Is acting with a positive map on a state not part of a larger system allowed?

Question

In the comments to a question I asked recently, there is a discussion between user1271772 and myself on positive operators.

I know that for a positive trace-preserving operator $\Lambda$ (e.g. the partial transpose) if acting on a mixed state $\rho$ then although $\Lambda(\rho)$ is a valid density matrix it mucks up the density matrix of the system it is entangled to - hence this is not a valid operator.

This and user1271772's comments, however, got me thinking. $\Lambda$ acting on a state which is not part of a larger system does indeed give a valid density matrix and there is no associated entangled system to muck it up.

My question is, therefore: Is such an operation allowed (i.e. the action of a positive map on a state which is not part of a larger system). If not, why not? And if so, is it true that any positive map can be extended to a completely positive map (perhaps nontrivially)?

Regarding the last sentence of the question, it may be helpful to note that any linear map $\Lambda$ from square matrices to square matrices, irrespective of being positive or completely positive, is uniquely determined by its action on pure state density matrices (simply because the pure state density matrices span the space of all matrices). So, there is no way to "extend" such a map to make it completely positive without changing its action on pure states. — John Watrous, May 15 '18 at 12:05
Why would the partial transpose acting on a pure state give a valid density matrix? Or do you just mean "acting on a state which is not part of a larger system"? (The former doesn't seem to make sense - any map will be "more positive" on mixed states than on pure states. The latter is simply called a "positive map".) — Norbert Schuch, May 15 '18 at 16:41
@NorbertSchuch I do mean "acting on a state which is not part of a larger system" - is this not one and the same as a pure state? — Quantum spaghettification, May 15 '18 at 16:48
@Quantumspaghettification No. (Well, it is a bit a matter of belief, but the way it is phrased it is highly misleading with regard to the usual language. I had to read it several times to guess what you mean. I would suggest to rephrase it accordingly. — Norbert Schuch, May 15 '18 at 18:00
@NorbertSchuch I have edited a post to remove the word 'pure state'. May I ask how you would define a 'pure state'? — Quantum spaghettification, May 15 '18 at 19:10
@Quantumspaghettification $\rho=|\psi\rangle\langle\psi|$: A pure state. Otherwise (i.e., the rank of $\rho$ is $>1$): mixed state. On either of them, the transpose yields a positive $\Lambda(\rho)$. Only if we apply $\Lambda\otimes I$ to a larger state (be it pure or mixed), we obtain a non-postive state. — Norbert Schuch, May 15 '18 at 20:22
@Quantumspaghettification: The paragraph of Eq. 6 and the paragraph after it explains that open system dynamics does not have to be completely positive, or even positive. The next paragraph makes sense of the fact that the dynamics in the equation leads to a non-positive density matrix. The conclusion explains why your thought-experiment on "a system entangled to a witness" is flawed: https://www.sciencedirect.com/science/article/pii/S0375960105005748. — user1271772 No more free time, May 17 '18 at 00:51
If a not completely positive map was physically meaningful, it would mean that we could build a device, a "black box", which produces meaningful output states for some kinds of inputs, but nonsensical results (non-positive states, that is, not states) for other inputs. That doesn't make much sense from a physical point of view, at least not within the framework of QM — glS, May 17 '18 at 14:13
the confusion for me is the concept "a state which is not part of a larger system". Unless there is some law of physics that prevents the small system from interacting with the rest of the Universe, there is the theoretical possibility that the two parts could be entangled, and that's what forces the map to be completely positive. And if it's not interacting with the rest of the Universe, note that you can't look at the state, interact with it, or do anything at all with it. Or perhaps I've misunderstood what you're after? — DaftWullie, Jun 15 '18 at 12:51

Niel de Beaudrap · Answer 1 · 2018-05-17T09:20:46.440

8

Any map which is not Completely Positive, Trace Preserving (CPTP), is not possible as an "allowed operation" (a more-or-less complete account of how some system transforms) in quantum mechanics, regardless of what states it is meant to act upon.

The constraint of maps being CPTP comes from the physics itself. Physical transformations on closed systems are unitary, as a result of the Schrödinger equation. If we allow for the possibility to introduce auxiliary systems, or to ignore/lose auxiliary systems, we obtain a more general CPTP map, expressed in terms of a Stinespring dilation. Beyond this, we must consider maps which may occur only with a significant probability of failure (as with postselection). This is perhaps one way of describing an "extension" for non-CPTP maps to CPTP maps — engineering it so that it can be described as a provocative thing with some probability, and something uninteresting with possibly greater probability; or at least a mixture of a non-CPTP map with something else to yield a total evolution which is CPTP — but whether it is useful to do so in general is not clear to me.

