Both $\hat{a}^\dagger\hat{a}$ and $\hat{a}\hat{a}^\dagger$ are Hermitian, how do we know which one represents the particle number?
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3 Answers
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We require that the number operator have the following property:
$$\hat n |0\rangle = 0.$$
We know that
$$\hat a |0\rangle = 0$$
and we know that
$$\hat a |1\rangle = |0\rangle$$
and we know that
$$\hat a^{\dagger} |0\rangle = |1\rangle. $$
Thus, it follows that
$$\hat a \hat a^{\dagger} \ne \hat n$$
since
$$\hat a \hat a^{\dagger} |0\rangle = \hat a|1\rangle \ne 0.$$
Now, it remains to be shown that
$$\hat a^{\dagger}\hat a = \hat n. $$
Can you take it from here?
answered Oct 10, 2014 at 23:28
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$\begingroup$ Thank you! You clearly solved my problem, but now I have another question: what's the difference between 0 and $|0\rangle$? (should I open a new question?) $\endgroup$– LePtCCommented Oct 11, 2014 at 4:05
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1$\begingroup$ @LePtC, $|0\rangle$ is a state - the state with 0 energy quanta present while 0 is a number. The product of the number 0 and any state is the number 0. Since $|0\rangle$ is an eigenstate of the number operator with eigenvalue 0, it must be the case that $\hat n |0\rangle = 0 |0\rangle = 0$. $\endgroup$ Commented Oct 11, 2014 at 16:56
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1$\begingroup$ @LePtC yes, that should be a new question, except that it's already been asked. $\endgroup$– David ZCommented Oct 12, 2014 at 7:30
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$\begingroup$ @AlfredCentauri you mean we can leave the ket when we have a 0 eigenvalue? when we get a zero spin for s=1 particle or zero energy for the Hamiltonian, why don't we leave the ket as well? $\endgroup$– LePtCCommented Oct 12, 2014 at 16:13
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$\begingroup$ @LePtC, I don't understand what you're trying to ask. A ket that is an eigenstate of some operator with eigenvalue zero has non-zero length, i.e., $\langle 0 | 0 \rangle \ne 0$. Only the $\varnothing$ (null) ket has zero length $\langle \varnothing | \varnothing\rangle = 0$ and it is not a state. $\endgroup$ Commented Oct 12, 2014 at 16:35
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Since you define e.g. in the bosonic case
$c_j^\dagger: H_N^S \rightarrow H_{N+1}^S,\quad c_j^\dagger | \ldots n_j \ldots \rangle := \sqrt{n_j+1} |\ldots n_j+1 \ldots \rangle$ $c_j: H_N^S \rightarrow H_{N-1}^S,\quad c_j | \ldots n_j \ldots \rangle := \sqrt{n_j} |\ldots n_j-1 \ldots \rangle$
it makes more sence to use $a^\dagger a$ which will give you $n_j$ (instead of $n_j+1$) as prefactor when acting on a state $| \ldots n_j \ldots \rangle $.
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$\begingroup$ But you are using the eigen state of $\hat{a}^\dagger\hat{a}$, maybe we can redefine the eigen state $|n'_i\rangle=|n_i+1\rangle$ to avoid that problem? $\endgroup$– LePtCCommented Oct 10, 2014 at 9:17
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2$\begingroup$ @LePtC: This would only "shift" your problem... $\endgroup$ Commented Oct 10, 2014 at 10:50
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The vacuum should have particle number $0$. In some detail: we would like $\hat{N}\ |0\rangle=0$ and $\hat{N}=a^\dagger a$ is the only ordering that does that. It follows from the usual commutation relations that $\hat{N} |n\rangle =n\ |n\rangle$ which is in sync with interpretation of $|n\rangle $ as an $n$-particle state in second-quantisation.
