Second quantization, creation and annihilation operators I found two notions of states for second quantization. 
One representation uses occupation numbers here, for example
Another one creates the n+1 th particle in a collection of n existent states. see for instance here.
Now, the problem is that in the first case the creation operator does 
$a_k^{\dagger} |N_1,N_2,..\rangle = \sqrt{N_k+1 } |N_1,N_2,..,N_{k}+1,..\rangle$
and in the latter case $a_k^{\dagger} |n\rangle = \sqrt{n+1 } |n+1 \rangle.$
So the action of this operator is very different depending on whether you write down the states in terms of their occupation number or whether you write them in terms of the ensemble of all the existing states.
Unfortunately, I just don't get how these two pictures are related to each other.
If anything is unclear, please let me know.
 A: @Xin Wang's last comment: In the first case you are simply, formally, looking at collection of k_max different, uncoupled oscillators. But you're only doing anything with the k'th one. k is an index in this case, nothing else but giving this specific oscillator a name. 
In the second case you only have one oscillator in your notation, so actually you don't need to give the annihilation operator an index, as it is implicitly fixed. It is acutally even clumsy, since you're not giving the corresponding occupation number variable n the same index.
Your question may be a semantic issue, but since you're not doing anything with all other but the k'th oscillator, their particle number will be fixed during the operation. It's just a definition to count the 'total particle number' by adding up all n_m.
A: Instead of $\sqrt{k+1}$ in your first expression, it should say $\sqrt{N_k+1}$. I think this should answer your question.
N.B. I edited your question so it might now say $\sqrt{N_k+1}$ in the first expression.
