OK, the problem with density operators is not as I had feared.  Although
I have found some illumination.

My current understanding:

* A density matrix is positive-definite with trace 1.

* A density matrix may always be obtained from an ensemble {p_i, psi_i}
  (i=1, ..., n) by rho = sum p_i |psi_i><psi_i|.  We discussed this in
  your office; also, Nielsen and Chuang prove it.

* Two different ensembles can give the same density matrix.  N&C give
  a theorem characterizing the conditions under which this can happen.

* "Pure" and "mixed" states have *no meaning* **except** with respect
  to a specified ensemble.  A state that is pure with respect to one
  ensemble may not be pure with respect to another.

* A state is pure with respect to an ensemble iff it is one of the state
  vectors in that ensemble.

* I proved that tr(rho) = 1 and tr(rho^2) <= 1, with equality iff 
  psi is one of the members of the ensemble.

* tr(rho) is basis-independent; tr(rho^2) is basis-dependent.  (I have
  specific examples.)

* Members of an ensemble apparently do not need to be orthogonal.  I find this
  bizarre.

* Whereas the notions of pure and mixed states involve a single vector space,
  "separable" and "entangled" involve tensor products.  A separable state is a
  decomposable tensor; an entangled state is an indecomposable tensor.

* Determination of pure vs. mixed, and separable vs. entangled, is utterly
  trivial for finite-dimensional state spaces.  I understand there are
  complications in the infinite-dimensional case.

* It is not at all clear to me how the density-operator representation is
  in any sense "more general" than the state-vector representation.  One may
  pass back *and* forth between them.