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Based on my limited and poor understanding, a quantum circuit in the "one clean qubit" model of quantum computation generally acts on a single pure (clean) qubit tensored with $n-1$ qubits in a maximally mixed state. The clean qubit can serve as a control to apply an arbitrary unitary $U$ to the maximally mixed qubits, and measuring this clean qubit in the $X$ or $Y$ basis enables characterizations of some properties of $U$ (such as its real and imaginary trace, which is likely classically difficult.)

I was inspired to consider the one clean qubit model of computation by a recent (2021) video of Yoganathan on her "Looking Glass Universe" channel, wherein she describes and summarizes the approach she and her colleagues took to answering an open problem on the power of such a model in the absence of entanglement. The headline is that the one clean qubit model without entanglement is classically tractable.

But as I begin to study their paper, it's not quite clear to me what's even meant by the one clean qubit model with and without entanglement.

What, in particular, is and is not entangled in the one clean qubit model?

The pure qubit would be entangled with mixed qubits after application of $U$, right? Or am I wrong, and even after the controlled application of $U$ the clean qubit need not be entangled with/is separable from the mixed qubits? If so, what is entangled in the simple circuit in the Wikipedia article?

EDIT: I am wrong! While nonetheless carrying in its amplitude some properties of $U$ (such as the real/imaginary trace), a (bi)partition between the clean qubit and the other qubits after application of $U$ has no entanglement, no matter what $U$ is. When one clean qubit is able to calculate the trace of $U$ efficiently, the the one clean qubit may have entanglement in other partitions wherein [the clean qubit $\otimes$ some other mixed qubits] is entangled with the remaining mixed qubits, but there is no entanglement in the "obvious" partition! This is what @DaftWullie has emphasized.

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This is a slightly messy topic which I often find to be misinterpreted.

I'm not sure you can really point at a specific bit of the system as say "the entanglement is here".

If you took the bipartition of (originally clean qubit) vs (everything else), which would seem like the obvious place to look, you will find that there is no entanglement according to that bipartition. Of course, that doesn't mean that there isn't entanglement anywhere else (although this statement has certainly been read in that way. I certainly did the first time I came across it). If you consider basically any other bipartition of the system, you will detect some entanglement across it (depending on the $U$). So, there can be some (relatively small amount of) entanglement everywhere throughout the system. I did a calculation of this at some point: https://arxiv.org/abs/1508.06474

I don't know if you could break the analysis down into finer grained details to localise what's going on a bit better. I suspect that would depend too heavily on the $U$ that you're tracing. Equally, you might ask about multi-partite entanglement. Bipartite certainly isn't the only option, just the easiest to calculate!

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  • $\begingroup$ Oh! So if you have, say $n=5$ qubits and, after application of $U$, choose to have Alice keep the pure qubit but let Bob take the $4$ remaining qubits, they can go their separate ways, nothing to it (although Alice can learn something classically difficult about $U$ such as its trace). But if you give $2$ of the $5$ to Alice and have Bob keep $3$ then there could be a (vanishingly small) amount of entanglement, depending on $U$. $\endgroup$
    – Mark Spinelli
    May 12, 2021 at 13:33
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    $\begingroup$ Yes. Of course, this scenario has to be repeated many times for Alice to actually learn something useful. You wouldn't even say the entanglement is vanishingly small at that point, there's a finite amount of it (depending on $U$). $\endgroup$
    – DaftWullie
    May 12, 2021 at 14:27
  • $\begingroup$ I see now upon re-watching her vid. The 1CQ model is interesting precisely because the obvious bipartition doesn't create entanglement, although the controlled $U$ might create other entanglement. Yoganathan and Cade showed that if $U$ doesn't create any entanglement in any other bipartition, then the circuit is classically efficient to simulate. Thus if $U$ is not classically efficient, the improvement in the 1CQ model could come from these other bipartitions. $\endgroup$
    – Mark Spinelli
    May 9, 2022 at 18:30

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