The number operator, counting the number of quanta is defined as follows:
$$ N = \int \frac{d^3 p}{(2\pi)^3} \hphantom{ii} a^{\dagger}_pa_p$$
with the momentum eigenstates being defined as $\lvert p_1, p_2, ...p_n \rangle = a^{\dagger}_{p_1}a^{\dagger}_{p_2}...a^{\dagger}_{p_n}\lvert0\rangle$.
The claim is that $N \lvert p_1, p_2, ...p_n \rangle = n \lvert p_1, p_2, ...p_n \rangle$.
Can anyone show this explicitly? I have no idea what the action of $a^{\dagger}_pa_p$ is on a multi-particle state such as $\lvert p_1, p_2, ...p_n \rangle$.
Answer
It all boils down to the fact that $[a_p,a_q^\dagger] = \delta_{p,q}1$. Consider as an example $|2_p\rangle = a_p^\dagger a_p^\dagger|0\rangle$. Then the operator $a_p^\dagger a_p$ on $|2\rangle$ gives $$\begin{align} a_p^\dagger a_p|2_p\rangle &= a_p^\dagger a_p a_p^\dagger a_p^\dagger|0\rangle\\ &=a_p^\dagger[a_p,a_p^\dagger] a_p^\dagger|0\rangle + a_p^\dagger a_p^\dagger a_p a_p^\dagger|0\rangle\\ &= |2_p\rangle + a_p^\dagger a_p^\dagger a_p a_p^\dagger|0\rangle\\ &= |2_p\rangle + a_p^\dagger a_p^\dagger [a_p, a_p^\dagger]|0\rangle +a_p^\dagger a_p^\dagger a_p^\dagger a_p|0\rangle\\ &= |2_p\rangle + |2_p\rangle + 0\\ &= 2|2_p\rangle \end{align}$$ If $q\neq p$, then $a_p^\dagger a_p$ commutes with $a_q^\dagger$ and therefore it goes straight to $|0\rangle$ in the expression of $|1_q\rangle = a_q^\dagger|0\rangle$, so it gives 0, because there are no particle of momentum $p$ in $|1_q\rangle$.
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