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I need help with some some notation I've not seen before. Is using the left-right arrow in this formula $$[P^μ,M^{ρσ}]=i\hbar(g^{\mu\sigma}P^\rho-(\rho\leftrightarrow\sigma))$$ equivalent to writing $$[P^μ,M^{ρσ}]=i\hbar(g^{\mu\sigma}P^\rho-g^{\mu\rho}P^\sigma)$$ if not, what does it mean?

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  • $\begingroup$ Where did you see the left-right arrow notation? Though I suspect that your interpretation is correct, it might be useful to see the original source. $\endgroup$ – Kyle Kanos Apr 29 '15 at 17:29
  • $\begingroup$ You are correct, this notation is relatively common (but pretty lazy), and usually means the same expression as given previously but with the two symbols exchanged. $\endgroup$ – Mark Mitchison Apr 29 '15 at 17:30
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Yes.

$$ f(A,B) \pm (A\leftrightarrow B) ~:=~ f(A,B)\pm f(B,A). $$

The notation is useful as a shorthand, or to convey a symmetry/antisymmetry that may often otherwise be less apparent.

See e.g. the last equation in my Phys.SE answer here for a nested example.

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I've never seen it either. It looks like it means "a term just like the one I just wrote, but with the indicies $\sigma$ and $\rho$ exchanged."

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    $\begingroup$ I've seen it regularly and I can confirm that's what it means. (I would post an answer but it'd say the same thing as yours) $\endgroup$ – David Z Apr 29 '15 at 17:30
  • $\begingroup$ @DavidZ Post it as an answer, and I'll delete mine since you have definitive information. $\endgroup$ – garyp Apr 29 '15 at 17:43
  • $\begingroup$ Ah, looks like Qmechanic already did so. $\endgroup$ – David Z Apr 30 '15 at 5:27

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