The conserved $U(1)$ current of the Dirac Lagrangian is given by $j^\mu = \bar{\psi} \gamma^\mu \psi$, where $\bar{\psi} = \psi^\dagger \gamma^0$. As this is interpreted as electric current I would expect it to flip sign under charge conjugation. Charge conjugation Of a spinor $\psi$ is defined as $\psi^c = C\psi^*$ where $C$ is the unitary charge conjugation matrix that satisfies $C^\dagger \gamma^\mu C = -(\gamma^\mu)^*$ for all gamma matrices.

If I calculate the $U(1)$ current under charge conjugation I find $$ j^\mu_c = \bar{\psi^c}\gamma^\mu \psi^c \\ = (C \psi^*)^\dagger \gamma^0 \gamma^\mu C \psi^* \\ = (\psi^\dagger)^* C^\dagger \gamma^0 C C^\dagger \gamma^\mu C \psi^* \\ = (\psi^\dagger)^* (\gamma^0)^* (\gamma^\mu)^* \psi^* \\ = (\bar{\psi} \gamma^\mu \psi)^*\\ = (j^\mu)^* $$

Which hasn’t flipped sign as I thought it would. Have I made an error in my analysis?

Any hints would be appreciated. Thanks!

  • 2
    $\begingroup$ This is one of those annoying things. Spinors are really Grassmann numbers, and in most QFT sources complex conjugation is defined to flip the order of Grassmann multiplication. So in your penultimate line, where you "pull out" the $*$, you need to pick up an overall minus sign because you didn't flip the order of the Grassmann multiplication, and Grassmann numbers anticommute. $\endgroup$ – knzhou May 4 '20 at 23:24
  • $\begingroup$ @knzhou Thank you for your comment. Do you mean $ (\psi^\dagger)^* (\gamma^0)^* (\gamma^\mu)^* \psi^* = - (\psi^\dagger \gamma^0 \gamma^\mu \psi)^* $? This is fine and I can show this with the rules of Grassman algebra now, but this still leaves that complex conjugate remaining. How do I account for that? If I used MannyC's answer from below this would tell me that $- (\psi^\dagger \gamma^0 \gamma^\mu \psi)^* = \psi^\dagger \gamma^0 \gamma^\mu \psi $ so I have shown $j^\mu_c = j^\mu$ which is not what I wanted to show. $\endgroup$ – Hermitian_hermit May 5 '20 at 20:13

For any fermion bilinear we have $$ \psi^T_\alpha A_{\alpha\beta} \chi_\beta = - \chi^T_\beta A^T_{\beta\alpha}\psi_\alpha\,. $$ So $$ \begin{aligned} (\bar\psi \gamma^\mu \psi)^* &= -\psi^* (\gamma^\mu)^\dagger(\gamma^0)^\dagger\psi \\&= -\psi^* \gamma^0\gamma^0(\gamma^\mu)^\dagger\gamma^0\psi \\&= -\bar\psi \gamma^\mu\psi\,. \end{aligned} $$ Where I used $(\gamma^0)^2 = 1$ and $\gamma^0(\gamma^\mu)^\dagger\gamma^0 = \gamma^\mu$. In the first line I applied the identity at the beginning with $\psi^T \to \bar\psi^*$ and $\chi \to \psi^*$.

  • $\begingroup$ So would this minus sign cancel with the minus sign picked up by pulling out the complex conjugate as Knzhou mentioned in his comment above? In this case the current would not change sign? $\endgroup$ – Hermitian_hermit May 5 '20 at 10:25
  • $\begingroup$ This is the minus sign that knzhou mentioned. $\endgroup$ – MannyC May 5 '20 at 10:27
  • $\begingroup$ Knzhou stated that pulling out the complex conjugate would mean $(\psi^\dagger)^* (\gamma^0)^* (\gamma^\mu)^* \psi^* = -(\psi^\dagger \gamma^0\gamma^\mu \psi)^*$.Then your answer states $-(\psi^\dagger \gamma^0\gamma^\mu \psi)^* =\psi^\dagger \gamma^0 \gamma^\mu \psi$ which means the current doesn't change sign. I apologise as I still haven't fully understood it $\endgroup$ – Hermitian_hermit May 5 '20 at 10:31
  • $\begingroup$ No, he meant that if you pull out the complex conjugate and you want to write it as a usual fermion bilinear (with $\bar\psi$ on the left), you have to commute the two spinors. That's what gives you a minus sign. $\endgroup$ – MannyC May 5 '20 at 10:41

Starting with your third to last line, we begin by rewriting \begin{equation} \begin{split} (\psi^\dagger)^*(\gamma^0)^* (\gamma^\mu)^* \psi^* &= \psi^T \big[(\gamma^0)^\dagger\big]^T \big[(\gamma^\mu)^\dagger\big]^T (\psi^\dagger)^T \\ &= \big[\psi^\dagger (\gamma^\mu)^\dagger (\gamma^0)^\dagger \psi \big]^T\\ &= \psi^\dagger (\gamma^\mu)^\dagger (\gamma^0)^\dagger \psi \end{split} \end{equation}

where in going from penultimate to last line we have used that the components of the current are complex numbers and thus not matrix valued, such that we may drop the transpose. We may then proceed in a way similar to my answer to this question, using the following properties of the gamma matrices

\begin{align} (\gamma^0)^\dagger &= \gamma^0, \\ (\gamma^\mu)^\dagger &= \gamma^0 \gamma^\mu \gamma^0, \\ (\gamma^0)^2 &= \mathbb{I}_{4}, \end{align}

where $\mathbb{I}_{4}$ is the identity to write

\begin{equation} \begin{split} \psi^\dagger(\gamma^\mu)^\dagger(\gamma^0)^\dagger \psi &= \bar{\psi}\gamma^\mu(\gamma^0)^2\psi\\ &= \bar{\psi} \gamma^\mu \psi. \end{split} \end{equation}

This is then the result $j^\mu_c = j^\mu$. This is a consequence of the charge conjugation symmetry of quantum electrodynamics.

  • $\begingroup$ How did you get the minus sign in your very first line? $\endgroup$ – knzhou May 5 '20 at 18:06
  • $\begingroup$ @knzhou originally through use of the relation $C^\dagger \gamma^\mu C = -(\gamma^\mu)$, although I realise that I have made a mistake in how I have used it. I will correct this and the conclusion now, of course we know that QED is invariant under charge conjugation so I should have expected as much. $\endgroup$ – nuLab May 5 '20 at 18:16
  • $\begingroup$ Actually, you've made two sign errors, so the final result is unchanged. First, you missed an extra sign flip in the first equality. Then you missed another in the second equality, because of the issue highlighted in my comment and in MannyC's answer. $\endgroup$ – knzhou May 5 '20 at 18:17
  • $\begingroup$ Surely there is no sign flip due to the spinors, as we have not commuted anything. The transpose takes care of that. $\endgroup$ – nuLab May 5 '20 at 18:24
  • $\begingroup$ That's the whole point of this question, though. "Surely" there can't be a sign flip, but because of the Grassmann nature of the spinors, there actually is. If you don't account for this then you get $j^\mu_C = j^\mu$, which is incorrect. $\endgroup$ – knzhou May 5 '20 at 18:27

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