The n-Category Café

Hi John,

You don’t see the extra 1/2 when you’re looking at the state space as a Heisenberg Lie group/algebra representation, so not in Duflo isomorphism there.

You do see it in the symplectic group representation, but it’s like the spin case: you have a Lie algebra representation which only exponentiates to a group representation up to sign.

This does show up in the Duflo isomorphism story, not for Heisenberg, but symplectic. There’s a mysterious factor there, one aspect of the usual “rho-shift” (rho=half sum of positive roots) that shows up in these formulas. This is the same 1/2 you’re thinking about.

Reading your question more carefully, I realize things to say are

1. H isn’t in the center of the universal enveloping algebra of the Heisenberg Lie algebra (if 1,q,p a basis of that Lie algebra, and 1 acts by the constant hbar, the center is the polynomials in hbar).

2. H also isn’t in the center of the universal enveloping algebra of the symplectic group. It is the action of the Lie algebra of a U(1) subgroup of the symplectic group (SL(2,R) in this case). The shift by 1/2 makes this exponentiate to a true representation of a double cover of SL(2,R). The center of SL(2,R) is polys in the Casimir, the rho shift shows up when you want to identify this algebra with Weyl-invariant polys on the Lie algebra of the Cartan (Harish-Chandra isomorphism) or ad-invariant polys on the Lie algebra (Duflo).

I would argue (somewhat unconventionally) that the $\tfrac{1}{2}$ is unnatural.

Let’s generalize this slightly to $n$ identical harmonic oscillators. Passing from the classical Hamiltonian

to the quantum Hamiltonian is inherently ambiguous since $p$ and $q$ don’t commute. Different orderings differ lead to quantum Hamiltonia that differ by a constant.

Classically, (1) has an apparent $O(2n)$ symmetry. The symplectic form $$\omega = \sum_{a=1}^n d p_a\wedge d q_a$$ has an $Sp(n)$ symmetry. The intersection $O(2n)\cap Sp(n) \simeq U(n)$. That symmetry is made manifest by writing (1) as

At the classical level, these are of course equal. But (2) makes the $U(n)$ symmetry manifest. Passing from (2) to the quantum Hamiltonian yields the one with vanishing ground state energy.