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P(n) & Q(n) as Generators, Irreducibility





   The preceding section suggests the following:

    <Theorem 6.1>:

   The operators P(n) and G(n) alone
   generate by commutation, the real Lie algebra so(n).

   Proof:


    <Theorem 6.2>:

   The operators Q(n), P(n) and G(n)
   generate by commutation, the real Lie algebra su(n),
   the complexification of which is sl(n, C).

   Proof:

   QED

    <Proposition 6.1>:


        [N(n), B!^k(n)]  =  + k B!^k(n)
        [N(n), B^k(n)]  =  - k B^k(n)


   for any integer k > 0.

   Proof:

   QED

   These two formulae also hold in standard QM.

    <Theorem 6.3>:

   i) The set of operators

        N^k(n)

   for k = 0, 1, 2, ..., n-1, are a basis for a maximal

   commutative subalgebra Alg( N(n) ) of Alg(n).
   Alg( N(n) ) is the set of operators that commute with N(n),
   i.e., those that are diagonal in the basis |n, k>.
   

   ii) Any operator A in Alg(n) such that

        [A, N(n)]  =  0

   is then a polynomial function of N(n) of degree at most n-1.

   iii) The only polynomial functions of B(n) and B!(n)
   that commute with N(n) are polynomials in N(n).
   This result also hold in standard QM.
   

   Proof:

   QED

    <Theorem 6.4>:

Any nxn Hermitean matrix is expressible as a polynomial in the operators Q(n) and P(n), or B(n) and B!(n) where the coefficients of the polynomial are real. Similarly, if the coefficients are permitted to be complex, then any nxn matrix is so expressible. In the first case, the polynomials span the Lie algebra u(n), and in the second case the Lie algebra gl(n, C).

The ordering of the factors in the polynomial must be fixed for a matrix to be represented by a unique polynomial.

Proof:

Irreducibility is physically associated with a simple system, that is not decomposable into noninteracting parts. A system composed of two noninteracting particles is reducible in terms of its observables. A measurement made on one particle has no effect on measurements made on the other, so measurements on the two particles are independent. If one views the three dimensions of space as being independent degrees of freedom, there is no room in Alg(n) to accommodate further spatial position operators since from theorem 5.1, the polynomials in B(n) and B!(n) span the full matric algebra Alg(n).




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Created: August 1997
Last Updated: May 28, 2000