The quantum pigeonhole principle and the nature of quantum correlationsby Yakir Aharonov and five less famous co-authors. Their paper claims to disprove the apparently trivial proposition

If you put 3 pigeons in 2 pigeonholes, at least one pair of pigeons (12/23/31) ends up in the same pigeonhole.It should be true because the number of pigeonholes is lower than the number of pigeons so they can't be hosted in accordance with the Pauli exclusion principle.

But is that statement right quantum mechanically? The authors claim that the seemingly obvious proposition isn't right which is why quantum mechanics assaults the "very core" of what numbers are and all the mathematics – lots of big words.

One must avoid prejudices. The technical claim in the paper may be right or wrong, and they may either say that they have found a problem with quantum mechanics (like hundreds of cranks love to claim these days) or not (just another example of the failures of the common sense). Which option is right?

First of all, one must understand the precise technical interpretation of the proposition "two pigeons are in the same pigeonhole", and so on. But informally, if you only look at the original slogan about the three pigeons, it is absolutely clear that the slogan holds quantum mechanically, too. After all, as my words below showed, it's nothing else than the reason why lithium, an atom with three electrons, cannot place all of its three electrons in a \(1s\) shell. There are only two pigeonholes in that shell!

So the claim actually holds both in classical physics and quantum mechanics when interpreted correctly.

All of chemistry – an application of atomic physics (quantum theory of atoms) – is based on the fact that the claim continues to hold in quantum mechanics and it's actually more important than it ever was in classical physics – because classical physics, due to its unavoidably continuous nature (and lack of discrete spectra) has never allowed us to discuss nice, sharply separated pigeonholes (discrete states for particles).

So what are they doing so that they end up claiming that the apparent tautology is violated in quantum mechanics? Consider three particles or pigeons with labels \(1,2,3\) that are prepared in the following unentangled (=simple factorized tensor product) state\[

\ket\Psi = \ket{+}_1 \ket{+}_2 \ket{+}_3

\] where each of the three tensor factors is a simple superposition of the particle's being in the "left pigeonhole" (L) and the "right pigeonhole" (R):\[

\ket{+} = \frac{1}{\sqrt{2}} \zav{ \ket L + \ket R }

\] They also told us that by the condition "pigeons 1,2 are in the same pigeonhole", they mean the simple pigeonholes L and R.

The problem with the paper is that the initial state and the definition of the observables we are going to measure already tell us

*everything*we need to know to calculate the probability that a pair of pigeons ends up in different pigeonholes. And let me tell you in advance that the probability that each of the three pairs \(12,23,31\) is found in different pigeonholes is zero, in agreement with the classical intuition.

How do we prove that? Define the operator that answers the question "are pigeons \(1,2\) found in two different pigeonholes?" By cyclical permutations, analogous operators may be written down for the pairs \(1,3\) and \(2,3\). For \(1,2\), we have the projection operator\[

\Pi_{1,2}^{\rm diff} = \ket{L}_1\ket{R}_2 \bra{R}_2\bra{L}_1

+\ket{R}_1\ket{L}_2 \bra{L}_2\bra{R}_1

\] It's simple: we are projecting to the subspace spanned by the basis \(\ket{LR}\) and \(\ket{RL}\). The operator above is a Hermitian projection operator. Its eigenvalues \(0\) and \(1\) correspond to the answers (to the aforementioned question) "No" and "Yes", respectively. If you aren't looking for a linear Hermitian projection operator whenever you are thinking about any "Yes/No" question about a physical system, then you haven't started to think quantum mechanically yet.

If you look carefully, you will see that in the paper, the equation 5b is screwed up. In all the bra vectors, the labels \(L\) and \(R\) are interchanged relatively to the correct form. We are talking about two pigeons' being in

*different*holes but the pairing to \(L1,R2\) or \(L2,R1\) must still be the same for ket vectors and bra vectors! (The errors probably occurred because they began to write the subscripts on the left side and got confused by them.)

But OK. They don't use this invalid formula, anyway. Instead, they also define the corresponding operator for the question "are \(1,2\) in the same pigeonhole?" which is obviously just "one minus the operator above", or, equivalently\[

\Pi_{1,2}^{\rm same} = \ket{L}_1\ket{L}_2 \bra{L}_2\bra{L}_1

+\ket{R}_1\ket{R}_2 \bra{R}_2\bra{R}_1

\] When this operator acts on the initial state \(\ket\Psi\), we get\[

\Pi_{1,2}^{\rm same}\ket\Psi =\frac 12 \zav{ \ket{L}_1\ket{L}_2 + \ket{R}_1\ket{R}_2 } \ket{+}_3

\] The state of the pigeon \(3\) that is found in \(\ket{+}\) wasn't changed. However, the mixed \(LR,RL\) states for pigeons \(1,2\) were eliminated.

