Hello, Mct mht, and
welcome to Wikipedia! Thank you for your contributions. I hope you like the place and decide to stay. Here are a few good links for newcomers:
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sign your name on talk pages using four tildes (~~~~); this will automatically produce your name and the date. If you need help, check out
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Oleg Alexandrov (
talk)
23:16, 2 April 2006 (UTC)reply
Math Information
Real-valued assumption in convergence theorems
You are removing the assumption that some measurable functions have real values from a lot of places. But then you can't talk about almost everywhere convergence, you need values at least in a metric space.
Do you have references when those theorems hold? I suggeest you go back to the real number-valued functions, and that's the classical case, and it is surely correct. Thanks, and you can reply here.
Oleg Alexandrov (
talk)
15:10, 12 April 2006 (UTC)reply
Yeah, in fact it is kind of weird that the previous contributors imposed the real-valued assumption. Surely something like the dominated convergence theorem, Egorov's, and Lusin's don't need that assumption. Should be in any reasonably modern introductory text on real analysis.
Mct mht19:22, 12 April 2006 (UTC)reply
Oleg, maybe I misunderstood your point. What I meant is the standard assumption, in this context, is that they're complex valued.
Mct mht19:26, 12 April 2006 (UTC)reply
But there is no "standard" assumption. A measurable function can theoretically take values in any measure space, as long as the inverse of a measurable set is measurable.
By the way, going to complex-valued function doesn't add much value in terms of generality, so I'd say we stick with real-valued functions. Most measure theory is done with such functions anyway.
Oleg Alexandrov (
talk)
23:09, 12 April 2006 (UTC)reply
In any case, considering metric spaces of metric-space valued functions on a measure space (X, M, μ) with an (extended) metric
A side note, apparently there are Lusin type results for Bochner integrals. Didn't know that.
Mct mht
Square root of a matrix
Hi. It seems you're still working on
square root of a matrix, so I won't edit that page myself at the moment. However, could you please bear in mind that the phrase "positive matrix" is often used to describe a matrix with positive entries? I gather that you're talking about a
positive-definite matrix. Great work, by the way. Cheers,
Jitse Niesen (
talk)
10:01, 17 August 2006 (UTC)reply
i just finished with that page. i appreciate the courtesy. re usage of positivity: maybe add a disclaimer somewhere?
Mct mht10:04, 17 August 2006 (UTC)reply
Yes, that's fine. I also came across
this edit which I don't understand: "a bounded operator on a complex Hilbert space is necessarily Hermitian, or self adjoint". Should there be an extra condition somewhere? Multiplication by i is not Hermitian, is it? --
Jitse Niesen (
talk)
10:12, 17 August 2006 (UTC)reply
not quite sure what you mean by mutiplication by i, Jitse. i think that edit is correct, no extra condition needed. for complex Hilbert spaces, if we have <x, Mx> ≥ 0 for all x, the polarization identity shows <x, My> = <Mx, y> for for x and y, so M is self adjoint.
Mct mht10:19, 17 August 2006 (UTC)reply
With "multiplication by i", I mean the operator which maps x to ix, where i is the square root of −1.
I think that instead of "a bounded operator on a complex Hilbert space is necessarily Hermitian, or self adjoint", you mean "a bounded operator M on a complex Hilbert space for which Re <x, Mx> ≥ 0 for all x, is necessarily Hermitian, or self adjoint"; that's the extra condition I'm talking about. --
Jitse Niesen (
talk)
10:53, 17 August 2006 (UTC)reply
damn, is that what i typed? lol, very sry, Jitse, saw it just now. it should be "a bounded positive operator on a complex Hilbert space is necessarily Hermitian, or self adjoint."
Mct mht11:08, 17 August 2006 (UTC)reply
Hello. You reversed a change I made to
vector space. Part of the reason why I was confident about that change was because the
algebra over a field article states that the bilinear multiplication operator must be such that (ax)(by)=(ab)(xy). However, now that I look into the formal definition of
bilinear, I do not see how this follows. That being said, I think that there are many important results (for example, results about commutable operators and their eigenspaces) that depend on this. It seems like there should be some stronger version of algebra that includes this property. I understand that in most cases this is a result of the commutivity of the field; however, strictly speaking the commutivity of the field should have nothing to do with ax=xa. Right? That is, a vector x need not have coordinates from the field. All I need with an algebra is some definition of scalar-vector multiplication; I can do that without having vector coordinates from a field. Thus, I do not see how commutivity of a field makes ax=xa true in general.
Thus, should
algebra over a field be changed? Or is there some stricter form of algebra that includes this special property. Any help?
Perhaps your point is that an algebra need not have this property; however, square matrix multiplication (and thus any linear operator) will naturally have this property in ADDITION to being an algebra. I'm willing to believe that. So if I'm asked to define an algebra, I should probably not state that (ax)(by)=(ab)(xy) without a note about the commutivity of fields. Right? --
TedPavlic20:42, 21 February 2007 (UTC)reply
hello. scalar multiplication is alway either left or right, strictly speaking. yes, among axioms of an algebra is that a(xy) = x(ay) where a is a scalar. what you added seems to mean that, if the algebra has a multiplicative identity I, then x(aI) = aI(x) gives some kind of justification to xa = ax. but not every algebra has an unit. consider the algebra of
compact operators on an infinite dimensional Hilbert space.
Mct mht01:15, 22 February 2007 (UTC)reply
I agree. Again, I was thinking of the (ax)(by)= (ab)(xy). However, I agree that stating that xa = ax is even stronger. That was definitely a big mistake on my part. --
TedPavlic10:57, 22 February 2007 (UTC)reply
also, the property (ax)(by)= (ab)(xy) you cited seems to me is part of the definition of an algebra. it is required, rather than a consequence of the underlying field be commutative.
Mct mht04:14, 22 February 2007 (UTC)reply
I would agree; however, I can find no credible reference that lists that as a property of an algebra except for the
algebra over a field article. That's why it seemed to me that it must be something special or a subtle consequence of one of the other more fundamental properties of an algebra. --
TedPavlic10:57, 22 February 2007 (UTC)reply
most linear algebra books should have something like that. e.g. the one by Roman. it's called Advanced Linear Algebra or something. but the property follows immediately from a(xy) = (ax)y = x(ay) (let's say we take this as part of the definition of an algebra), and field multiplication being commutative, as you said. the converse is also true. so i suppose to simply require (ax)(by)= (ab)(xy) in the definition, as the WP algebra article has done, is reasonable.
