# What physical significance has the Heisenberg Group?

I read that the canonical commutation relation between momentum and position can be seen as the Lie Algebra of the Heisenberg group. While I get why the commutation relations of momentum and momentum, momentum and angular momentum and so on arise from the Lorentz group, I don't quite get where the physical symmetry of the Heisenberg group stems from.

Any suggestions?

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–  Qmechanic Jun 23 '13 at 23:53

You might like to see:

http://www.math.columbia.edu/~woit/QM/heisenberg.pdf

I.e. the lectures "Quantum Mechanics for Mathematicians: The Heisenberg group and the Schrodinger Representation" by Peter Woit, wherein the significance of the Heisenberg group is discussed in detail. But its physical significance is NOT as a group of symmetries of the physical situation. So be careful about tight analogies between the canonical commutation relation and the finite (say $n$) dimensional Hiesenberg Lie group $\mathfrak{H}_n\left(\mathbb{R}\right)$. The thing on the RHS of the relationship $\left[\mathbf{x},\,\mathbf{p}\right] = i \, \hbar \,\mathbf{i}$ in the finite dimensional algebra $\mathfrak{h}_n\left(\mathbb{R}\right)$ is NOT the identity matrix - it is simply something that commutes with everything else in the Lie algebra. It was Hermann Weyl who pointed out that the canonical commutation relation cannot be referring to a finite dimensional Lie algebra: in such algebras, a Lie bracket $\left[\mathbf{x},\,\mathbf{p}\right]$ (between square matrices) has zero trace but the identity matrix (or a scalar multiple, as on the RHS of the CCR) does not. One has to pass to operators on infinite dimensional Hilbert spaces ($e.g.$ $p = i \,\hbar \,d/dx$) to find full realisation of the canonical commutation relationship.

Another way to understand that the behaviour of the finite dimensional matrix Heisenberg Lie algebra is radically different from the CCR is the uncertainty principle itself. The product of RMS uncertainties for simulataneous measurements from two non-commuting observables $\hat{a}, \hat{b}$ given a quantum state $\psi$ is bounded from below by the positive real number $\frac{1}{2}\left|\left<\psi|c|\psi\right>\right|$ where $\left[\hat{a},\hat{b}\right] = i c$ (see section 10.5 of edition 3 of Merzbacher "Quantum Mechanics"). If $c$ is a finite square matrix, and, as in the Heisenberg algebra, it is not of full row rank, there are certain states (those in $c$'s nullspace) where the uncertainty product can be nought. So the finite dimensional matrix algebra can't model Heisenberg's physical postulate.

Minor comment to the answer (v2): The sign in the displayed Schroedinger representation of $p$ is not the conventional sign. –  Qmechanic Jun 24 '13 at 7:58