Talk:No-cloning theorem

message: Dosnt this rule break the second law of thermodynamics ie A is destroyed in a case of quantum teleportation?  — Preceding unsigned comment added by 124.170.0.157 (talk) 13:28, 5 April 2012 (UTC)
replies: text: The state of A becomes |0> or |1>. By destroyed, we mean the state changes to one of these values. Does that address your concern? My knowledge of physics is limited. Skippydo (talk) 14:20, 5 April 2012 (UTC)
thread_title:  Second law of thermodynamics

message: The article states, "The no cloning theorem prevents superluminal communication via quantum entanglement, as cloning is a sufficient condition for such communication."  Logically, this makes no sense:  to prevent superluminal communication, cloning would have to be a _necessary_ condition.  As stated, there is nothing to say that some condition other than cloning might also imply superluminal communication.  Indeed, if cloning implies superluminal communication, then a proof of no superluminal communication would imply no-cloning as a corollary.
:yes, you're right. feel free to modify the article. Mct mht 02:31, 26 June 2007 (UTC)
replies: text: In your contrapositive the consequent is missing a negation. You are correct in stating that the no-cloning theorem is a corollary of no superliminal communication. Skippydo 05:24, 26 July 2007 (UTC)
replies: text: The no cloning theorem indeed prevents superluminal communication via quantum entanglement. Following the thought experiment in that paragraph, no matter Alice perform measurements or not, Bob alway got an ensemble of mixed state. The reason is because if Alice really performs an experiment, Bob will randomly get |+z> or |-z> state, he does not know which state he has, so following no cloning theorem, he can not clone enough states to perform a coherent experiment to see if he really gets a pure state. So at the end overall he only has a bunch of equal mixed states which is equivalent to the situation Alice does not make any measurement. So Bob can not distinguish whether Alice performs an experiment or not.  — Preceding unsigned comment added by 128.42.156.137 (talk) 20:21, 1 July 2013 (UTC)
replies: text: Superluminal's got nothing to do with it. You can't communicate sub-luminally, either, that's the no-communication theorem.  User:Linas (talk) 21:15, 18 November 2013 (UTC)
thread_title: No-cloning prevents superluminal communication?

message: Shouldn't the title be No-cloning theorem, as opposed to No cloning theorem? --Robin (talk) 20:02, 5 September 2009 (UTC)
replies: text: It would be great if someone moved this, and the otehr no-xxx articles, from short-dash to long-dash (en-dash to em-dash). I no longer have that ability :-( User:Linas (talk) 21:03, 18 November 2013 (UTC)
thread_title:  Article title 

message: For example:
: | \phi \rangle = {1 \over \sqrt{2}} \bigg( | 0 \rangle + | 1 \rangle \bigg)  
and
: | \psi \rangle = {1 \over \sqrt{2}} \bigg( | 0 \rangle - i| 1 \rangle \bigg) 
Can someone clarify on what exactly these are examples of? Coz they're neither normalized nor orthogonal to each other.--Dhatsavan (talk) 03:56, 11 January 2011 (UTC)
replies: text: I am not sure, but I think that these (well, their replacement in the present article), are meant to be a demonstration of the particular states which may be cloned. I will attempt to clarify this section, as it confused me as well, even though the theorem is extremely simple.zipz0p (talk) 18:01, 1 August 2011 (UTC)
replies: text: The example is kind-of-bogus; it doesn't really show anything pertinent to the theorem, and mostly serves to confuse and distract the reader. It should probably be removed. User:Linas (talk) 20:54, 18 November 2013 (UTC)
thread_title:  the "example" state vectors 

message: Why does this article have so many references to category theory --- and so close to the top of the article!  This is really a pretty fringe view of quantum mechanics.   Most people working in quantum could lecture for two hours and never mention the word category.  I think the average reader is probably more baffled by the references than informed.   (Was it inserted by the authors? Probably!)  — Preceding unsigned comment added by 163.1.246.246 (talk) 09:59, 22 October 2020 (UTC)
thread_title:  Why so much category theory 

message: I think even one knows the state in system B and erases the information in B, one can still not clone A into B. The argument still holds even one does not require U to apply for any |e> state. Fangzhangmnm (talk) 00:47, 28 June 2021 (UTC)
thread_title:  The initial state |e> does not necessary be unknown 

message: The no cloning thm is basically trivial and does not need any lengthy proof. Any Unitary operators is linear acting on the state vectors
If you have U(a psi1 +b psi2) phi, (a b complex constants, psi's and phi are states) then 
that equals a (U psi1 phi)+ b (U psi2 phi) But the state (a psi1 +b psi2)(a chi1+b chi2) zeta , ie, the originla state plus a clone,  is clearly not linear in a, or b. 
Thus it cannot arise from a unitary transformation. 
This is no matter what the states psi1, psi2, chi1, chi2, phi, or zeta are. (the dimensionality of total Hilbert space containing chi times zeta must of course have the same dimensionality as phi). Ie, linearity implies
no cloning. It also cannot arise from a measurement. The projection operator onto a measurement outcome is also a linear operator. (assuming that the projection operator is independent of a,b. )
 This wiki page should be drastically simplified.  — Preceding unsigned comment added by 75.155.165.57 (talk) 05:03, 28 January 2022 (UTC)
thread_title:  No-cloning basically trivial