Mct mht12:31, 22 February 2007 (UTC)reply
Not to beat a dead horse here, but it looks like the reason why (ax)(by)= (ab)(xy) is not because the underlying field is commutative. It is because the vector multiplication defining xy must be bilinear. I didn't think that had enough meat, but if xy is linear in its second argument, then (x)( a y ) = a (x)(y). It's interesting to me that this also means that (a x)(b y)=a (x)(b y)=ba (x)(y) AND (a x)(b y)=b (a x)(y)=ab (x)(y). This implies that scalar ab=ba, and so it's very important that the underlying field is commutative. If it weren't, this would be a contradiction. However, I think the reason why (ax)(by)= (ab)(xy) is simply that vector multiplication is bilinear. --
TedPavlic01:20, 23 February 2007 (UTC)reply
I see you've pretty much written this article so I hope it's ok to ask you some stuff about it. Is the radical mentioned the same as
Jacobson radical (there are a lot of equivalent formulations of the Jacobson radical but none on the page obviously matched what you had as the definition). If it is the same can we use that definition for general (not necessarily finite dimensional) algebras. If it isn't the Jacobson radical I think we need to disambiguate because semisimple algebra is definitely used in this sense in Banach algebra theory. Thanks.
A Geek Tragedy21:13, 25 May 2007 (UTC)reply
hi, A Geek Tragedy, :-). i believe what's currently on the page is more or less valid: a finite dimensional semisimple algebra is a Cartesian product of simple (matrix) algebras. on the other hand, what you say is also right (semisimple <=> trivial Jacobson radical, yes?). in the finite dimensional case, IIRC they coincide. anyway, no objection from me if you wanna modify/add stuff, with the suggestion/request that discussion on the finite dimensional case in the current version be preserved, if possible. maybe Banach algebra considerations could have its own section.
Mct mht18:57, 26 May 2007 (UTC)reply
Cool, I'll get onto it. I don't think it needs to mention Banach algebras at this article (it was the purely algebraic definition I needed) just that it wasn't a different "semisimnple".
A Geek Tragedy21:21, 26 May 2007 (UTC)reply
Hi, I came across this article which you wrote in March. My familiarity with Wold's decomposition is from signal processing, where it is stated as saying that any
stochastic process can be represented as the sum of a deterministic process and a purely indeterministic process
[1]. Is this a special case of your article? If so, do you think it would be worthwhile to write this special case as a section in the article, so that people (like me) who know nothing about operator theory could still make some sense of it? Cheers,
Zvika18:45, 17 June 2007 (UTC)reply
hi, Zvika thanks for the comment. there's a (rather rich, IIRC) connection between operator theory and stochastic processes in general but i don't know all the details. at first glance the two Wold decompositions do seem similar (not just in the statement but i suspect in their proofs), if we think of "deterministic" as "unitary operator" and "regular" as "pure isometry". i am sorry that i cannot give a better answer at the moment.
Mct mht21:53, 17 June 2007 (UTC)reply
Zvika, i went back and looked at it again. i think the Wold decomposition for stochastic processes is indeed a special case.
Mct mht02:09, 19 June 2007 (UTC)reply
I still think it makes sense to have a section on the special case of stochastic processes, since right now the article is close to useless for people like me. Perhaps I will add this someday. A quick Google search seems to show that many (if not most) people interested in Wold decomposition really care about the stochastic process application. --
Zvika12:16, 19 June 2007 (UTC)reply
i ain't disagreeing that having a section on the special case of stochastic processes is a good idea, :-). to be more precise, it is a special case of the Wold decomposition for a countable family of isometries, indexed by Z. this result is alluded to in the stub section "A sequence of isometries". so a full implementation would probably mean filling out that stub, laying out some background for stochastic processes, and state's Wold's result in that context. perhaps we should advertise somewhere and get some knowledgable attention.
Mct mht01:57, 21 June 2007 (UTC)reply
Thanks for your message! I am not totally satisfied with my edit of the proof though. When I have time, I'll change it slightly, concentrating from the beginning on the orthogonal of the kernel. This would fit better with the statement of the theorem, as it stands now.
With best wishes, --
Bdmy (
talk)
17:58, 15 March 2009 (UTC)reply
The spectrum
Hi, in your edits for
Spectrum (functional analysis) a couple of years ago, you changed the section on bounded operators to say that bijectivity is not required for invertibility (you suggested that the surjectivity requirement could be dropped in the edit text). I feel fairly confident that this is not correct; my understanding of the true situation is as follows (note: T is assumed to be linear, which implies all other maps mentioned below are):
A point λ is said to be regular (i.e. not in the spectrum) if λ - T has a bounded inverse (i.e. a two-sided inverse that is a bounded linear map).
λ - T is bijective is a necessary condition for λ to be regular.
In the case that T is a closed (possibily unbounded) map, λ - T is bijective is also a sufficient condition for λ to be regular (so overall λ is regular iff λ - T bijective).
In the case that T is a bounded map, λ - T is bijective is again a sufficient condition for λ to be regular (so again, overall λ is regular iff λ - T bijective).
The only case where it is not sufficient is when T is an unbounded map that is not closed, when bijectivity of λ - T does imply that an inverse exists but not that it is necessarily bounded.
It seems that you disagree with my point (2) above, but only in the bounded case. I think it is right for the following reasons:
I gave a direct mathematical argument on the
talk page. Having done more research into your edits, I guess the dense range case is the one that is most relevant.
Your viewpoint is contradictory: bounded linear maps are just special cases of (possibily unbounded) closed linear maps, where you agree bijectivity is equivilent to regularity.
W/o looking at the article, the point you're making seems trivially true. Any algebraic map with a two-sided inverse is bijective; so that gives necessity. Open mapping theorem gives sufficiency. So you're completely correct.
Don't know what happened with the "surjectivity not necessary" thing. Certainly not right. An operator is invertible iff it's bounded below and has dense range. But then it's surjective. I very much appreciate your looking into this and putting me right.
Mct mht (
talk)
22:51, 20 June 2009 (UTC)reply
Wide sense stationarity and Hilbert space techniques
I would like to understand the section you added in
stationary process. I don't understand why any stationary process can be considered as a Fourier transform while the Bochner's theorem is only applied on the autocovariance function.
Why do you consider treating the heat bath as made up of a large number of loosely-couple copies of the main system is incorrect? (It enables you to then the results from the MCI for a large number of coupled discrete systems with a prescribed total energy.)
Linuxlad16:13, 19 May 2006 (UTC)reply
if you put it that way, then yes fine. if a comment to that effect is to be added, should be phrased like that as well in the article, not as it previously was.
