# What is the transpose of Lorentz transformation under spinor representation?

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?

• In this question, $\gamma^0$ is hermitian Dec 1, 2018 at 23:55
• @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 $S^T$. Dec 2, 2018 at 0:08

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.

• Why does eq.(2) satisfy eq.(3)? Dec 2, 2018 at 3:37
• @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). Dec 2, 2018 at 3:58
• I see. Could you explain that how did you get the equation (4) ? Dec 2, 2018 at 5:38
• $\gamma^\mu\gamma^\nu C=-(\gamma^\mu C)(\gamma^\nu)^T=-(-C(\gamma^\mu)^T(\gamma^\nu)^T=C(\gamma^\mu)^T(\gamma^\nu)^T$ Is it like this ? Dec 2, 2018 at 5:48
• why could you get eq.(4) from $(\gamma^\mu\gamma^\nu)^TC=C(\gamma^\mu\gamma^\nu)^{-1}$ ? Dec 2, 2018 at 7:18