# How can a Hamiltonian determine the Hilbert space?

Sometimes, when discussing quantum field theory, people speak as if a Hamiltonian determines what the Hilbert space is. For example, in this answer AccidentalFourierTransform says

Imagine an $H_0$ that depends on the phase space variables $P$,$X$. [...] If you add the perturbation $\vec{L} \cdot \vec{S}$, with $\vec{S}$ the spin of the particle, then you change the Hilbert space, because the new space has three phase space variables $P,X,S$, and you cannot span the latter with a basis of the former.

This kind of language also pops up when introducing the free scalar field -- lots of lecture notes and textbooks speak of 'building' or 'constructing' the Hilbert space, or 'finding' the 'Hilbert space of the Hamiltonian'.

This kind of reasoning seems exactly backwards to me. How can one possibly define a Hamiltonian, i.e. an operator on a Hilbert space, if we don't know the Hilbert space beforehand? Without a Hilbert space specified, isn't $H = p^2/2m + V(x)$ just a meaningless string of letters with no mathematical definition? I find this shift of perspective so bewildering that I feel like I missed a lecture that everybody else went to.

For example, when dealing with the harmonic oscillator, it is possible to show that the Hilbert space must contain copies of $\{|0 \rangle, |1 \rangle, \ldots \}$ using only the commutation relations. But there's no way to pin down how many copies there are unless we use the fact that the Hilbert space is actually $L^2(\mathbb{R})$ which shows that $a |0 \rangle = 0$ determines a unique state. Similarly I would imagine for quantum fields we should start with a Hilbert space where the individual states are classical field configurations, but I've never seen this done in practice -- there seems to be no input but the Hamiltonian itself. How could that possibly be enough?

• All Hilbert spaces are isomorphic, so you can use whatever Hilbert space you want. What the Hilbert space determines is what Hilbert space is easiest to work with. – Slereah Oct 7 '17 at 10:25
• @Slereah That is completely false. $\mathbb{C}^2 \neq \mathbb{C}^3$, for instance. This is a good question in my opinion. – gj255 Oct 7 '17 at 10:32
• Oh yes, separable infinite dimensional, specifically – Slereah Oct 7 '17 at 10:53
• @Slereah while that is technically true, it is profoundly unphysical. An isomorphism between Hilbert spaces spanned by functions on spaces of different dimension will generally require you to choose a fixed basis on each, and that is without further information only granted by the axiom of choice. To choose an actually usable isomorphism, you pretty much need to start with an eigenbasis, which you only have if the Hamiltonian is already defined in the right space! – leftaroundabout Oct 7 '17 at 17:41

The point is that sometimes one starts from a more or less explicit algebraic formalism where only algebraic manipulations of algebra elements are initially used. Here operators are not operators in a precise Hilbert space, but just elements of a unital $^*$-algebra and only compositions (multiplication with scalars, sum and algebra product) and the involution operation (formal adjoint) are used. Next one sees if this algebra, with further technical conditions (some operator must be self-adjoint, some representation must be irreducible) or physical requirements (a suitable state exists) uniquely determine a Hilbert space where this algebra is faithfully represented in terms of operators with suitable domains.
For instance, the algebra of $a,a^*$ completely determines the standard harmonic oscillator representation in $L^2(\mathbb R, dx)$ if assuming that $a^*a$ is essentially self-adjoint on a dense common invariant domain and the arising representation is irreducible.
In summary, the Hamiltonian itself, viewed as an element of a unital $^*$-algebra is not enough to determine a Hilbert space where the theory can be implemented in the standard way, the whole algebra should to be fixed and more information is usually necessary.