I'm taking a course on quantum mechanics and I'm getting to the part where some of the mathematical foundations are being formulated more rigorously. However when it comes to Hilbert spaces, I'm somewhat confused.

The way I understand it is as follows: The Hilbert space describing for example a particle moving in the $x$-direction contains all the possible state vectors, which are abstract mathematical objects describing the state of the particle. We can take the components of these abstract state vectors along a certain base to make things more tangible. For example:

  • The basis consisting of state vectors having a precisely defined position. In which case the components of the state vector will be the regular configuration space wave function
  • The basis consisting of state vectors having a precisely defined momentum. In which case the components of the state vector will be the regular momentum space wave function
  • The basis consisting of state vectors which are eigenvectors of the Hamiltonian of the linear harmonic oscillator

It's also possible to represent these basis vectors themselves in a certain basis. When they are represented in the basis consisting of state vectors having a precisely defined position, the components of the above becoming respectively Dirac delta functions, complex exponentials and Hermite polynomials.

So the key (and new) idea here for me is that state vectors are abstract objects living in the Hilbert space and everything from quantum mechanics I've learned before (configuration and momentum space wave functions in particular) are specific representations of these state vectors.

The part that confuses me, is the fact that my lecture notes keep talking about the Hilbert space of 'square integrable wave functions'. But that means we are talking about a Hilbert space of components of a state vector instead of a Hilbert space of state vectors?

If there is anybody who read this far and can tell me if my understanding of Hilbert spaces which I described is correct and how a 'Hilbert space of square integrable wave functions' fits into it all, I would be very grateful.


4 Answers 4


This is a good question, and the answer is rather subtle, and I think a physicist and a mathematician would answer it differently.

Mathematically, a Hilbert space is just any complete inner product space (where the word "complete" takes a little bit of work to define rigorously, so I won't bother).

But when a physicist talks about "the Hilbert space of a quantum system", they mean a unique space of abstract ket vectors $\{|\psi\rangle\}$, with no preferred basis. Exactly as you say, you can choose a basis (e.g. the position basis) which uniquely maps every abstract state vector $|\psi\rangle$ to a function $\psi(x)$, colloquially called "the wave function". (Well, the mapping actually isn't unique, but that's a minor subtlety that's irrelevant to your main question.)

The confusing part is that this set of functions $\{\psi(x)\}$ also forms a Hilbert space, in the mathematical sense. (Mumble mumble mumble.) This mathematical Hilbert space is isomorphic to the "physics Hilbert space" $\{|\psi\rangle\}$, but is conceptually distinct. Indeed, there are an infinite number of different mathematical "functional representation" Hilbert spaces - one for each choice of basis - that are each isomorphic to the unique "physics Hilbert space", which is not a space of functions.

When physicists talk about "the Hilbert space of square-integrable wave functions", they mean the Hilbert space of abstract state vectors whose corresponding position-basis wave functions are square integrable. That is:

$$\mathcal{H} = \left \{ |\psi\rangle\ \middle|\ \int dx\ |\langle x | \psi \rangle|^2 < \infty \right \}.$$

This definition may seem to single out the position basis as special, but actually it doesn't: by Plancherel's theorem, you get the exact same Hilbert space if you consider the square-integrable momentum wave functions instead.

So while "the Hilbert space of square integrable wave functions" is a mathematical Hilbert space, you are correct that technically it is not the "physics Hilbert space" of quantum mechanics, as physicists usually conceive of it.

I think that in mathematical physics, in order to make things rigorous it's most convenient to consider functional Hilbert spaces instead of abstract ones. So mathematical physicists consider the position-basis functional Hilbert space as the fundamental object, and define everything else in terms of that. But that's not how most physicists do it.

  • $\begingroup$ Nice answer: +1. Isn't there a further subtly though when it comes to the fact that a state in H.S is really a ray? Or can we essentially just map the phase of the abstract ket to that of the function to make it bijective (at least w.r.t the phase)? (I'm still getting my mind around projective Hilbert spaces) $\endgroup$ Dec 28, 2018 at 2:01
  • $\begingroup$ @InertialObserver I'm not sure I quite understand your question. It's true that a (pure) state is a ray in Hilbert space; there is no "Hilbert space of states", as there's no way to add states, only state vectors. (There is a projective Hilbert space of states, but confusingly, a projective Hilbert space is not a Hilbert space.) But note that my answer never mentioned anything about states, only state vectors. The distinction between states and state vectors is completely independent of the relation between abstract state vectors and wave functions. $\endgroup$
    – tparker
    Dec 28, 2018 at 3:20
  • 1
    $\begingroup$ All separable, infinite dimensional, Hilbert spaces are isomorphic. However, there are inequivalent Hilbert representations of the canonical commutation and anticommutation relations. It is the choice of such representation, for a quantum theory with given physical parameters such as mass and spin, that determines the physical hilbert space of the theory. This is why the hilbert space for one non-relativistic particle is different from the hilbert space for $N$ non-relativistic spin one-half fermions that is again different from the space of a scalar free relativistic massive field. $\endgroup$
    – yuggib
    Dec 28, 2018 at 8:02
  • $\begingroup$ Great answer. Just to add: the abstract ket vectors representing a state are actually time-independent (no Schrödinger equation for them), its the basis 'x' moving in time (via Heisenberg equation), leading to the Schrödinger equation for $$\psi (x)$$. This leads to a lot of confusion if one wants to gloss over the math and physics approach differences. $\endgroup$
    – lalala
    Dec 28, 2018 at 8:35
  • $\begingroup$ @lalala Isn't that only true in the Heisenberg picture, not in the Schrodinger picture? $\endgroup$
    – tparker
    Dec 28, 2018 at 19:01

Your description of an abstract Hilbert space is correct, and that is how we usually think of quantum states. The problem is that to actually define a Hilbert space you need to give a concrete representation: you need to say what its vectors actually are, mathematically. And in our case, one way to define the space is as the set of square integrable functions1. This defines one basis, the basis of position eigenstates (though you could also define the basis to consist of momentum eigenstates), and hence all other basis.

