Interpret the canonical commutation relations as a Lie algebra

lie-algebrasmathematical physicsquantum mechanics

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:

  1. The vector $c$ is not a scalar multiple of the identity, unlike $\mathrm i \hbar I$
  2. The Heisenberg group is finite-dimensional while $P, Q, I$ are infinite-dimensional
  3. 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?

Best Answer

The physics CCR group is the Heisenberg group. I'd follow WP, to fix concepts and notation, as you seem to confuse representations with the Lie algebras they represent.

The three-dimensional Lie algebra $\mathfrak h$ of the Heisenberg group H (over the real numbers) is known as the Heisenberg algebra. (Three generators, so three parameters, a,b,c.)

It may be represented using the space of 3×3 upper-triangular matrices of the form $$\begin{pmatrix} 0 & a & c\\ 0 & 0 & b\\ 0 & 0 & 0\\ \end{pmatrix} , $$ with $a, b, c\in\mathbb R $; its exponential is the generic group element, $$\begin{pmatrix} 1 & a & c+ab/2\\ 0 & 1 & b\\ 0 & 0 & 1\\ \end{pmatrix} , $$

Note that, in this (defining) representation, the algebra basis elements are not hermitean, and hence the group elements are not unitary. You'd be very wrong if you imagined the group is U(3).

The following three elements form a basis for $\mathfrak h$, $$ X = \begin{pmatrix} 0 & 1 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0\\ \end{pmatrix};\quad Y = \begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0\\ \end{pmatrix};\quad Z = \begin{pmatrix} 0 & 0 & 1\\ 0 & 0 & 0\\ 0 & 0 & 0\\ \end{pmatrix}. $$

The basis elements satisfy the commutation relations, $$ [X, Y] = Z;\quad [X, Z] = 0;\quad [Y, Z] = 0. $$

In physics, it is also represented by infinite-dimensional hermitean matrices/operators, $$\left[\hat x, \hat p\right] = i\hbar I;\quad \left[\hat x, \hbar I\right] = 0;\quad \left[\hat p, \hbar I\right] = 0. $$

Note the obligatory physics i upon transition to the hermitean basis! The group elements result from exponentiation with an i and a parameter, resulting into unitary infinite-dimensional matrices.

Further note in this rep that the trace of the central element Z, proportional to the infinite-dimensional identity here, is not zero, even though it amounts to a commutator. This is a standard famously frequent question on the PSE, with a subtle limit resolution.

The unitarity of the representations or not is not a feature of the Group, but, instead a feature of the representation.

Finally, the identity is never a feature of the Lie algebra, but of the universal enveloping algebra. In this representation, it coincides with the center Z.

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