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$\vec{P}=\int \frac{d^3p}{(2\pi)^2}\vec{p}a_\vec{p}^{\dagger}a_\vec{p}$ is the total momentum operator.
How does this operator act on the state $\lvert\vec{p}\rangle=a_\vec{p}^\dagger \lvert0\rangle$?
Note: I know that this should give $P \lvert \vec{p} \rangle = \vec{p} \lvert \vec{p}\rangle$, but I don't see how this should follow.
Use the anticommutation relations and be careful with what you call $\vec{p}$ .
