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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?

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  • $\begingroup$ 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

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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.

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  • $\begingroup$ 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

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