On a higher level — while we may consider entanglement a strange phenomenon, and in some way special to quantum mechanics, the laws of quantum mechanics themselves make no distinctions between entangled states and product states. There is no sense in which quantum mechanics is delicate or sensitive to the mere presence of nonlocal correlations (which are correlations in things which we are concerned with), which would render impossible some transformation on entangled states merely because it might produce an embarrassing result. Either a process is impossible — and in particular not possible on product states — or it is possible, and any embarrassment about the outcome for entangled states is our own, on account of the difficulty in understanding what has happened. What is special about entanglement is the way it challenges our classically-motivated preconceptions, not how entangled states themselves evolve in time.

edited May 17 '18 at 09:20

answered May 15 '18 at 10:49

Niel de Beaudrap

12,052
1
31
69

What physics law requires that subsystems of the universe must evolve this way? If we only assume that the universe evolves according to the Schroedinger equation, can we prove that all subsystems must evolve in a CPTP way? I have never seen such a proof, and others agree: https://www.sciencedirect.com/science/article/pii/S0375960105005748. I asked the question here: https://quantumcomputing.stackexchange.com/questions/2073/only-assuming-the-universe-evolves-according-to-a-positive-trace-preserving-map. – user1271772 No more free time May 17 '18 at 00:28
After more reading, I have found a counter-example to your claim that dynamics must be CPTP. When the initial density matrix is given by Eq. 6 of https://www.sciencedirect.com/science/article/pii/S0375960105005748, and the Hamiltonian is given in that same paragraph, $e^{-iHt}\rho e^{iHt}$ leads to a "total" density matrix where the subsystem density matrix is not even positive. The key idea is that the system and its bath are entangled even at time $t=0$. I believe you have to assume no entanglement between system and bath at $t=0$ in order to force CPTP in Choi's way or Alicki's way. – user1271772 No more free time May 17 '18 at 00:56
@user1261772: if you are not allowed to assume no entanglement between system and bath, then in what respect is it even meaningful to consider a map on the system alone? The pre-existing entanglement makes a nonsense of the idea that we're even trying to provide a "more-or-less complete account" of how the system evolves. And --- finally --- if the subsystem operator is not even positive, how on earth do we interpret the possibility of obtaining negative probabilities (or supernormalised probabilities) of some of the eigenstates? – Niel de Beaudrap May 17 '18 at 07:53
1

