In a lot of literature the notion of different Hilbert spaces has been mentioned. In QFT, for non-interacting theories the Hilbert space is a called a 'Fock space', this is different from the interacting theory 'Physical Hilbert Space'.

When the notion of Hilbert spaces was introduced to me it was described as the space of all continuous functions defined over a domain. Since states for both the non-interacting and interacting theories are described by continuous wavefunctions it seems like really the 'Fock space' and 'Physical Hilbert Space' are subspaces of the much larger space, space of all continuous functions.

Such a space would be way too large. Instead what I am thinking is rather the Hilbert space of our theory is described by the spectrum of the Hamiltonian. This subspace spun by the Hamiltonian eigenbasis would give us all the states relevant to our particular theory. This is because any state in this eigensubspace can be written as a linear combination of energy eigenstates, it obviouly has vector space structure (ie any linear combination of states in the spectrum is also a state in the spectrum) and since states evolve according to the Hamiltonian such a state written in this spectrum eigensubspace stays in the eigensubspace as time evolves.

This also explains why we always can say the Hamiltonian resolution of the identity is complete over the 'Hilbert space', since the Hilbert space is defined as the spectrum of the Hamiltonian (often I would get the answer that to know if the resolution of the identity is complete over the Hilbert space was guess work).

My question is that is this the correct way of understanding 'different Hilbert spaces' in Quantum Mechanics? Vague use of the term Hilbert spaces without explaining properly has left me confused about it for quite some time and I'm hoping to get to the bottom of my confusion.

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    $\begingroup$ You may find useful commentary about this from a lecture given by P. Dirac: youtu.be/zoKWlOmMDIY?t=431 (starts at 7:11). I hope this helps. $\endgroup$
    – ad2004
    Commented Sep 26, 2020 at 21:57

3 Answers 3


I'll start with a generic perspective, and then I'll apply it to the question about Hilbert space. Here's the generic perspective:

  • Sometimes we use one mathematical thing to represent another mathematical thing. Here, a representation is a mapping from $A$ to $B$ that preserves the essential structure of $A$, where $A$ and $B$ are both mathematical things.

  • In physics, we also use another type of representation: a mapping from physical things to mathematical things. To define a mathematical model of a physical system, we need to provide this second type of representation.

Hilbert spaces and their (math-to-math) representations

A Hilbert space is a vector space over the complex numbers $\mathbb{C}$, equipped with a positive-definite inner product and satisfying a completeness condition. That's all. In fact:

  • For any given finite $N$, all $N$-dimensional Hilbert spaces over $\mathbb{C}$ are isomorphic to each other: they're all the same as far as their abstract Hilbert-space-ness is concerned.

  • When the number of dimensions is not finite, quantum theory requires the Hilbert space to be separable, meaning that it has a countable orthonormal bases. Once again, all infinite-dimensional separable Hilbert spaces over $\mathbb{C}$ are isomorphic to each other: they're all the same as far as their abstract Hilbert-space-ness is concerned.

We can represent (math-to-math) a Hilbert space using matrices, or using Fock spaces, or using single-variable functions, or using seventeen-variable functions, or whatever. Those representations introduce extra structure that is superfluous as far as the Hilbert-space-ness is concerned, but such representations can still be useful. In particular, different representations can simplify the task of describing different linear operators on the Hilbert space. In quantum theory, the linear operators on a Hilbert space are more important than the Hilbert space itself.

The physics-to-math representation

The thing that makes a given quantum model interesting is how it represents measurable things in terms of linear operators on a Hilbert space. The word observable is used for both sides of this physics-to-math mapping.

Consider these two models:

  • The usual quantum mechanics of a single non-relativistic spinless particle, with a Hamiltonian of the form $H=P^2/2m + V(X)$, where $P$ is the momentum observable and $X$ is the position observable.

  • Quantum chromodynamics (QCD). By the way, QCD can be rigorously well-defined by treating space as a discrete lattice, so it's mathematically legit.