Mct mht16:17, 19 May 2006 (UTC)reply
Yep - (I thought I wasn't misremembering my old lecture notes even though they are now rather old!) I'll see if I can reinstate the point more to your liking...Bob aka
Linuxlad16:30, 19 May 2006 (UTC)reply
I apologize if my quick and harsh responses have felt like I was
biting you, but deletion is a serious matter. However, I advise you to refrain from commenting on other editors' intelligence as you did, which is at the least
uncivil and at worst a
personal attack. In fact, I advise you to stop commenting so much in the AfD in general: you really haven't added anything new to the discussion since your nomination, and you should let the community speak. As you are a newcomer, though, let me direct you to
Wikipedia:Introduction to Deletion Process, where you can learn about how deletion works and what the processes are, and where to read more, as well as
Wikipedia:Moving and merging, where you can learn about how to deal with redundant or badly named articles. And if this is a question of not being sure you could edit out parts of an article by yourself, wikipedia encourages you to
be bold.
Mangojuicetalk18:06, 22 May 2006 (UTC)reply
i thought the discussion was over as far as i was concerned, and i wasn't going to comment further. since you decide to come to my talk page i will respond, one last time. my question re your ability to read the article somewhat intelligently has become a question whether you can read it critically at all. in fact it has become obvious you're not competent to judge, from your comments. inability to recognice and distinguish relevant material is pretty clear. after initiating the process, i had assumed whatever objections encountered by the proposal would be well-informed and educated. whatever the deletion policy is, if all AfD's solicit such ignorant responses, hopefully not too many articles will go thru this nonsense in the future.
Mct mht18:26, 22 May 2006 (UTC)reply
Comment on deletion policy
i hope it's not inappropriate to have a comment such as this on the talk page. as can be seen above, i got into it a bit regarding an AfD. the full exchange is
here. while that is over and done with, it seems to bring about questions regarding the deletion policy for articles of a particular nature.
my position was that the article
derivation of the partition function is technically worthless. i assumed that whatever objections raised would be technically sound. when the justification of the first 2 votes to keep seemed to be superficial and had no apparent technical merit, i mentioned the question whether one is actually "required to read the article somewhat intelligently to enter the discussion" (my words), the responses of the two voters were:
1. one flat out stated that such requirement is unnecessary.
2. the second voter claimed to be insulted and offered further justifications/explanations which reinforced the suspicion that there's no real understanding behind his/her comments.
There is nothing uncivil about calling an ignorant comment what it is. it's unpleasant but it needs to be done. i am very happy, and did, listen to
objections/suggestions raised that from those who understand the context, be they disagreements on the specific issues i raised or the overall value in retaining the article. it seems funny that a discussion on the deletion proposal for such an article would attract untrained attention, whose only interest, apparently, is to ensure that some "policy" is followed, as they understand that policy, sensibly or otherwise. the discussion then becomes pointless. as the risk of being overly dramatic, it would be similarly ridiculous to have an amateur sitting on the editorial board of, say, Phy. Lett. X (does it even go to X? :)), and decides what gets in the journal. again at the risk of overdramatizing, this reminds of the story when the physicist Alan Sokal, as a prank, submitted some gibberish to a sociology(?) journal and got accepted. it would be funny to see whether one can duplicate that here, write up some garbage filled with technical jargon, put it on AfD, and see whether it gets defended.
if that's the WP policy, so be it, but then one needs to be extra careful with actions which could bring un-knowledgable attention to articles, such as AfD.
Mct mht22:39, 22 May 2006 (UTC)reply
De facto, no requirement of specific familiarity with an article is needed, as the first voter said (and as Jitse Niesen also said). The ultimate goal of the debate on AfD is to establish, if possible, a consensus for whether to keep or delete the article. The quality of the debate matters... it matters a lot, actually, while people simply piling on votes matters much less. However, the qualification of the debators does not matter. The way that deletion ultimately works is that after the debate has been open for several days, an administrator will arrive and try to determine whether a consensus exists, and act on it if one does exist. It's impossible for these people to fully understand who is an expert and who isn't, but they can look at the debate and evaluate it.
Another problem is the availability of experts. For articles on an obscure subject, we presume that those who edit the article will notice if it becomes deleted (via watchlists), but in practice, people aren't always active within any given week, and AfD debates only last for a week (not that this would be fixed if they lasted two weeks instead; inactivity can persist for a long time).
Okay, that's the bad news. The good news is that nothing that happens on WP is permanent, which means that mistakes can be undone. If an article is deleted and should be recreated, it can be recreated via
deletion review. If an article isn't deleted once, it can still be nominated for deletion again, and can be deleted that time. I completely agree with your last point, but let me rephrase. One must be extra-careful when seeking input from unknowledgeable editors... it's a lousy way to improve articles. This is why I generally try to stop articles from being deleted when the main reasoning in the nomination has to do with concerns that should be addressed by editing, which is exactly why I voted the way I did at the time.
I see your point in disliking this situation... but how else could Wikipedia really operate? This is an encyclopedia that anyone can edit, after all, so we have to accept that for the most part, it's written by amateurs.
Mangojuicetalk20:35, 24 May 2006 (UTC)reply
PS -- here is a fine place to get into it, except that only people who want to talk specifically to you will come here. If you want to try to get Wikipedia to actually work differently, you might want to get involved at the talk pages of some of the policy pages, like, for instance,
Wikipedia talk:Deletion policy.
Mangojuicetalk20:35, 24 May 2006 (UTC)reply
here you come again.
1. the vast majority of the technical articles on WP, certainly all the ones i've encountered, along with discussions on their talk pages, are very professional. certainly almost everyone on
Wikipedia talk:WikiProject Mathematics seems to be a professional mathematician. that's why i was initially surprised the AfD attracted attention such as yours.
2. first you claim to possess adequate understanding and now say it's an obscure subject. as far as i am concerned, that renders your already zero credibility (in the present discussion) below zero. in any case, there's nothing obscure about statistical mechanics, millions of college kids know it, obscuredness is not relevant anyhow. don't excuse the ignorance.
3. i will leave the policy to the (self-appointed?) WP bureaucrats. however, as stated above, i will certainly be careful with actions that might attract undesirable interest.
Greetings! Since you said the worthwhile information from
Derivation of the partition function is present elsewhere in the encyclopedia, I infer that you know where it is. I don't, or I would do the following task: please change or remove the links broken by the deletion, which may be found
here. (Only the links in actual article pages need fixing.) Ideally, the admin who did the deleting should have done this, but he may not have known to do so or may not have felt qualified to judge what changes should be made. --
Cyan22:14, 17 June 2006 (UTC)reply
thanks for the heads up. both "cabin" and "elevator" can be found in the literature. they convey the same idea. no objections from me if you strongly prefer elevatorvand wanna put it back.
