{"id":2028,"date":"2016-01-23T10:24:33","date_gmt":"2016-01-23T17:24:33","guid":{"rendered":"http:\/\/www.wall.org\/~aron\/blog\/?p=2028"},"modified":"2018-09-05T13:09:13","modified_gmt":"2018-09-05T20:09:13","slug":"quantum-mechanics-ii-decoherence-states","status":"publish","type":"post","link":"https:\/\/www.wall.org\/~aron\/blog\/quantum-mechanics-ii-decoherence-states\/","title":{"rendered":"Quantum Mechanics II: Decoherence &#038; States"},"content":{"rendered":"<p>Strictly speaking, most of the other rules about QM are already implicit in what <a title=\"Quantum Mechanics I: Interference\" href=\"http:\/\/www.wall.org\/~aron\/blog\/quantum-mechanics-i-interference\/\">I&#8217;ve already said<\/a>.\u00a0 But a few implications of this setup are worth pointing out.<\/p>\n<p>First note that, in QM, the &#8220;state&#8221; includes information about every single object in the system.\u00a0 So, when you add up the different histories, they only interfere if the final states are exactly the same in every respect.\u00a0 If even one tiny particle is in a different place than it otherwise would be, then they don&#8217;t interfere.\u00a0 In that case, you just add up the probabilities normally.<\/p>\n<p>This is why measurement is such a significant thing in QM.\u00a0 If you try to catch out Nature by explicitly measuring which slit the particle went through, then <em>YOU<\/em> are now different as a result of you knowing which slit it went through.\u00a0 As a result, the two histories don&#8217;t interfere.\u00a0 But it needn&#8217;t be a person which does the &#8220;measurement&#8221;.\u00a0 Even if you refuse to look at it, the detector being different still prevents the interference from happening.\u00a0 As far as we know experimentally, there is no special relationship between consciousness and QM (although some people have proposed interpretations of QM in which there is a connection between the two.).<\/p>\n<p>Usually, once histories become sufficiently different from each other, for a long enough period of time, their random interactions with the environment will tend to be different, so that the chances of getting everything perfectly the same become tiny, and the histories won&#8217;t interfere anymore.\u00a0 This phenomenon is called <em>decoherence<\/em>.\u00a0 People argue about what this tells us about the interpretation of QM, but the phenomenon itself can be studied in the laboratory, so my use of this word should not be regarded as an endorsement of any particular interpretation.<\/p>\n<p>Secondly, if you have two or more distinct states, then it&#8217;s possible to take a <em>quantum superposition <\/em>of the two states, formed by adding them up with complex coefficients.\u00a0 For example, if <strong>X<\/strong> and <strong>Y<\/strong> are two distinct states, then $$(\\mathbf{X} + \\mathbf{Y}) \/ \\sqrt{2}$$ or $$(\\mathbf{X} &#8211; \\mathbf{Y}) \/ \\sqrt{2}$$ or $$(2\\mathbf{X} +i \\mathbf{Y}) \/ \\sqrt{5}$$ are all equally valid states!\u00a0 (The reason for the square root in the denominator, is to make it so that, by the <a title=\"Quantum Mechanics I: Interference\" href=\"http:\/\/www.wall.org\/~aron\/blog\/quantum-mechanics-i-interference\/\">Born Rule<\/a>, the total probability of the state is still 1.)\u00a0 These states are just as much valid states as <strong>X<\/strong> or <strong>Y<\/strong> themselves would be.<\/p>\n<p>The possibility of quantum superpositions is implicit in the <a title=\"Quantum Mechanics I: Interference\" href=\"http:\/\/www.wall.org\/~aron\/blog\/quantum-mechanics-i-interference\/\">quantum probability rules<\/a>, since if you start with a particular state <strong>A<\/strong>, in general it will evolve to a superposition of different states as time passes.\u00a0 And there&#8217;s no particularly good reason you couldn&#8217;t also have <em>started out<\/em> the experiment with a quantum superposition.