## Weyl form of the Heisenberg commutation relations

Consider the Heisenberg commutation relations, for simplicity, for a free particle with one degree of freedom, is $$ PQ-QP = -i\hbar $$ where $P,Q$ should be self-adjoint operators on a Hilbert space $\mathcal{H}$.

Several observations:

- $\mathcal{H}$ should be finite-dimensional if $\hbar\neq 0$, since the trace of any commutator vanish, while $\operatorname{Tr}(-i\hbar)$ doesn’t.
- Both $P,Q$ should be unbounded.

*Proof:*First of all, $P$ and $Q$ should be both bounded or unbounded, since $P$ and $Q$ play symmetric roles via the fact that interchanging them amounts to replacing $i$ by $-i$ or to selecting a different square root of $-1$. Since $P$ and $Q$ are self-adjoint, by Stone's theorem, they can be exponentiated to one-parameter unitary groups $$ U_t = \exp(itP), V_t=\exp(itQ). $$ Then by expansion (which can be justified pointwise, applied to a vector in the range of a spectral projection of $Q$), $$ U_t Q U_{-t} = \operatorname{Ad}(\exp(itP))Q = Q+t\hbar $$ meaning that $Q$ is unitarily equivalent to $Q+t\hbar,\forall t\in \mathbb{R}$. The spectrum of $Q$ must consist of the whole real line, and hence $Q$ is unbounded. □

Using the notation introduced in the proof, in a similar manner, functional calculus yields
$$
U_t f(Q) U_{-t} = f(Q+t\hbar)
$$
for any real analytic function $f$ on the spectrum of $Q$. In particular,
$$
\begin{equation}\label{weyl-form}
U_t V_\sigma U_{-t} = U_t e^{i\sigma Q}U_{-t} = \exp(i\sigma(Q+t\hbar))=e^{i\sigma t}V_\sigma
\end{equation}
$$
This is the multiplicative form of the commutation relations, discovered by Weyl, which is given the name *Weyl integrated form* of the commutation relation.

## The Stone-von Neumann Theorem

Let $P_j,Q_k$ be self-adjoint operators, $1\leq j,k\leq n$, satisfying the canonical comutation relations $$ [P_j,P_k] = [Q_j,Q_k]=0, [P_j,Q_k] = -i\delta_{jk}\hbar, $$ and ignore the difficulties presented by the unboundedness of the operators by going to the Weyl integrated form, that is, assume that there are representations $U,V$ of $\mathbb{R}^n$, then the theorem of Stone and von Neumann is

*Proof:*This is a sketch. Exchanging $U,V$, it is easy to see that $\omega$ is skew-symmetric. Without lose of generality, the relations can be reduced to $$ U(x)V(y) = e^{i\langle x,y\rangle}V(y)U(x) $$ where $\langle\cdot,\cdot\rangle$ is the usual Euclidean inner product on $\mathbb{R}^n$ since any nondegenerate bilinearity form is equivalent to its standard form. Given any representation of the Weyl integrated relation on a Hilbert space $\mathcal{H}$, there's a large number of self-adjoint projections on $\mathcal{H}$ of the form $$ P_\psi = \iint U(x)V(y)\psi(x,y)dxdy $$ for $\psi\in\mathcal{S}(\mathbb{R}^2n)$ (the Schwartz space). It can be proved that $P_\psi$ vanish only if $\psi\equiv 0$. In particular, if one put $$ \psi(x,y)=\frac{1}{(2\pi)^n} e^{-i\langle x,y\rangle/2}e^{-(|x|^2+|y|^2)/4} $$ then $P_\psi$ is a self-adjoint projection. For this $\psi$, $P_\psi U(x) P_\psi$ and $P_\psi V(y) P_\psi$ agree with $P_\psi$ up to scalar factors depending on $x,y$. Thus if the representation of $U,V$ is irreducible, $P_\psi$ must be of rank one, otherwise $P_\psi$ would be able to be written as the sum of two proper subprojections that generate proper invariant subspaces of $\mathcal{H}$. Given $\mathcal{H},\mathcal{H}'$ and two irreducible representations of the Weyl integrated relations, correspondingly there are $P_\psi$ and $P_{\psi'}$, and a map sending a unit vector in $\operatorname{Range}P_\psi$ to that of $\operatorname{Range}P_{\psi'}$ extends uniquely to a unitary intertwining operator, and thus intertwining the representations of $U,V$. □

## Mackey’s version

Recall that a *covariant representation* of a covariant system is a pair $(\pi,\rho)$ where $\rho$ is a unitary representation of a locally compact group $G$ on $\mathcal{H}$ and $\pi$ is a $\ast$-representation of a $C^\ast$-algebra $A$ on $\mathcal{H}$. There is furthermore a homomorphism $\alpha: G\to \operatorname{Aut}(A)$, s.t.
$$
\rho(g)\pi(a)\rho(g)^\ast = \pi(\alpha_g(a)).
$$
When $A=C_0(G)$, there is a canonical $\alpha$ given by $\alpha_g(f(h)) = g\cdot f =f(g^{-1}h)$ where $f\in C_0(G)$. Then the covariance relation becomes
$$
\rho(g)\pi(f)\rho(g)^\ast = \pi(g\cdot f)
$$

### Generalization: Morita Equivalence

This can be generalized to the theorem

This in a more modern language means the following. The covariant pairs of representations of $G$ and that of $C_0(G/H)$, or *systems of imprimitivity* based on $G/H$, can be identified with representations of a crossed product algebra $C_0(G/H)\rtimes G$. The content of the theorem says that $C_0(G/H)\rtimes G$ is (strongly) Morita equivalent to the group $C^\ast$-algebra $C^\ast(H)$. The correspondence of representations matches induced representations $\operatorname{Ind}(\sigma)$ of $G$ with the inducing representations $\sigma$ of $H$.

### Generalization: Takai Duality

Stated in terms of crossed product, Theorem 3 can be read as

This turns out to be a special case of *Takai duality* and its noncommutative generalization.

This can be generalized to nonabelian groups and even quantum groups.