The Interpretation of $|\psi|^2$ and the No-Clone Theorem - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-09-21T20:07:31Z https://physics.stackexchange.com/feeds/question/310075 https://creativecommons.org/licenses/by-sa/4.0/rdf https://physics.stackexchange.com/q/310075 0 The Interpretation of $|\psi|^2$ and the No-Clone Theorem Dvij Mankad https://physics.stackexchange.com/users/20427 2017-02-06T12:39:33Z 2018-09-08T02:18:11Z <p>The standard interpretation of $|\psi|^2$ is taken as the <em>probability</em> density of the wave-function collapsing in the given infinitesimal small region. The way this <em>probability</em> is interpreted (at least in the text-book by Griffiths) is that if I prepare a large number of identical states and then perform a measurement on each of them then the <em>probability</em> associated with $|\psi|^2$ actually represents the statistical results of the measurements made (individually) on all these wave-functions of the entire ensemble. This seems like a good enough argument to accept $|\psi|^2$ as the probability density of the wave-function collapsing in a given infinitesimal small region even if I have only one wave-function.</p> <p>But the <a href="http://en.wikipedia.org/wiki/No-cloning_theorem" rel="nofollow noreferrer">No-Clone Theorem</a> suggests that it is fundamentally impossible to make two states that are completely identical and thus, to my understanding, it makes absolutely no sense to talk about those identical wave-functions that were used to project $|\psi|^2$ as the probability density. Put in other words, if I can't replicate a wave-function at all then how do I confirm in my laboratory that the probabilities obtained by $|\psi|^2$ actually represents the <em>likely-hood</em> of the collapse happening in a given region? </p> <p><strong>Edit</strong> Is it the catch that the No-Clone Theorem suggests only that a given state can't be evolved to a state identical to another state and it allows the two states being identical if they are so from eternity? </p> https://physics.stackexchange.com/questions/310075/the-interpretation-of-psi2-and-the-no-clone-theorem/310106#310106 8 Answer by ACuriousMind for The Interpretation of $|\psi|^2$ and the No-Clone Theorem ACuriousMind https://physics.stackexchange.com/users/50583 2017-02-06T14:46:15Z 2017-02-06T14:46:15Z <blockquote> <p>But the No-Clone Theorem suggests that it is fundamentally impossible to make two states that are completely identical[...]</p> </blockquote> <p>No, it doesn't say that. It says that it is impossible to build a <em>quantum apparatus</em> (modeled by a unitary operator and a state space for that apparatus) that takes <em>any</em> quantum state as input and then outputs the exact same state, i.e. "clones" it. It does not prohibit the existence of an apparatus that produces a never-ending stream of copies of a <em>fixed</em> state, and it say nothing at all about two states being "accidentally" the same. </p> <p>If you consider that the crucial feature of bosons is that more than one of them can exist in the same state, the no-cloning theorem in your phrasing would be in outright contradiction to what quantum mechanics actually says.</p> https://physics.stackexchange.com/questions/310075/the-interpretation-of-psi2-and-the-no-clone-theorem/310114#310114 1 Answer by alanf for The Interpretation of $|\psi|^2$ and the No-Clone Theorem alanf https://physics.stackexchange.com/users/28512 2017-02-06T15:13:19Z 2017-02-06T15:15:44Z <p>You have misunderstood the no-cloning theorem. The theorem says the following. Suppose you have a quantum system S in an arbitrary state $|\psi\rangle_S$. There is no physical process that can take that system and copy its state to another system in a blank state $|0\rangle_C$. That is, no physical system can perform the operation $|\psi\rangle_S|0\rangle_C \to |\psi\rangle_S|\psi\rangle_C$ for an arbitrary state.</p> <p>The no-cloning theorem is perfectly consistent with being able to prepare two different systems in the same state to an arbitrarily good approximation since this doesn't involve copying. In addition, if you have many systems prepared in the same way you can perform measurements to find the state in which the system was prepared to any desired degree of accuracy - this is called quantum tomography:</p> <p><a href="https://arxiv.org/abs/1508.01907" rel="nofollow noreferrer">https://arxiv.org/abs/1508.01907</a></p> <p>You are also confused about what the amplitudes mean in quantum mechanics. The square amplitude $|\psi|^2$ only acts as a probability when the system concerned has undergone decoherence. Before the system has undergone decoherence, the square amplitude can change in ways that break the rules of probability:</p> <p><a href="https://arxiv.org/abs/math/9911150" rel="nofollow noreferrer">https://arxiv.org/abs/math/9911150</a></p> <p><a href="https://arxiv.org/abs/quant-ph/0306072" rel="nofollow noreferrer">https://arxiv.org/abs/quant-ph/0306072</a></p>