Introduction
In quantum mechanics, a commutator is $$ [A, B] := AB - BA \: , $$ where $A$, $B$ are observables (ie. self-adjoint operators on a Hilbert space). Their interpretation is usually the following:
Each observable corresponds to an infinitesimal transformation, whose “full form” is $U_{\!A}(t) = \exp t \, \mathrm iA$.
A commutator is a measure of the local non-commutativity of two such transformations.
For example, the momentum operator corresponds to an infinitesimal translation, angular momentum to a rotation, etc.
This is exactly how Lie algebras relate to Lie groups – more precisely, given a set of observables $A_k$ the transformations $U_{\!A_k}(t)$ form one-parameter subgroups of some unitary group $\mathrm U(n)$, their corresponding vectors from the Lie algebra $\mathfrak u(n)$ being $\mathrm i A_k$. Therefore it seems natural to think about commutator relations in QM as vectors from the Lie algebra $\mathfrak u(n)$, or its infinite-dimensional counterpart.
Canonical Commutation Relations
The canonical commutation relations (or CCR for short) of quantum mechanics read $$ [ Q, P ] = \mathrm i \hbar \, I \: , $$ where $Q$ and $P$ are observables and $I$ is the identity. Interpreting each operator as $(-\mathrm i)$-times a generic vector, we get the Lie algebra $$ [a, b] = c \: , \qquad [a, c] = 0 \: , \qquad [b, c] = 0 \: . $$ This is the Lie algebra of the Heisenberg group, a 3×3 non-unitary matrix group. There are three big differences between the quantum-mechanical operators and the Heisenberg group:
- The vector $c$ is not a scalar multiple of the identity, unlike $\mathrm i \hbar I$
- The Heisenberg group is finite-dimensional while $P, Q, I$ are infinite-dimensional
- The Heisenberg group is not a subgroup of the unitary group, while the transformations $U_P, U_Q$ are unitary
The Question
Is there a way to interpret the quantum-mechanical realization of CCR in terms of Lie groups and Lie algebras? In particular, does the algebra of $Q = x, \; P = \mathrm i \partial_x$ correspond to some Lie group, and if so, how does that group relate to the Heisenberg group?
Finally, I struggle to rigorously formulate the requirement that $c$ be a scalar multiple of the identity operator, as there is no “identity” in an abstract Lie algebra. Is there a way to formulate such a requirement in the language of Lie algebras?