Timeline for Fermionic Wick Theorem
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Aug 4, 2016 at 23:09 | comment | added | Igor Khavkine | Many authors in the physics literature use this technique implicitly. It is sometimes made somewhat more explicit, for instance in mathematical treatments of supergeometry. Authors like Berezin, DeWitt, Henneaux & Teitelboim all have books, in a rather hands-on way with specific applications in mind, where you can see this technique of enlarging scalars to include formal anticommuting parameters. There is no limit to the mathematical sophistication with which these things can be stated and more abstract treatments can be found by looking up mathematical references on supergeometry. | |
| Aug 4, 2016 at 23:05 | comment | added | Igor Khavkine | Sorry for the delayed reply. If you are comfortable with constructing algebras/rings from generators and relations, that's really all you need. The hint in talking about the $\epsilon$'s as c-numbers is that you need to extend the scalars of all the algebras you are working with from $\mathbb{C}$ to $\mathbb{C}[\epsilon]$/relations. This makes scalar products valued in the extended scalars instead of simple complex numbers, which should address the question in your last comment. | |
| Aug 1, 2016 at 8:38 | comment | added | Robert Rauch | Why? Although we have not yet explained what those c-numbers actually are, they are not just complex numbers, because they're required to anticommute among themselves. If I unsterstood you correctly, these $\epsilon_i$ are elements of some extension $A$ of my CAR C*-algebra $A_0$, so in an equation like $\langle\epsilon\cdots\rangle=\epsilon\langle\cdots\rangle$ the l.h.s. should be a complex number, whereas the r.h.s. is an element of $A$. Some hints on how to make your argument rigorous would be helpful. | |
| Jul 28, 2016 at 9:07 | comment | added | Igor Khavkine | As c-numbers, they can be freely moved in and out of expectation values, $\epsilon \langle {\cdots} \rangle = \langle \epsilon \cdots \rangle$ and $\langle \cdots \epsilon \rangle = \langle {\cdots} \rangle \epsilon$. | |
| Jul 28, 2016 at 7:56 | comment | added | Robert Rauch | By the way, can you recommend any reference on this technique? | |
| Jul 28, 2016 at 7:54 | comment | added | Robert Rauch | I'm now figuring out how to "factor out $\prod_{j=1}^n \epsilon_{k_j}$" from the right hand side, but as far as I see we have not yet explained how the $\epsilon_k$ act on $\mathcal{H}$, so expressions like $\langle \epsilon_kc_j\rangle$ are kind of meaningless so far... | |
| Jul 28, 2016 at 7:22 | comment | added | Igor Khavkine | @RobertRauch, yes and sorry about the typo! Thanks for fixing it. | |
| S Jul 28, 2016 at 7:21 | history | suggested | Robert Rauch | CC BY-SA 3.0 |
Fix typo in commutator computation
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| Jul 28, 2016 at 7:11 | review | Suggested edits | |||
| S Jul 28, 2016 at 7:21 | |||||
| Jul 28, 2016 at 7:10 | comment | added | Robert Rauch | Ok, I see now. $\eta$ is "just another $\epsilon$", so it makes sense. I was a bit confused because you wrote $[\epsilon c_i, \epsilon c_j^\dagger]$. | |
| Jul 28, 2016 at 7:05 | comment | added | Robert Rauch | From what I have understood, I coumpute as $[\epsilon c_i, \epsilon c_j^\dagger]=\epsilon c_i\epsilon c_j^\dagger - \epsilon c_j^\dagger\epsilon c_i=\epsilon^2[c_i,c_j^\dagger]=0$. And what is $\eta$? | |
| Jul 27, 2016 at 12:24 | comment | added | Theo Johnson-Freyd | This is not just a trick but in fact a universal recipe. One can prove in various ways (e.g. it is a basic exercise in categorical reasoning) that any fermionic anything --- Berezinian integrals, statements about supergeometry, etc. --- can be recovered from including c-numbers and then only working with bosonic combinations. | |
| Jul 27, 2016 at 10:07 | history | answered | Igor Khavkine | CC BY-SA 3.0 |