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.
Best Answer
Reference: Fetter and Walecka, Quantum Theory of Many Particle Systems, Ch. 1
The Hamiltonian for a SHO is:
$$ H = \sum_{i = 0}^{\infty}\hbar \omega ( a_i^{+} a_i + \frac{1}{2} ) $$
where $\{a^+_i, a_i\}$ are the creation and annihilation operators for the $i^\textrm{th}$ eigenstate (momentum mode). The Fock space $\mathbf{F}$ consists of states of the form:
$$ \vert n_{a_0},n_{a_1}, ...,n_{a_N} \rangle $$
which are obtained by repeatedly acting on the vacuum $\vert 0 \rangle $ by the ladder operators:
$$ \Psi = \vert n_{i_0},n_{i_1}, ...,n_{i_N} \rangle = (a_0^+)^{i_0} (a_1^+)^{i_1} \ldots (a_N^+)^{i_N} \vert 0 \rangle $$
The interpretation of $\Psi$ is as the state which contains $i_k$ quanta of the $k^\textrm{th}$ eigenstate created by application of $(a^+_k)^{i_k}$ on the vacuum.
The above state is not normalized until multiplied by factor of the form $\prod_{k=0}^N \frac{1}{\sqrt{k+1}}$. If your excitations are bosonic you are done, because the commutator of the ladder operators $[a^+_i,a_j] = \delta_{ij}$ vanishes for $i\ne j$. However if the statistics of your particles are non-bosonic (fermionic or anyonic) then the order, in which you act on the vacuum with the ladder operators, matters.
Of course, to construct a Fock space $\mathbf{F}$ you do not need to specify a Hamiltonian. Only the ladder operators with their commutation/anti-commutation relations are needed. In usual flat-space problems the ladder operators correspond to our usual fourier modes $ a^+_k \Rightarrow \exp ^{i k x} $. For curved spacetimes this can procedure can be generalized by defining our ladder operators to correspond to suitable positive (negative) frequency solutions of a laplacian on that space. For details, see Wald, QFT in Curved Spacetimes. Now, given any Hamiltonian of the form:
$$ H = \sum_{k=1}^{N} T(x_k) + \frac{1}{2} \sum_{k \ne l = 1}^N V(x_k,x_l) $$
with a kinetic term $T$ for a particle at $x_k$ and a pairwise potential term $V(x_k,x_l)$, one can write down the quantum Hamiltonian in terms of matrix elements of these operators:
$$ H = \sum_{ij} a^+_i \langle i \vert T \vert j \rangle a_i + \frac{1}{2}a^+_i a^+_j \langle ij \vert V \vert kl \rangle a_l a_k $$
where $|i\rangle$ is the state with a single excited quantum corresponding the action of $a^+_i$ on the vacuum. (For details, steps, see Fetter & Walecka, Ch. 1).
I hope this helps resolves some of your doubts. Being as you are from math, there are bound to be semantic differences between my language and yours so if you have any questions at all please don't hesitate to ask.