Confusion about ladder operators Let´s consider a system, that consists out of $N$ bosonic particles, that are not interacting with each other. The Hamiltonian of this system would be given as
$$H = \sum_{i=1}^N \frac{\hbar^2}{2m}\hat{p}^2_i \quad .$$
This Hamiltonian acts on the Hilbert space of $N$ particles $\mathcal{H_N}$. We can rewrite the Hamiltonian using ladder operators
$$H = \sum_{\vec{k}} \frac{\hbar^2 \vec{k}^2}{2m}c^\dagger_{\vec{k}} c_{\vec{k}} \quad . $$
What bothers me is the following: the Hamiltonian is an operator, that acts on the Hilbert space $\mathcal{H_N}$. By taking a look at the Hamiltonian using second quantization, it looks like the ladder operators "connect" different Hilbert spaces:
$$c_{\vec{k}}: \mathcal{H_N} \rightarrow \mathcal{H_{N-1}}$$
and
$$c^{\dagger}_{\vec{k}}: \mathcal{H_{N-1}} \rightarrow \mathcal{H_{N}}$$
However using the commutation relations, we can re-express the given Hamiltonian:
$$H = \sum_{\vec{k}} \frac{\hbar^2 \vec{k}^2}{2m} (c_{\vec{k}}c^\dagger_{\vec{k}}+\mathbb{1})\quad $$
In this case it looks like the operators switched the spaces, they act on:
$$c^{\dagger}_{\vec{k}}: \mathcal{H_{N}} \rightarrow \mathcal{H_{N+1}}$$
and
$$c_{\vec{k}}: \mathcal{H_{N+1}} \rightarrow \mathcal{H_{N}} \quad.$$
It appears to me very wrong, that we change the nature of the ladder operators simply by taking advantage of commutation relation. Therefore I hope for some clarification about this problem.
 A: I think the problem lies in the fact that although one can define the creation and annihilation operators for a fixed $N$, their well-known commutation relations only make sense if the operators are extended to the Fock space:
$$\mathcal F = \bigoplus\limits_{N=0}^\infty \mathcal H_N \quad . $$
Indeed, writing
$$[c,c^\dagger] =  c c^\dagger - c^\dagger c \tag{1} $$
is ill-defined if both $c$ and $c^\dagger$ are the creation and annihilation operators defined for an arbitrary but fixed $N$, e.g.  as maps
\begin{align}
c^\dagger&: \mathcal H_N \longrightarrow \mathcal H_{N+1} \\
c&:\mathcal  H_{N+1} \longrightarrow \mathcal H_N \quad .
\end{align}
This is rather easy to see, since first term on the RHS in equation $(1)$ is a map from $\mathcal H_N$ to $\mathcal H_N$, but the second term cannot act on any state $|\psi\rangle \in\mathcal  H_N$, as $c$ is only defined for states in $\mathcal H_{N+1}$. So the commutation relation you've used is actually not valid.
However, if the operators are extended to $\mathcal F$ (cf. this), then we can define their corresponding commutator and obtain the usual commutation relations. So we have $c,c^\dagger :\mathcal  F \longrightarrow\mathcal  F$ and we can identify a $N$-particle state $|\psi\rangle \in \mathcal H_N$ with $ \mathcal F \ni \psi = (0,\ldots,|\psi\rangle,0,\ldots)$. Note that it still holds that e.g. $c^\dagger$ maps a state with $N$ particles to a state with $N+1$ particles. But we only have one $c^\dagger$ (per mode) on $\mathcal F$ and not one operator (per mode) for each $N$.
Consequently, your Hamiltonian is an operator $H:\mathcal  F\longrightarrow \mathcal F$, with the property that if $\psi \in\mathcal F$ is a $N$-particle state, then $H\psi \in F$ is a $N$-particle state too, i.e. $H$ is number conserving: $[H,N]=0$, where $N = \displaystyle \sum\limits_k c^\dagger_k c_k$ is the number operator on $\mathcal F$.
A: One of the problems is that your two Hamiltonians aren't actually equal.
The second one,
$$H = \sum_{\vec{k}} \frac{\hbar^2 \vec{k}^2}{2m} c^†_{k}c_k$$
is the Hamiltonian for a single boson, not for multiple bosons.
The creation and annihilation operators here create and annihilate particular frequency modes for the single particle. They are not creating and annihilating particles. To really make the two Hamiltonians match your second version should be
$$H = \sum_{i}\sum_{\vec{k}} \frac{\hbar^2 \vec{k}^2}{2m} c^†_{(i),k}c_{(i),k}$$
Now each $c_{(i),k}$ is an operator that removes the mode $\vec{k}$ from the state of the $i$th particle.
The other part of your question is how changing the order of the $c$ operators appears to change which Hilbert space they act on. Your mistake is the part where you write
$$c_k : \mathcal{H}_N \rightarrow \mathcal{H}_{N-1}$$
This notation implies that $c_k$ is an operator that only acts on $\mathcal{H}_N$. But that's definitely not true. The $c$ operators act on the entire Hilbert space, which is made up of subspaces with different occupation numbers. Put in math notation:
$$\mathcal{H} = \mathcal{H}_1\oplus\mathcal{H}_2\oplus \dots$$
$$c_{k} : \mathcal{H}\rightarrow \mathcal{H}$$
$$c_k(\mathcal{H}_N) \subseteq  \mathcal{H}_{N-1}\text{ for any $N$}$$.
Put this way, $c^†_kc_k$ and $c_k c^†_k$ clearly act on the same Hilbert space, and there is no problem with transposing them.
A: I believe it is probably incorrect to be thinking about creation/destruction of particles for a seemingly non-relativistic system or even more generally a system without interactions. Second quantization in this context still means a fixed number of particles while the interpretation of the ladder operators is to excite or relax the modes of a given particle.
Having said that, you must understand that the operator $c_i$ always acts on the portion (factor) of the full Hilbert space (Fock Space) ${\cal F} = {\cal H}_1 \otimes \cdots\otimes {\cal H}_i \otimes \cdots\otimes {\cal H}_N$ that corresponds to the $i$-th particle, namely only on ${\cal H}_i$. So within ${\cal H}_i$ you have all the excitations. And the operator $c_i$ is always going from ${\cal H}_i$ to itself. As long as there are no interaction terms between the particles the ladder operators by themselves do not connect different one-particle Hilbert spaces.
PS: For relativistic systems particles can actually decay into other particles or radiation or vice versa be created in certain processes. This only happens when the energy is high enough that relativistic speeds are involved.
