With this choice the distinguished state| 0 〉will be an eigenstate ofHwith
eigenvalue^12.
- The choice of the coordinatezgives a decomposition
M⊗C=C⊕C (24.10)
where the first subspaceChas basis vectorz, the second subspace has
basis vectorz.
- The decomposition 24.10 picks out a subgroupU(1)⊂SL(2,R), those
symplectic transformations that preserve the decomposition. In terms of
the coordinatesz,z, the Lie bracket relations 16.15 giving the action of
sl(2,R) onMbecome
{zz,z}=−iz, {zz,z}=iz
{
z^2
2
,z
}
= 0,
{
z^2
2
,z
}
=iz
{
z^2
2
,z
}
=−iz,
{
z^2
2
,z
}
= 0
The only basis element ofsl(2,R) does not mix thezandzcoordinates
iszz. We saw (see equation 24.1) that upon exponentiation this basis
element gives the subgroup ofSL(2,R) of matrices of the form
(
cosθ sinθ
−sinθ cosθ
)
- Quantization of polynomials inz,zinvolves an operator ordering ambigu-
ity sinceaanda†do not commute. This can be resolved by the following
specific choice, one that depends on the choice ofzandz:
Definition.Normal ordered product
Given any productPof theaanda†operators, the normal ordered product
ofP, written:P:is given by re-ordering the product so that all factorsa†
are on the left, all factorsaon the right, for example
:a^2 a†a(a†)^3 : = (a†)^4 a^3
For the case of the HamiltonianH, the normal ordered version
:H: = :
1
2
(aa†+a†a): =a†a
could be chosen. This has the advantage that it acts trivially on| 0 〉and has
integer rather than half-integer eigenvalues onF. Upon exponentiation
one gets a representation ofU(1) with no sign ambiguity and thus no
need to invoke a double covering. The disadvantage is that :H: gives a
representation ofu(1) that does not extend to a representation ofsl(2,R).
