Appropriate space in second quantization

Question

The appropriate space for the study of a system of identical bosons, for instance, is something like

\begin{equation} \tag{1} \mathbb{C}\oplus\mathcal{H}\oplus(\mathcal{H}\otimes\mathcal{H})_S\oplus(\mathcal{H}\otimes\mathcal{H}\otimes\mathcal{H})_S\oplus \ldots \end{equation}

where $\mathcal{H}$ is the one-particle Hilbert Space and the subscript $S$ denotes the symmetric sector of the products.

My question is: how is this different from

\begin{equation} \tag{2} (\mathcal{H}\otimes\mathcal{H}\otimes\mathcal{H}\otimes\mathcal{H}\ldots)_S~? \end{equation}

That is, intuitively my impression is that the first space 'converges' to this one.

For example: suppose a state of 3 bosons. Regarding (1), this state is specifically in the subspace

$$ (\mathcal{H}\otimes\mathcal{H}\otimes\mathcal{H})_S $$

but regarding (2), it is also in the sector that is 'non-vacuum' in 3 of the coordinates of the tensor product.

Why is (1) and not (2) the Fock Space?

Answer 1

I'm not sure whether I understand your misunderstanding correctly.

The space $$ \mathbb{C}\oplus\mathcal{H} \oplus(\mathcal{H}\otimes\mathcal{H})_S \oplus(\mathcal{H}\otimes\mathcal{H}\otimes\mathcal{H})_S\oplus \ldots \tag{1}$$ contains states with all possible numbers of particles.

Saying it more explicitly:
$\mathbb{C}$ contributes the zero-particle state (the vacuum state), $\mathcal{H}$ contributes the one-particle states, $(\mathcal{H}\otimes\mathcal{H})_S$ contributes the 2-particle states, and so on.

Your space $ (\mathcal{H}\otimes\mathcal{H}\otimes\mathcal{H}\otimes\mathcal{H}\ldots)_S $ contains only states with infinitely many particles (what ever this means). It does not contain, for example, the 3-particle states, or states with any other finite number of particles.

Your space $ (\mathcal{H}\otimes\mathcal{H}\otimes\mathcal{H})_S $ contains only states with exactly 3 particles.

Only space (1) is the correct Fock space, because only in this space you are able to describe processes with changing particle number. For example, think of creation and annihilation of photons.

Answer 2

You're mixing up two different ways of constructing the Fock space.

The first way is to focus on particles. You can have any integer number $n$ of particles, represented by the Hilbert space $\mathcal{H}^{\otimes n}$. But this should be symmetrized since the particles are identical. Summing over the possibilities gives $$\mathcal{H}_{\text{Fock}} = \mathbb{C} + \mathcal{H} + (\mathcal{H}^{\otimes 2})_S + (\mathcal{H}^{\otimes 3})_S + \ldots.$$ The second way is to focus on fields. Starting from a quantized free field, each field mode $i$ is associated with a Hilbert space $\mathcal{H}_i$, describing the excitations in that mode. The Fock space is $$\mathcal{H}_{\text{Fock}} = \bigotimes_i \mathcal{H}_i.$$ The main point of the first couple weeks of a quantum field theory class is that these two procedures physically give the same result, though of course there are always more mathematical subtleties. (Of course, if one works in a finite box with a UV cutoff, the number of factors is finite, and there are no mathematical subtleties at all.)

You are mixing up these two. In particular, the subspaces $\mathcal{H}_i$ are not the same thing as $\mathcal{H}$. Physically, they represent modes and particles, respectively. Mathematically, $\mathcal{H}_i$ is formally the Hilbert space for a 1D harmonic oscillator in field space, which is $L^2(\mathbb{R})$, while $\mathcal{H}$ is $L^2(\mathbb{R}^3)$. You need the sum in the first expression because you have to sum over the different possible numbers of particles. You don't have a sum in the second expression because there's always the same number of modes.

Answer 3

The space in (2) simply does not exist as it is described there. It can be defined but it needs arbitrary choices. Therefore the correct notion of Hilbert space when dealing with infinite identical systems (here bosons) is (1).

See Tensor product of Hilbert spaces at the item "Infinite tensor products". You see that to start with a definition of tensor product of an infinite set of Hilbert spaces $H_n$ you have to arbitrarily pick out the preferred unit vectors $\xi_n$. There is no physical way to choose them. This idea can be traced back to von Neumann.

Some attempts to use this weird structure in physics exist (I remember a paper by Streater and a discussion in Haag's book). However, equipped with the usual mathematical structure arising from elementary QFT, where no such choices are made, the unique tensor product structure is that indicated by (1) in the OP. There, differently form what happens if dealing with a precise notion as (2), the physically crucial notions of vacuum and subspace of states with a finite number $N$ (arbitrarily large) of particles can be defined.