Spin Chronicles Part 38: Inside a representation

 Suppose ρ is a *-representation of a *-algebra A (with unit 1) on a Hilbert space H. We will assume that both A and H are finite-dimensional. Then one of two things can happen: either ρ is irreducible, or ρ is reducible. Let us examine the case of irreducible ρ first.

Inside a representation

The case of ρ irreducible: every non-zero vector is cyclic.

Suppose ρ is irreducible. That means H has no nontrivial invariant subspaces. Choose any non-zero vector x in H. Consider ρ(A)x:

ρ(A)x = {ρ(a)x: a∈A}.

Then ρ(A)x is a linear subspace of H (Why?). Moreover ρ(A)x is an invariant subspace (Why?). By definition it must be trivial. Since it is different from {0} (Why?) - it mast be the whole H. We say that the vector x is cyclic for ρ(A). A vector is cyclic for an algebra if acting with the algebra elements on this vector you get the whole space (in general a dense subspace, but we are in finite dimensions, so every dense subspace is the whole space).

The case of ρ reducible.

Now suppose ρ is reducible. That means there is a nontrivial invariant subspace F. But then F^⟂ - the orthogonal complement of F in H, is also invariant (Why?). Now we have two representations of A, one on F, and one in F^⟂. We choose a non-zero vector in each of them and repeat the reasoning. Since we assume the dimension of H is finite, this procedure must end after a finite number of steps with all irreducible representations.  So, we have a decomposition

H = F₁⊕F₂⊕...⊕F_k,

and vectors x₁,x₁,...,x_k, with x_i being a cyclic vector for ρ(A) acting on F_i.

Vectors and states.

Let us go back to a general case. We have A, H, and ρ, with H being different from {0}. Let us choose a nonzero x vector in H, and let us suppose that x is normalized, wit ||x||=1. The x defines a state functional f on A by

f(a) = (x,ρ(a)x).

It is easy to check (check it!) that f(a*)=f(a)*, f(a*a) ≥ 0 for all a in A, and f(1) = 1.

Notice that vectors x and ix, define the same state f. In fact, vectors x and cx, where c i a complex number with |c|=1, define the same f. So there is some redundancy where replacing states (positive normalized functionals on A) by norm 1 vectors in H. Perhaps the source of this redundancy is already contained in A itself. In quantum theory A is assumed to be an algebra over complex numbers. What is the physical meaning of the product ca,where c is a complex number and a is an "observable" in A? Nobody knows it in a general case. For our Clifford algebra the meaning is geometric, but what it has to do with quantum theory? Good question.

In the next post we will finally see the GNS construction: starting from a state f, and constructing H, ρ, and a cyclic vector x. This representation sometimes, for some states f,  will be irreducible. We will see that this construction gives us a particular x, not just a class of physically "equivalent" cx. There is some mystery here.