Mct mht19:12, 23 September 2006 (UTC)reply
basis function
FYI, I checked a couple of mathematical references and did not find the term.
I don't know enough about the subject to have an opinion,
but I did find the
where functional analysis was added to the article. It was in there for three years!
I find problems all the time in articles and not just technical ones.
Sometimes it is helpful review the edit history to find out just when a problematic change was made.
It might then be possible to contact the original editor and ask what he had in mind.
If the editor was anonymous and left no edit summary, I feel comfortable removing the material.
--
Jtir19:04, 24 September 2006 (UTC)reply
seems to me that no one bothered to clean up that page, for three years, as you said. i think most mathematicians will agree that the presentation and content is not mathematical. there are a few entries in the edit history that are from mathematicians, with edit summaries like "more work needs to be done". the discussion is in the same vein as what can be found in some physics texts, see for example Intro to Quantum Mechanics (i believe that's the title) by David Griffiths. it's certainly misleading to call it functional analysis. as i said on the talk page, the stuff the article seems to purport to cover is discussed in
Hamel basis and
Hilbert space in much better fashion.
Mct mht03:52, 25 September 2006 (UTC)reply
would you mind taking it a little less personal? what's your problem? it is clear from your own talk page that you attack viciously anthing not to your liking. that might me fair. you also have to agree from time to time that you might have missed a detail and might also not be omnipotent. so i find it fair to ask wether you have other ways of response than labelling something general bs when you don't see all it includes. there is no point or reason in questioning your intelligence but you can slow down insulting mine. thank you. you can reply here, to my talk page or the quantum hamiltonian and i do expect some substance. thank you. andrej.westermann 04:06, 17 October 2006 (UTC)
why didn't you mention that you contribute to
Mathematical formulation of quantum mechanics. This of course explains your vigour. I actually tend to that side, for the record. If we want to get there in a convincing manner, it might help not to get too agressive towards the so called classical community. Not to have a point of view (or to ponder many) can be the soundest scientific position. Wikipedia is as much about points of view as about verification of those points of view. It is not a a bible and does not try to be. so thanks for your efforts, keep it up and take it easy when people ask you questions, as well as it might help to use a softer style yourself. but hey, this is a free world. that has never meant it should be free of all rules. civilized behavior saves a lot of energy for the things that really matter, like getting a deeper understanding of our universe and how we deal with it. take care.
84.227.129.10215:35, 17 October 2006 (UTC) oops, was not logged in. andrej.westermann 15:55, 17 October 2006 (UTC)reply
i thought i was done with you. you spout nothing but nonsense bs and then try to hide behind pretense of civility. you're a crank and should be treated accordingly. cranks are about the only people not welcome on my talk page. you keep posting here and i am gonna report you.
Mct mht17:27, 17 October 2006 (UTC)reply
This article is hopeless. Material gets added to this article that I have trouble even parsing. Can you make any sense out of the following graf?
However, it should be noted that the Robertson-Schrödinger uncertainty relation, is not the uncertainty as stated in the Heisenberg uncertainty principle. That is, this derivation is inherently only applicable to measurements on a statistical ensemble of systems, and says nothing, contrary to most popular statements, about the simultaneous measurements of individual systems.
hey, CSTAR. not really sure, i've only heard of, very vaguely, Robertson-Schrödinger states. those are, i believe, states that minimize the uncertainty relation in some way, i.e. when equality is achieved. also, the part "...applicable to measurements on a statistical ensemble of systems..." seems to me is true about the Heisenberg uncertainty principle (not just particular to Robertson-Schrödinger principle, whatever that is), since the Heisenberg uncertainty principle can be viewed as a statement about the standard deviations of measuring different observables. what you think?
Mct mht17:47, 26 October 2006 (UTC)reply
i just looked at the article and agree it's not in good shape. seems to me a separate article of mathematical nature, where imprecision and verbosity is less likely, could be useful here.
Mct mht18:02, 26 October 2006 (UTC)reply
CSTAR, if you still watching that page, things don't look much better. that statement you pasted above is still there. i agree that the uncertainty relation is a statement about measurement on statistical ensembles, that part about "Robertson-Schrödinger uncertainty relation, is not the uncertainty as stated in the Heisenberg uncertainty principle..." looks like b.s. the derivation given for the "Robertson-Schrödinger uncertainty relation" doesn't look very different from those in Griffiths or Sakurai for the Heisenberg uncertainty principle, IIRC.
Mct mht11:23, 16 January 2007 (UTC)reply
hi, Mike. i believe they can be easily merged again, as the current stub doesn't say anything new and it's probably better to discuss pure states in the context of density matrices.
Mct mht10:57, 27 October 2006 (UTC)reply
Hi there, i apologise for the last-paragraph problem with the Enthalpy article, there's browser problems here which are causing textbox data loss mid-edit. I was wondering if you'd be interested in creating a collaboration project with regards to the Enthalpy article; there's a good basis of information, but it needs verifying and tone-changing.
thanks for the comment. no need to apologize there. that paragraph happens to have been added by me; it seems to flow better without breaking, is all. i'd be glad to contribute what contents i can and have discussions, although i must emphasize that i don't pretend to be a chemist or experimental physicist.
Mct mht14:37, 27 November 2006 (UTC)reply
Math help
Mct mht, thank you for answering my questions on a variety of math talk pages. I've been learning a lot, and I think the process has improved the Wikipedia pages. Thanks.
—Ben FrantzDale11:26, 1 May 2007 (UTC)reply
Hi, I'm a little bit new to wikipedia so I'm not sure if this is the right way to contact you.
Although I'm on a different IP, I was the guy who edited the quantum teleportation article regarding the collapse of the qubits into a bell state. I trust that you know more about the subject than I do, but I wanted to point out that what I added was actually a direct quote from the article on bell states:
http://en.wikipedia.org/wiki/Bell_state
hey. i wouldn't worry too much about being a little bit new, eh. the remarks in dispute are the following:
Note that because the qubits were not in a Bell state before, they get projected into a Bell state (according to the projection rule of
quantum measurements), and as Bell states are
entangled, a Bell measurement is an entangling operation.
it is the "because" in the first sentence that makes it misleading. the system collapses to a Bell state because that's the eigenbasis of your measurement not "because the qubits were not in a Bell state before."
