Timeline for Derivation of the quadratic form of the Dirac equation

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13 events

when toggle format	what		by	license	comment
May 16, 2020 at 10:01	history	edited	Qmechanic♦		edited tags
May 16, 2020 at 9:39	answer	added	Cuntista		timeline score: 4
Dec 2, 2013 at 13:21	history	tweeted			twitter.com/#!/StackPhysics/status/407499735642439680
Dec 2, 2013 at 11:42	answer	added	Trimok		timeline score: 4
Dec 2, 2013 at 6:13	history	edited	jinawee		edited tags
Dec 2, 2013 at 6:12	comment	added	mikefallopian		so after a bit of algebra I get $\sigma^{\mu\nu} \left( i \hbar \partial_{\nu} - \frac{e}{c} A_{\nu} \right) \left( i \hbar \partial_{\mu} - \frac{e}{c} A_{\mu} \right) = \frac{i \hbar e}{2c} \sigma^{\mu \nu}F_{\mu \nu}$, which leads me to be off by a factor of 2 in the end. Also, the sign of the first term is still wrong...
Dec 2, 2013 at 5:54	comment	added	mikefallopian		Oh of course, I see what you're getting at. $\sigma^{\mu \nu}=-\sigma^{\nu \mu}$, so the diagonal terms of the expansion vanish
Dec 2, 2013 at 5:43	comment	added	joshphysics		Re-examine your "well-certainly..." assertion; recall that $\sigma^{\mu\nu}$ is proportional to the commutator of $\gamma$ matrices.
Dec 2, 2013 at 5:38	comment	added	mikefallopian		Well certainly $\sigma^{\mu\nu}$ should be symmetric but I'm not sure how that helps. Also, I looked more closely at my work and the sign of the $\left(i \hbar \partial - \frac{e}{c} A \right)^2$ term appears to come out to be opposite what I'd expect.
Dec 2, 2013 at 5:35	history	edited	mikefallopian	CC BY-SA 3.0	corrected sign
Dec 2, 2013 at 4:01	review	First posts
Dec 2, 2013 at 5:18
Dec 2, 2013 at 4:00	comment	added	Zo the Relativist		What do you know about the symmetry of $\sigma^{\mu\nu}$? What does this tell you about which of those terms in your expansion should vanish?
Dec 2, 2013 at 3:44	history	asked	mikefallopian	CC BY-SA 3.0