Particle mixing and indistinguishability Neutral kaons have two flavor combinations: $\mathrm{d}\bar{\mathrm{s}}$ and $\mathrm{s}\bar{\mathrm{d}}$. They can also be weak eigenstates: $\mathrm{\frac{d\bar{s} \pm s\bar{d}}{\sqrt{2}}}$. 
But are neutral kaons of different compositions indistinguishable? If yes, how is it so, since they are composed of different fundamental particles? If not, what will happen when neutral kaon oscillates? They are suddenly indistinguishable when one oscillates into another? That indistinguishability can vary over time?
 A: What are the two particles about which we may safely say that they're indistinguishable? They're indinstinguishable if they're particles of the very same species. 
Imagine that two particles are of the type $K_0$ or $\bar K_0$. How can you say whether they are indistinguishable? Well, a $K_0$ is indistinguishable from another $K_0$ but it is distinguishable from a $\bar K_0$.
However, it's more natural to choose a different basis in the Hilbert space. The fact that $K_0$ oscillates into a $\bar K_0$ really means that there is an off-diagonal element of the Hamiltonian that is able to evolve one to the other. Because there is an off-diagonal element of the Hamiltonian, it means that the Hamiltonian isn't diagonal in this basis. In other words, $K_0$ and $\bar K_0$ aren't the energy (or mass) eigenstates. That's why oscillate.
However, there exists a linear superposition of the states in the Hilbert space that is a mass eigenstate (if the particle is in the rest frame, it's the same thing as an energy eigenstate). The corresponding linear superpositions are known as $K_{0S}$ and $K_{0L}$ where L,S stand for long-lived and short-lived. If a kaon is a $K_{0S}$, then it will be a $K_{0S}$ until the end of its (short) life: it will never become a long-lived one. And similarly $K_{0L}$ will remain a $K_{0L}$ until the end of its (a bit longer) lifetime; it will never be the short-lived one. So they don't oscillate; however, they still have an exponentially decaying probability of staying alive because they decay, with different decay rates (which really determine the imaginary part of their mass).
Again, for the mass eigenstates, it's true that two $K_{0S}$ particles are indistinguishable but $K_{0L}$ is distinguishable from a $K_{0S}$.
If you have two kaons that oscillate and each of them is exactly in some phase of the oscillation, then they're partly distinguishable, partly indistinguishable. Just write the two-particle state as a linear superposition of some basis vectors, e.g. 
$$ |K_{0S},K_{0S}\rangle, \quad  |K_{0S},K_{0L}\rangle, \quad  |K_{0L},K_{0S}\rangle, \quad  |K_{0L},K_{0L}\rangle, \quad $$
Well, the second and third ket vectors are actually identical if the particles are coincident. But if the particles are at two positions (or, more generally, in two different states specified by some other quantum numbers than positions), they're effectively distinguishable, anyway, as I will reiterate below. So the second and third vectors are identical but there are twice as many states of this kind so each of the 4 "templates" really gives you the same number of states.
If you write the most general two-kaon state vector, it will be a linear superposition of the ket vectors of the 3 or 4 types above. The state vector must also remember the information about the position of the kaons (and/or other quantum numbers). This position part of the wave function is symmetric (i.e. reflecting indistinguishability) for the first and last components in the state vector (SS and LL) but it is a general wave function for the middle (mixed) terms in the decomposition. By comparing the integrated squared probability amplitudes, you may even quantify the probability that the two particles you have are indistinguishable. 
Much like almost all questions in quantum mechanics, the answer to the question "are the two kaons indistinguishable?" depends on the state and if you have some general state, you may only answer probabilistically and the probability is a number between 0 and 100 percent.
