Total angular momentum in QFT

Can anyone show explicitly how the QFT total angular momentum operator $$\hat{\vec{J}} = - i \int \frac{d^3p}{(2 \pi)^3} \hat{a}^{\dagger}_{\vec{p}} ( \vec{p} \times \nabla_{\vec{p}}) \hat{a}_{\vec{p}}$$ gives $$\hat{\vec{J}} |\vec{0} \, \rangle = 0~?$$

Derivation of all the above is here. The question is essentially the same but I can't really implement the existing answer mathematically.

$|\vec{0} \, \rangle$ being a momentum eigenstate.

• you have an annihilation operator to the very right, and the integral is roughly a sum of many such terms. Dec 31 '14 at 14:55
• Is it that simple? Dec 31 '14 at 14:56
• hang on you changed the question while I was typing my comment. So the ket you put there is not the vacuum, right? Dec 31 '14 at 14:57
• It's the vacuum for the momentum. $a_p |0\rangle=0$. What about the action og $J$ on a state of generic momentum $q$, $|q\rangle$? Dec 31 '14 at 15:00
• There's no "vacuum for the momentum". Either it is the vacuum, i.e. annihilating it gives just zero, or it is a one-particle state with zero momentum, i.e. annihilating it with $a_0$ gives the vacuum, or, rather, it is created from the vacuum by $a^\dagger_0$. Which do you mean? Dec 31 '14 at 15:09

If the state you're using is just the vacuum, then my comment to your question applies. If otherwise $|\mathbf 0\rangle = a_{\mathbf 0}^\dagger|0\rangle$, then just use that $[a_{\mathbf p},a_{\mathbf q}^\dagger] = \delta_{\mathbf p,\mathbf q}1$, with $\mathbf q=\mathbf 0$ to fix the integral at the term with $\mathbf p = \mathbf 0$ through the Dirac delta $\delta_{\mathbf p,\mathbf 0}$.
• Those kets arise from the Fock Hilbert space, which is constructed out of single particle Hilbert spaces, which you can take to be $L^2$ spaces, that is functions. Jan 3 '15 at 13:07