Any quantum state is an eigenstate of some complete observable? $\newcommand{\ket}[1]{\left|#1\right>}$
Quantum Mechanics and Experience by David Albert states (p. 63):

$$\ket{A} = \tfrac{1}{\sqrt{2}} \ket{\text{black}}_1  \ket{\text{white}}_2 - \tfrac{1}{\sqrt{2}} \ket{\text{white}}_1  \ket{\text{black}}_2$$
$\ket{A}$, like any quantum state, is necessarily an eigenstate of some complete observable of this state of electrons.

Note that two-valued “colour” is just a metaphor for the spin state of some electron in some particular basis.
My question is, how can I understand the highlighted statement? It actually came as a bit of a surprise. I suppose it follows from some simple maths, though my QM maths is pretty rusty. Also, how should I understand it from a physical perspective? I am used to equating observables with measurable physical quantities, but how can an entangled state be observable/measurable in this sense?
 A: This is true only for pure states but given you have a ket (which is a pure state), the statement can be taken as true.
The point is: given a (normalized) ket $\vert \psi\rangle$, construct the operator $$
\hat {\cal O}=\vert \psi \rangle \langle \psi\vert + \vert \phi\rangle \langle \phi\vert
$$
where $\langle \psi\vert\phi\rangle=0$.  Clearly
$$
\hat {\cal O}\vert \psi\rangle= 
\vert \psi \rangle \langle \psi\vert \psi\rangle+ \vert \phi\rangle \langle \phi\vert \psi\rangle=
\vert \psi\rangle \tag{1}
$$
so $\vert \psi\rangle$ is an eigenstate of $\hat {\cal O}$ with eigenvalue $+1$.
That's the basic result.  It needs to be augmented to account for possible degeneracies.  Suppose you have $\hat T$ so that $[\hat T,\hat{\cal O}]=0$, then $\vert \psi\rangle$ is either an eigenstate of $\hat T$ or a linear combination of eigenstates of $\hat T$ with the same eigenvalues.   You can in this way refine (1) until your $\vert \psi\rangle$ is actually an eigenstate of a set of commuting operator (possibly taking combos of eigenstates of $\hat T$ in the process).
It has to be a complete set else there could be another different state
$\vert \psi'\rangle$ for which $\hat{\cal O}$ etc would have the same eigenvalues (i.e. same outcomes), and you'd have no way of differentiating it from $\vert \psi\rangle$ given a set of outcomes of your commuting operators.
In your case,  using the shorthand notation
\begin{align}
\vert bb\rangle &= \vert \hbox{black}\rangle \vert \hbox{black} \rangle\, ,\\
\vert ww\rangle &= \vert \hbox{white}\rangle \vert \hbox{white} \rangle\, \\
\vert bw;\pm\rangle &= \frac{1}{\sqrt{2}}
\left(\vert \hbox{black}\rangle \vert \hbox{white} \rangle\pm 
\vert \hbox{white}\rangle \vert \hbox{black} \rangle \right)
\end{align}
your state is an eigenstate of
\begin{align}
\hat{\cal O}= \vert bb\rangle\langle bb\vert +  \vert ww\rangle\langle ww\vert +  \vert bw;+\rangle\langle bw;+\vert
+  \vert bw;-\rangle\langle bw;-\vert\, . \tag{2}
\end{align}
Clearly $\hat{\cal O}$ is not unique,  but your state is an eigenstate of
$\hat{\cal O}$, and that's all you need.
Note that there is nothing to suggest that $\hat{\cal O}$ needs to be realizable in practice: it could be a purely formal operator and it could be impossible to construct an apparatus to measure this operator, but formally it exists and that's good enough for the argument.
A: Here is  another answer. Suppose the space is finite-dimensional or countably infinite dimensional. We concentrate on the latter   case only since the former is a trivial subcase. Let $\{|n\rangle\}_{n \in \mathbb N}$ be a Hilbert basis such that $|0\rangle = |A\rangle$.
Define the observable
$$H = \sum_{n\in \mathbb N} n |n\rangle \langle n|\:.$$
This observable forms a complete observable without adding further observables, and
$H|A\rangle =0\:.$
As in the other answer, in general $H$ has no direct physical meaning.
The procedure can be extended to the case of an infinte dimensional space whose cardinality is the one of $\mathbb{R}$.
