What is the relation between joint measurability and common refinement (pure state decomposition) of density operators? Here page 13, the author states "...just as two quantum observables are often not jointly measurable, two
decompositions of  mixed states often have no common refinement (Actually, in
the formalism of quantum theory these are two variants of the same theorem)". My question is two-fold:
1.How is non-commutativity related to impossibility of joint measurement (for the non-commutative observables). In particular, what is definition of joint-measurability?
2. How does one relate indistinguishability of particles in Quantum statistics (Fermi-Dirac or Bose-Einstein) to the above properties (impossibility of joint measurement and/or impossibility of determining a preferred pure state decomposition for mixed states)?
I appreciate comments on any of the above questions!
 A: They are probably just talking about the uniqueness of the eigendecomposition of an Hermitian operator.
Two observables being jointly measurable means that they are mutually diagonalisable. Given any observable (Hermitian operator) $\mathcal O$, measuring $\langle\mathcal O\rangle$ can always be understood as performing a projective measurement in the (a) basis of eigenvectors of $\mathcal O$, and then performing some classical post-processing on the associated measured probabilities. A pair of observables is jointly measurable when there is a single projective measurement which allows to retrieve the expectation values of both. In other words, two observables $A$ and $B$ are jointly measurable iff there is some orthonormal basis (for whatever underlying space we are dealing with) $\{|i\rangle\}_i$ such that $A|i\rangle,B|i\rangle\propto|i\rangle$.
Regarding states, I assume when you talk of "decompositions" you are referring to decompositions of a density matrix $\rho$ as a convex combination of pure states. I'm not quite sure what a "common refinement" would mean in this context though. Given a (finite-dimensional) state $\rho$, one can write it as
$$\rho = \sum_i p_i \mathbb P(u_i)
= \sum_i q_i \mathbb P(v_i)$$
for some $p_i,q_i\in[0,1]$ with $\sum_i p_i=\sum_i q_i=1$, and $u_i,v_i$ unit vectors, and $\mathbb P(u)\equiv uu^\dagger\equiv|u\rangle\!\langle u|$ the projection onto the vector, if and only if there is some unitary operator $V$ such that
$$\sqrt{p_i} u_i = \sum_j V_{ij} \sqrt{q_j} v_j.$$
This is shown thinking about the possible purifications of a density matrix.
If the text is talking about using a similar theorem as for the observables, I suppose they are talking about decompositions in terms of orthogonal pure states? In fact, such decompositions are nothing but eigendecompositions of the density matrix: if there are orthonormal vectors $\{u_i\}_i$ such that
$$\rho = \sum_i p_i \mathbb P(u_i),$$
for some probability vector $(p_i)_i$, then $u_i$ are eigenvectors of $\rho$ with eigenvalues $p_i$.
Note that eigendecompositions are uniquely determined by the operator itself (albeit there is some freedom in the choice of eigenvectors when the operator is degenerate).
So in this sense, decompositions of a state in terms of orthogonal pure states are "essentially" unique.
The question about particle indistinguishability is quite different, and should probably be asked separately.
