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?

  • $\begingroup$ I'm not 100% sure how to answer your question, but quite often when the notion of countable vs. continuous comes up in quantum mechanics the answer lies in the fact that we are using a rigged Hilbert space. $\endgroup$
    – Charlie
    Mar 1, 2021 at 11:49
  • 1
    $\begingroup$ @Charlie you don’t need rigged Hilbert spaces to formulate quantum mechanics — it can be done with just the normal Hilbert space if you quantize only rapidly decaying functions on the phase space. The problem is that physically interesting functions such as $x$ and $p$ are not rapidly decaying. To define their quantizations rigorously, the Hilbert space is usually extended to the rigged Hilbert space, but it is mostly a mathematical trick — the physics still takes place on the Hilbert space. The answer to OPs question is different and IMO beautiful, PTAL at my answer which outlines this $\endgroup$ Mar 1, 2021 at 13:13
  • $\begingroup$ @Prof.Legolasov I see, that's interesting! $\endgroup$
    – Charlie
    Mar 1, 2021 at 13:37
  • $\begingroup$ related, as it also inquires about the notion of uncountably many basis vectors: physics.stackexchange.com/q/359965/50583 $\endgroup$
    – ACuriousMind
    Mar 1, 2021 at 18:04

2 Answers 2


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.


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 $$ {\psi}=\prod_{i}^N(a_i^{\dagger})^{\lambda_i}|0>, $$ is countable (Hilbert space), whereas the same discrete basis (with $N \rightarrow \infty $) of $$ {\psi}=\prod_{i}^\infty(a_i^{\dagger})^{\lambda_i}|0>, $$ 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.