This result isn't a multiple of \(\ket\Psi\), so \(\ket\Psi\) isn't an eigenvector of \(\Pi_{1,2}^{\rm same}\). Instead, you may calculate the expectation value of the projection operator\[

\bra{\Psi}\Pi^{\rm same}_{1,2}\ket{\Psi} = P^{\rm same}_{1,2}

\] which is the right formula for the probability that the answer is "Yes". Because \(\Pi^2=\Pi\) (it is the projection operator), this probability is the same thing as the squared norm of the state \(\Pi\ket\Psi\), and it is \((1/2)^2+(1/2)^2=1/2\). There are many ways to quickly see this result. Without the operator in the middle, the expression above would be a sum of eight terms equal to \(1/8\). The projection operator erases four (one-half) of them, so the probability is \(4/8=1/2\) that the pigeons \(1,2\) are found in the same state.

Now, can you find \(1,2\) in different states, \(2,3\) in different states, and \(1,3\) in different states? What is the probability of that (namely that all these conditions are obeyed at the same moment)? Well, if we are naive, this question may be associated with the projection operator\[

\Pi_{\rm paradox}=\Pi_{1,2}^{\rm diff}\Pi_{2,3}^{\rm diff} \Pi_{3,1}^{\rm diff}.

\] You should check whether the three operators \(\Pi_{i,j}^{\rm same}\) for \(ij=12,23,31\) commute with each other. This is an important thing to check because non-commuting observables cannot be measured simultaneously. The paper is completely sloppy about all such things – in fact, you won't find the word "commute" in the paper at all!

The operators \(\Pi_{i,j}^{\rm same}\) (and "diff") commute with each other for a simple reason: all of them are diagonal in the basis containing vectors \(\ket{LLL},\ket{LLR}\), and six more vectors like that. And diagonal (in the same basis) operators simply commute with each other. You may prove the vanishing commutator in many other ways, too.

The correct claim that the operators \(\Pi_{i,j}^{\rm same}\) commute with each other was included in the first version of this blog post. In the second, revised version that was posted for a few minutes, I mixed the operators with other, seemingly "similar" but ultimately different projection operators "capital upsilon"\[

\Upsilon_{1,2}^{\rm same} = \frac{ 1+\pi_{1,2} }{2}

\] where \(\pi_{1,2}\) is the self-explanatory operator permuting the pigeons \(1,2\) – or, equivalently, the operator switching the complex amplitudes in front of \(\ket{LR}_{12}\) and \(\ket{RL}_{12}\). It's easy to see that \(\Upsilon_{1,2}^{\rm same}\) doesn't commute with \(\Upsilon_{2,3}^{\rm same}\) because \(\pi_{1,2}\) doesn't commute with \(\pi_{2,3}\). When the product of these two \(\pi\) unitary operators is written in one order, you get the operator corresponding to the cyclic permutation \(123\to 231\) while if you write it in the other order, you get the operator corresponding to \(123\to 312\).

These are different unitary operators for cyclic permutations because they are different permutations. So \(\Upsilon_{1,2}^{\rm same}\) and \(\Upsilon_{2,3}^{\rm same}\) cannot be measured simultaneously. They can't be simultaneously diagonalized.

Because these projection operators don't commute, the triple product \(\Upsilon_{12}^{\rm same}\Upsilon_{23}^{\rm same}\Upsilon_{31}^{\rm same}\) isn't the right operator for the hypothetical "combined" question. It actually isn't a projection operator; and it isn't necessarily Hermitian, either. So this is would be a wrong way to operationally define the question whether each pair of pigeons occupies different pigeonholes (or a similar question with \(\Pi\) replaced by \(\Upsilon\)). If you measure the three pairs one by one, the three measurements just won't commute with each other, so you can't interpret the result of the second and third measurement as telling you something about the initial state before the first measurement!

