4
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ What does the multi-particle state symbolize? In particular, how many particles of each momentum $p_j$ are there? $\endgroup$
    – PPR
    Commented Dec 30, 2014 at 21:01
  • $\begingroup$ Apparently it means that there are $n$ particles, each with a momentum $p_i$ $\endgroup$ Commented Dec 30, 2014 at 21:03
  • $\begingroup$ Good, so now you know how the number operator acts on such a state. $\endgroup$
    – PPR
    Commented Dec 30, 2014 at 21:04
  • $\begingroup$ But this is just a definition. I would like to see explicitly how the $N$ written above fulfills this task $\endgroup$ Commented Dec 30, 2014 at 21:10

2 Answers 2

7
$\begingroup$

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 $|1_q\rangle$.

$\endgroup$
3
  • $\begingroup$ 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 $\endgroup$ Commented Dec 31, 2014 at 17:55
  • 1
    $\begingroup$ 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
    Commented Dec 31, 2014 at 18:06
  • $\begingroup$ cool, thanks. They always just define it with its action on the vacuum :) $\endgroup$ Commented Dec 31, 2014 at 18:52
4
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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? $\endgroup$ Commented Dec 30, 2014 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
    Commented Dec 30, 2014 at 23:15
  • 1
    $\begingroup$ I guess the physical meaning is "an annihilation operator must annihilate something" $\endgroup$
    – rob
    Commented Dec 30, 2014 at 23:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.