Stone-von Neumann Theorem

Prerequisites: Canonical Commutation Relations, Hilbert Space. Full path (19 pages)

Intuitive
Concrete
Abstract
Why is it interesting?
History

Intuitive

The Stone-von Neumann theorem roughly says that for any operators $A$ and $B$ satisfying the canonical commutation relation, we can get away with using the standard representation $A \rightarrow u,\ B \rightarrow -i \hbar \frac{d}{du}$ without loss of generality. (More precisely, it says that any representation of the *exponentiated* canonical commutation relation on a sufficiently smooth Hilbert space is unitarily equivalent to the standard representation, so any other representation basically just describes the same physics in a different coordinate system.)https://physics.stackexchange.com/a/264587/37286

In typical physics quantum mechanics textbooks, one often sees calculations made just using the Heisenberg commutation relations, without picking a specific representation of the operators that satisfy these relations. This turns out to be justified by the remarkable fact that, for the Heisenberg group, once one picks the constant with which Z acts, all irreducible representations are unitarily equivalent. In a sense, the representation theory of the Heisenberg group is very simple: there’s only one irreducible representation. This is very different from the theory for even the simplest compact Lie groups (U(1) and SU(2)) which have an infinity of inequivalent irreducibles labeled by weight or by spin. Representations of a Heisenberg group will appear in different guises (we’ve seen two, will see another in the discussion of the harmonic oscillator, and there are yet others that appear in the theory of theta-functions), but they are all unitarily equivalent, a statement known as the Stone-von Neumann theorem. […] In the case of an infinite number of degrees of freedom, which is the case of interest in quantum field theory, the Stone-von Neumann theorem no longer holds and one has an infinity of inequivalent irreducible representations, leading to quite different phenomena. […]It is also important to note that the Stone-von Neumann theorem is formulated for Heisenberg group representations, not for Heisenberg Lie algebra representations. For infinite dimensional representations in cases like this, there are representations of the Lie algebra that are “non-integrable”: they aren’t the derivatives of Lie group representations. For such non-integrable representations of the Heisenberg Lie algebra (i.e., operators satisfying the Heisenberg commutation relations) there are counter-examples to the analog of the Stone von-Neumann theorem. It is only for integrable representations that the theorem holds and one has a unique sort of irreducible representation.https://www.math.columbia.edu/~woit/QM/qmbook.pdf

Concrete

In the algebraic approach, the fundamental structure of a quantum theory is given by an abstract algebra of the canonical commutation relations, which can be given various different representations in terms of subalgebras of the bounded operators on a Hilbert space. Two such representations, each consisting of a Hilbert space $H$ and the set of bounded operators $\hat O$ defined on it, are unitarily equivalent if there is a (one-to-one, linear, norm preserving) map $U:G\to H'$ such that $(\forall i)(U^{-1}\hat{O}'_iU=\hat{O}_i)$ . The Stone–von Neumann theorem guarantees that in the finite-dimensional case all irreducible representations of the abstract algebra are unitarily equivalent, but in the infinite-dimensional case there are unitarily inequivalent representations of the algebra.http://publish.uwo.ca/~csmeenk2/files/HiggsMechanism.pdf

Theorem (Stone-von Neumann). Any irreducible representation π of the group H3 on a Hilbert space, satisfying $π 0 (Z) = -i1$ is unitarily equivalent to the Schrödinger representation (ΓS, L2 (R)).https://www.math.columbia.edu/~woit/QM/qmbook.pdf

Abstract

The commutation relation $[Q,P] = i\hbar$ cannot hold for bounded operators: taking the trace on a finite-dimensional space gives $0 = i\hbar\, n$ , and an infinite-dimensional argument shows that at least one of $Q$ , $P$ must be unbounded. Statements about unbounded operators depend on domains, so the theorem is formulated for the exponentiated operators. Define the one-parameter unitary groups $U(a) = e^{iaQ}$ and $V(b) = e^{ibP}$ . The commutation relation becomes the Weyl relation $U(a)\, V(b) = e^{-i\hbar a b}\, V(b)\, U(a), \qquad a,b \in \mathbb{R} ,$ which follows from $e^{A}e^{B} = e^{B}e^{A}e^{[A,B]}$ for central $[A,B]$ .

Theorem. Let $U$ and $V$ be strongly continuous one-parameter unitary groups on a separable Hilbert space $\mathcal H$ satisfying the Weyl relation. Then $\mathcal H$ decomposes into a direct sum of invariant subspaces, on each of which $(U,V)$ is unitarily equivalent to the Schrödinger representation on $L^2(\mathbb{R})$ , where $(U(a)\psi)(x) = e^{iax}\psi(x)$ and $(V(b)\psi)(x) = \psi(x + \hbar b)$ . In particular, any irreducible representation is unitarily equivalent to the Schrödinger representation. The same holds for $n$ degrees of freedom.

Group-theoretic form. The Weyl relations define a unitary representation of the Heisenberg group, the central extension of $\mathbb{R}^{2n}$ by $\mathbb{R}$ with group law $(z, t)\cdot(z', t') = \left( z + z',\; t + t' + \tfrac{1}{2}\omega(z,z') \right) ,$ where $\omega$ is the standard symplectic form on $\mathbb{R}^{2n}$ . The theorem says that for every nonzero value of the central character (fixed by $\hbar$ ) there is exactly one irreducible unitary representation up to equivalence. Equivalently, for a finite-dimensional symplectic vector space the C*-algebra generated by the Weyl operators has a unique regular irreducible representation.

Consequences. Since a linear symplectic transformation $S \in Sp(2n,\mathbb{R})$ preserves the Weyl relations, the representation $W(Sz)$ is equivalent to $W(z)$ . The intertwining unitaries are fixed up to a phase and give a projective representation of $Sp(2n,\mathbb{R})$ , which lifts to a true representation of its double cover, the metaplectic representation. The Fourier transform is the intertwiner for the quarter turn in the $(Q,P)$ plane, which exchanges $Q$ and $P$ up to a sign, and the free and harmonic oscillator time evolutions arise this way.

Where it fails. Each hypothesis matters:

For infinitely many degrees of freedom, as in quantum field theory or the thermodynamic limit, there are uncountably many inequivalent irreducible representations of the canonical commutation relations. Different vacua of a spontaneously broken symmetry, free fields of different mass, and interacting fields live in such inequivalent representations, which underlies Haag's theorem.
If the configuration space is not $\mathbb{R}^n$ , the relevant group changes. For a particle on a circle, the position is only defined through $e^{i\theta}$ , and the inequivalent irreducible representations are labelled by an angle $\vartheta \in [0, 2\pi)$ , which appears physically as an Aharonov-Bohm flux or as the $\theta$ -parameter in a periodic potential.
There are pairs of self-adjoint operators satisfying $[Q,P] = i\hbar$ on a common dense invariant domain whose exponentials do not satisfy the Weyl relations, and which are not equivalent to the Schrödinger pair. The Weyl form is needed for the theorem.

Why is it interesting?

The Stone-von Neumann theorem is important, for example, to understand why symmetry breaking is possible in quantum field theory.

History

A Selective History of the Stone-von Neumann Theorem by Jonathan Rosenberg