Timeline for How to evaluate traces of Dirac matrices with slashed momentum in between?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Oct 8, 2020 at 16:08 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
edited tags; edited tags
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| S Oct 8, 2020 at 15:38 | history | suggested | Urb | CC BY-SA 4.0 |
Fixed capitalization
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| Oct 8, 2020 at 15:34 | review | Suggested edits | |||
| S Oct 8, 2020 at 15:38 | |||||
| Oct 8, 2020 at 15:32 | vote | accept | Manas Dogra | ||
| Oct 8, 2020 at 15:25 | answer | added | aitfel | timeline score: 1 | |
| Oct 8, 2020 at 15:08 | comment | added | jkb1603 | Yes, $\gamma^{\mu} p^{\nu} = p^{\nu} \gamma^{\mu}$ for any $\mu, \nu \in \{0,1,2,3\}$, since $\gamma^{\mu}$ is a matrix and $p^{\nu}$ (I mean the components here, not the vector) is a number (therefore $p^{\nu}$ can be pulled out of the trace). | |
| Oct 8, 2020 at 14:59 | comment | added | Manas Dogra | So, the $\gamma$ s and $p$ s commute? | |
| Oct 8, 2020 at 14:47 | comment | added | jkb1603 | Yes. It is a four vector, but its components $p_0,p_1,p_2,p_3$ are numbers in the internal (four-dimensional) vector space in which the Dirac matrices act. The trace is to be taken with respect to this vector space. | |
| Oct 8, 2020 at 14:45 | comment | added | Manas Dogra | Isn't $p_{\mu}$ a covariant 4-vector because it has one index $\mu$ attached to it? | |
| Oct 8, 2020 at 14:43 | comment | added | jkb1603 | Recall that $\not{p} = p_{\mu} \gamma^{\mu}$ (with the implicit sum over $\mu$ of course). The factor $p_{\mu}$ can be pulled out of the trace, since it is just a number, and you can use the usual trace formulas (you will also need formulas for the contraction of gamma matrices). | |
| Oct 8, 2020 at 14:38 | comment | added | Manas Dogra | Related: physics.stackexchange.com/q/255243 | |
| Oct 8, 2020 at 14:38 | history | asked | Manas Dogra | CC BY-SA 4.0 |