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Jul 8, 2022 at 12:27 comment added Quantumwhisp @flippiefanus this isn't the usual complex dirac field, but instead a real fermionic field. If you want, you can take $\psi(x)$ to be the real part of one component of the usual dirac field. See for example the article by Jackiw (inis.iaea.org/collection/NCLCollectionStore/_Public/22/056/…, p133, V Fermionic systems)
Jul 8, 2022 at 12:17 comment added flippiefanus Shouldn't $\{\psi(x), \psi(y) \} = 0$?
S Jul 8, 2022 at 11:28 history suggested Brendan Darrer CC BY-SA 4.0
improved grammar
Jul 8, 2022 at 10:57 review Suggested edits
S Jul 8, 2022 at 11:28
Jul 8, 2022 at 10:49 history edited Qmechanic CC BY-SA 4.0
added 1 character in body; edited tags; edited title; edited tags
Jul 8, 2022 at 8:53 history asked Quantumwhisp CC BY-SA 4.0