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?

  • 2
    $\begingroup$ Where is the vacuum state in (2)? $\endgroup$
    – Qmechanic
    Commented Jul 9, 2019 at 18:13

3 Answers 3


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.

  • $\begingroup$ Thank you for the answer! My problem is that it seems to me that $\mathcal{H}\otimes\mathcal{H}\otimes\mathcal{H}\ldots$ does contain lesser numbers of particles. Can't, for instance, the subspaces corresponding to all $\mathcal{H}$ 'be' in its vacuum state but one which is in a particle-state? $\endgroup$
    – GaloisFan
    Commented Jul 9, 2019 at 18:59
  • $\begingroup$ No, the space $\mathcal{H}\otimes\mathcal{H}\otimes\mathcal{H}$ does not contain the space $\mathcal{H}\otimes\mathcal{H}$. $\endgroup$ Commented Jul 9, 2019 at 19:02
  • $\begingroup$ @GaloisFan You're getting confused between particles and modes again. It doesn't make sense for a particle Hilbert space $\mathcal{H}$ to "be in the vacuum state". It only makes sense for modes, as covered in my answer. $\endgroup$
    – knzhou
    Commented Jul 9, 2019 at 19:04
  • $\begingroup$ @GaloisFan If you're still confused, just think about one particle, like in a first undergraduate quantum mechanics course. It's described by a wavefunction $\psi(x)$. What wavefunction corresponds to not having any particle at all? $\endgroup$
    – knzhou
    Commented Jul 9, 2019 at 19:05
  • $\begingroup$ @knzhou What was confusing me was the impression that there could be 'coordinates' or quantum numbers which could set the component of a single existing $\mathcal{H}$ to be the vacuum, so that, for instance, I thought that something like $|r_1, 0>$ could be a state in $\mathcal{H}\otimes\mathcal{H}$. $\endgroup$
    – GaloisFan
    Commented Jul 9, 2019 at 22:29

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.

  • $\begingroup$ Thank you for the answer! I understand that each element in the sum of the first space is the space of a different number of particles. What I don't understand is why the tensor product of an infinite number of $\mathcal{H}$ also doesn't contain arbitrary number of particle states. $\endgroup$
    – GaloisFan
    Commented Jul 9, 2019 at 19:02
  • $\begingroup$ @GaloisFan A tensor product is an "AND". If you tensor produce two Hilbert spaces, it means you have something in the first one and something in the second one. So $\mathcal{H}^{\otimes n}$ means you have specifically $n$ particles, not an arbitrary number. $\endgroup$
    – knzhou
    Commented Jul 9, 2019 at 19:03
  • $\begingroup$ It is wrong that "they give the same result", simply the second construction is not mathematically consistent when the number of factors is infinite. There are several proofs of this very well known msthematical result. $\endgroup$ Commented Jul 9, 2019 at 19:44

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.


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