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I want to prove that the particle number operator $$N = \sum_i a^\dagger_i a_i $$ commutes with a general product of creation / annihilation operators, e.g., $$(a^\dagger_1)^2 (a^\dagger_2) (a_1)^7 (a^\dagger_3)^2 ...$$ as long as the number of creation operators is equal to the number of annihilation operators.

So I plan to compute the below commutator $$ [N, \prod_j\, (a^\dagger_j)^{n_j} (a_j)^{k_j}] $$

using the commutator basic identities and $$[a_i,a^\dagger_j]=\delta_{ij}$$

expecting to be zero only when the number of $a^\dagger$ is the same of $a$.

Is this logically correct and sufficient to prove my question or am I missing something?

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    $\begingroup$ Why are there factorials on $n$ and $k$? Going off the line above, I would expect you'd want an operator $\prod_{j} (a_{j}^{\dagger})^{n_{j}} (a_{j})^{k_j}$ in your commutator (for some collection of numbers $\{n_{j}, k_{j}\}_{j}$, where your line above would have $n_{1}=2$, $k_{1} = 7$, then $n_2=1$, $k_2 =0$ and so on) $\endgroup$ Commented Oct 26, 2020 at 1:11
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    $\begingroup$ It does makes more sense, on the contrary to my assumption. I'm going to edit it $\endgroup$
    – EinRock
    Commented Oct 26, 2020 at 1:39

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The result you seek is $$ \left[ N , \prod_{j} (a_j^{\dagger})^{n_j} (a_{j})^{k_j} \right] \ = \ \left( \sum_{i} ( n_{i} - k_{i} ) \right) \prod_{j} (a^{\dagger})^{n_{j}} (a_{j})^{k_j} \ . $$ As you guessed, this commutes with a RHS of $0$ when $n_j = k_j$ for all $j$. However, interestingly there are other combinations of operators which make this commute (for example, if you have two oscillators, then picking $n_1=2$, $n_2 = 0$ and $k_1=1$, $k_2=1$ makes the above commute ie. $N = a_1^\dagger a_1 + a_2^\dagger a_2$ would commute in this example with the operator $(a_1^\dagger)^2 a_1 a_2$).

To see how this result follows, note that this commutator has the form $$ [ \sum_{j} A_j , \prod_{j} X_{j} ] \ = \ \sum_{j} [ A_{j}, X_j ] \prod_{\ell \neq j} X_\ell \ , $$ which follows when you are careful about the tensor product structure of the operators (see Appendix A below). This means that our commutator becomes $$ \left[ N , \prod_{j} (a_j^{\dagger})^{n_j} (a_{j})^{k_j} \right] \ = \ \sum_{j} \left[ a_j^\dagger a_j, (a_j^{\dagger})^{n_j} (a_{j})^{k_j} \right] \; \prod_{\ell \neq j} (a_\ell^{\dagger})^{n_\ell} (a_{\ell})^{k_\ell} $$ It follows that $\left[ a_j^\dagger a_j, (a_j^{\dagger})^{n_j} (a_{j})^{k_j} \right] = ( n_j - k_j ) (a_j^{\dagger})^{n_j} (a_{j})^{k_j}$ (see Appendix B below). With this we get the quoted result.

Appendix A: To see why this is true, imagine there are only two oscillators, then we have $$ [A_1 + A_2, X_1 X_2] \to [ A_{1} \otimes \mathbb{I} + \mathbb{I} \otimes A_2 , X_1 \otimes X_2 ] = [ A_{1}, X_1 ] \otimes X_2 + X_1 \otimes [ A_2 , X_2 ] , $$ and the generalization of this is quoted above.