But back to the \(\Pi\) operators. With these operators, the triple product \(\Pi_{\rm paradox}\) defined above is identically equal to zero. Like the three individual \(\Pi\) operators, their product is diagonal in the basis \(\ket{LLL},\ket{LLR},\dots ,\ket{RRR}\). That's why the three qubits behave as three classical bits and you may see that each of the eight diagonal elements is zero as well. They're zero for the usual classical reason. There is no basis vector among these eight – an arrangement of three letters L/R – that would guarantee that each pair of the letters among the three are two different letters. ;-)

**So in contrast with the claim of the paper, the paradoxical situation proposed by Aharonov et al. has a vanishing probability to occur, regardless of the initial state. The paradox is impossible!**

There are lots of valid operators to measure. For example, you may define the number operators \(N_L,N_R\) for the number of pigeons in the left and right pigeonhole, respectively. On this 8-dimensional Hilbert space, they will identically obey \(N_L+N_R=3\) which is why they obviously commute with one another, too. This isn't directly relevant for the question whether "each pair is in different pigeonholes".

You may also try to construct the Hermitian projection operator expressing the possibility that "each of the three pairs is found in different pigeonholes". If you construct it as a sum of simple projection operators on the subspaces spanned by \(\ket{LLL}\) or its 7 counterparts, you will find out that none of these eight basis vectors really obeys the condition, so none of the corresponding projection operators will be added to yours. Your operator corresponding to the paradox will therefore obey \(\Pi_{\rm paradox}=0\) identically. The paradox cannot occur.

There is another way to mathematically express the fact that the paradox cannot occur. If there were a state in which all the three pairs \(12,23,31\) are found in different (\(L\) vs \(R\)) pigeonholes, such a state \(\ket\Phi\) would have to be a tensor product of a Bell-like state for \(1,2\) and a state for the third pigeon, but it would have to admit two similar tensor decompositions related to the first one by cyclic permutations. No such (nonzero) state exists. This claim is really equivalent to the widely discussed (in the quantum gravity community) "monogamy of entanglement". In this sense, the well-known principle of "monogamy of entanglement" is what shows that they are wrong when they claim that the paradox may occur.

**Measuring something else than what you measured**

To make their paper really messed up, they simultaneously talk about a different measurement, well, three more measurements, whether the pigeons 1,2,3 are found in the state \(\ket{+i}\) or in the state \(\ket{-i}\) where\[

\ket{\pm i}_{j} = \frac{1}{\sqrt{2}} \zav{ \ket{L}_{j}+i\ket{R}_{j} }, \quad j=1,2,3

\] The measurements of these three particular quantum bits may be done. But they cannot be done simultaneously with the measurements of \(\Pi_{1,2}^{\rm same}\) and its two cyclic counterparts!

Just like in all the textbook situations of quantum mechanics, the operators also refuse to commute with each other, i.e. \(\Pi_{1,2}^{\rm same}\) doesn't commute with the projection operator on \(\ket{+i}_1\) for the first particle,\[

\Pi_{1,+i} = \ket{+i}_1 \bra{+i}_1,

\] just check it, so the measurements of the \(\pm i\) qubits for the three particles modify the values of the observables \(\Pi_{i,j}^{\rm same}\). They modify the answer to the question whether a pair of pigeons is in the same state. It's exactly the same principle as the fact that the measurement of the position inevitably modifies the momentum and vice versa.

So if pigeons 1,2 are found in different pigeonholes after you measured their \(\pm i\) qubit, it may be a consequence of the \(\pm i\) measurement (and its unavoidable effect on the system), not a property of the initial state.

To make things really hopeless, they also talk about "post-selected" states but they're actually different states than the states you get by the projection corresponding to the measurement you just did!

Undergraduate students who would write such a preprint as their term paper should get an F. It's a pile of hopeless rubbish showing that they have misunderstood pretty much all the basic principles of quantum mechanics, especially the fact that any measurement – any procedure that finds some nontrivial information about any physical system – inevitably modifies the state of the physical system (and in particular, changes the distribution of all observables that don't commute with the just measured one).

Note that this mistaken opinion that you may circumvent the uncerttainty principle – to find a value of a measurable without appreciating that this "finding out" (=measurement) generally changes the state and the distribution of other observables not commuting with the just measured one – is also hiding in the foundations of the "weak measurement" pseudoscience. And yes, Aharonov has been writing "influential" papers about that rubbish, too. The fresh new paper doesn't use the term "weak measurement" at all but the elementary misconceptions behind this paper and the "weak measurement" papers are pretty much isomorphic.

Mr Aharonov and his fellow crackpots whose names I refuse to memorize: Why don't you pick a credible basic textbook of quantum mechanics, like Dirac's textbook, and learn the basics of quantum mechanics? You have violated pretty much all of them.