Both of these models use the same abstract Hilbert space, namely the one-and-only infinite-dimensional separable Hilbert space over the complex numbers. The two models are different, though, because they describe different (simplified) worlds that have different types of measurable things. QCD does not have position observables, and single-particle quantum mechanics does not have Wilson-loop observables. Even if we only consider the association between observables and spacetime, QCD and single-particle QM are still very different: the pattern of observables in QCD is Lorentz-symmetric (to a good approximation at resolutions much coarser than the lattice spacing) and the pattern of observables in single-particle QM is not.

Sometimes physicists use the term "Hilbert space" to mean a particular representation (math-to-math) of the Hilbert space along with a particular set of observables (physics-to-math) suggested by that representation. I personally prefer to reserve the term "Hilbert space" for the abstract mathematical thing (neither type of representation), because I think that's more clear. Preferences aside, the important message is that models are not distinguished from each other by their abstract Hilbert spaces or by how those Hilbert spaces are represented in the math-to-math sense. Instead, different models are distinguished from each other by their observables — by the physics-to-math mapping from measurable things to linear operators on the Hilbert space.


The single-particle Hilbert space is the space of functions $\psi: \, \mathbb{R} \to \mathbb{C}$ [with value $\psi(\vec{r})$] with finite $L^2$ norm (normalizable functions). Furthermore boundary conditions may limit this set. In any case it is indeed not the space of all continuous functions. Fock space is the many-particle generalization, the space of normalizable $N$-body wavefunctions, $\psi(\vec{r}_1,\dots \vec{r}_N)$ which is now a function from $\mathbb{R}^N \to \mathbb{C}$. Furthermore, for identical bosons or fermions we require that this function is symmetric or antisymmetric in the arguments. I'm not sure if this answers your questions but hopefully it clarifies some things.


As was mentioned in the other answer, a Hilbert space is not necessarily a space of continuous functions. This is just one example of a Hilbert space. The term Hilbert space applies to any set of mathematical objects that satisfy the axioms of a vector space over either $\mathbb{R}$ or $\mathbb{C}$, that have an inner product defined on the space, and that form a complete metric space with the inner product serving as the distance function (this last requirement is often not explicitly mentioned in physics textbooks).

This means that "normal" vector spaces, like $\mathbb{R}^n$, are also Hilbert spaces. As another example, we can form a Hilbert space from continuous, normalizable functions by defining the inner product to be $\psi\cdot\phi\equiv\int dx \psi^*(x)\phi(x)$. We can also form a Hilbert space using continuous, normalizable functions of two variables by defining the inner product to be $\psi\cdot\phi\equiv\int\int dxdy \psi^*(x,y)\phi(x,y)$. The possibilities are endless. We could even take two different Hilbert spaces and form another Hilbert space by taking their outer product.

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    $\begingroup$ "a Hilbert space is not necessarily a space of continuous functions." In no case the space of continuous function is a Hilbert space: the limit of a sequence of continuous function is not a continuous function, in general. $\endgroup$ Commented Sep 27, 2020 at 9:13
  • $\begingroup$ @GiorgioP actually in the finite-dimensional case the space of all continuous functions from a (discrete) n-element set to your ground field coincides with the space of all functions to the ground field, and forms a Hilbert space. $\endgroup$ Commented Sep 27, 2020 at 14:09
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    $\begingroup$ @MarttiKarvonen Is it possible to get any continous function by using a basis made by a finite number of continous functions? I have some difficulty in understanding how. Any reference? $\endgroup$ Commented Sep 27, 2020 at 14:21
  • $\begingroup$ My point is simpler: you can identify the Hilbert space $\mathbb{C}^n$ with the space of functions $\{1,\dots,n\}\to\mathbb{C}$ which coincides with the space of continuous such functions. For more interesting spaces the space of complex-valued continuous functions is unlikely to be finite-dimensional. $\endgroup$ Commented Sep 27, 2020 at 14:27
  • $\begingroup$ When $\{1,\dots, n\}$ is given the discrete topology (i.e. the topology it inherits as a subset of the real line) all functions out of it are continuous. $\endgroup$ Commented Sep 27, 2020 at 14:38

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