"his is perhaps one way of describing an "extension" for non-CPTP maps to CPTP maps — engineering it so that it can be described as a provocative thing with some probability, and something uninteresting with possibly greater probability" -- do you have any example for that? It seems to me that this would with some probability produce an output which is non-positive, which cannot be. – Norbert Schuch May 17 '18 at 08:24
@Neil: I never said you are not allowed to assume no entanglement between system and bath. The paper said that the arguments made for CPTP maps by Choi and Alicki both assumed no initial correlation, then gave an example of how an OQS that is initially correlated with its bath, can have non-positive evolution when the system+bath are evolved using $e^{-iHt}\rho e^{iHt}$ and then the bath is traced out. You say that the pre-entanglement idea is "nonsense", but if you search "initial correlations" you will find a huge body of literature on OQSs that are initially correlated with their baths. – user1271772 No more free time May 17 '18 at 08:24
@NieldeBeaudrap: About the negative probabilities. I agree with you on that one, though I don't think you had to go so far as to say "how on earth..." Can't we be a little nicer? Believe it or not, you were my T.A. in Waterloo more than a decade ago. In the original comments that started this question, I was arguing that the map has to be positive (but not completely positive). But the paper I found today gave an example where $e^{-iHt}\rho e^{iHt}$ on the system+bath gives non-positive evolution for the system. The example is simple (just 4x4 matrix) so maybe it's better to find something – user1271772 No more free time May 17 '18 at 08:31
wrong with it rather than asking how on earth we interpret it. I honestly don't know either, since I also thought the map would have to be positive (but not necessarily completely positive). I must be mis-understanding the paper, but it seems to say quite clearly that the map doesn't have to be CPTP (or even positive). – user1271772 No more free time May 17 '18 at 08:35
@Norbert: there are techniques for realising Kraus operators which are themselves linear combinations of unitary operators, while not being unitary themselves. They're used in improvements to the precision of HHL. But they involve postselection, which of means that (a) they can't be performed deterministically, and (b) in particular we rely on special properties of the state we apply them to, in order to realise them with high probability. Which is to say that they are only partial and effective descriptions of something which is actually CPTP. – Niel de Beaudrap May 17 '18 at 08:46
@NorbertSchuh: as regards non-positivity, I certainly wouldn't expect that we could under any conditions realise a transformation which realises a negative (or indefinite) marginal. However, we could consider a map which looks like a convex combination of the partial transpose, and a completely depolarising map; or "universal negation" and a completely depolarising map. – Niel de Beaudrap May 17 '18 at 08:53
@user1271772: I never said that entanglement with the bath is nonsense (this is obviously not a reasonable thing to say): I said that the presence of this entanglement indicates that what comes after, and is described as an evolution of the system, is not a complete account of that evolution. What about the part that came before which brought about that entanglement in the first place? – Niel de Beaudrap May 17 '18 at 08:59
@user1271772: I mean no offense by saying "how on earth" --- I mean to emphasise the importance of the claim, specifically: one thing wrong with the claim is that the states it admits as output lie completely outside of the mathematical formalism of quantum theory. It is as bad as predicting a negative probability of rain in the weather report: it doesn't have a sensible interpretation in terms of what outcomes you can obtain. It does suggest that there is an analytical misstep, which I will attempt to pinpoint when I can get around/through the paywall for the article. – Niel de Beaudrap May 17 '18 at 09:06
@Niel My only point was that it is impossible to realize sth. non-CP even probabilistically. As far as I understand, we agree on this point. – Norbert Schuch May 17 '18 at 09:14
@user1271772:$\def\ket#1{\lvert#1\rangle}\def\bra#1{\langle#1\rvert}$ In that paper, around page 52--53, they consider $\rho_S = \tfrac{1}{2}(1 + a_1 \sigma_1) = \tfrac{1+a_1}{2} \ket+!\bra+ + \tfrac{1-a_1}{2} \ket-!\bra-$, meant to be the marginal of $\rho_{ST}$ depending on parameters $a_j$, $b_k$, and $c_{j,k}$ as in Eqn.(6). They note an evolution yielding a state with a negative eigenvalue when $a=-1$ and $c_{13} = 1$. But for $a=-1$, $\rho_S = \ket-!\bra-$, which is pure. Then $\rho_{ST}$ must not be a positive operator: they have used an invalid initial condition. That's the problem. – Niel de Beaudrap May 17 '18 at 10:43
@NieldeBeaudrap: (1) If all particles in the universe began in an entangled state, then why do we need "evolution" in order to have initial entanglement? Entanglement between system and bath at $t=0$ where $t=0$ is the big bang, does not seem to violate any laws of physics or mathematics. (2) I think the point about negative probabilities could have bee made exactly the same, but without the "how on earth" ... I was also surprised when they said the map was not even positive. – user1271772 No more free time May 17 '18 at 18:16
@NieldeBeaudrap: (3) Sorry the article had a paywall, I couldn't find an arXiv version. Now it is me that suffers from the paywall since I'm at home and I want to look at the paper again to understand your critique of it. What is the invalid initial condition? Is it that they start off by saying $\rho_S(t)$ is not pure, but their argument for the map being non-positive relies on it being pure, meaning there's a contradiction? – user1271772 No more free time May 17 '18 at 18:19