Mct mht04:34, 3 May 2007 (UTC)reply
Hilbert space
Hello, I was a little puzzled by your edit summary in
Hilbert space. If you believe that the sentence that you've removed should be replaced with a more accurate one, are you going to replace it yourself? I actually thought that the sentence was reasonable, and don't quite understand how (and mostly, why) are you going to project onto a non-closed subspace. For example, sequences with only finitely many non-zero terms are dense in l2 and form a linear subspace, but however you define the projection onto it, this projection will not have any good properties.
Arcfrk22:50, 3 May 2007 (UTC)reply
hi, Arcfrk. well, you're right about unbounded projections not being as nice, but i hope we agree that they do arise on occasion. for instance, one might have a non-closed subspace affiliated to a von Neumann algebra. projection can defined purely algebraically, without topological requirements, if one wishes. that's why i thought that sentence was somewhat misleading. as for replacement, that section being about motivation, maybe something physical could be cited instead. or the article could remark that completeness is crucial whenever algebraic conditions is translated to topological ones and vice versa. but i got lazy there.
Mct mht09:33, 4 May 2007 (UTC)reply
I am not sure I understand your comment about von Neumann algebras, but I agree that motivational section should provide good motivations, not massage technical details. Unfortunately, I can't find the right words to express the idea of algebraic vs topological consideration that you've alluded to. If you can overcome your laziness, it would be great if you fill in a sentence or two there.
Arcfrk04:22, 5 May 2007 (UTC)reply
hello, Mr. Hardy. good to see ya, :-). i don't know enough about infinite monkey theorem's role in popular culture to make a sufficiently well-informed comment. but now there are some well-considered opinions from knowledgable folks, whichever way they vote. this does not seem to be the case for the most part in the first AfD. (but man, did you step on some toes and got folks circling their wagons, eh.)
Mct mht19:27, 11 August 2007 (UTC)reply
Triangular matrix
Please note that i undid your undo, of a change that I made to the
Triangular matrix article. The claim I removed is clearly false. As you sensibly recommended, I put a remark in the discussion page. Please do not revert the false section again.
Tom Lougheed01:25, 12 August 2007 (UTC)reply
read the article carefully. not sure how notable it is; maybe people familiar with the terminology can supply a reference. but the part you removed was trivially correct.
Mct mht01:27, 12 August 2007 (UTC)reply
Analytic or geometric properties of groups?
Various properties of groups - amenability, a-T-menability, property (T) - are studied by geometric group theorists, ergodic theorists and operator algebraists alike. Each one can be defined by a geometric condition, as Gromov has pointed out. I am not so sure that these pages should automatically be classified as
analysis, rather than geometry or topology. I think many geometric group theorists might take issue. The field of "algebra", however, for amenable groups was obviously wrong. This page by the way does not mention amenable actions, which I intend to add in the near future. The action on the Gromov boundary is amenable; it's hard to say whether that's geometry or analysis, because it's probably both.
Mathsci08:02, 29 September 2007 (UTC)reply
hello, Mathsci. i know nothing about geometric group theory at the moment (something i hope to remedy in the future). so if someone wants to list property (T) under topology or geometry instead of analysis, i am in no position to object. seems to me that it'd be good in general if more than one subfields can be put in the banner.
Mct mht03:51, 30 September 2007 (UTC)reply
Hello -- After I added a brief section on the von Neumann equation to
density matrix, you put in an edit note suggesting that it might be merged into
mathematical formulation of quantum mechanics. I'm not sure exactly what you mean by this: If you mean that the section should be removed from the density matrix article, I would disagree -- it's certainly important information about the density matrix that belongs in an encyclopedia article about the density matrix, and in fact there's plenty in the "density matrix" article is also in the "math. formulation" article. If, on the other hand, you mean that there's content in one presentation but not the other that should be copied, it sounds like a great idea, although I'm not sure I'll get a chance to do so myself. Happy holidays! --
Steve (
talk)
06:32, 18 December 2007 (UTC)reply
hi, Steve. unfortunately i meant it in the first sense, :-). anyway, a "see also" on top of the section you added couldn't hurt. happy holidays and see you around.
Mct mht (
talk)
12:14, 18 December 2007 (UTC)reply
sry, Steve, i was a bit sloppy with my statement there.
Mathematical formulation of quantum mechanics talks about unitary time evolution for pure states. extending by linearity gives the Liouville-von Neumann equation. the unitary time evolution of observables in the Heisenberg picture seems to me should be the same for both pure and mixed states.
Mct mht (
talk)
20:54, 18 December 2007 (UTC)reply
Tensor product
Hi,
I notice that in May last year you removed the definition of the tensor product from the article on the
tensor product of vector spaces. The edit summary indicates that you thought the definition was confusing; for this I apologize, as I gave that definition there, and perhaps I was less than clear. However, it was the only actual definition of the tensor product that the article contained! (which is why I took the time to add it! Yet this is typical of the shameful state of WP articles.) If you read the current version of the article carefully, you will notice that while it uses the otimes notation liberally, it never actually ever defines what otimes means, except by general dancing about and examples using finite algebra! Please consider restoring the deleted text, and/or modifying it so that otimes is actually defined. Thanks.
linas (
talk)
04:12, 8 January 2008 (UTC)reply
Never mind. The whole thing irked me enough that I just reverted back to the correct definition. I am displeased by the fact that many of my WP edits are driven by irritation rather than by pleasure. Worse, this is soo far off teh mark of what I'm trying to study that its ridiculous. Oh well.
linas (
talk)
05:11, 8 January 2008 (UTC)reply
hi, linas. i am not quite sure which edit you are referring to. i never touched that section, Tensor product of vector spaces, you're talking about. i thought it was too messy to mess with. anyway, seems to me that section is now correct after
User:Silly rabbit's edits.
Mct mht (
talk)
09:28, 8 January 2008 (UTC)reply
I saw some answers you replied at
Talk:Quantum entanglement.
I think you know pretty much about the quantum entanglement. Could you
forgive me for making some discussion here?
I have studied a bit quantum mechanics because of my interest in the ability of parallelism of quantum computer.
It seems that quantum entanglement plays a critical role to achieve the parallelism. I believe the entanglement
phenomenon can be derived from the Schrödinger equation. The thought experiment of EPR paradox would be my evidence.
But after the study, I can not find an article which explains the relations between the Schrödinger equation and
quantum entanglement so far. As below: (Part I: Quantum entanglement, Part II: Schrödinger equation and Part III: Question)
Part I: Quantum entanglement
As I know, the entanglement only happens in composite system. For example, there are three systems , , (each of which consists of a single particle) with respective Hilbert space , , . The Hilbert space of composite system of the three is the tensor product of respective Hilbert spaces:
If each Hilbert space has two basis in the set:
and
If I am right, the composite Hilbert space will have basis (but not basis)
in the set:
Therefore, is an 8-dimensional Hilbert space.
