= Normal-ordered free scalar four-momentum
{title2=$P^\mu=\int\frac{d^3p}{(2\pi)^3}p^\mu a_{\mathbf p}^\dagger a_{\mathbf p}$}
For a canonically quantized <real scalar field>, the quadratic Hamiltonian contains $\frac12\sum_{\mathbf p}E_{\mathbf p}$ in a regulated box. <Normal ordering> subtracts this <zero-point energy>, leaving
$$
P^\mu=\int\frac{d^3p}{(2\pi)^3}p^\mu a_{\mathbf p}^\dagger a_{\mathbf p}.
$$
The vacuum momentum is zero with an inversion-symmetric regulator. The <canonical commutation relation> then gives $[P^\mu,a_{\mathbf p}^\dagger]=p^\mu a_{\mathbf p}^\dagger$, identifying the momentum and energy added by one particle. The un-subtracted energy differs by a constant and is not literally the same operator.
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