What are the adequate Hilbert spaces for Schrödinger, Schrödinger–Pauli, Dirac equations, and QFT?

Question

In quantum mechanics (both non-relativistic and relativistic), it is possible to study physical systems by looking for solutions of PDEs, whose solutions belong to suitable Hilbert spaces:

• Schrödinger equation for spinless particles* provides solutions which are $C^2$-wavefunctions, which form a dense subset of the Hilbert space $\mathcal{H} = L^2(\mathbb{R}^3)$ of functions square-integrable over $\mathbb{R}^3$.

• The Schrödinger-Pauli equation is a generalization for 1/2-spin particles, whose natural Hilbert space seems to be $\mathcal{H} = L^2(\mathbb{R}^3)\otimes \mathbb{C}^2$.

• Dirac equation, include spin and antiparticles, and its solutions can be considered a dense subset of $\mathcal{H} = L^2(\mathbb{R}^3)\otimes \mathbb{C}^4$.

• For "collision states" which are non-square-integrable wavefunctions, rigged Hilbert spaces can be used, for a mathematically rigorous formulation.

However, in the ordinary QFT approach we do not use evolution equations as above, nor do we make explicit to which type of Hilbert space the physical states belong. My question is:

Why is it not easy or useful to generalize the above approach to QFT?

Answer 1

Ordinary QM has essentially a fixed Hilbert space of $L^2(\mathbb{R}^n)\otimes S$, where $n$ is the number of spatial dimensions and $S$ some representation of the rotation group. This is due to the Stone-von Neumann theorem that tells us that any representation of $n$ position/momentum pairs with $[x_i,p_i] = \mathrm{i}$ is isomorphic to $L^2(\mathbb{R}^n)$, and we rarely have independent observables other than position, momentum and spin, so $L^2(\mathbb{R}^n)\otimes S$ with position as multiplication and momentum as differentiation on $L^2(\mathbb{R}^n)$ is the default setting for a lot of ordinary QM. In particular, we don't have to change this space when we change the Hamiltonian from a free to an interacting one - the SvN theorem guarantees all these Hamiltonians can live on the same space.

In principle, QFT does have a Schrödinger representation analogous to $L^2(\mathbb{R})$ where the states are "wavefunctionals" (see e.g. this question and its answers or this question), but this space is difficult to explicitly write down or to make rigorous.

E.g. in Glimm and Jaffe's "Quantum Physics", this Schrödinger representation is a subspace $L^2(D'(\mathbb{R}^{n-1},\mathrm{d}\nu))\subset L^2(D'(\mathbb{R}^n),\mathrm{d}\mu)$, where $\mathrm{d}\mu$ is the measure in the path integral physicists usually would write more like $\mathrm{e}^{\mathrm{i}S[\phi]}\mathcal{D}\phi$ and the subspace is arrived at by restricting to $t=0$. $D'(\mathbb{R}^n)$ is a suitable space of distributions, e.g. the tempered distributions. You'll note that this space depends on the action/Hamiltonian/Lagrangian because the measure depends on it, in contrast to the nice independent $L^2(\mathbb{R}^n)$ of QM with finitely many d.o.f., so even if you managed to study this space with explicit methods for one particular case, they might be of no use at all for a different system.

Abstractly, this is a manifestation of Haag's theorem, which says that the representations of the field operators on our Hilbert space cannot be isomorphic between free and interacting theories, so something like ordinary QM where both a free and an interacting Hamiltonian live on $L^2(\mathbb{R}^n)$ with $x$ and $p$ acting in the same way in all theories is impossible in QFT.

So, indeed, it is neither easy to write down a Schrödinger-like representation (because the spaces involve measures on infinite-dimensional function spaces that change from system to system) nor very useful because systems don't "share" a Hilbert space as they do in ordinary QM thanks to the SvN theorem.

Answer 2

One way to proceed is to start with the classical observable and find some suitable Lie algebra containing these as generators. Upon quantization, the observables are promoted to operators, now acting on representations of the Lie algebra. In most cases, one can choose the representations to be unitary and irreducible (or indecomposable, which is the next best thing if you don't need unitary).

Thus, as pointed out by others, if you start with $x$ and $p$ you get the “standard” Hilbert space.

On the other hand, you might be interested in (nuclear) quadrupole deformation, containing the quadrupole moments, which are the traceless symmetric combinations of $x_ix_j$ in 3D, and include angular momenta to describe the associated rotations. You would then naturally “recover” the semi-direct sum $\mathbb{R}^5\rtimes so(3)$ and states in your Hilbert space now span unitary irreps of the associated group.

You might then care to add the kinetic energy as an observable, in which case you must expand $\mathbb{R}^5\rtimes so(3)$ to $\mathfrak{sp}(6,\mathbb{R})$ as the minimal model to include the above. In fact there is a series of such nuclear physics models where the construction process is clear. Some key references are:

• Ui, Haruo. "Quantum Mechanical Rigid Rotator with an Arbitrary Deformation. I: Dynamical Group Approach to Quadratically Deformed Body." Progress of Theoretical Physics 44.1 (1970): 153-171.
• Weaver, L., and L. C. Biedenharn. "Nuclear rotational bands and SL (3, R) symmetry." Nuclear Physics A 185.1 (1972): 1-31,
• and finally Rowe, D. J. "Microscopic theory of the nuclear collective model." Reports on Progress in Physics 48.10 (1985): 1419.

In this way, the Hilbert space emerges quite naturally from the observables in your model as spanned by states in irreducible representations of a Lie algebra.

Answer 3

Here is another approach. Instead of defining a state space $L^2(\text{something})$ and then defining observables as the self-adjoint operators on that space, you can go the other way around:

Choose an abstract (unital, $C*$)-algebra $A$ of observables, and then define states as a certain class of linear functionals $A\to\mathbb{C}$. It is important that these states are not vectors in any Hilbert space yet.

After choosing a fixed state $\rho$, you can always find a concrete representation of $A$ (meaning a representation as operators on a Hilbert space), such that $\rho$ is just a vector in that Hilbert space (GNS construction). But this is not unique. Different $\rho$ can give you different representations with different Hilbert spaces. While one representation contains many states (both vector states and other "mixed" states), typically it is impossible to captured all states in a single representation.

In this view, the (algebra of) observables is the fundamental object, not any particular Hilbert space. This approach is common for example in "algebraic quantum field theory".