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

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