Commutator question: constraint on operator coefficients I'm trying to understand for which constants $c_{ijkl}$ the two operators $\sum_{n\in\mathbb{Z}}a^\dagger_na_n$ and $\sum_{ijkl\in\mathbb{Z}}c_{ijkl}(a^\dagger_ia_ja_ka_l + a^\dagger_ja^\dagger_ka^\dagger_la_i)\delta_{i,j+k+l}$ commute. An obvious option is $c_{ijkl}=1$, but can we be more general? I would guess that the answer should be in a form of some restriction (symmetry or antisymmetry) on $c_{ijkl}$.
I also have a similar question on $\sum_{n\in\mathbb{Z}}a^\dagger_na_n$ and $\sum_{ijkl\in\mathbb{Z}}d_{ijkl}(a^\dagger_i a^\dagger_ja_ka_l + a^\dagger_ka^\dagger_la_ia_j)\delta_{i+j,k+l}$. Here my guess is that $d_{ijkl}=d_{jikl}$ and $d_{ijkl}=d_{ijlk}$ would be sufficient. Is this correct?
CLARIFICATION
The operators are bosonic, $[a_m,a_n^\dagger]=\delta_{m,n}$.
 A: The operator
$$N=\sum_n a_n^\dagger a_n$$
is the total number of particles.
We will need several commutators between $N$ and various products of $a^\dagger_.$ and $a_.$
By using the bosonic commutator relations it is easy show that
$$\begin{align}
[N,a_i^\dagger] &= +a_i^\dagger \\
[N,a_i] &= -a_i
\end{align} \tag{1}$$
With a some more effort you can prove
$$\begin{align}
[N,a_i^\dagger a_j^\dagger] &= +2a_i^\dagger a_j^\dagger \\
[N,a_i^\dagger a_j] &= 0 \\
[N,a_i a_j] &= -2a_i a_j
\end{align} \tag{2}$$
and then
$$\begin{align}
[N,a_i^\dagger a_j^\dagger a_k^\dagger] &= +3a_i^\dagger a_j^\dagger a_k^\dagger \\
[N,a_i^\dagger a_j^\dagger a_k] &= +a_i^\dagger a_j^\dagger a_k \\
[N,a_i^\dagger a_j a_k] &= -a_i^\dagger a_j a_k \\
[N,a_i a_j a_k] &= -3a_i a_j a_k
\end{align} \tag{3}$$
and then
$$\begin{align}
[N,a_i^\dagger a_j^\dagger a_k^\dagger a_l^\dagger] &= +4a_i^\dagger a_j^\dagger a_k^\dagger a_l^\dagger \\
[N,a_i^\dagger a_j^\dagger a_k^\dagger a_l] &= +2a_i^\dagger a_j^\dagger a_k^\dagger a_l \\
[N,a_i^\dagger a_j^\dagger a_k a_l] &= 0 \\
[N,a_i^\dagger a_j a_k a_l] &= -2a_i^\dagger a_j a_k a_l\\
[N,a_i a_j a_k a_l] &= -4a_i a_j a_k a_l
\end{align} \tag{4}$$
Notice especially the zero result in the middle equation of (4).
This was to be expected because the operator $a_i^\dagger a_j^\dagger a_k a_l$
annihilates 2 particles and creates 2 particles,
and therefore doesn't change the total number of particles.
Hence it commutes with $N$.

With the results from (4) we can calculate the commutator:
$$\begin{align}
 &\left[N,\sum_{ijkl}d_{ijkl}(a^\dagger_i a^\dagger_ja_ka_l + a^\dagger_ka^\dagger_la_ia_j)\delta_{i+j,k+l}\right] \\
=&\sum_{ijkl}d_{ijkl}\left([N,a^\dagger_i a^\dagger_ja_ka_l]+[N,a^\dagger_ka^\dagger_la_ia_j]\right)\delta_{i+j,k+l} \\
=&\sum_{ijkl}d_{ijkl}(0+0)\delta_{i+j,k+l} \\
=&0
\end{align}$$
Obviously the result is zero completely independent of the $d_{ijkl}$.
The other commutator is more difficult to calculate:
$$\begin{align}
 &\left[N,\sum_{ijkl}c_{ijkl}(a^\dagger_i a_j a_k a_l + a^\dagger_j a^\dagger_k a^\dagger_l a_i)\delta_{i,j+k+l}\right] \\
=&\sum_{ijkl}c_{ijkl}\left([N,a^\dagger_i a_j a_k a_l]+[N,a^\dagger_j a^\dagger_k a^\dagger_l a_i]\right)\delta_{i,j+k+l} \\
=&\sum_{ijkl}c_{ijkl}\left(-2a^\dagger_ia_ja_ka_l+2a^\dagger_ja^\dagger_ka^\dagger_la_i\right)\delta_{i,j+k+l} \\
=&...
\end{align}$$
I leave it to you to finish the calculation
and see for which $c_{ijkl}$ this reduces to zero.