It means (again if I am right) any state of the composite system can be represent by an 8-dimensional vector.
Part II: Schrödinger equation
Now, we try to described the composite system of three particles above by a 2-dimensional time-independent Schrödinger equation:
where , , , are position vectors.
The position vector of each particle is in the Cartesian coordinate (or Cartesian space). The composite wave function of the three is in the the space of cartesian product of respective Cartesian spaces:
where:
: set of all real numbers.
The composite space will have basis (but not basis) in the set:
Therefore, is a 6-dimensional Cartesian space. And .
Part III: Question
1. My explanations in Part I & Part II look so similar to me. The wave function in Part II may be rewritten as
does that mean the composite system in Part II can be written as a ket
with six basis
?
2. Why the same composite system can have two explanations (Part I & Part II) but with different number of basis (8 versus 6)?
3. How to connect Schrödinger equation with quantum entanglement?
4. Why the entangled state is in the space yielded from tensor product rather than from cartesian product?
hi, Justin. thanks for dropping by. i am not enough of a physicist (not much of one at all, in fact) to give a good answer for 3) and 4). 4) in particular. but i will try to very quickly answer 1) and 2) for now. i might also pass 3) and 4) to some physics folks around here if that's ok. a better way to formulate 3) would be to ask whether entanglement can be broken by unitary time evolution. (IIRC, the answer is no?)
in part I above, you were correct. all three spaces were finite dimensional. in part II, you're talking about tensor product of two infinite dimensional Hilbert spaces, which of course is itself infinite dimensional. so what you did doesn't quite make sense there. the wave function ψ is viewed as a state vector in that tensor product, which seems to me should be L2(R6), for three particles moving in the plane. hope to see you around.
Mct mht (
talk)
01:43, 26 January 2008 (UTC)reply
Looking forward to the answers from folks for 3) & 4). Thanks for passing.
>> 3) would be to ask whether entanglement can be broken by unitary time evolution.
Sorry, I don't really get it. Does that mean 3) relates to eigenvalue problem? If you are correct, entanglement is irrelevant with Schrodinger. Then there must be some way to prove or derive the result mathematically that the composite Hilbert space is tensor product of the three spaces, right?
>> in part II, you're talking about tensor product of two infinite dimensional Hilbert spaces...
Why two but not three? I think there are three particles... And what does L2(R6) mean?
Well, I think the questions I posted here are too tough to be answered. Or it would take much effort to just answer such questions. :) And I should have posted the questions somewhere else rather than personal talk. Hope I didn't bother you too much because of my rudeness... I think the discussion should stop here. Anyway, thank you both
Mct mht and
Steve for reading my questions and attempting to solve them. :) --
Justin545 (
talk)
01:47, 28 January 2008 (UTC)reply
A section in "Separable states" is removed?
You removed the entire section of "Separable wave function and tensor product" in article "Separable states" at 01:43, 26 February 2008 (UTC). Could you give me the reasons? -
Justin545 (
talk)
02:42, 26 February 2008 (UTC)reply
I am not sure have you read the
Q&A that I put in your talk page a few days ago. The section I added in "Separable states" is based on the knowledge I got from the Q&A. If you have not read it, could you please take a look? I would like to know what's wrong with the Q&A. Or do you have some references which disapprove my edit? If the "Separable states" is for advanced physicist only. I would like to find another place to put the removed section. I believe there must be an appropriate place to put it. Could you give some suggestions? Create a new page or add it into an existing article? I am pretty curious about why you removed it. Please give me some clue, Thanks! -
Justin545 (
talk)
06:31, 26 February 2008 (UTC)reply
the edits i reverted were confused and unclear. non-"positive definite" and "non-degeneracy" are not the same. a metric tensor is a smooth assignment of a non-denegerate bilinear form <,>x to each tangent space TxM at x. non-denegeracy means <v, v>x ≠ 0 for all x and all v in TxM. if <v, v>x > 0 for all x and all v in TxM, the metric is called positive definite, a Riemannian metric. Otherwise, it's a pseudo-Riemannian metric. in both Riemannian and pseudo-Riemannian cases, one can easily have <v, w>x = 0 for v ≠ w in TxM; this gives a notion of orthogonality on each tangent space.
Mct mht (
talk)
00:27, 22 May 2008 (UTC)reply
Thanks for the reply. However, not being a mainstream mathematician I have a few problems with it. Firstly, as I understand it the real projective line is itself a one point compactification of the Euclidean line - that is, frankly, the whole reason why it was developed. Secondly, you do not answer (but merely counter) the particular point I raised on the article's discussion page, in that I would call it a topological loop (1-manifold) but not a "circle". And there is another difference which I thinks helps illustrate the point: a circle in the plane is orientable and shrinkable - it has an inside and an outside and can be shrunk to a point without affecting the topology of the surrounding plane, while a projective line is not shrinkable and nor does it have an inside and an outside, since it does not divide the plane. Please could you explain a little how come such an object can be described as a "topological circle"? I should like to explain the difference in the article, but I need to understand your point of view before I can do so in a way that would be acceptable to you. -- Cheers,
Steelpillow19:10, 28 May 2008 (UTC)reply
well, what's your precise definition of a topological manifold? there seems to be confusion as to what a topological manifold is. up to homeomorphism, there are only 4
topological 1-manifolds: (0,1), [0,1), [0,1], and the circle S1. which one is your "loop"?
as for your remarks starting with "a circle in the plane is orientable and shrinkable...", they make no mathematical sense. there is a connection between the
fundamental group of a manifold and its
orientability but it's not anything close to what you're describing. just to cover all the objects you mentioned:
the fundamental group of the circle, i.e. the projective line, is the integers Z. the circle is orientable.
the fundamental group of the plane is trivial. the plane is orientable.
the fundamental group of the projective plane is the multiplicative group {-1,1}. the projective plane is not orientable.
let me elaborate a bit and respond more directly, if you're still watching. the definition of a topological manifold makes no reference to an embedding (it wouldn't make much sense if it did). so it makes no sense to talk about how the projective line sits inside a "plane" (what plane?) unless you specify an embedding into some Rn. such an embedding always exists, if one makes n sufficiently large; see
Whitney embedding theorem. but, again, the definitions of topological properties intrinsic to the manifold makes no reference to an embedding.
the projective line is the set of lines in R2, with a "natural" topology. it is homeomorphic to the circle. similarly, the
one point compactification of R is homeomorphic to the circle.
let me take apart your statement "a circle in the plane is orientable and shrinkable..." again, the "in the plane" makes no sense. no embedding is needed to consider orientability (the circle can be obviously embedded in Rn for any n = 2, 3..., but that's not relevant). yes, any loop in the plane is, to use your language, shrinkable. that's a topological property of the plane; one says the plane is
simply connected. it has nothing to do with the circle, the 1-manifold. (a circle embedded in the
torus need not be "shrinkable." that's a topological property of the torus. it has nothing to do with the circle.)