This is not only possible definition: taking the basis of eigenstates of the harmonic oscillator, we see that this space is also the space of infinite sequences $a_n$ of complex numbers with $\sum |a_n|^2 < \infty$, so we could also have defined it like that: it's just that the $L^2$ definition is more obvious. In fact, all infinite dimensional separable Hilbert spaces are isomorphic, and since the Hilbert spaces we use in QM are pretty much always separable, we don't actually need the $L^2$ definition, since there's only one possible infinite dimensional Hilbert space. However, since physics books usually don't want to bother with all this, they just define the space as $L^2$ because it's simpler.

1 Technically the space $L^2$ is the set of square integrable functions modulo functions with zero norm, so not exactly a space of functions.

  • $\begingroup$ -1: The space of square integrable functions form a Hilbert space. It doesn't 'define a basis.' You can also work out a lot of the mathematics of Hilbert spaces without giving any kind of concrete presentation - mathematicians do this all the time. It's required for the physics of the problem to hand. $\endgroup$ Dec 27, 2018 at 23:24
  • $\begingroup$ "Define a basis" is not the greatest wording, but $L^2$ does come with a preferred "basis": that of Delta functions (which of course is not really a basis). This is analog to $\mathbb{R}^n$ having a special basis even if the abstract concept of Euclidean space does not. And I know that you don't need a presentation to do things, but AFAIK you do need one to show that this space actually exists. $\endgroup$
    – Javier
    Dec 27, 2018 at 23:43

On one side we have the (very specific) Hilbert space of square integrable functions is a vector space where the vector is the function $f$ (i.e., and application from $\mathbb R\ni [a,b]\rightarrow \mathbb C$) with the property

$$\int_a^b|f|^2dx <\infty$$

On the other side we have Quantum Mechanics, which postulates that the states of a physical system may be represented by a vector in some Hilbert space. This state, being a vector, doesn't need components to be defined to retain its identity, but if we want, we can expand it on some suitable basis.

Who provides basis?

A basis could be built from scratch, or we could use Hermitian operators, which are renown to possess a complete and orthonormal/orthonormalizable basis $\phi_a$ function of the parameter $a$ which represents the specific eigenvalue associated to the eigenvector $\phi_a$ and such that $(\phi_a,\phi_b)=\delta_{ab}$ ($(\,,\,)$ being the inner product in our Hilbert space). Our generic state $\psi\in\mathcal H$ will be expanded as


If the operator has a continuous spectrum, let's call it $\hat R$, we now have a basis of vectors $\phi_r$ labeled by a continuous eigenvalue $r$ and normalized as $(\phi_r,\phi_{r'})=\delta(r-r')$, we could write

$$\psi=\int_a^b dr(\phi_r,\psi)\phi_r$$

The complex number $(\phi_r,\psi)$ is a complex function of the variable $r$, and if we add the requirement that all physical states must have norm squared equal to some finite number (for example 1),this brings the following consequence:

$$(\psi,\psi)=\left(\int_a^b dx(\phi_r,\psi)\phi_r ,\int_a^b dr'(\phi_{r'},\psi)\phi_{r'}\right)=\int_a^b dr(\phi_r,\psi)^*\int_a^b dr'(\phi_{r'},\psi)\underbrace{\left(\phi_r^* ,\phi_{r'}\right)}_{\delta(r-r')}=$$

$$=\int_a^b dr(\phi_r,\psi)^*(\phi_{r},\psi)=\int_a^b dr|(\phi_r,\psi)|^2=1$$

Then, $(\phi_r,\psi)$ must be a square integrable function. What is $(\phi_r,\psi)$? It is basically the component of $\psi$ in the "direction" described by $r$ in the Hilbert space.

If we change operator (still with continuous spectrum) $\hat Q$, we change basis $\xi_q$, and we have a new expansion for the state $\psi$, but the component of $\psi$ in the new basis, $(\xi_q,\psi)$ will still be a square integrable function of the new variable $q$.

Concluding, the "mathematical" Hilbert space of square integrable functions provides a prototype for Quantum Mechanics, to describe physical states:

  • The physical state $\psi$ is the analogous of the generic function $f$,

  • The components $(\phi_r, \psi)$ or $(\xi_q,\psi)$ are the analogous of the function $f$ evaluated in some point $r\in Range[\hat R]$ or $q\in Range[\hat Q]$: it's worthwhile to notice that the 2 Hilbert spaces resulting, $L^2(Range[\hat R])$ and $L^2(Range[\hat Q])$, are in principle different (regardless of the possibility of them being eventually isomorphic).

So, in programming jargon I would say that the mathematical Hilbert space is the class, while the various physical possibilities are the instances of that class.


Paul Dirac managed to help invent QM without knowing and probably not caring what a Hilbert space is, and Pauli told von Neumann who invented the notion of Hilbert spaces:

If mathematics was physics, then you'd make a great physicist.

The implication being that physics is not the same thing as mathematics. It's worth noting that Hilbert spaces are not enough in QM. For example, Diracs delta function is not even a function and for this we need the theory of distributions. For this you need the theory of rigged Hilbert spaces.

It's possible to represent basis vectors themselves in a certain basis.

This is just as true for ordinary, garden-variety vector spaces. If you're having trouble with Hilbert spaces it might be worth revisiting how they work as much of Hilbert space theory is merely generalising results from the finite-dimensional context to the infinite-dimensional context.


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