Is the Fock space, defined in the context of the quantum field theory, really a quantum mechanical Hilbert space?

Question

It can be proven that if one considers a separable Hilbert space to define the theory on, then the orthonormal basis for this space is countable.

If one considers a discrete set of harmonic oscillators with a Hamiltonian: $$ H=\sum_{i=1}^{n}\hbar w\left(a^\dagger_ia_i+\frac{1}{2}\right). $$ Then one can define for such theory a bosonic Fock space such that: $$ \mathcal{F}=\oplus_{n}H_n, $$ where $H_n$ is the Hilbert space of $n$ excitations. In this context, for this space one can consider the countable basis with elements:

$$ {\psi}=\prod_{i,\lambda_i}(a_i^{\dagger})^{\lambda_i}|0\rangle, $$ where $i$, indicates the $i$th oscillator and $\lambda_i$ is the occupation number of that oscillator.

So far everything looks fine with me because the states above are discrete and countable and I see no contradiction with the claim that $\mathcal{F}$ is a Hilbert space, but when one generalizes the notions above to the quantum field theory then one has an uncountable set of oscillators ($\hbar=1$): $$ H=\int \:{d^3p} ~ w_p \left(a^\dagger_pa_p+\frac{1}{2}\right) $$

Then the basis for the said Fock space is continuous accordingly. Now how is this generalized Fock space a separable Hilbert space?

Answer 1

You bring up a very interesting point.

In constructive QFT, which is a mathematically sound formulation of QFT on the Hilbert space (which is separable), the Fock space has a different definition.

Start by choosing a countable basis on a space of first quantized states (one particle sub space). That is a Hilbert space in QM, so one such basis must exist. Elements of this basis are called modes.

Then define the Fock space in the following way. First, consider a space of all (sums of) polynomials in creation operators for modes acting on the vacuum. There must be a finite number of creation operators in each polynomial.

Finally, take the closure of the space defined above by adding to it the limits of sequences.

You will end up with a Hilbert space (which is separable) which is the Fock space.

The naive definition from your question is demonstrably different: for example, a state that has occupation number 1 for each of the modes belongs to your space but not to the Fock space (as it would require a product of infinite number of creation operators, and no sequence in the Fock space converges to it).

The existence of these exotic states can be traced back to the failure of the Stone von Neumann theorem for QFT (and in general, for systems with an infinite number of degrees of freedom). These belong to different, unitarily inequivalent representations of the CCR.

Interestingly this isn’t just a mathematical detail. Physical operators such as eg the Hamiltonian have well defined action only on the Fock space defined in my answer, try computing them on one of the exotic states that don’t belong to it and you’ll get infinity.

Finally, this gets super interesting for interacting QFT. I can’t possibly tell the whole story here so I encourage you to read up on constructive QFT, Wightman axioms, Haags theorem etc.

Essentially, for interactive QFT it is possible to prove that the Fock space can’t realize the QFT operators. One needs a different representation of the CCR.

Answer 2

You are conflating "discrete" with "countable".

QM Hilbert space is countable, but QFT Fock space is NOT countable.

More specifically, the discrete basis (with N being finite) of $$ {\psi}=\prod_{i}^N(a_i^{\dagger})^{n_i}|0>, $$ is countable (Hilbert space), whereas the same discrete basis (with $N \rightarrow \infty $) of $$ {\psi}=\prod_{i}^\infty(a_i^{\dagger})^{n_i}|0>, $$ is not countable (Fock space).

Why is that? Let's look at an infinite length chain of quantum spins (corresponding to QFT Fock space), characterized by up and down states at discrete locations $ i =1, 2, 3$, etc. The state at the location $i$ can be described as $n_i = 0$ or $n_i= 1$, corresponding to spin down or up.

Are the states of this quantum chain countable? let's construct a binary number (rather than a decimal number) $$ x = 0.n_1n_2n_3... = \frac{1}{2}n_1 + \frac{1}{4}n_2 + \frac{1}{8}n_3 + ... $$ The number $x$ is actually a real number ranging from 0 to 1! Unlike the natural number, real number is NOT countable according to Cantor's categorization of infinite sets. Hence QFT Fock space is uncountable.

On the other hand, if the length of the chain is finite (corresponding to QM Hilbert space), say N quantum spins, we can always move (N positions to right) the decimal point of above $x$ to make it into an integer number. Therefore, QM Hilbert space is countable in Cantor's sense.

Answer 3

One of the most common misunderstandings in quantum mechanics is the notion that the position (or momentum) Hilbert spaces in QM have uncountable dimension. This is very false. The source of this confusion is the common abuse of language in which the """position eigenvectors""" are treated as a linear algebraic basis for the Hilbert space. But they are emphatically no such thing. Let us focus on quantum mechanics for simplicity.

The Hilbert space of any one dimensional quantum mechanical system with no spin is $L^2(\mathbb{R})$, the Hilbert space of normalizable wavefunctions on the real line. The distributions $\delta(x-x_0)$ do not belong to this space. Furthermore. linear combinations of these distributions do not belong to this space either, so they certainly cannot be a basis. (Integrals are often analogized as being continuous linear combinations, but this is only an analogy). To see that the dimension of $L^2(\mathbb{R})$ is countable, consider the quantum harmonic oscillator Hamiltonian $H=-\partial_x^2 + x^2$. It is a standard result that $H$ has a purely discrete spectrum with no degeneracy, so there is an orthonormal basis $|n\rangle$ given by its eigenstates. This is an honest basis (these are essentially perturbations of Gaussians after all!), and they are spanning. So the QHO eigenstates form a basis for $L^2(\mathbb{R})$, allowing us to conclude that we have a countable Hilbert space.

Now to anticipate another common confusion that I have encountered related to this issue, the Hilbert space is not an object that depends on the Hamiltonian. The $L^2(\mathbb{R})$ in the QHO is the same $L^2(\mathbb{R})$ for the free particle. In fact we can represent everything about the free particle, including the free particle Hamiltonian, in terms of this new basis $|n\rangle$.

Finally, to address how this extends to Fock space, the 1-particle Hilbert space $H$ is always going to be of countable dimension. They will essentially always look like $L^2(X)$. This time $X$ is something like a mass shell, but the fact remains true that the dimension is countable (this is a pretty general fact for reasonable $L^2$ spaces).

The Fock space is $$\mathcal{F}(H)=\bigoplus_{n=0}^\infty H^{\otimes n}$$

An orthonormal basis for this can be constructed by starting with an orthonormal basis $\{e_i\}$ for $H$.

There is an associated basis on every $H^{\otimes n}$ which I will express as $e_{i_1\dots i_n} = e_{i_1}\otimes \dots \otimes e_{i_n}$.

This finally gives us the basis $\{e_{i_1, \dots, i_n}\}$ indexed by sequences of indices of arbitrary but finite length. You can write this as a union of the bases for the $n$-particle states. A countable union of countable sets is countable, so you can conclude.