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

Question

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.

Answer 1

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 \tag{see below} \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.

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.

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

Answer 2

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

Answer 3

It's easy to see from the anticommutation relations alone that $γ^0 γ^μ γ^0 = (γ^μ)^{-1}$, so $γ^0 γ^μ γ^0 = (γ^μ)^\dagger$ iff $γ^μ$ is unitary. Generators satisfying the anticommutation relations needn't be unitary, so the claim as stated is false. E.g., take $$γ^μ = \begin{cases} Γ^μ \cosh α + Γ^{1-μ} \sinh α & μ\in\{0,1\} \\ Γ^μ & μ\in\{2,3\} \end{cases}$$ where the $Γ^μ$ are standard unitary generators. These matrices and their adjoints both satisfy the anticommutation relations, but if $α\ne 0$ then for $μ\in\{0,1\}$ $$(γ^μ)^\dagger = \pm(Γ^μ \cosh α - Γ^{1-μ} \sinh α) \ne γ^0 γ^μ γ^0 = \pm γ^μ.$$