I hope that the text above has also answered the question whether it's possible to distribute three big pigeonholes into the brains of six physicists so that the paper that they write together is as brainless as a paper written by one physicist with one big pigeonhole in his skull. The answer is Yes.

Thanks, the 2-vs-3 mistake was fixed before I read your comment but after you posted it. ;-)

ReplyDeleteWell, lithium has Z=3 and "3" is still pretty important number (a low integer) so you can't be surprised that Lisa and others sometimes talk about it!

I think that they still talk about lithium less frequently than about Z=1 hydrogen and Z=2 helium. ;-)

Lubos - Nice take down. I'm glad to see that it hasn't disturbed your equanimity after the recent flood of nonsense.

ReplyDeleteFor some time it has seemed to me that Aharonov has created a cottage industry which produces these sleight-of-hand tricks for the amateur prestidigitators of the philosophy of quantum mechanics. I still find it a bit strange since he wasn't always such a dope.

Pure assholery with a pinch of condescendence.

ReplyDeleteA proper spelling should be Lisium.

ReplyDeleteI had not heard that phrase, Lubos, but it makes perfect sense. Closely related is, "There's no such thing as a victimless crime."

ReplyDeleteThe concluding paragraph: "a new quantum effect"(Read: a great discovery) "Requires us" (Read: All Physicists) "to revisit basic notions of quantum mechanics" (Read: Rewrite the Textbooks). I love how the next sentence admits it's "early" but then characterizes their own paper as "major." To hell with peer review!

ReplyDeleteTRF is the place where crappy quibbles with QM end up and get cleaned out by its chief janitor! :)

ReplyDeleteCarbon gives you three pigeons,

ReplyDeletehttp://physicsworld.com/cws/article/news/2014/jul/08/carbon-nucleus-seen-spinning-in-triangular-state

If one reacts Meldrum's acid anion with Hydride Sponge, does there obtain a stable five-coordinate carbon? "8^>)

This is sort of like Obamacare. Even if you agree with the ideology, which is crazy, the implementation and details are so deeply flawed and awful that it can never work.

ReplyDeleteI think Francis should be congratulated for the discovery that RFP was actually human! I had long believed he was really some shape-shifting lizard alien come to earth to give us the benefit of his superior extra-terrestrial intellectual powers.

ReplyDeleteBut no, we learn that, along with a not insignificant fraction of the world's population, he would sleep with a pretty girl, given half a chance.

The astonishing insight that Francis displays is to be rewarded, not mocked, surely :-)

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ReplyDeleteDear Lubos,

ReplyDeleteWhat I find strange while glancing at the article is that I cannot find a remark that indicates whether or not the pigeons are indistinguishable/distinguishable and if indistinguishable whether they are bosons or fermions. They say for example “consider three particles”. Is that identical or non-identical particles? That’s sloppy isn’’t it?

In the paper, they're definitely a priori distinguishable, non-identical particles a priori, even though many of the states they consider are compatible with the Bose-Einstein statistics.

ReplyDeleteTo have the "default" expectation that particles are identical is a "professional deformation" of people who mostly work with quantum field theory or its extensions, and people like the authors of this paper simply aren't that far.

Svik,

ReplyDeleteI understand now and take it back.

It's very easy to get the wrong impression with the written word where there's no accompanying body language to suggest the remark is made in jest, even if one reads carefully.

Unfortunately, the only way in general to avoid doubt—as far as I can see anyway—is to use smileys. I don't particularly like them. I think they're twee. But they're by far the lesser of two evils, if one remembers to stick them in, that is. :)

No. Smolin has already been vindicated. There has only ever been one testable prediction of string theory, supersymmetry. And that's considered falsified now that the LHC has failed to find any partner particle for so long.

ReplyDelete"An undergraduate in spirit for life" - I like that description Dr. Motl!

ReplyDeletehttp://www.youtube.com/watch?v=qWabhnt91Uc&list=PL9B57112BE4DA7090

Dear Lubos,

ReplyDelete"There is no basis vector among these eight – an arrangement of three letters L/R – that would guarantee that each pair of the letters among the three are two different letters. ;-)"

This is all you have to say, haven't you. By linearity you can infer the required property from the basis vectors to all vectors in the Hilbert space. A nice exercise for a beginners course in quantum mechanics. Really scary.

I will remember to put on lots of smilies ;--)

ReplyDeleteBy the way I searched for the most common 50000 words websites in English. One site has 42000 from common tv scripts and I recognized all on the last block except 1 which was barbitus. Which is Latin for vearded . So it looks like our vocabulary is at least 50k words.

