Let $S$ be the Lorentz transfortmation under spinor representation, and from any quantum field theory textbooks, we know that $$ S^\dagger=\gamma^0S^{-1}\gamma^0 \\ S^{-1}=\gamma^0S^\dagger\gamma^0 $$

where $\gamma^\mu$ is Dirac matrices ($\mu=0,1,2,3$).

The question that confused me is that how are $S^T$ and {$S, S^\dagger, S^{-1}, \gamma^\mu$} related?

  • $\begingroup$ In this question, $\gamma^0$ is hermitian $\endgroup$
    – Wang Yun
    Commented Dec 1, 2018 at 23:55
  • $\begingroup$ @Dan Yand I want to analyze the properties of $\psi^T \gamma^\mu C \psi$ under Lorentz tramsformation, where $C=i\gamma_0\gamma_2$, and $\psi$ is spinor. At this situation, we are going to encounter $S^T$. $\endgroup$
    – Wang Yun
    Commented Dec 2, 2018 at 0:08

1 Answer 1


Short answer: it depends on which $4\times 4$ matrix representation is used.

In any representation, as far as its action on a Dirac spinor $\psi$ is concerned, the connected part of the Lorentz group is generated by transformations of the form $$ \psi\rightarrow S \psi \hskip2cm S = \exp\left(\frac{\theta}{2}\gamma^\mu\gamma^\nu\right) \tag{1} $$ with $\mu\neq \nu$. This is a Lorentz boost if $\mu=0$ or $\nu=0$, and it is an ordinary rotation if $\mu\geq 1$ and $\nu\geq 1$.

Now suppose that we use a representation in which $\gamma^\mu$ is

  • hermitian for $\mu=0$ and anti-hermitian for $\mu =1,2,3$

  • symmetric for $\mu=0,2$ and anti-symmetric for $\mu=1,3$

The representation used in Peskin and Schroeder's An Intro to QFT satisfies these conditions. In this representation, the matrix $$ C\propto \gamma^0\gamma^2 \tag{2} $$ satisfies $$ (\gamma^\mu)^T C = -C\gamma^\mu, \tag{3} $$ which implies $$ S^T C = C S^{-1}. \tag{4} $$ Therefore, under the Lorentz transformation $\psi\rightarrow S\psi$, the quantity $\psi^T C\psi$ transforms as a scalar: $$ \psi^T C\psi\rightarrow \psi^T S^T CS\psi = \psi^T C\psi \tag{5} $$ and the quantity $\psi^T C\gamma^\mu\psi$ transforms as a vector: $$ \psi^T C\gamma^\mu\psi\rightarrow \psi^T S^T C\gamma^\mu S\psi = \psi^T C (S^{-1}\gamma^\mu S)\psi. \tag{6} $$ By the way, the similar-looking quantity $\psi^T \gamma^\mu C\psi$ transforms like this: $$ \psi^T \gamma^\mu C\psi\rightarrow \psi^T S^T \gamma^\mu CS\psi = \psi^T S^T \gamma^\mu (S^T)^{-1}C\psi, \tag{7} $$ which is not the way a vector (or any other tensor) should transform.

In any representation, we can think of (4) as the defining condition for a matrix $C$. The matrix $C$ that satisfies this condition depends on the representation; but if such a matrix exists, then equations (5)-(6) hold automatically. We can think of equations (5)-(6) as the motive for the condition (4).

Similarly, if we choose a matrix $A$ so that $$ S^\dagger A = A S^{-1}, \tag{8} $$ then the quantity $\psi^\dagger A\psi$ transforms as a scalar: $$ \psi^\dagger A\psi\rightarrow \psi^\dagger S^\dagger AS\psi = \psi^\dagger A\psi \tag{9} $$ and the quantity $\psi^\dagger A\gamma^\mu\psi$ transforms as a vector: $$ \psi^\dagger A\gamma^\mu\psi\rightarrow \psi^\dagger S^\dagger A\gamma^\mu S\psi = \psi^\dagger A (S^{-1}\gamma^\mu S)\psi. \tag{10} $$ We can think of equations (9)-(10) as the motive for the condition (8). In the representation described above, the familiar choice $A\propto\gamma^0$ satisfies the condition (8), but this depends on the representation.

  • $\begingroup$ Why does eq.(2) satisfy eq.(3)? $\endgroup$
    – Wang Yun
    Commented Dec 2, 2018 at 3:37
  • $\begingroup$ @StephenWong In the given representation, eq (2) says that $C$ is the product of the two symmetric $\gamma$-matrices. If $\gamma^\mu$ is one of these matrices, then it anti-commutes with $C$ (because it commutes with itself and anti-commutes with the other $\gamma$-matrix in $C$); and since it is symmetric, this $\gamma^\mu$ satisfies eq (3). If $\gamma^\mu$ is one of the anti-symmetric matrices, then it anti-commutes with both of the $\gamma$-matrices in $C$ because it is distinct from both of them, so it commutes with $C$. Since this $\gamma^\mu$ is anti-symmetric, it again satisfies (3). $\endgroup$ Commented Dec 2, 2018 at 3:58
  • $\begingroup$ I see. Could you explain that how did you get the equation (4) ? $\endgroup$
    – Wang Yun
    Commented Dec 2, 2018 at 5:38
  • $\begingroup$ $\gamma^\mu\gamma^\nu C=-(\gamma^\mu C)(\gamma^\nu)^T=-(-C(\gamma^\mu)^T(\gamma^\nu)^T=C(\gamma^\mu)^T(\gamma^\nu)^T$ Is it like this ? $\endgroup$
    – Wang Yun
    Commented Dec 2, 2018 at 5:48
  • $\begingroup$ why could you get eq.(4) from $(\gamma^\mu\gamma^\nu)^TC=C(\gamma^\mu\gamma^\nu)^{-1}$ ? $\endgroup$
    – Wang Yun
    Commented Dec 2, 2018 at 7:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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