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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?

Thank you in advance for your help!

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    $\begingroup$ 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 $|\Psi\rangle\langle \Psi|$. $\endgroup$ Oct 6, 2023 at 18:00
  • $\begingroup$ 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). $\endgroup$
    – pcalc
    Oct 7, 2023 at 17:19
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    $\begingroup$ 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 $\endgroup$ Oct 7, 2023 at 22:00

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For applications such as BB84, it really doesn't matter which of the two descriptions you use, because from everybody's perspective apart from Alice, they do not have access to the first system, and the best description they can have of what is travelling to Bob is that of tracing out Alice's system. In either description, that partial trace gives the same answer.

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.

Why might you go the other route and use a pure state? Well, one possible reason is that you need a source of randomness. Particularly for a crypto application, you want a good source of randomness. There is no true randomness in classical probabilities. You have to generate it from quantum by, for example, preparing a system in superposition and measuring it. But if you measure and then do something based off the measurement, you could just as easily do a controlled-unitary based off the un-measured system. Another reason is just for the purpose of using a pure state for as long as possible, which is a much more concise description than a density matrix.

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  • $\begingroup$ Thanks for your answer @DaftWullie ! Ok, I see that from outside Alice's lab it looks the same. Nevertheless, it seems to make a big difference when we look at the security analyses (I think the argument is called "source-replacement scheme", where the P&M version is then turned into an entanglement-based version of the same protocol) as then the structure of Alices share of the state (so the trace over the Bob system) becomes relevant in the argument. As the matrices then do not agree at all, I would expect different key rates. Thus, it seems to me like there is a "right" way to look at it(?) $\endgroup$
    – pcalc
    Oct 7, 2023 at 17:28
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    $\begingroup$ then you use whichever way gives you the easiest analysis/best parameters ;-) $\endgroup$
    – DaftWullie
    Oct 9, 2023 at 6:22

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