 '13 at 14:31
• 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 $\gamma^0 \gamma^0 \gamma^0=\gamma^0 =(\gamma^0)^{\dagger}$ – user10001 Dec 9 '13 at 14:46
• 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 $h\in G$, $(hx,y)'=\displaystyle\sum_{g\in G} (ghx,gy) =\displaystyle\sum_{g\in G} (ghx,gh h^{-1}y)=(x,h^{-1}y)'$. – user10001 Dec 9 '13 at 15:29
• You can look at any of the references mentioned in this post – user10001 Dec 9 '13 at 15:30

How about just testing the two different cases?

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

$$(\gamma^\mu)^\dagger= (\gamma^i)^\dagger= -\gamma^i \tag{see below}$$ while the RHS becomes

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

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

EDIT: because of the comment by the OP I will add the following to the answer:

The properties of the gamma matrices can be derived from the properties of the $\vec{\alpha},\beta$-matrices. Some of the properties of the $\vec{\alpha},\beta$ matrices are imposed upon them motivated by physics arguments, such that the Dirac hamiltonian must be hermitian which implies $\vec{\alpha},\beta$ hermitian etc.

Notice that $$\gamma^\mu := (\beta, \beta\vec{\alpha}).$$

For instance: $$(\gamma^i)^\dagger = (\beta\alpha^i)^\dagger = (\alpha^i)^\dagger\beta^\dagger=\alpha^i\beta=-\beta\alpha^i=-\gamma^i\tag{QED}$$

See page 10 in this PDF for more on the Dirac equation, or see the old book "Quarks and Leptons..." by Halzen and Martin.

• 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. – Danu Dec 9 '13 at 14:41
• that page 10 of pdf halzen martin doesn't exist anymore. – MycrofD Mar 15 '19 at 10:40

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 $(\gamma^0)^\dagger= \gamma^0 = \gamma^0 \gamma^0 \gamma^0$

• 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. – Danu Dec 9 '13 at 20:48