I'm reading chapter 3 of Peskin and Schroeder and am stuck on page 43 of P&S. They have defined the Lorentz generators in the spinor representation as: \begin{equation} S^{\mu \nu} = \frac{i}{4}[\gamma^\mu,\gamma^\nu] \end{equation} such that a finite transformation is given by: \begin{equation} \Lambda_{1/2}=e^{-\frac{i}{2} \omega_{\mu \nu} S^{\mu \nu}} \end{equation} where $\gamma^\mu$ are the gamma matrices and $\omega_{\mu \nu}$ are the elements of a real and antisymmetric matrix. According to P&S on page 43 (between equation (3.32) and (3.33), they say that the adjoint of a Dirac spinor transforms as follows: \begin{equation} \psi^\dagger \rightarrow \psi^\dagger \left(1+\frac{i}{2} \omega_{\mu \nu}(S^{\mu \nu})^\dagger \right) \end{equation} However, I would expect the transformation to be: \begin{equation} \begin{aligned} \psi^\dagger & \rightarrow \psi^\dagger \left(1-\frac{i}{2} \omega_{\mu \nu}S^{\mu \nu} \right)^\dagger \\& = \psi^\dagger \left(1+\frac{i}{2} (\omega_{\mu \nu})^\dagger (S^{\mu \nu})^\dagger \right) \\& = \psi^\dagger \left(1-\frac{i}{2} \omega_{\mu \nu} (S^{\mu \nu})^\dagger \right) \end{aligned} \end{equation} where in the last line I made use of the fact that $\omega$ is a real and antisymmetric matrix: \begin{equation} (\omega_{\mu \nu})^\dagger = (\omega_{\mu \nu})^T = \omega_{\nu \mu} = - \omega_{\mu \nu} \end{equation} This implies that according to my calculations, equation (3.33) of P&S should actually be: \begin{equation} \overline{\psi} \rightarrow \overline{\psi} \Lambda_{1/2} \end{equation} This equation must be wrong because it means that $\overline{\psi} \psi$ does not transform as a scalar and therefore the Dirac Lagrangian is not correct. However, I do not know where my mistake is and was hoping someone could help me out?
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2$\begingroup$ I fixed the "daggers"; you just need to use the command \dagger instead of \dag. $\endgroup$– joshphysicsOct 29, 2013 at 17:32
2 Answers
The mistake you are making is in "daggering" the object $\omega_{\mu\nu}$. For each $\mu, \nu = 0,\dots 3$, the symbol $\omega_{\mu\nu}$ is a real number, so its dagger (which is really just complex conjugation in this case) does nothing; $(\omega_{\mu\nu})^\dagger = \omega_{\mu\nu}$.
When we say that $\omega_{\mu\nu}$ is an antisymmetric real matrix, we really mean that the matrix with these numbers as components is such a matrix, not that $\omega_{\mu\nu}$ is a matrix for each $\mu$ and $\nu$.
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$\begingroup$ Is this only true because $\omega_{\mu \nu}$ (i.e. the matrix elements) is contracted with a proper matrix $S^{\mu \nu}$? Or is it always true what you say? $\endgroup$– HunterOct 29, 2013 at 17:35
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2$\begingroup$ @Hunter It is always the case that each $\omega_{\mu\nu}$ is a real number for each $\mu$ and $\nu$, and as such, it is always true that the dagger of each of these guys is itself. $\endgroup$ Oct 29, 2013 at 17:37
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The last step you performed is incorrect. $$\begin{equation} \begin{aligned} \psi^\dagger & \rightarrow \psi^\dagger \left(1-\frac{i}{2} \omega_{\mu \nu}S^{\mu \nu} \right)^\dagger \\& = \psi^\dagger \left(1+\frac{i}{2} (\omega_{\mu \nu})^\dagger (S^{\mu \nu})^\dagger \right) \\& = \psi^\dagger \left(1-\frac{i}{2} \omega_{\mu \nu} (S^{\mu \nu})^\dagger \right) \end{aligned} \end{equation}$$ $\omega $ is real, simply means that $$\begin{equation} (\omega_{\mu \nu})^\dagger = \omega_{\mu \nu} \end{equation}$$ So we have $$\begin{equation} \begin{aligned} \psi^\dagger & \rightarrow \psi^\dagger \left(1-\frac{i}{2} \omega_{\mu \nu}S^{\mu \nu} \right)^\dagger \\& = \psi^\dagger \left(1+\frac{i}{2} (\omega_{\mu \nu})^\dagger (S^{\mu \nu})^\dagger \right) \\& = \psi^\dagger \left(1+\frac{i}{2} \omega_{\mu \nu} (S^{\mu \nu})^\dagger \right) \end{aligned} \end{equation}$$ To find the Lorentz invariance of $\bar\psi\psi$ you are missing $\bar\psi = \psi^\dagger\gamma^0$. When this $\gamma^0$ passes through $S^{\mu\nu}$ it fixes the problem. Solve it and share with us again.