@Niel The sentence where they said $a_1$=-1 was preceded by a statement saying "The new state is not positive for ALL possible choices of $a_1$ and $c_{12}$. For instance if $a_1$=-1...." The example that you refute, was just an example they gave, but before that they said that for ALL possible values of $a_1$, the density matrix has negative eigenvalues. Honestly this paper isn′t written in a language I′m very familiar with, I can assume that $\sigma_1$ means means X, but I had to go much further back to figure out that $\tau$ is a Pauli matrix too. I do have respect for Sudarshan though. – user1271772 No more free time May 18 '18 at 01:02
@user1271772: The point of contention is not if all particles in the universe start in an in entangled state: it is whether, when you wish to perform a map, the system A on which you wish to act is entangled with the bath B in precisely the way which matters for describing the evolution as CPTP, at the moment at which we wish to consider the map. Presuming that the entanglement is at all likely to be present in a way that matters, is tantamount to supposing that not only is "a map on a state" not necessarily CPTP, it is not even well-defined for practical purposes. – Niel de Beaudrap May 21 '18 at 14:29
@user1271772: I cannot confirm precisely what operator describes their initial condition: Eqn. (6) reads $R=\tfrac{1}{4}(1+a_j\sigma_j +b_k\tau_k +c_{jk}\sigma_j\tau_k)$, with some unusual summation convention, where $\sigma_j,\tau_k$ are presumably spin-1/2 Pauli operators. The point is that they presume that, for any marginal on the input system, the bath somehow conspires to 'purify' it in precisely this way (but why?), requiring in some cases that the joint state have negative eigenvalues (meaning what?), without clearly explaining either idea. I do not find this to be meaningful physics. – Niel de Beaudrap May 21 '18 at 14:41
@NieldeBeaudrap: $\sigma_i$ and $tau_i$ are indeed defined as Pauli matrices earlier in the paper, and I mentioned this in a comment on the other question. I don't know what you mean by the bath conspiring to purify the system, but with their Hamiltonian and initial state, they claim that unitary evolution on the system+bath (in this case just 2 qubits) results in negative eigenvalues for the system density matrix. Maybe there's an arithmetic error, but no one has been able to pinpoint it. The author Sudarshan published the 2nd most cited paper of all time in CPTP literature, maybe he made – user1271772 No more free time May 21 '18 at 17:46
a mistake in this paper, but there seems to be no proof that a subsystem of the universe must evolve in a CPTP way. If there is one, I'd be keen to see an answer here: https://quantumcomputing.stackexchange.com/questions/2073/only-assuming-the-universe-evolves-according-to-a-positive-trace-preserving-map – user1271772 No more free time May 21 '18 at 17:47
@user1271772: Here is what I mean by the bath "conspiring to purify the system". First, I will concede that 'purify' is not a good choice of word, as it seems likely that $R$ will very often not be a rank 1 operator --- but the point is the following. If the evolution is to be considered "a map" (CPTP or otherwise) on the system S, the input state $\rho_S$ should be a free choice. The system+bath state $R$ should also be a valid quantum state, in which the marginal state of the bath does not depend on the marginal on S; or else in what sense is this "a (single) map" to be applied to states? – Niel de Beaudrap May 21 '18 at 18:16
@NieldeBeaudrap: I agree that the system+bath state must be a valid quantum state. Now if $\rho_{\textrm{total}}(t=0)$ is valid, then $e^{-iHt}\rho_{\textrm{total}}(t=0)e^{iHt}$ is valid. I see no proof that $\textrm{Tr}B\rho{\textrm{total}}$ has to evolve according to a CPTP map though, just that closed systems have to be PTP, and the only closed system is the universe, so I see no proof that a subsystem of the universe must have CPTP evolution. – user1271772 No more free time May 21 '18 at 19:29
@NieldeBeaudrap: Finally it seems someone has come up with an explanation for the non-positive map seen for the case where the system and bath are initially correlated: https://arxiv.org/pdf/1806.02381.pdf, though for me it is still too early to tell if this refute's the 2005 paper that we discussed. – user1271772 No more free time Jun 12 '18 at 20:15
@user1271772: from the abstract -- "we argue that the correct definition of the evolution map is obtained by considering a counterfactual scenario wherein the system is reprepared independently of any systems in its causal past while the rest of the circuit remains the same" sounds a lot like what I was arguing; "we highlight two distinct mistakes that retrospectively become evident [...] (ii) in a (retrospectively unnecessary) attempt to introduce variability in the input state, it inadvertently introduced variability in the inference map itself" is not terribly surprising. – Niel de Beaudrap Jun 12 '18 at 20:32
@NieldeBeaudrap: Thanks for pointing that out. "scenario wherein the system is reprepared independently of any systems in its causal past " seems to contradict the definition of "initially correlated" though -- I think I have to read the Spekkens paper in more detail. – user1271772 No more free time Jun 12 '18 at 20:35
@user1271772: My point was that if the auxiliary system was correlated to the input, then in particular either the input is not truly free to choice, or the idea that the auxiliary system is in any way actually "auxiliary" is not very sensible. In either case I would wonder why we artificially ignore the past interaction of the input and auxiliary systems (their 'common cause' in the language of [arXiv:1806.02381]). By not doing so, the evolution is like a poorly programmed procedure which has significant side-effects --- ie., not a 'function' or 'map' in the mathematician's sense. – Niel de Beaudrap Jun 12 '18 at 20:40

