How should I interpret the expectation value $\langle x p\rangle$ in quantum mechanics? $xp$ is not a hermitian operator and hence doesn't represent an observable. Then, how can we interpret the expression
$$
\langle x p \rangle \text{,}
$$
the expectation value of position times momentum?
How to interpret the expectation value of a hermitian operator, say $p$, is clear: make several measurements on identically prepared systems and get the expectation value from the measured quantities.
But now we are talking about the expectation value of an operator that doesn't represent an observable. The discussion above hence isn't valid. However, computing $\langle x p \rangle$ knowing $\Psi$ according to the mathematical definition can be done, so $\langle x p \rangle$ can clearly be deduced if we know the wave function - but how can we interpret the quantity we get?
(The question arises from a line in my textbook - "in a stationary state, $\frac{d}{dt} \langle x p \rangle = 0$, ..." without further explanation - why is this obvious?)
 A: Here is how you interpret: the expectation value in a state $|\Psi\rangle$ of products of Hermitian operators like $x$ and $p$ (corresponding to observables) quantifies how correlated (quantum mechanically entangled) the two observables are in that state.
Therefore, the statement $\frac{d}{dt}\langle xp\rangle=0$ in English reads: "the extent to which the two observables 'position' and 'momentum' are entangled correlated does not change with time".  Now since stationary states are independent of time (up to a phase), it should be pretty clear that the correlation should not change with time (i.e. it is fixed).
A: The line from your textbook you're puzzling over is in fact quite a bit easier: in a stationary state nothing changes apart from phases that cancel out in any expectation value, and therefore $\frac{d}{dt}\left[\textrm{anything} \right]=0.$
You're right to point out that $xp$ is not a hermitian operator, but that does not mean that the expectation value $\langle xp \rangle$ is meaningless. Specifically, take the uncertainty relation $xp-px=i\hbar$ and substract $xp$ twice: you get $$xp+px=-i\hbar+2xp,$$ which you can rearrange into $$\langle xp\rangle = \left\langle \frac{xp+px}{2}\right\rangle+\frac{i\hbar}{2}.$$
The expectation value is now of the hemitian operator $\frac{1}{2}(xp+px)$, and you can see that your original expectation value has a trivial imaginary part. 
If it's a physical interpretation for this quantity you're after, try this question.
