# How to correctly describe state preparation?

I came across a rather basic state-preperation description that I didn't fully understand. So far, I thought I got the basics of density matrices, but now I am confused. I hope you can help me finding out if I just misinterpret what's written or if there is indeed a problem in what I thought is true in QM.

The problem is motivated by the BB84 protocol in Quantum Communication, but the underlying question is even decoupled from that specific application. According to the protocol description of the prepare and measure variant of the protocol, Alice prepares one of the following states $$|\phi_1\rangle:=|H\rangle, |\phi_2\rangle:=|V\rangle, |\phi_3\rangle:=|D\rangle, |\phi_4\rangle:=|A\rangle$$ with equal probability and sends them to Bob, who performs measurements.

According to literature, the prepared state is $$|\Psi\rangle:= \frac{1}{2}\sum_{x=1}^{4} |x\rangle \otimes |\phi_x\rangle$$, consequently the density matrix of the prepared state $$\rho = |\Psi\rangle\langle\Psi| = \frac{1}{4}\sum_{x,y=1}^{4} |x\rangle\langle y|\otimes |\phi_x\rangle\langle\phi_y|$$.

Why do we mix state vectors and not density matrices, i.e., $$\tilde{\rho} := \frac{1}{4}\sum_{x=1}^{4} |x\rangle\langle x| \otimes |\phi_x\rangle\langle\phi_x|$$, which, at least to me, sounds equally "valid"?

What determines if I first take the sum and then build the density matrix of the obtained state or if I can first stick bra and ket together and then sum over it, seeing it as a classical mixture of quantum states? At the moment, both seem to be somehow valid processes, with drastically different results, of course.

What am I missing?

• Good catch - you are absolutely correct that these correspond to different preparations. If one flips two classical coins to choose what state to prepare, you get the classical mixture. If you prepare a superposition of four states, then you get the $|\Psi\rangle\langle \Psi|$. Oct 6, 2023 at 18:00
• Thanks for your answer @QuantumMechanic ! Ok, and what is it that is actually happening/being prepared in prepare&measure BB84? On the one hand, what is happening sounds like "flipping a quantum coin (i.e., generating true random numbers) and preparing the mixture. On the other hand, it feels like this cannot be what we need to obtain quantum correlations, i.e., we actually prepare a superposition of four states (at least that is what it looks like for an observer from outside). Oct 7, 2023 at 17:19
• For BB84 classical randomness is sufficient. Or better yet, Alice might send a string of bits with no randomness, but then the average state of any of those photons chosen at random will be a mixed state Oct 7, 2023 at 22:00

If you had asked me to write down the density matrix of what's going on, I would probably have written down the $$\tilde\rho$$ that you did.