In relativistic QFT, Haag's theorem states that the Hilbert space of an interacting theory is generally not the same as the Hilbert space of the associated free theory.
I thought this wasn't a problem in nonrelativistic QFT or nonrelativistic QM. For example, the Hilbert space for a single spinless particle should just be $L^2(\mathbb{R}^3)$ no matter what the interactions are. However, in the comments in this question, it is claimed that even in the context of single particle nonrelativistic QM, the interacting Hilbert space can be different!
Is this claim correct? If so, can one give a concrete example where this happens?
Answer
Haag's theorem is about degrees of freedom. More precisely, about the fact that a quantum theory with an infinite number of degrees of freedom is either free or ill-defined. In this sense, you would only encounter it in QFT, either relativistic or non-relativistic. You cannot clash with Haag's result if you are studying the Dirac equation for a finite number of particles; and you will clash with it if you are studying the Schrödinger field. Haag's theorem has little to do with whether the system is relativistic or not; it has to do with the number of degrees of freedom in it.
As for a concrete example we will follow Itzykson and Zuber. Let us consider a lattice of $N$ half-integers spins. The phase-space variables are $\vec\sigma(i)$, where $i$ labels the site on the lattice. We label the states through their eigenvalue under $\sigma_3(i)$: $$ |\pm\pm\cdots\pm\rangle $$ which are generated by the action of the $\sigma_-$ upon the vacuum $$|0\rangle=|++\cdots+\rangle $$
We can make the unitary transformation $\vec\tau(i)\equiv U^\dagger(\theta) \vec\sigma(i)U(\theta)$, with $$ U=\exp\left[i\frac{\theta}{2}\sum_{j=1}^N\sigma_2(j)\right] $$ under which $$ \begin{aligned} \tau_1&=\sigma_1\cos\theta-\sigma_3\sin\theta\\ \tau_2&=\sigma_2\\ \tau_3&=\sigma_1\sin\theta+\sigma_3\cos\theta \end{aligned} $$
The operators $\vec\sigma(i)$ and $\vec\tau(i)$ are unitary equivalent. They satisfy the same algebra (that of $\mathfrak{su}(2)$).
Let us now consider the ground state of $\vec\tau$: $$ |\theta\rangle=U^\dagger(\theta)|0\rangle $$
It is straightforward to check that $$ \langle 0|\theta\rangle=\cos^N\frac\theta2 $$
The main point is the following: We now try to conside the same system, in the limit $N\to\infty$. The Hilbert space of states is constructed from the ground state $|0\rangle$ by the action of a finite number of creation operators $\sigma^-$ (and Cauchy completion). We can perform the same rotation as before, $$ \begin{aligned} \tau_1&=\sigma_1\cos\theta-\sigma_3\sin\theta\\ \tau_2&=\sigma_2\\ \tau_3&=\sigma_1\sin\theta+\sigma_3\cos\theta \end{aligned} $$ and ask ourselves if the $\vec\tau$'s are unitary equivalent to the $\vec\sigma$'s. For one thing, they satisfy the same algebra, and as such, they physically represent the same system.
To see that no such unitary transformation can exist, we note that all the nonrotated states are orthogonal to the rotated ones. For example, $\langle 0|\theta\rangle\to0$ in the $N\to\infty$ limit. Other states are orthogonal to the nonrotated vacuum as well, $\langle 0|\theta,i\rangle=0$ where $i\in\mathbb N$ labels the different rotated states, with $i=0$ corresponding to the rotated vacuum.
These vectors being orthogonal implies that there is no unitary transformation $\vec\tau =U^\dagger(\theta)\vec\sigma U(\theta)$ such that $|\theta\rangle=U(\theta)|0\rangle$. Indeed, assume that such an operator exists, and write \begin{equation} 1=\langle 0|U U^\dagger|0\rangle=\sum_{i\in\mathbb N}\langle 0|U|i\rangle\langle i|U^\dagger|0\rangle=0 \end{equation} a contradiction. Thus, no such $U$ exists, despite the fact that the systems are physically equivalent.
The take-home message is that in theories with an infinite number of degrees of freedom, physically equivalent observables are not necessarily unitary equivalent.
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