What's the exact connection between bosonic Fock space and the quantum harmonic oscillator?

Let's suppose I have a Hilbert space $K = L^2(X)$ equipped with a Hamiltonian $H$ such that the Schrödinger equation with respect to $H$ on $K$ describes some boson I'm interested in, and I want to create and annihilate a bunch of these bosons. So I construct the bosonic Fock space

$$S(K) = \bigoplus_{i \ge 0} S^i(K)$$

where $S^i$ denotes the $i^{th}$ symmetric power. (Is this "second quantization"?) Feel free to assume that $H$ has discrete spectrum.

What is the new Hamiltonian on $S(K)$ (assuming that the bosons don't interact)? How do observables on $K$ translate to $S(K)$?

I'm not entirely sure this is a meaningful question to ask, so feel free to tell me that it's not and that I have to postulate some mechanism by which creation and/or annihilation actually happens. In that case, I would love to be enlightened about how to do this.

Now, various sources (Wikipedia, the Feynman lectures) inform me that $S(K)$ is somehow closely related to the Hilbert space of states of a quantum harmonic oscillator. That is, the creation and annihilation operators one defines in that context are somehow the same as the creation and annihilation operators one can define on $S(K)$, and maybe the Hamiltonians even look the same somehow.

Why is this? What's going on here?

Assume that I know a teensy bit of ordinary quantum mechanics but no quantum field theory.

Answer

Let's discuss the harmonic oscillator first. It is actually a very special system (one and only of its kind in whole QM), itself being already second quantized in a sense (this point will be elucidated later).

First, a general talk about HO (skip this paragraph if you already know them inside-out). It's possible to express its Hamiltonian as $H = \hbar \omega(N + 1/2)$ where $N = a^{\dagger} a$ and $a$ is a linear combination of momentum and position operator). By using the commutation relations $[a, a^{\dagger}] = 1$ one obtains basis $\{ \left| n \right >$ | $n \in {\mathbb N} \}$ with $N \left | n \right > = n$. So we obtain a convenient interpretation that this basis is in fact the number of particles in the system, each carrying energy $\hbar \omega$ and that the vacuum $\left | 0 \right >$ has energy $\hbar \omega \over 2$.

Now, the above construction was actually the same as yours for $X = \{0\}$. Fock's construction (also known as second quantization) can be understood as introducing particles, $S^i$ corresponding to $i$ particles (so HO is a second quantization of a particle with one degree of freedom). In any case, we obtain position-dependent operators $a(x), a^{\dagger}(x), N(x)$ and $H(x)$ which are for every $x \in X$ isomorphic to HO operators discussed previously and also obtain base $\left | n(x) \right >$ (though I am actually not sure this is base in the strict sense of the word; these affairs are not discussed much in field theory by physicists). The total hamiltonian $H$ will then be an integral $H = \int H(x) dx$. The generic state in this system looks like a bunch of particles scattered all over and this is in fact particle description of a free bosonic field.