Timeline for Spinor field normalisation from poles in the propagator
Current License: CC BY-SA 3.0
17 events
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| Jul 6, 2017 at 14:15 | vote | accept | AccidentalFourierTransform | ||
| Jul 6, 2017 at 14:15 | answer | added | AccidentalFourierTransform | timeline score: 5 | |
| Apr 20, 2016 at 13:40 | history | edited | AccidentalFourierTransform | CC BY-SA 3.0 |
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| S Oct 29, 2015 at 21:55 | history | bounty ended | CommunityBot | ||
| S Oct 29, 2015 at 21:55 | history | notice removed | CommunityBot | ||
| Oct 21, 2015 at 23:48 | answer | added | Void | timeline score: 5 | |
| Oct 21, 2015 at 23:13 | history | tweeted | twitter.com/StackPhysics/status/656971508749869056 | ||
| Oct 21, 2015 at 21:53 | history | edited | AccidentalFourierTransform | CC BY-SA 3.0 |
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| S Oct 21, 2015 at 20:23 | history | bounty started | AccidentalFourierTransform | ||
| S Oct 21, 2015 at 20:23 | history | notice added | AccidentalFourierTransform | Improve details | |
| Oct 17, 2015 at 16:26 | history | edited | AccidentalFourierTransform | CC BY-SA 3.0 |
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| Oct 16, 2015 at 17:39 | comment | added | Javier | Here's something that might help. $\not{p}=m$ is impossible, at least in the usual representation of the gamma matrices. What happens is that if $p$ goes on shell (that is, $p^2=m^2$), $\not p -m$ doesn't have a matrix inverse. Also, remember that the propagators are functions of $p^\mu$; writing them as $S(\not p)$ is just a way of saying that $p^\mu$ only appears contracted with $\gamma^\mu$. | |
| Oct 9, 2015 at 2:36 | answer | added | user2309840 | timeline score: 2 | |
| Oct 8, 2015 at 22:59 | comment | added | gented | $\not p = p^{\mu}\gamma_{\mu}$, where $\gamma_{\mu}$ are Clifford algebras operators, having one representation as Dirac matrices. However, a matrix (if you prefer) in the denominator means that you have to multiply the numerator by the inverse. Exploiting the contraction of the $\gamma$ matrices you will end up with something like $f(p)\cdot 1_{\textrm{Cliff}}$, where $f(p)$ has residues as complex function. | |
| Oct 8, 2015 at 22:48 | comment | added | AccidentalFourierTransform | Im not sure what you are referring to as an operator... $\not p$ is a matrix, not an operator. The operators are $\psi$ and $\phi$, but the propagators are just c-numebers, $S(x-y)=\langle \Omega|T \psi^\alpha(x)\psi^\beta(y)|\Omega\rangle$. As these are expectation values of operators, I dont know where I am supposed to place the identity operator? | |
| Oct 8, 2015 at 22:40 | comment | added | gented | Don't forget that any of the above expressions need to be multiplied by the identity operators in the respective spaces where they live. This said, you of course cannot bring $\not p$ in the denominator, that being an operator, unless you think of it as multiplication by $(\not p \pm m)^{-1}$; when so, multiplying by the other corresponding sign and contracting the Dirac matrices leaves back a function (whose residues do make sense) times the identity operator. | |
| Oct 8, 2015 at 22:09 | history | asked | AccidentalFourierTransform | CC BY-SA 3.0 |