Mct mht (
talk)
10:49, 30 May 2008 (UTC)reply
Phew! Some of that is way above my head, but let's see if I get this right: those four 1-manifolds, (0,1) etc., are just lines with various combinations of open or closed or joined ends? In which case, OK topologists use the word "circle" in a broader sense than ordinary geometers do. I take your point about orientability - that's probably me not knowing the formal definition. There is a confusion over terms of reference when embedding one manifold in another - for example, is a "line" to be taken as a geodesic of the one manifold or the other. Also, as a geometer, one tends to distinguish "lines" from "circles" even when the line is projective - for example using homogeneous coordinates in the projective plane we find that "lines" and "circles" behave quite differently (e.g. the shrinkability thing, or the number of intersections between two lines vs. two circles). But I think this is a bit away from the article. Anyway, I don't think we will understand each other fully in the near future, so I will try to think up a simple edit that helps explain these different viewpoints. Meanwhile, many thanks for your patience. -- Cheers,
Steelpillow11:38, 31 May 2008 (UTC)reply
it's pretty safe to say everything i said so far is completely standard. there is no ambiguity. confusing various notions, which i must say your remarks indicate you are, is not a good idea. for example, talking about
embedding one manifold into another and the
geodesics on a manifold (which requires a Riemannian structure) the way you did makes no sense. i suggest consult the mathematicians watching the page while considering the edits you wanna make.
Mct mht (
talk)
21:01, 31 May 2008 (UTC)reply
My understanding of geometry also comes from standard, even classic, works in its own field - such as Hilbert & Cohn-Vossen's Geometry and the imagination, Coxeter's Projective geometry and also Edwards' book of the same name. These works are all synthetic or practical in nature, and quite unlike the analytical tradition common among the majority of mathematicians today. Nevertheless, I am sure you will recognise my remarks on homogeneous coordinates - even if you do not choose to answer the point I made. The unfamiliarity to you of way I use words, and the lack of sense you perceive in them, could not more clearly illustrate the gulf in terminology and culture between our two, equally respectable, mathematical disciplines. The more I study maths, and especially polyhedra, the greater I realise this gulf to be. The projective line is an important, even fundamental, concept in modern synthetic geometry and in perspective drawing. Anybody versed more in these disciplines than the analytic will be confused by the present wording of the article, and I believe that this wording would benefit from some clarification. Clearly, the present import of it should not be perverted in the process. -- Cheers,
Steelpillow08:42, 1 June 2008 (UTC)reply
somehow i doubt you will find any professional mathematician, a geometer especially, who agrees with the way you phrase things. since you cite Coexter, one can look at his Introduction to Geometry, which covers projective geometry and also a bit of differential geometry and topology. it contains no great inconsistency in language and terminology.
Mct mht (
talk)
09:35, 1 June 2008 (UTC)reply
BTW, as an anecdote on Coxeter, I am told by one eminent geometer that Coxeter was from time to time prone to small errors and inconsistencies in his writings. He was usually delighted when these were pointed out to him. -- Cheers,
Steelpillow09:57, 2 June 2008 (UTC)reply
Hi again. I have modified my edit in the light of various recent discussions/comments, and because "finite" means different things in different contexts and I just realised that I had not made my usage clear. If you are still unhappy with it, please check out the
Talk page before reverting again. Apologies for inviting comment in the edit summary, but you had not replied to my proposal and I didn't want to blank you. -- Cheers,
Steelpillow10:48, 7 June 2008 (UTC)reply
Howdy, I undid a big deletion with no edit summary at
Hermitian adjoint. I guess it was probably accidental. The deleted material wasn't shakespeare, but it was relevant and looked correct. If you did want to get rid of it, definitely include an edit summary, and for so much reasonable text maybe also on the talk page, so we know what's up.
JackSchmidt (
talk)
03:04, 3 July 2008 (UTC)reply
it's either completely trivial or funnily put ("...similar properties."? it's exactly the same thing, in a fixed basis.)
Mct mht (
talk)
03:09, 3 July 2008 (UTC)reply
Definitely trivial, but important to mention. The only difference is matrix versus operator. The fact that the same definition is the topic of a different article seems worthwhile to mention. It is in the leads of both articles, but leads are
supposed to be summaries of the articles, so it is good to say it again. The kernel section was also not terribly profound, but seemed reasonable to leave in (maybe that part was accidental?). Something a little more direct for the matrix thing would be fine too. "The matrix analogue of the hermitian adjoint is the
conjugate transpose."
JackSchmidt (
talk)
03:26, 3 July 2008 (UTC)reply
you recently
reverted my recent
over at PD matrix. While in general it makes sense to include an edit summary, when you RV an registered user (especially one making a change discussed on the discussion page), I'd highly recommend you (a) write an edit summary and (b) write something in the discussion.
Pdbailey (
talk)
04:03, 3 July 2008 (UTC)reply
Quantum Teleportation
Hi there, I noticed you deleted a bunch of stuff from the introduction of Quantum Teleportation. I'm neither a mathematician nor a physicist so I'm can't vouch for the validity of the information, however I understand enough to be the last person before you to try and make any sense of that page. The paragraph you deleted was the introduction before I added something more sensible. Could you add something to the Talk Page to explain that it was gibberish? Otherwise I fear whoever wrote it will just revert.
Master z0b (
talk)
09:08, 17 July 2008 (UTC)reply
Your recent revert in the article Homotopy
Please study the confusion at
Talk:Homotopy, and comment there, if you have a better idea on how to make the text clearer!
As you can see, some users definitely have got the wrong idea that an homotopy equivalence would be an homotopy. Also view the confusing text in the section
Homotopy#Homotopy equivalence and null-homotopy, from which you removed the {{Contradict}}; this text shows that at least one editor has mixed up the concepts! The text
Intuitively, two spaces X and Y are homotopy equivalent if they can be transformed into one another by bending, shrinking and expanding operations
would make clear sense if "are homotopy equivalent" were replaced by "are homotopic"; but right now it looks rather confused.
(The fact that the two compositions of the equivalence and its "quasi-inverse" are homotopic to identites does not mean that an homotopy equivalence is some kind of homotopy. It isn't.)