<\/p>\n<p>(Note that if we take any state like \\((\\mathbf{X} + \\mathbf{Y}) \/ \\sqrt{2}\\), and we multiply it by a <em>phase<\/em> (a number on the unit circle of complex numbers, e.g. \\(i\\), or \\(-1\\), or \\((1+i)\/\\sqrt{2}\\)) then we can&#8217;t tell the difference between that and the original state in any way!\u00a0 That&#8217;s because, when we work out the patterns of <a title=\"Quantum Mechanics I: Interference\" href=\"http:\/\/www.wall.org\/~aron\/blog\/quantum-mechanics-i-interference\/\">interference<\/a>, we only care about the <em>relative<\/em> phases between different histories, not the absolute phase of the whole system.\u00a0 So it&#8217;s good to remember that there is a slight redundancy in our description here: two states that differ by a phase are really the same state.)<\/p>\n<p>Now if we have a system with N possible states, then we can imagine a higher dimensional geometry consisting of all possible superpositions of these N possible states (including, for mathematical convenience<em><\/em>, those for which the probability doesn&#8217;t add to 1).\u00a0 This is called the <em>Hilbert Space<\/em> of that system.\u00a0 It is a kind of <a href=\"https:\/\/en.wikipedia.org\/wiki\/Vector_space\">vector space<\/a> with N <a href=\"https:\/\/en.wikipedia.org\/wiki\/Complex_number\">complex<\/a> dimensions, which means in terms of real numbers it&#8217;s a 2N-dimensional space.\u00a0 But don&#8217;t worry about these details for the moment.<\/p>\n<p>(It&#8217;s kind of hard to visualize a Hilbert space when N is greater than about 2, but it&#8217;s still very useful mathematically!)<\/p>\n<p>The simplest nontrivial Hilbert Space is the one with N = 2 states.\u00a0 (I&#8217;ll give a physical example in a moment.)\u00a0 This would normally involve a 4-dimensional space, but to keep things as simple as possible, I give you permission to ignore the bit about complex numbers and just think about a 2-dimensional plane.\u00a0 (This is the space of all states of the form $$a\\mathbf{X} + b\\mathbf{Y}$$where \\(a\\) and \\(b\\) are now real numbers.)\u00a0 Then we can think of <strong>X<\/strong> as a unit vector pointing along the x-axis, and\u00a0<strong>Y<\/strong> as a unit vector pointing along the (wait for it&#8230;) y-axis.<\/p>\n<p>Perhaps a picture will help:<\/p>\n<p style=\"text-align: center;\"><a href=\"http:\/\/www.wall.org\/~aron\/blog\/wp-content\/uploads\/2016\/01\/HS.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-4661 aligncenter\" title=\"HS\" src=\"http:\/\/www.wall.org\/~aron\/blog\/wp-content\/uploads\/2016\/01\/HS.jpg\" alt=\"\" width=\"371\" height=\"225\" \/><\/a><strong><em>The Hilbert space for a system with 2 states.<br \/>\n<\/em><\/strong><\/p>\n<p>As you can see, the Hilbert space has an origin, which is the point in the middle which represents &#8220;zero&#8221;.\u00a0 Each state is a represented by a <em>vector<\/em> coming out of the origin, pointing in some direction.\u00a0 (But remember that \\(-X\\) is really the same state as \\(+X\\), since they differ by a -1 phase.\u00a0 I didn&#8217;t draw \\(-X\\) on the picture, but if I had it would be 180\u00ba around from \\(X\\).)\u00a0 The Born Rule tells us that length = total probability squared.\u00a0 That means that in order for a vector to be a state-in-good-standing, it needs to be length 1.\u00a0 (In other words, by the Pythagorean Theorem, the sum of the squares of its \\((x,y)\\) coordinates needs to add up to 1).\u00a0 So don&#8217;t ask me what the physical meaning of the &#8220;zero&#8221; vector is, since it doesn&#8217;t have one.<\/p>\n<p>A physical example of an \\(N = 2\\) state system would be the <em>polarization <\/em> of a photon coming straight at you from your computer screen.\u00a0 Light can be either horizontally polarized (the <strong>X<\/strong> state, corresponding to an electric field that points in the \\(x\\) direction) or it can be vertically polarized (the <strong>Y<\/strong> state, corresponding to an electric field that points in the \\(y\\) direction).