score 0 · Answer 2 · answered Jun 15 '18 at 12:39

The situation of non-completely positive maps (or more generally non-linear maps) is controversial partly due to the precise definition of how you should construct the map. But it is easy to come up with an example of something that would seem to be NCP or even not linear.

Non linear map.

Consider a preparation device that can create a qubit in an arbitrary state $\rho$ (this device has 3 dials). Now let this device be constructed so that it also prepares a second state $\rho$ in the environment. I.e, you think you prepared a one qubit state $\rho$ but actually you prepared a two qubit state $\rho\otimes\rho$. The second qubit is the environment (which you cannot access), so if you perform tomography on your qubit, everything seems ok.

No imagine that you also have the following black box - it has (as far as you can tell) one input and two outputs. In reality (unknown to you) it has two inputs and two outputs and it simply spits out both the system qubit and the environement qubit. As far as you can tell, this black box is a cloning machine, violating linearity.

NCP

Similar to the idea above, but the preparation device prepares $\rho\otimes\rho^T$ (clearly this could be done in the lab). The black box will now be a one rail box (one qubit input one qubit output as far as the user is concerned), which swaps the system and environement. To you, it seems like a trasposition map.

Note that both preparation devices are physical, but the way you construct the map might depend on how you use them. In the example above I assumed that a mixed state $\rho$ would only be constructed by using the three dials in the machine. In principle, I could try to construct a mixed state by flipping coins and preparing pure states with the right probability. Tomorgraphy would show that the processes are equivalent, but the environment would be different, and the map you would construct for the black boxes would be different.

user1271772 No more free time · Answer 3 · 2018-05-15T23:57:22.280

No law of physics states that we must be able to evolve a sub-system of the universe on its own.

There would be no way to definitively test such a law.

The density matrix of the universe must have a trace of 1 and be positive semi-definite, by the mathematical definition of probabilities¹. Any change in the universe must¹ preserve this, for mathematical reasons and due to definitions. If $\rm{Tr}(\rho_{\rm{universe}})\lt1$, you just haven't included the whole universe in $\rho_{\rm{universe}}$. If it's more than 1, or if $\rho_{\rm{universe}}<0$, what you have is not actually a density matrix, by the definition of probability¹.

So the map: $\rho_{\rm{universe}}(0)\rightarrow\rho_{\rm{universe}}(t)$ must¹ be positive and trace-preserving.

For convenience, we like to model sub-regions of the universe, and introduce complete positivity for that. But one day an experiment might come along that we find impossible to explain², perhaps because we have chosen to model the universe in a way that's not compatible with how the universe actually works.

If we assume gravity doesn't exist, and we can magically compute anything we want, we believe that evolving $\rho_{\rm{universe}}$ using the right positive trace-preserving map, then doing a partial trace over all parts of the universe not of concern, will give accurate predictions. Introducing the notion of modeling only a sub-system of $\rho_{\rm{universe}}$, using a CPT map, is also something we believe will work, but we might bet slightly less on this, because we've added the assumption that sub-systems evolve this way, not just the universe as a whole.

1: Even this is debatable because the relationship between a wavefunction or density matrix and probabilities comes from a postulate of quantum mechanics called the Born rule, which until fewer than 10 years ago was never tested at all, and still has only been confirmed to be true within an $\epsilon$, and for a particular system: If Born's rule isn't true, Eq. 6 of this would not be zero. To test if Born's rule is true for a particular system (in this case, photons coming from some particular source), you would have to do an infinite number of instances, of all 7 of these experiments, or come up with a different way to test Born's rule (and I don't know of any). In 2009 we published this saying that Born's rule was true (for this system) to within an $\epsilon$ that was smaller than the experimental uncertainty, so we only know Born's rule is true for this system, and to within a precision limited by the experiment.

2: This is actually already the case, but let's pretend that gravity does not exist and that quantum mechanics (QED+QFD+QCD) is correct, and we still find it impossible to explain something, despite having (somehow) magical computer power to compute anything we want instantly.

You're bringing up field theories, and there the notion of traces is much more subtle. But it was unnecessary for the question. No need to say anything like $Tr \rho_{universe}$ — AHusain, May 16 '18 at 06:35
@AHusain: The question was about trace-preserving maps, which involves the trace. The question was directed at me. Let me decide how I would like to answer the question. — user1271772 No more free time, May 16 '18 at 19:16
Just wanted to point out that finite and infinite dimensional Hilbert spaces have some substantial differences. States on different sorts of VonNeumann algebras. That is all. — AHusain, May 16 '18 at 20:52
@AHusain: Okay. The Hilbert space of a single particle can be uncountably infinite dimensional too, so these substantial differences don't just occur for $\rho_{\rm{universe}}$. Anyway the point I was trying to make in my answer was that quantum mechanics (QED+QFD+QCD) requires that $\rho_{\rm{universe}}$ evolves in a way that preserves trace and positivity (assuming the Born's rule axiom to be true). Does this mean all subsystems of the universe need to evolve by a CPT map? I have never seen a proof of this. — user1271772 No more free time, May 16 '18 at 23:58
If you're going to downvote an answer that took a whole morning (maybe 3-4 hours?) to write and format, would it not be fair to explain what you didn't like about it? — user1271772 No more free time, May 17 '18 at 08:43

Is acting with a positive map on a state not part of a larger system allowed?

3 Answers3

Linked