In Messiah's Quantum Mechanics Vol. II, properties of the Dirac matrices are derived. There is so-called fundamental theorem, which states that,

Let $\gamma^\mu$ and $\gamma^{'\mu}$ be two systems of 4 fourth-order unitary matrices satisfying the relations $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu} I_{4\times 4}$ (similarly for primed gamma matrices). There exists a unitary matrix $U$, defined to within a phase, such that $\gamma^{'\mu}=U\gamma^\mu U^\dagger (\mu=0,1,2,3)$.

The unitarity of gamma matrices is defined as $\gamma^{\mu\dagger}=\gamma^0\gamma^\mu\gamma^0$. Then, it is easy to check that $\gamma^{\mu *}$ are also unitary and obey the relations $\{\gamma^{\mu *},\gamma^{\nu *}\}=2g^{\mu\nu} I_{4\times 4}$. Therefore, there is a unitary matrix, called $B$, such that

$\gamma^\mu=B\gamma^{\mu *}B^\dagger,\; \gamma^{\mu *}=B^*\gamma^\mu B^T$ ----- Eq.(1)

I can show that $BB^*=cI_{4\times 4}$ with $c$ a constant. I cannot show that $c=-1$. Can you guys help me out? Thank you very much!

The way to show $BB^*=cI_{4\times 4}$ is that by the first expression in Eq.(1), solve for $\gamma^{\mu *}=B^\dagger\gamma^\mu B$. Comparing this with the second expression in Eq.(1), we have

$B^\dagger\gamma^\mu B=B^*\gamma^\mu B^T \Rightarrow \gamma^\mu BB^*=BB^*\gamma^\mu$

So $BB^*$ commutes with $\gamma^\mu$ and by the basic properties of gamma matrices, $BB^*=cI_{4\times 4}$.

There is a hint in Messiah's book. It is that $BB^*$ is the same for whatever the system of 4 unitary matrices $\gamma^\mu$ used to define $B$.

So I choose a different system of $\gamma^{'\mu}$ and so $\gamma^{'\mu}=U\gamma^\mu U^\dagger$. Taking complex conjugates of both sides and applying Eq.(1), finally I get

$B'=c'UBU^T$ where $B'$ is for the system of $\gamma^{'\mu}$ and $c'$ is a second constant.

Simple calculation shows that $B'B'^*=UBU^TU^*B^*U^\dagger=UBB^*U^\dagger$, assuming $c'=1$ (normalization). But I cannot prove that $B'B'^*=BB^*$.


Notice that both $B$ and $B^\star$ are unitary matrices and, therefore, $B B^\star$ must be unitary as well, which implies that its determinant must be either $1$ or $-1$. Since you have already determined that $B B^\star = c\,I$, this narrows the possible values of $c$ down to $c = \pm 1$.

About the expression $B^\prime B^{\prime\star} = U B B^\star U^\dagger$, having also proved that $B B^\star = c\,I$ implies that $B^\prime B^{\prime\star} = c\, U U^\dagger = c\,I$. Therefore $B^\prime B^{\prime\star} = B B^\star$.

I still miss a proof to exclude the $c=1$ solution. Sorry.

  • $\begingroup$ Thanks @jordix all the same. You solved one of my puzzles. $\endgroup$ – Drake Marquis Nov 6 '14 at 15:28

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