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I am computing the commutator of the Lorentz generators, from the Eqn (3.16) to Eqn (3.17) in Peskin & Schroeder. $$ \begin{aligned} J^{\mu\nu} &= i(x^\mu \partial^\nu - x^\nu \partial^\mu ) &(3.16)\\ [J^{\mu\nu}, J^{\rho\sigma}] &= i (g^{\nu\rho}J^{\mu\sigma} - g^{\mu\rho}J^{\nu\sigma} - g^{\nu\sigma}J^{\mu\rho} + g^{\mu\sigma}J^{\nu\rho}) &(3.17) \end{aligned} $$

However, my answer differs from Eqn (3.17) by a factor of $i$, as I pulled out of the factors of both terms in the commutator and got $-1$ instead of $i$. I wonder where it went wrong here?

Also, when computing it, I had: $$ \begin{aligned} &x^\mu \partial^\nu(x^\rho \partial^\sigma)\\ =& x^\mu \partial^\nu x^\rho \partial^\sigma + x^\mu x^\rho \partial^\nu \partial^\sigma \\ =& g^{\nu\rho}x^\mu\partial^\sigma + 0, \end{aligned} $$ where I assumed the second term is zero. But I don't have a solid proof why $x^\mu x^\rho \partial^\nu \partial^\sigma = 0$.

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    $\begingroup$ If the second term is zero, why is $J^{\mu\nu}$ itself not also zero? $\endgroup$
    – Oбжорoв
    Commented Jan 19 at 15:34
  • $\begingroup$ These are operators, they are hungry to act on something (some function $f(x)$ in your case). So: $x^\mu \partial^\nu( x^\rho \partial ^\sigma f(x))=\ldots$. At the end of your calculation, you write your result again without the dummy function $f(x)$. $\endgroup$
    – Hyperon
    Commented Jan 19 at 15:40
  • $\begingroup$ @Hyperon Thanks. But I don't think this is why my answer was wrong. I computed that $[x^\mu \partial^\nu, x^\rho \partial^\sigma]$. Say, adding the dummy function, it would be $([x^\mu \partial^\nu, x^\rho \partial^\sigma])f(x)$. All calculations happens before applying to that function. What I meant here is $x^\mu \partial^\nu$ acting on $x^\rho \partial^\sigma$. $\endgroup$
    – user174967
    Commented Jan 19 at 15:51
  • $\begingroup$ @Oбжорoв I thought that was the two partial differentiations that makes the term zero. But I just reviewed my calculations. Those terms involving $\partial^\mu \partial^nu$'s cancels out if assuming $x^\mu$'s and $\partial^\mu$'s commute with themselves. But I'm not sure if this is always true. $\endgroup$
    – user174967
    Commented Jan 19 at 15:57
  • $\begingroup$ Are you able to reproduce the operator relation $x^\mu \partial^\nu x^\rho \partial^\sigma = x^\mu g^{\nu \rho} \partial^\sigma + x^\mu x^\rho \partial^\nu \partial^\sigma$? Good that you don't have a "solid proof" for $x^\mu x^\rho \partial^\nu \partial^\sigma=0$, because this is simply wrong. As you should also know, $[x^\mu ,\partial^\nu]= -g^{\mu \nu}$. $\endgroup$
    – Hyperon
    Commented Jan 19 at 16:08

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