Appendix B: We just need to evaluate a commutator of the form $[ a^\dagger a, (a^\dagger)^n a^k]$ here. One way to do this is consider the cases of $n>k$, $n=k$ and $n<k$ separately. Clearly for $n=k$ we have find that $[ a^\dagger a, (a^\dagger)^n a^k] = [ N, N^n ] = 0$. For $n>k$, we can set $n = k+m$ and the commutator has the form $$ [ a^\dagger a, (a^\dagger)^n a^k] = [ N, (a^\dagger)^m N^k] $$ which is easy to evaluate case-by-case for $m=1,2,3,\ldots$ once you know that $[N,a^{\dagger}] = a^{\dagger}$. It then follows that the above is equal to $$ \cdots = m (a^{\dagger})^m N^k = (n-k) (a^{\dagger})^n a^k \ . $$ The remaining case for $n<k$ follows similarly and gives the same result.

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  • $\begingroup$ Remarkable. However I didn't quite understood the explanation related to appendix A and the appendix A itself. Care to explain or indicate me some resources so I can profound my knowledge on it? $\endgroup$
    – EinRock
    Commented Oct 26, 2020 at 3:25
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    $\begingroup$ For two oscillators as in Appendix A: There's a separate Hilbert space for each oscillator. Note that you label states for two oscillators with two labels $| n_1, n_2 \rangle$, where the operators with "$1$" subscripts only talk to the $n_1$ in the above, and similarly for the "$2$" subscripts. For example, you have operations like $a_1^{\dagger} a_{1} | n_1, n_2 \rangle = n_1 | n_1, n_2 \rangle$ or also $a_2^{\dagger} a_{2} | n_1, n_2 \rangle = n_2 | n_1, n_2 \rangle$. This should convince you that the overall Hilbert space is a tensor product of each of the two oscillators Hilbert spaces. $\endgroup$ Commented Oct 26, 2020 at 20:37
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    $\begingroup$ It is annoying to include the $\otimes$ symbols, so for brevity people write $| n_1 \rangle \otimes | n_2 \rangle \to | n_1,n_2 \rangle$, or $N = a_1^{\dagger} a_1 \otimes \mathbb{I} + \mathbb{I} \otimes a_2^{\dagger} a_2 \to a_1^\dagger a_1 + a_2^\dagger a_2$. So you've secretly been using a tensor product structure (there's lots of good literature out there about tensor products). Appendix A is just being explicit calculating $[ a_1^{\dagger} a_1 \otimes \mathbb{I} + \mathbb{I} \otimes a_2^{\dagger} a_2, (a_1^{\dagger})^{n_1} a_1^{k_1} \otimes (a_1^{\dagger})^{n_2} a_2^{k_2} ]$ $\endgroup$ Commented Oct 26, 2020 at 20:44
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Actually it does not commute. What you have is $[\hat N,a_i^\dagger]=a_i^\dagger$ so that \begin{align} [\hat N,(a_i^\dagger)^2]=[\hat N,a_i^\dagger]a_i^\dagger + a_i^\dagger [\hat N,a_i^\dagger]= 2a_i^\dagger \end{align} by the derivative rule. The induction hypothesis is that $$ [\hat N,(a_i^\dagger)^p]=p (a_i^\dagger)^p $$ so that \begin{align} [\hat N,(a_i^\dagger)^p]&= [\hat N,a_i^\dagger (a_i^\dagger)^{p-1}]\\ & =[\hat N,a_i^\dagger](a_i^\dagger)^{p-1}+ a_i^\dagger[\hat N,(a_i^{\dagger})^{p-1}]\\ &= a_i^\dagger(a_i^\dagger)^{p-1} + a_i^\dagger (p-1)(a_i^\dagger)^{p-1}\\ &= p (a_i^{\dagger})^p \end{align} As similar argument can be made for $[\hat N, (a_i)^p]$.

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  • $\begingroup$ You are absolutely right! However I am studying the case of the N operator commuting with a general product of $a^\dagger$s and $a$s as in my example above. Said that, actually this question helped me getting the result $\endgroup$
    – EinRock
    Commented Oct 26, 2020 at 3:20

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