# Positive maps on pure states?

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
• @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

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.

• 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: sciencedirect.com/science/article/pii/S0375960105005748. I asked the question here: quantumcomputing.stackexchange.com/questions/2073/…. – user1271772 May 17 '18 at 0:28
• 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 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 May 17 '18 at 0:56
• @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 7:53
• "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 8:24
• @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 May 17 '18 at 8:24

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.

1. 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.

1. 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.

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 6:35
• @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 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 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 May 17 '18 at 8:43