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Momentum is usually defined in QFT as: $$P^i = \int d^3x T^{0i}$$ where $T^{uv}$ refers to energy-momentum tensor, and I believe $P$ refers to linear momentum. In fermionic field, canonical energy-momentum tensor is no longer symmetric, and we define Belinfante-Rosenfeld stress-energy tensor, which is equivalent to stress-energy tensor we use in general relativity.

Creation operator of quantum field theory is $a^{\dagger}_{k,s}$, where $k$ refers to momentum and $s$ referring to spin. Is $k$ referring to $P$ defined with $T$ being Belinfante-Rosenfeld stress-energy tensor or $P$ defined with canonical Noether energy-momentum tensor?

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    $\begingroup$ The crucial thing about the Belinfante construction is that the charge built out of the canonical or symmetric stress tensor are both the same! It's a good exercise to prove this. So the answer to your question is $P$ is the same no matter which stress tensor we use. $\endgroup$
    – Prahar
    Commented Apr 10, 2018 at 21:36
  • $\begingroup$ @Prahar Oh. But canonical Noether energy-momentum tensor is not symmetric, right? (for fermionic field) So is it that $T^{0i}$ of canonical tensor is different from $T^{0i}$ of symmetric tensor, but "charge" $P$ turns out to be same? Or is it that $T^{0i}$ is same for both symmetric and canonical, but $T^{i0}$ is different? Just wishing some clarification on my understanding. $\endgroup$
    – Brion Brion
    Commented Apr 10, 2018 at 21:43
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    $\begingroup$ $T^{0i}$ is different in both cases (generically) but the charge is the same. As I said previously - the proof of this is quite easy, you should try it. $\endgroup$
    – Prahar
    Commented Apr 10, 2018 at 21:45
  • $\begingroup$ This is explained in Section V of my Phys.SE answer here. $\endgroup$
    – Qmechanic ♦
    Commented Apr 11, 2018 at 10:32

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