How can a single quantum system be prepared in a particular mixed state? Let's consider an electron in some sort of potential well. Suppose $|i\rangle$ is an $i$th eigenstate, and we want to prepare the electron in state with state vector (up to normalization) $|\psi_1\rangle=|1\rangle+|2\rangle$.
We can prepare such a state arbitrarily precisely by e.g. making the barrier finite (but still large) in size, and shedding two quasimonochromatic wave packets from left and from right — one with frequency corresponding to state $|1\rangle$ and the other with frequency of state $|2\rangle$. After some time, because of resonance the waves will noticeably tunnel through barriers inside the well, and we have our pure state.
Now, suppose we want to prepare a mixed state from same eigenstates: with density operator $\hat\sigma=|1\rangle\langle1|+|2\rangle\langle2|$. How could we go to prepare our electron in such a state? Do we have to prepare many such electrons, randomly selecting whether they'll be in $|1\rangle$ or in $|2\rangle$ and then measure randomly selected one, or can we just make that single electron in such a mixed state?
 A: This is, to the best of my knowledge, not doable without some sort of 'forgetfulness' on your part, or on the part of the physical systems you use to prepare your state.
The most obvious way is to flip a coin and decide which of $|1⟩$ and $|2⟩$ you want to prepare, and not record the outcome of the coin, and this is a perfectly fine way to generate
$$
\hat\sigma=\tfrac12\left(|1⟩⟨1|+|2⟩⟨2|\right).
$$
Of course, with a single realization this sort of feels like cheating, but the fact is that there is no way to measure, or corroborate, the full quantum state of a single realization of a system.
If this feels inelegant, let me propose another way. If you are able to produce the pure state
$$|+⟩=\tfrac{1}{\sqrt{2}}\left(|1⟩+|2⟩\right),$$
then you will also be able to generate its orthogonal conjugate
$$|-⟩=\tfrac{1}{\sqrt{2}}\left(|1⟩-|2⟩\right),$$
for example by changing the phase relationship between the two waves you mention. As it happens, the mixed state that has 50% chance of being either in $|+⟩$ or in $|-⟩$ is also $\hat\sigma$, so that flipping a coin to decide on the phase relationship will also generate $\hat\sigma$. More elegantly, though, you can set that phase to drift naturally throughout the course of the experiment, and this will also result in a mixed state. (You need to be careful, on the other hand, to ensure that this phase drift does not affect the resonant capture of your particles.)
Finally, a third method is to use one half of an entangled state, which will locally look like a mixed state even though the global state is a pure one. Assume, then, that you have some other two-level system, with states $|\!\uparrow⟩$ and $|\!\downarrow⟩$, and that your preparation procedure is able to generate $|1⟩\otimes|\!\uparrow⟩$ or $|2⟩\otimes|\!\downarrow⟩$, or arbitrary pure-state superpositions of those two. One simple way to do this is to spin-polarize the plane waves of your proposal.
For a system like this, any experiment that is local to the $\{|1⟩,|2⟩\}$ half of the Hilbert space will be sensitive to the partial trace over $\{|\!\uparrow⟩,|\!\downarrow⟩\}$, and this will be exactly the mixed state $\sigma$. To observe the pure state, you need an entangling measurement between the two degrees of freedom.
It is important to note that this third method is actually a generalization of the previous two, and it is unclear whether it is the only way to generate mixed states or whether 'really' mixed states are possible in nature. From a many-worldsian perspective, you can substitute the spin degree of freedom for a more macroscopic system, like
$$
|\!\uparrow⟩=|\text{Ruslan measures a quantum coin to give heads and runs experiment}⟩,
$$
$$
|\!\downarrow⟩=|\text{Ruslan measures a quantum coin to give tails and runs experiment}⟩,
$$
and you get identical predictions for your experimental outcomes.
A: If you can find and measure an observable that has $\vert \psi \rangle = \vert 1 \rangle  + \vert 2 \rangle$ as an eigenstate, you could measure the value of this operator on random electrons and select those that are in the desired linear combination of states. Otherwise, I think you cannot fabricate a pure state.
