Quantum Mechanics for Mathematicians

Note that this is a differentSO(2) action than the one with moment map the
Hamiltonian, it acts separately on positions and momenta rather than mixing
them.
To see what happens if one instead uses the Bargmann-Fock representation,
using

qj=

1

√

2

(zj+zj), pj=i

1

√

2

(zj−zj)

the moment map is

μL=

i

2

((z 1 +z 1 )(z 2 −z 2 )−(z 2 +z 2 )(z 1 −z 1 ))

=i(z 2 z 1 −z 1 z 2 )

Quantizing, the operator

UL′ =a† 2 a 1 −a† 1 a 2 = Γ′(2X 2 )

gives a unitary representation ofso(2). The factor of two here reflects the fact
that exponentiation gives a representation ofSO(2)⊂Sp(4,R), with no need
for a double cover.

25.4.2 Three degrees of freedom andSO(3)

The cased= 3 corresponds physically to the so-called isotropic quantum har-
monic oscillator system, and it is an example of the sort of central poten-
tial problem we studied in chapter 21 (since the potential just depends on
r^2 =q^21 +q^22 +q^23 ). For such problems, we saw that since the classical Hamilto-
nian is rotationally invariant, the quantum Hamiltonian will commute with the
action ofSO(3) on wavefunctions, and energy eigenstates can be decomposed
into irreducible representations ofSO(3).
Here the Bargmann-Fock representation gives an action ofH 7 oMp(6,R)
on the state space, with aU(3) subgroup commuting with the Hamiltonian
(more precisely one has a double cover ofU(3), but by normal ordering one
can get an actualU(3)). The eigenvalue of theU(1) corresponding to the
Hamiltonian gives the energy of a state, and states of a given energy will be
sums of irreducible representations ofSU(3). This works much like in thed= 2
case, although here our irreducible representations are on the spacesHnof
homogeneous polynomials of degreenin three variables rather than two. These
spaces have dimension^12 (n+ 1)(n+ 2). A difference with theSU(2) case is that
one does not get all irreducible representations ofSU(3) this way.
The rotation groupSO(3) will be a subgroup of thisU(3) and one can ask
how theSU(3) irreducibleHndecomposes into a sum of irreducibles of the
subgroup (which will be characterized by an integral spinl= 0, 1 , 2 ,···). One
can show that for evennone gets all even values oflfrom 0 ton, and for odd
none gets all odd values oflfrom 1 ton. A derivation can be found in some
quantum mechanics textbooks, see for example pages 456-460 of [60].