\u00a0 Now since physics is rotationally symmetric, it&#8217;s obvious that if light can be horizontal or vertical, it can also be <em>diagonal<\/em>.\u00a0 So you might have na\u00efvely thought the photon would have infinitely many possible states.\u00a0 And in a sense this is true, but each of these diagonal states is really just a <em>quantum superposition <\/em>of the <strong>X<\/strong> and <strong>Y<\/strong> states.<\/p>\n<p>Yet on a plane, the choice of axes is arbitrary.\u00a0 You can rotate the coordinate system by 45\u00ba, and it would be just as good as the original coordinate axis.\u00a0 In the same way, we are currently thinking of <strong>X<\/strong> and\u00a0<strong><strong>Y<\/strong><\/strong> as the two possible states of the system (with every other state being a superposition of <strong>X <\/strong>and\u00a0<strong><strong>Y<\/strong><\/strong>)\u2014but this is an arbitrary choice!\u00a0 We could just as well say that every state is a superposition of \\((\\mathbf{X} + \\mathbf{Y}) \/ \\sqrt{2}\\) and \\((\\mathbf{Y} &#8211; \\mathbf{X}) \/ \\sqrt{2}\\)!\u00a0 So actually every state is a quantum superposition, of certain other states.<\/p>\n<p>Although the choice of coordinate axis is arbitrary, it is important that the states you pick are all &#8220;orthogonal&#8221; to each other (i.e. at right angles in the Hilbert space).\u00a0 That is what tells you that it represents a set of\u00a0 mutually exclusive possibilities.\u00a0 Any such set of N orthogonal states is called a <em>basis <\/em>of the Hilbert space.\u00a0 (The plural of &#8220;basis&#8221; is &#8220;bases&#8221;, pronounced BASE-EES.\u00a0 Just like the plural of &#8220;index&#8221; is &#8220;indices&#8221;.)\u00a0 A basis gives the possible set of outcomes for some particular way to measure the system.<\/p>\n<p>For example, suppose we start with a diagonal photon in the \\((\\mathbf{X} + \\mathbf{Y}) \/ \\sqrt{2}\\) state, and we <em>measure it<\/em> to see whether it is horizontally or vertically polarized.\u00a0 (Maybe by passing it through some kind of material in which these two polarizations follow different trajectories.)\u00a0 What happens?<\/p>\n<p>Well, people disagree about interpretation (what is <em><\/em><em>ultimately<\/em> going on), but everyone agrees on the practical set of rules you&#8217;d use in the laboratory.\u00a0 We just look at the state \\((\\mathbf{X} + \\mathbf{Y}) \/ \\sqrt{2}\\).\u00a0 It has an amplitude of \\(1\/\\sqrt{2}\\) to be \\(X\\), and also \\(1\/\\sqrt{2}\\) to be \\(Y\\).\u00a0 By the Born Rule, we&#8217;ve got to square these numbers, so we get a 1\/2 chance for it to be horizontal, and a 1\/2 chance for it to be vertical.<\/p>\n<p>Let&#8217;s suppose it turns out to be vertical (the <strong>Y<\/strong> state).\u00a0 Then from now on, the particle behaves just as if it had been in the <strong>Y <\/strong>state all along.\u00a0 (This is called &#8220;projection&#8221; or sometimes &#8220;collapse of the wavefunction&#8221;; but see my remarks on decoherence earlier in this post.)\u00a0 For example, if we measure it a second time to see if it is in the <strong>Y<\/strong> state.\u00a0 If we check to see whether it is in the\u00a0<strong>X<\/strong> state, it is definitely not.<\/p>\n<p>But now we can ask a separate question: is it in the \\((\\mathbf{X} + \\mathbf{Y}) \/ \\sqrt{2}\\) state, or the \\((\\mathbf{Y} &#8211; \\mathbf{X}) \/ \\sqrt{2}\\) state?\u00a0 This corresponds to sending it through a different kind of filter, which discriminates between the two 45\u00b0 diagonal polarization choices.\u00a0 We would then find a 1\/2 chance of it being the former, and a 1\/2 chance of it being the latter.<\/p>\n<p>Supposing it turns out to be \\((\\mathbf{Y} &#8211; \\mathbf{X}) \/ \\sqrt{2}\\), this is a bit paradoxical.\u00a0 Since if we had just started off asking whether the \\((\\mathbf{X} + \\mathbf{Y}) \/ \\sqrt{2}\\) photon was in the \\((\\mathbf{Y} &#8211; \\mathbf{X}) \/ \\sqrt{2}\\) state, Nature&#8217;s answer would have been &#8220;Nope.