QuantumPhysics.dvi

This immediately gives the real dimension dimSU(N) =N^2 −1. [Note thatSU(N) itself

still contains non-trivial elementsMC which are proportional to the identityMC =εI as

long asεN= 1. The set of all such elements forms a finite group,ZNwhich is referred to as

thecenter ofSU(N) since they commute with all elements ofSU(N).]

9.6.3 The groupSp(2N)

IsU(N) the most general symmetry group leaving the Hamiltonian invariant? In general, it

is a difficult question to answer whether one has found the most general symmetry of a given

system. In particular, for the harmonic oscillator Hamiltonian, it would seem that we can

have a symmetry still larger thanU(N). Introduce the following 2N-dimensional column

matrixX,

H=

1

2 m

XtX X=

(

mωQ

P

)

(9.51)

The Hamiltonian H is now clearly invariant under arbitrary orthogonal transformations

X→X′=MX, whereM∈SO(2N). This group is larger thanU(N), since we have

SO(N)⊂U(N)⊂SO(2N) (9.52)

Actually, even thoughSO(2N) leaves the Hamiltonian invariant, it is not a symmetry of

the full quantum system, because the canonical commutation relations are not invariant.

Expressed on the componentsxα withα = 1,···, 2 N of the matrixX, the commutation

relations [qi,pj] =i ̄hδijbecome,

[xα,xβ] =i ̄hmωJαβ J=

(

0 +I

−I 0

)

(9.53)

Real linear transformationsX → X′ = MX which leave these canonical commutation

relations invariant must obey

MtJM=J (9.54)

The set of all real matricesM satisfyingMtJM=J forms thesymplectic groupSp(2N),

for which the inverse is given byM−^1 =−JMtJ.

The Hamiltonian is invariant underSO(2N) while the canonical commutation relations

are invariant underSp(2N). The entire system will thus be invariant under the largest

common subgroup ofSO(2N) andSp(2N). This is the set of real 2N× 2 N matricesM

which satisfy bothMtM= 1 andMtJM=J, which requires that [M,J] = 0. As a result,

Mmust be of the form

M=

(

A B

−B A

)

MtM=I (9.55)