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 $\lvert p_1, p_2, ...p_n \rangle$.

  • $\begingroup$ What does the multi-particle state symbolize? In particular, how many particles of each momentum $p_j$ are there? $\endgroup$ – PPR Dec 30 '14 at 21:01
  • $\begingroup$ Apparently it means that there are $n$ particles, each with a momentum $p_i$ $\endgroup$ – SuperCiocia Dec 30 '14 at 21:03
  • $\begingroup$ Good, so now you know how the number operator acts on such a state. $\endgroup$ – PPR Dec 30 '14 at 21:04
  • $\begingroup$ But this is just a definition. I would like to see explicitly how the $N$ written above fulfills this task $\endgroup$ – SuperCiocia Dec 30 '14 at 21:10

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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  • $\begingroup$ Thanks. That's exactly what I wanted. Is there a shortand to this that allows not not expand the state $|2_p\rangle$ as ^$ a_p^\dagger a_p^\dagger|0\rangle$? I.e., what is $a_p |kp\rangle$ in general? With $k$ integer $\endgroup$ – SuperCiocia Dec 31 '14 at 17:55
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    $\begingroup$ yes the rule is exactly that of the number operator: set $N_p = a_p^\dagger a_p$, then $N_p|k_p\rangle = k|k_p\rangle$. $\endgroup$ – Phoenix87 Dec 31 '14 at 18:06
  • $\begingroup$ cool, thanks. They always just define it with its action on the vacuum :) $\endgroup$ – SuperCiocia Dec 31 '14 at 18:52

The operator $a_p$ is an annihilator: on a state with momentum $p$ it gives the vacuum, $$ a_p \left| p\right> = \left|0\right>,$$ but on any other state it gives zero. (N.B. the vacuum state is quite different from the number zero!)

The operator $a_p^\dagger$ is a creator: on a state without a particle with momentum $p$, such as the vacuum, it creates one: $$ a_p^\dagger \left| 0 \right> = \left| p \right>.$$ What the creation operator does if there's already a particle with $p$ depends on what type of particles you're talking about. For fermions you're not allowed to have two particles in the same state, so $a_p^\dagger\left|p\right>=0$ (though you might properly need to specify some other quantum numbers, too). For bosons you may have multiple particles in the same state e.g. in lasers or condensates, so $a_p^\dagger \left|p\right>$ would not vanish.

This means that the product operator $a_p^\dagger a_p$ that it vanishes when acting on states which contain no particles with momentum $p$, $$ a_p^\dagger a_p \left| q \right> = 0, $$ but has eigenvalue 1 on states which contain particles with momentum $p$, $$ a_p^\dagger a_p \left| p \right> = \left| p \right>. $$ That's essentially what a Dirac delta function does. Your number operator, $$ N = \int \frac{d^3 p}{(2\pi)^3} \, a^{\dagger}_pa_p,$$ integrates over all momenta; it will leave the state unchanged, but pull out one term for every particle in the state.

You can be more explicit using commutators, as shown by Phoenix87.

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  • $\begingroup$ Thanks for your answer. I did the maths and I found that indeed $a_p^\dagger a_p \left| q \right> = 0$ when $q$ is different from $p$. But what is the physical meaning of this? I mean, if the state has no particle with momentum $p$, shouldn't $a_p just do nothing? $\endgroup$ – SuperCiocia Dec 30 '14 at 22:30
  • $\begingroup$ I think you're asking why $a_p|\text{vacuum}\rangle \neq |\text{vacuum}\rangle$? That would mean that the vacuum is an eigenstate of $a_p$, which would screw up the Hermitian-conjugate relationship between $a_p$ and $a_p^\dagger$. You can see it explicitly if you construct raising and lowering operators for a simple harmonic oscillator and try to lower the ground state; whatever state you might end up with has a normalization of zero. $\endgroup$ – rob Dec 30 '14 at 23:15
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    $\begingroup$ I guess the physical meaning is "an annihilation operator must annihilate something" $\endgroup$ – rob Dec 30 '14 at 23:16

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