Is acting with a positive map on a state not part of a larger system allowed? - Quantum Computing Stack Exchange most recent 30 from quantumcomputing.stackexchange.com 2020-01-28T19:56:28Z https://quantumcomputing.stackexchange.com/feeds/question/2058 https://creativecommons.org/licenses/by-sa/4.0/rdf https://quantumcomputing.stackexchange.com/q/2058 11 Is acting with a positive map on a state not part of a larger system allowed? Quantum spaghettification https://quantumcomputing.stackexchange.com/users/2015 2018-05-15T07:33:05Z 2019-10-23T20:09:54Z <p>In the comments to a <a href="https://quantumcomputing.stackexchange.com/questions/2047/two-notions-of-general-quantum-operators">question</a> I asked recently, there is a discussion between <a href="https://quantumcomputing.stackexchange.com/users/2293/user1271772">user1271772</a> and myself on positive operators. </p> <p>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.</p> <p>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.</p> <p>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)?</p> https://quantumcomputing.stackexchange.com/questions/2058/-/2064#2064 5 Answer by Niel de Beaudrap for Is acting with a positive map on a state not part of a larger system allowed? Niel de Beaudrap https://quantumcomputing.stackexchange.com/users/124 2018-05-15T10:49:44Z 2018-05-17T09:20:46Z <p>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. </p> <p>The constraint of maps being CPTP comes from the physics itself. Physical transformations on closed systems are unitary, as a result of the Schr&ouml;dinger 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 &mdash; 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 &mdash; but whether it is useful to do so in general is not clear to me. </p> <p>On a higher level &mdash; while <em>we</em> 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 <em>we</em> 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 &mdash; and in particular not possible on product states &mdash; 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. </p> https://quantumcomputing.stackexchange.com/questions/2058/-/2066#2066 -3 Answer by user1271772 for Is acting with a positive map on a state not part of a larger system allowed? user1271772 https://quantumcomputing.stackexchange.com/users/2293 2018-05-15T19:01:26Z 2018-05-15T23:57:22Z <p><strong>No law of physics states that we must be able to evolve a sub-system of the universe on its own.</strong> </p> <p>There would be no way to definitively test such a law. </p> <hr> <p>The density matrix of the universe must have a trace of 1 and be positive semi-definite, by the mathematical definition of probabilities<sup><strong><a href="https://arxiv.org/pdf/0811.2068.pdf" rel="nofollow noreferrer">1</a></strong></sup>. Any change in the universe must<sup><strong><a href="https://arxiv.org/pdf/0811.2068.pdf" rel="nofollow noreferrer">1</a></strong></sup> 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 $\rho_{\rm{universe}}&lt;0$, what you have is not actually a density matrix, by the definition of probability<sup><strong><a href="https://arxiv.org/pdf/0811.2068.pdf" rel="nofollow noreferrer">1</a></strong></sup>.</p> <p>So the map: $\rho_{\rm{universe}}(0)\rightarrow\rho_{\rm{universe}}(t)$ must<sup><strong><a href="https://arxiv.org/pdf/0811.2068.pdf" rel="nofollow noreferrer">1</a></strong></sup> be positive and trace-preserving. </p> <p>For convenience, we like to <strong><em>model</em></strong> sub-regions of the universe, and introduce complete positivity for that. But one day an experiment might come along that we find impossible to explain<sup><strong>2</strong></sup>, perhaps because we have chosen to model the universe in a way that's not compatible with how the universe actually works.</p> <p>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 <strong>positive</strong> 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 <strong>only</strong> a sub-system of $\rho_{\rm{universe}}$, using a CPT map, is also something we believe will work, but we might bet <strong><em>slightly</em></strong> less on this, because we've added the <em>assumption</em> that <strong><em>sub-systems</em></strong> evolve this way, not just the universe as a whole.</p> <p><hr> <strong><a href="https://arxiv.org/pdf/0811.2068.pdf" rel="nofollow noreferrer">1</a></strong>: 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 <a href="https://arxiv.org/pdf/0811.2068.pdf" rel="nofollow noreferrer">this</a> would not be zero. To test if Born's rule is true for a <em>particular</em> 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. </p> <p><strong>2</strong>: 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.</p> https://quantumcomputing.stackexchange.com/questions/2058/-/2357#2357 1 Answer by Aharon Brodutch for Is acting with a positive map on a state not part of a larger system allowed? Aharon Brodutch https://quantumcomputing.stackexchange.com/users/1864 2018-06-15T12:39:07Z 2018-06-15T12:39:07Z <p>The situation of non-completely positive maps (or more generally non-linear maps) is controversial partly due to the precise definition of <a href="https://arxiv.org/abs/1212.4387" rel="nofollow noreferrer">how you should construct the map</a>. But it is easy to come up with an example of something that would seem to be NCP or even not linear. </p> <ol> <li>Non linear map. </li> </ol> <p>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. </p> <p>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. </p> <ol start="2"> <li>NCP</li> </ol> <p>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.</p> <p>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. </p>