# Does the issue of a distinct 'interacting Hilbert space' occur in nonrelativistic single particle QM?

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?

• I think you (and/or the commenters there) need to clarify the notion of "different Hilbert spaces" for this question to be well-defined. Since all separable Hilbert spaces are isomorphic as Hilbert spaces, the notion of a "different Hilbert space" is, taken literally, non-sensical. Haag's theorem talks about unitarily inequivalent representations of the QFT CCR. For CCR in finitely many d.o.f., the Stone-von Neumann theorem shows all representations are unitarily equivalent, so the precise analogue of Haag's theorem does not hold in QM. – ACuriousMind Feb 15 '17 at 21:41
• @ACuriousMind If I knew enough math to make this question 'well-defined', I would already know the answer, since 99% of the challenge here is formulating it right. So I guess my question is, what are those comments really saying, and is there a way for them to be correct? – knzhou Feb 15 '17 at 21:44
• You might be interested in the recent comments below AFT's answer; we are both no longer sure that his/her example is quite correct. – tparker Aug 14 '18 at 15:17

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 $$1=\langle 0|U U^\dagger|0\rangle=\sum_{i\in\mathbb N}\langle 0|U|i\rangle\langle i|U^\dagger|0\rangle=0$$ 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.

• Thanks for the answer! Just to make sure, does your first comment imply that the comments I linked to are incorrect, and that this issue never occurs in QM (non-QFT)? – knzhou Feb 15 '17 at 21:56
• @knzhou well, not necessarily. Haag's theorem implies non-equivalent Hilbert spaces, but this doesnt mean that non-equivalent Hilbert spaces imply Haag's theorem. In a system with a finite number of d.o.f., Haag's theorem doesnt apply, but in principle you may have non-equivalent Hilbert spaces for other reasons. That being said, I agree with Prahar's first comment, but not quite with the rest (and specially with the third one: Prahar says that the Hilbert spaces of $H_i=\frac{P^2}{2m_i}$ are non-equivalent for $m_1\neq m_2$, which is clearly false: both Hilbert spaces are $L^2(R)$, and – AccidentalFourierTransform Feb 15 '17 at 22:03
• cont. you can expand any wave function in the former in terms of plane waves constructed in the latter). But he has a point in the 1st comment: imagine an H_0 that depends on the phase space variables P,X. If you add the perturbation $V=P^4$, then you dont change the Hilbert space. But if you add the perturbation $\vec L\cdot \vec S$, with $S$ the spin of the particle, then you do 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 (because S represents a new d.o.f., which is not present in the former) – AccidentalFourierTransform Feb 15 '17 at 22:06
• Going back to this after a while, I'm confused about the claim that "there is no $U$ mapping $|0 \rangle$ to $|\theta\rangle$ because $\langle 0 | \theta \rangle = 0$". Why can't you have a unitary take a state to an orthogonal state? – knzhou Oct 14 '17 at 14:04
• @tparker number of degrees of freedom = number of phase-space variables (or = dimension of the phase-manifold). For a system of $N$ particles in 3d space, $|\text{d.o.f.}|=|\{\vec x_i\}_{i=1,\dots,N}|+|\{\vec p_i\}_{i=1,\dots,N}|=6N$. In field theory with field $\{\phi^a\}_{a=1,\dots,n}$, the phase-space variables are $\{\phi^a(\vec x),\pi_a(\vec x)\}_{a=1,\dots,n}^{x\in\mathbb R^3}$, that is, there are $3n$ for each point $\vec x$ in space $\mathbb R^3$. That is, an infinite number of d.o.f. In the example in my post, the phase-space variables are $\{\vec\sigma(i)\}_{i=1,\dots,N}$ (1/2) – AccidentalFourierTransform Aug 13 '18 at 13:23