Timeline for Is the expectation value of a Fermi field operator a Grassmann number?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Dec 3, 2019 at 21:45 | comment | added | knzhou | You could choose to introduce fermionic coherent states involving Grassmann coefficients, in which case all bets are off, but you don't have to. | |
| Dec 3, 2019 at 21:42 | comment | added | knzhou | Forget all about path integrals and Grassmann numbers. In good ol' canonical quantization, you have a standard Hilbert space over the complex numbers. Some of the degrees of freedom in this Hilbert space happen to represent particles with half-integer spin. In no case do you need a Grassmann number, all expectation values are manifestly complex. | |
| Dec 3, 2019 at 21:27 | answer | added | MadMax | timeline score: 2 | |
| Dec 2, 2019 at 3:32 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
added 1 character in body; edited tags; edited title
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| Dec 2, 2019 at 3:30 | comment | added | Qmechanic♦ | Possible duplicate: physics.stackexchange.com/q/269699/2451 | |
| Dec 2, 2019 at 2:45 | review | First posts | |||
| Dec 2, 2019 at 3:44 | |||||
| Dec 2, 2019 at 2:44 | history | asked | Abe Levitan | CC BY-SA 4.0 |