Timeline for Does the $U(1)$ vector current flip under charge conjugation?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 5, 2020 at 20:00 | vote | accept | Hermitian_hermit | ||
| May 5, 2020 at 20:05 | |||||
| May 5, 2020 at 10:41 | comment | added | MannyC | 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. | |
| May 5, 2020 at 10:31 | comment | added | Hermitian_hermit | 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 | |
| May 5, 2020 at 10:27 | comment | added | MannyC | This is the minus sign that knzhou mentioned. | |
| May 5, 2020 at 10:25 | comment | added | Hermitian_hermit | 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? | |
| May 5, 2020 at 10:05 | history | answered | MannyC | CC BY-SA 4.0 |