What is more fundamental: Fock space or Hilbert space? And why?

Question

Consider the following state for some bosons represented in Fock space:

$$|2\rangle_{k_1}|1\rangle_{k_2}$$

where $k_i$ is some distinguishing index. You may think of these as the two different wavevectors for photons.

Now, if we use the Hilbert space representation of each individual boson, the same normalized state is written as

$$\frac{1}{\sqrt{3}} \left( |k_1,k_1,k_2\rangle+|k_1,k_2,k_1\rangle+|k_2,k_1,k_1\rangle \right)$$

Is this correct? Are the two expressions above equivalent? Now, consider that the particle in mode $k_2$ is transferred to the mode $k_1$ (photon addition from one mode into another, for example). This would give the state

$$\frac{3}{\sqrt{3}} |k_1,k_1,k_1\rangle$$

which is clearly not normalized. How is this possible? Does it mean Fock space is more fundamental? However, the Hilbert space picture appears more intuitive at first glance.

In any case, how do we make sense of the prefactor of $\sqrt{3}$ intuitively?

I am trying to somehow relate it to the prefactors that appear due to the action of creation operator in mode $k_1$:

$\hat{a}^{\dagger}|2\rangle=\sqrt{3}|3\rangle$

Answer 1

Fock space is the Hilbert space of multiparticle systems.

The difference you're alluding to is a difference of representation. The one you're naturally associating with Fock space is the occupation number representation, which is obviously more natural to describe multiparticle systems due to the indistinguishability of particles.

However, notice that the basis states of the occupation number representation are the same basis states that you get when you construct the full Hilbert space as a product of single-particle Hilbert spaces and write down the full basis as a product of bases of single-particle Hilbert spaces (see, for example, $|1\rangle_k|1\rangle_K$ is the same as $\frac{1}{\sqrt{2}}(|k\rangle |K\rangle + |K\rangle |k\rangle)$.

Hence, the occupation number representation just writes all of the same basis states in a new notation, a much helpful one, of course). So, not only is Fock space the same Hilbert space, the occupation number basis is not even some new basis used to span the same space.

The $k_1$ going to $k_2$ trick is misleading for the reason that we're talking about $\underline{\text{basis}}$ and you're mixing them with $\textit{states}$. One can perform some unitary transformations on the basis to get to a new basis but keeping everything else the same and just letting $k_1$ go to $k_2$ is not such a transformation.

If you want to talk about a physical state labeled by arbitrary $k_1$ and $k_2$, then you can do such a trick, and say that when this happens (in the expansion of the state into basis vectors), the coefficient in front of the basis vector with different $k_1$ and $k_2$ become zero and those coefficient which have all three particles having the same momentum i.e. ($k_1$ = $k_2$) goes to one.