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This question is based on a scenario that is partly hypothetical and partly based on the experimental features of molecule-based quantum devices, which often present a quantum evolution and have some potential to be scalable, but are generally extremely challenging to characterize in detail (a relevant but not unique example is a series of works related to this electrical control of nuclear spin qubits in single molecules).

The scenario: Let us say we have a variety of black boxes, each of which is are able to process information. We don't control the quantum evolution of the boxes; in the language of the quantum circuit model, we do not control the sequence of quantum gates. We know each black box is hardwired to a different algorithm, or, more realistically, to a different time-dependent Hamiltonian, including some incoherent evolution. We don't know the details of each black box. In particular, we don't know whether their quantum dynamics are coherent enough to produce a useful implementation of a quantum algorithm (let us herein call this "quantumness"; the lower bound for this would be "it's distinguishable from a classical map"). To work with our black boxes towards this goal, we only know how to feed them classical inputs and obtain classical outputs. Let us here distinguish between two sub-scenarios:

  1. We cannot perform entanglement ourselves: we employ product states as inputs, and single qubit measurements on the outputs. However, we can choose the basis of our input preparation and of our measurements (at minimum, between two orthogonal bases).
  2. As above, but we cannot choose the bases and have to worked on some fixed, "natural" base.

The goal: to check, for a given black box, the quantumness of its dynamics. At least, for 2 or 3 qubits, as a proof-of-concept, and ideally also for larger input sizes.

The question: in this scenario, is there a series of correlation tests, in the style of Bell's inequalities, which can achieve this goal?

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    $\begingroup$ So is it coherence specifically that should be witnessed or not? In any case, maybe you will find this preprint interesting. $\endgroup$ – Kiro Apr 8 '18 at 13:44
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    $\begingroup$ Indeed, 1212.0194 by Nori et al seems very interesting, I'll check it with some further care. In any case, I edited the question trying to be more clear both in the goals and the conditions. $\endgroup$ – agaitaarino Apr 8 '18 at 14:17
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Let's assume that your black box processes classical inputs (i.e. a bit string) to classical outputs in a deterministic way, i.e. it defines a function $f:x\mapsto y$.

If you can only prepare and measure separable states in that basis, all you can determine is what that function $f$ is. Assuming that all the outputs are different, it could have been computed either by a reversible classical computation or a quantum computation, and you wouldn't be able to tell.

So, let's assume you can prepare product states and measure in two different bases, $X$ and $Z$ for the sake of argument. One thing that you could do (which may be hopelessly inefficient for all I know, but it's somewhere to start) is first determine the function $f(x)$ using the $Z$ basis. Then, for any pair of bit string $x_1$ and $x_2$ that differ in only one position, prepare the state $(\left|x_1\right\rangle\pm\left|x_2\right\rangle)/\sqrt{2}$. This is a product state, using the $Z$ basis on all but one site. Let's assume that the outputs $y_1=f(x_1)$ and $y_2f(x_2)$ differ on $k>0$ sites. (If $k=0$, the evolution wasn't coherent anyway.) For the bits where $y_1$ and $y_2$ are supposed to be equal, just measure them in the $Z$ basis to make sure you get what you expect to get. On the remaining $k$ sites, if the black box is coherent, you receive a GHZ state of $k$ qubits, $$ \frac{1}{\sqrt{2}}(\left|y_1\right\rangle\pm\left|y_2\right\rangle). $$ If it were completely incoherent, you'd get a rank two mixed state $$ \frac{1}{2}\left(\left|y_1\right\rangle\left\langle y_1\right|+\left|y_2\right\rangle\left\langle y_2\right|\right). $$ If $k=1$, you can distinguish these directly by measuring that qubit in the $X$ basis (repeating a few times to get statistics). For $k>1$ you have a few options. Either you can peform a Bell test ($k=2$) or equivalent for GHZ states (such as all versus nothing proofs), or apply an entanglement witness (there are some based on single-qubit observables). Alternatively, measure every qubit in the $X$ basis and record the outcomes. In the case of the entangled state, the last outcome should be entirely predictable based on the previous outcomes. For the mixed state, the answer will be completely unpredictable. If you want to make a more quantitative statement, you could use something like an entropy, $H(X|Y)$ where $X$ is the random variable describing the output of the last measurement, and $Y$ is the random variable describing the outcome of all the previous measurements.

One possible issue is that by testing only inputs with a single site prepared in the $X$ basis, there are a lot of options you're not testing, so I don't know whether testing all of these coherences is enough, or whether one ought to start analysing what happens if you prepare pairs of sites in the $X$ basis, and so on.

Of course, while this tells you something about how coherent the implementation of the black box is, whether or not that coherence contributes to the speed of operation of the black box is a completely different matter (for example, that's the sort of thing people want to know about with transport processes in photosynthetic bacteria, or even something like D-Wave).

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Why not input one half of a maximally entangled state as the input to the black box (so that half has the same dimension as the input dimension)? Then you could test your favourite measure, such as the purity, of the full output state. If the oracle corresponds to a unitary evolution, the purity is 1. The less coherent the smaller the purity. Incidentally, the output state describes the map that the black box implements, via the Choi-Jamiołkowski isomorphism.

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  • $\begingroup$ To the material you suggested I'd add for readability this intuitive explanation of Choi-Jamiolkowski isomorphism. I rewrote the question, motivated by your answer and by the other answer that was offered. In particular, I was assuming that one is not able to prepare an entangled state as input, and if I understand your suggestion correctly, this is a critical problem. $\endgroup$ – agaitaarino Apr 8 '18 at 14:06
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    $\begingroup$ I clearly hadn’t understood the classical input/output specification of your question on first reading. If you have a single fixed basis of input and output, I guess there’s very little you can do except to count the number of different answers you can get over all possible inputs (perhaps the distribution of answers would also be relevant). If the black box is coherent, mapping product states in that chosen basis into other product states in that basis, each output should be unique. But it also would be for a classical reversible computation, and I don’t see how you’d distinguish the two. $\endgroup$ – DaftWullie Apr 8 '18 at 15:31
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I'm not exactly sure what you mean by quantumness of your black box. So maybe there are some more sophisticated approaches (similar to the other answer you could use an entanglement witness to show that your black box is not entanglement breaking). However, in general you could perform quantum process tomography (see e.g. arXiv:quant-ph/9611013).

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  • $\begingroup$ Thanks! Likewise, let me check the paper with some care before getting back to you, but on a first vire this seems to be much closer to what I asked, since in arXiv:quant-ph/9611013 one can use a product state as initial state (more compatible with "We know how to feed them classical inputs"). $\endgroup$ – agaitaarino Apr 8 '18 at 12:25
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    $\begingroup$ @agaitaarino Apparently I did not pay much attention to that sentence. If by classical inputs and outputs you mean a single basis, than process tomography will not work. With this restriction there is no way to distinguish the box from a classical map. $\endgroup$ – M. Stern Apr 8 '18 at 13:15
  • $\begingroup$ I tried to improve the question to distinguish between the two sub-scenarios. If I understand your answer, for sub-scenario 1 the problem is solved (at least in quant-ph/9611013 for a particular case) while for sub-scenario 2 it is unsolvable. Is that correct? $\endgroup$ – agaitaarino Apr 8 '18 at 13:53

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