According to the definitions in our article (and incidently according to widespread usage), spaces are never homotopic; but continuous functions may be homotopic. I personally would only talk about "homotopic spaces" if there are impicit continuous functions involved, in such a manner that both I and my "audience" in a simple manner may replace the spaces with these functions.
E.g., let X be a subspace of the real plane R2, and consider two positively oriented "crossing free" loops L1 and L2 in X. Now, I might use this, presenting the Li as subspaces of X; but what this means formally is that there are "parametrisations" and , which are injective and continuous, and give the same "counter clockwise" orientation to the images as to the unit circle S1. Formally, the fi are the loops, not the Li. In this case, I might speak about an homotopy between L1 and L2, if both I and those I speak to understand that I'm actually speaking about homotopies between chosen parametrisations f1 and f2. The reason for this abuse of language making sense is that all different parametrisations of L1, which have the properties I just decribed, are homotopic to each others; and the same goes for L2. Therefore, in this particular case, it is not that important which parametrisation I choose.
In a situation where I'm not sure that all others understand this, I'd be more careful not to call the subspaces themselves "homotopic" or "non-homotopic". I think this also holds true for introductory books. I just picked out M. A. Armstrong, Basic Topology, and I note that he follows this consistently. E.g., he defines a map (i.e., a continuous function) from X to Y as null homotopic, if the map is homotopic to a constant map (sending the whole of X to some point in Y; see Problem 5 to Chapter 5); and he essentially defines a space X as contractible if the identity map 1X is null homotopic (in Section 5,4). He also is very careful in stressing that "loops" are maps, not spaces.
Of course, if you can find some literature with a different usage of the terms "homotopy" and "homotopic", then our article should be extended with information of this usage. Some alternatives do exist. E.g., Armstrong employs "being of the same homotopy type" as a synonym to "homotopy equivalent". However, he never calls homotopy equivalent spaces "homotopic". In this way, he avoids contradictions. E.g., consider the "figure 8" or "figure " space, consisting of two "loops" meeting in one point. Now as spaces, these "loops" are homeomorphic and thus a fortiori homotopy equivalent. However, the two "loops" as parametrisations are not homotopic; there is no "continuous deformation" of one of them to the other (where the intermediate image all the time is contained in the "figure 8" space).
I noticed that you weakened one of the statements in
Eberlein–Šmulian theorem on the claim that there are infinite-dimensional normed spaces whose weak topology is metrizable. This is actually not true. The counterexample you gave was a separable Hilbert space. Although it is true that the weak topology is metrizable on norm-bounded subsets, it is not metrizable on the whole space. In fact, more generally the weak and weak-* topologies on a reflexive Banach space coincide, and it is easy to show using the Hahn-Banach theorem that a weak-* topology is never metrizable.
More generally the weak topology on a normed space X is not metrizable. Otherwise, by first countability there would be a sequence xn∗ in X∗ such that the weak neighborhoods
are a basis of neighborhoods at 0. But then for any fixed x∗∈X∗, the weak neighborhood must contain Wn for some n. It follows that x∗ is a linear combination of x1∗,...,xn∗. That is, the Banach space X* is a union of countably many finite-dimensional subspaces, in violation of the Baire category theorem.
siℓℓy rabbit (
talk)
15:46, 16 December 2008 (UTC)reply
yes, my mistake. thank you for the heads-up and correction. following is another counter example, from Lax: take n- dimensional subspaces Vn and finite sets Sn from the sphere of radius n, so that the sphere is covered by 1/n-balls centered at elements of Sn. then 0 lies in the weak closure of ∪ Sn.
Mct mht (
talk)
05:54, 17 December 2008 (UTC)reply
Differential form of Maxwell's equations (in Stokes' theorem)
You recently deleted a note on SI units in the section of
Stokes' theorem where Maxwell's equations are presented, with the comment that it is "absurd to imply a physical formula is valid only in certain units". Whether you like it or not, the differential forms presented are in fact for the electromagnetic fields expressed in SI units. In other systems of units, the differential forms as such look similar but may involve different scaling factors. For example, in
Gaussian units, Faraday's law of induction takes the form , to be compared with the same law expressed in SI units, which yields . As a reference you may consult the standard textbook "Classical Electrodynamics" by
J.D. Jackson, in which an entire chapter is devoted to the subject of Maxwell's equations expressed in different systems of units.
Hakkasberra (
talk)
23:38, 10 January 2009 (UTC)reply
*-homomorphisms
Why the revert on C*-algebras? It's more useful to mention that non-zero *-homomorphisms have norm 1 than to simply say they're non-expansive. And why not use the discussion page?
76.126.116.54 (
talk)
04:47, 16 March 2009 (UTC)reply
Hi, Mct mht. I saw that you wrote most of the article on
Choi's_theorem_on_completely_positive_maps, which current is the destination for a redirect if anyone searches for "completely positive map". I was thinking it would be more logical to move the contents of the page to "Completely Positive Map" with a large section on Choi's theorem (or, if the page got to large in the future, I would move the material on Choi's theorem to its own page, separate from the CPM page).
Good suggestion. That page needs a lotta trimming and editting. Maybe there should be a page tieing together some basic notions/results that involve CP maps (nuclear C*-algebras, Choi's theorem, Choi-Effros lifting theorem, etc. Maybe we can advertise on math project page to get expert inputs there.) But anyway, your suggestion sounds good.
Mct mht (
talk)
22:56, 20 June 2010 (UTC)reply
Hello there, thanks for dropping a note. 2-positive just means that the map remains positive after passing to 2 by 2 matrices. Same goes for n-positivity. I have to confess, in hindsight, that whole section (written by me) is probably not well-written. The POV that ought to be emphasized, IMHO, is that
completely positive maps directly generalize vector states. Mentioning 2-positivity is probably not appropriate.
Mct mht (
talk)
10:11, 2 December 2010 (UTC)reply
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First, thanks for your work on the article on
quantum relative entropy. My comment relates to the divergence property of S(ρ||σ) in the subsection Non-finite relative entropy. The demonstration of the divergence of the relative entropy diverges is a good one I think. However, only positive comments follow (i.e. "This makes physical sense. Informally, ..."). The negative repercussions of the divergence are not mentioned. Specifically, S(ρ||σ) can diverge when ρ and σ differ by an epsilon amount (as measured by some norm). An example: let σ have the diagonal representation with for and for , and let for a small positive number . As ρ has support (i.e. ) in the null space of σ, S(ρ||σ) is divergent even though the trace norm of the difference ρ-σ is 2ε. I think this should be pointed out as a negative feature of the relative entropy. What's your take on this?
DireNeed (
talk)
06:54, 8 April 2017 (UTC)reply