Timeline for Does the $U(1)$ vector current flip under charge conjugation?
Current License: CC BY-SA 4.0
14 events
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| May 5, 2020 at 19:51 | comment | added | Hermitian_hermit | @nuLab Wait a second, so your answer is that the current density does not flip sign under charge conjugation? knzhou stated in his previous comment that $j^\mu_C = j^\mu$ is incorrect. I would expect it to flip sign too. | |
| May 5, 2020 at 19:36 | comment | added | nuLab | Added some clarification, thank you. Please let me know if you still spot a mistake. | |
| May 5, 2020 at 19:35 | history | edited | nuLab | CC BY-SA 4.0 |
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| May 5, 2020 at 18:58 | comment | added | knzhou | As currently written, on line $-1$. You start by pulling out the complex conjugation in line $4$ of OP, but that should have incurred a minus sign. | |
| May 5, 2020 at 18:56 | comment | added | nuLab | Apologies if I am misunderstanding. On which line do you believe I should have reversed the order? | |
| May 5, 2020 at 18:53 | comment | added | knzhou | That's precisely the point. Complex conjugation is defined to reverse the order. But you didn't reverse the order. Hence, in order to undo the reversal, you had to implicitly perform an additional commutation, which is where the sign flip comes from. | |
| May 5, 2020 at 18:52 | comment | added | nuLab | I accept that if we were commuting the spinors then we would pick up a minus sign through a relation of the form $\{a,b\}=0$ for Grassmann variables $a$ and $b$. But here no commutation has taken place. | |
| May 5, 2020 at 18:27 | comment | added | knzhou | 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. | |
| May 5, 2020 at 18:24 | comment | added | nuLab | Surely there is no sign flip due to the spinors, as we have not commuted anything. The transpose takes care of that. | |
| May 5, 2020 at 18:18 | history | edited | nuLab | CC BY-SA 4.0 |
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| May 5, 2020 at 18:17 | comment | added | knzhou | 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. | |
| May 5, 2020 at 18:16 | comment | added | nuLab | @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. | |
| May 5, 2020 at 18:06 | comment | added | knzhou | How did you get the minus sign in your very first line? | |
| May 5, 2020 at 17:57 | history | answered | nuLab | CC BY-SA 4.0 |