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The composition property of Lorentz transformations forces the commutation relation:

$$[P^{\mu},P^{\rho}]=0$$

where $P$ is the four-momenta. The above seems to imply that Energy and 3-momentum operators must commute with each other for the operator of Lorentz transformations to have the form that we usually use.

Being new to QFT, I am having a hard time wrapping my head around this seeing that all through QM we almost always had Hamiltonians that didn't commute with momentum. I understand that this is a different ball game altogether but still, pretty weird.

So could anyone tell me what I am missing in all of this?

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In quantum mechanics, Hamiltonian usually acts as a function of momentum $\hat{p}$ and coordinates $\vec{x}$. It is not commutative with momentum $\hat{p}$ usually due to the coordinates or some other operators like angular momentum. But for quantum field theory, Hamiltonian operator is no longer the function of coordinates and momentum, it's now the function of fields and fields' derivative $H(\phi,\partial_\mu\phi)$. Here operator $P^i=\int dx^3T^{0i}$ is the whole system's momentum, which is different from the fields' canonical momentum $\pi(x)=\frac{\partial L}{\partial \phi}$. Pictorially we may see $H$ and $P$ are now at the same level, independently, which is not the case in QM. They form the four-vector $P^\mu$, which shows the spirit of QFT making time and space at the same level.

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    $\begingroup$ This answer is kind of misleading. In QM, $H$ and $P$ commute as well. There is no fundamental difference between QFT and non-relativistic QM when it comes to the symmetry algebra: in both cases we have $[P^\mu,P^\nu]\equiv 0$. $\endgroup$ Aug 6, 2017 at 9:20
  • $\begingroup$ Non-relativistic quantum mechanics doesn't need to obey Lorentz symmetry right? $P$ the canonical momentum won't always commute with $H$, I think. The simplest example: $H=\frac{\hat{p}^2}{2m}+\frac{1}{2}\omega x^2$ $\endgroup$
    – Kangle
    Aug 6, 2017 at 16:19

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