\u00a0 Definitely not.\u00a0 Those states are orthogonal and therefore if it&#8217;s the one, it&#8217;s not the other!&#8221;<\/p>\n<p>But somehow, merely by answering a series of questions about the photon&#8217;s polarization, we managed to trick Nature into converting the photon from its original polarization to one 90\u00b0 away, which is inconsistent with the first.\u00a0 <em>By measuring the photon we have affected it<\/em>!<\/p>\n<p>So we see that, somehow, we can get the photon to be definitely &#8211; or | polarized, or definitely \/ or \\ polarized.\u00a0 But we can&#8217;t get both of these things to be definite simultaneously.\u00a0 This is an <em>uncertainty relationship<\/em>.\u00a0 It&#8217;s analogous to the &#8220;Heisenberg uncertainty principle&#8221; where you can&#8217;t measure position and momentum at the same time; so that measuring one makes the other uncertain.\u00a0 (Although it&#8217;s not exactly the same, since position and momentum are continuous variables, while each polarization choice is a yes-no question.)<\/p>\n<p>In the case we are considering, we&#8217;re been lucky that the Hilbert space is directly related to two dimensions of the physical space.\u00a0 That means that the rotation of axes in the Hilbert space is the same thing as a rotation of physical space.\u00a0 In general, however, we are not so lucky and the Hilbert space is more abstract.\u00a0 But it is still true that there are a bunch of different possible bases of the Hilbert space, that are related by rotations in the Hilbert space.\u00a0 (Since the Hilbert space is complex, we are really only interested in those rotations that don&#8217;t mess with the notion of &#8220;multiplying-by-\\(i\\)&#8221;.\u00a0 These are called <em>unitary transformations<\/em>.)<\/p>\n<p>As long as I&#8217;m talking about complex numbers, I should mention that there&#8217;s also such a thing as <em>circularly <\/em>polarized photons, which involve complex superpositions like \\((\\mathbf{X} \\pm i \\mathbf{Y}) \/ \\sqrt{2}\\).\u00a0 But most of the bizarreness of superpositions can be illustrated without thinking about complex numbers.<\/p>\n<p><a title=\"Quantum Mechanics III: Wavefunctions\" href=\"http:\/\/www.wall.org\/~aron\/blog\/quantum-mechanics-iii-wavefunctions\/\">Continue to the final post<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Strictly speaking, most of the other rules about QM are already implicit in what I&#8217;ve already said.\u00a0 But a few implications of this setup are worth pointing out. First note that, in QM, the &#8220;state&#8221; includes information about every single object in the system.\u00a0 So, when you add up the different histories, they only interfere [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"class_list":["post-2028","post","type-post","status-publish","format-standard","hentry","category-physics"],"_links":{"self":[{"href":"https:\/\/www.wall.org\/~aron\/blog\/wp-json\/wp\/v2\/posts\/2028","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.wall.org\/~aron\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.wall.org\/~aron\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.wall.org\/~aron\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.wall.org\/~aron\/blog\/wp-json\/wp\/v2\/comments?post=2028"}],"version-history":[{"count":105,"href":"https:\/\/www.wall.org\/~aron\/blog\/wp-json\/wp\/v2\/posts\/2028\/revisions"}],"predecessor-version":[{"id":4690,"href":"https:\/\/www.wall.org\/~aron\/blog\/wp-json\/wp\/v2\/posts\/2028\/revisions\/4690"}],"wp:attachment":[{"href":"https:\/\/www.wall.org\/~aron\/blog\/wp-json\/wp\/v2\/media?parent=2028"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.wall.org\/~aron\/blog\/wp-json\/wp\/v2\/categories?post=2028"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.wall.org\/~aron\/blog\/wp-json\/wp\/v2\/tags?post=2028"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}