Now if you know Latin and Greek it should be easy to top 100k.

Dear Lubos: As usual, you have solid arguments on your side. Would you be willing to invite Aharonov or some one from his group to defend their case on your blog? The second point is that their previous work on weak measurement and post selection did get through referees presumably because of Aharonov's name. There was a physics today article also. Do you see any connection between that paper and the present article?

ReplyDeleteSorry. I noticed you do mention weak measurement in the last paragraph. But I was wondering about more direct connection which might give them a loophole.

ReplyDeleteDear Kashyap, the weak measurement (first) paper not only got through the peer review, It created a huge industry of followup papers.

ReplyDeleteYes, of course that I see connections between the weak measurement papers and the present one. It's based on the same wrong assumption that Heisenberg was wrong and you can measure things without affecting them.

The weak measurement, as previously defined them, isn't a measurement. It's something that falls out of the spectrum and produces an error whose exact behavior depends on the precise way how the non-measurement was done. That's why you could never interpret the result of a single "weak measurement" as reliably telling you something, like whether two pigeons are in the same pigeonhole.

In this new paper, just like if it were another weak measurement paper, they ignore this fact and pretend that they measure the Yes/No questions *and* manage not to change the state of the pigeons. But that's not possible.

"

ReplyDeleteNow if you know Latin and Greek..."I don't!

I simply happened to be sitting in the class for a number of years when it was forced on all us young boys, and I never 'officially' did Greek. I have no linguistic tendencies at all. Canings were frequent. :)

Yes, I was the consummate linguistic duffer

par excellence, mit oak leaf cluster, and bar. I never enjoyed learning foreign languages—after all, English was bad enough—beyond finding out how their basic structures worked. The grammars were interesting but I found the prose exercises tediousin extremis. :)It occurred to me once that I might try reading Newton's

Principiain the original Latin. A couple of milliseconds later it occurred to me what a dumb idea that was! So I never did. :)On a more serious note, I have long since been grateful that I had no choice in the matter, and I have a good deal of affection for those old masters, bastards though I thought they were at the time. Hindsight through rose-tinted glasses? :) No, not really — it's genuine enough. I think people lose out by not having done it. Boys these days don't get the chance.

Yes we need more teachers with big yard sticks ;--). And nun schools too with fast rulers.

ReplyDeleteThe best people are trained is schools like this. Romanians are trained like this and make very good employed.

By the way are you awailable to work. ;----)

Probably I misunderstood your post or the paper (nonexclusive "or"), but it seems to me that you criticize something else. If I understand the paper correctly, what the do and state is: prepare the system in the |+>|+>|+> state and measure all three particles for each you get +-i and the state is one of 8 possible. In all cases something interesting happens. Say it is in the |i>|i>|i> state. Then ask the question prior the measurement can we say that two particles were in the same box. No measurement is involved here and no question whether they are in the left or in the right box. Then the answer is that you cannot say that they were in the same box. Because that would mean that the state of the system was in a state orthogonal to the state you've just found it in. This is interesting but in a way not so surprising. It is reminiscent to the which slit did the particles go through if you observe an interference pattern on the screen. You cannot say that they went through exactly one of the slits. The difference is that you need many trials.

ReplyDeleteDear BBB, it's physically meaningless - strictly forbidden by the most basic laws of quantum mechanics - to ask what the value of an observable is before the measurement.

ReplyDeleteAn observable only has a well-defined value when it's actually measured.

Dear Lyubosh, here is an example. Suppose I have a qubit in the state spin down. Can I say that it is not in the state spin up without measuring? Yes, I can. Because if I do measure, in the up-down direction, I will not find spin up 100% of the time. If I had just measured spin down, can I say that prior to the measurement it wasn't in spin up state? Yes, I can. Because had it been in the up state I wouldn't have measured down. Right? There is nothing controversial here.

ReplyDeleteThe problem is in the word "suppose". The only way to "suppose" that an observable has a particular value is to measure it. Quantum mechanics just doesn't allow to "suppose" values of observables that are not measured.

ReplyDeleteSo you can't "suppose" that the spin is not spinning up because you can't even "suppose" that it is spinning down. Spinning up and not spinning down is exactly the same thing for the j=1/2 system so this whole game of yours with negation is completely vacuous and has nothing whatever to do with your mistake.

Dear Lubos, if you read more carefully you will see that I never mentioned observables, so you could've saved your 5000 shouts. I said suppose the system is in down state. Of course that means that it has been prepared in that state, which involves measurement. What was unclear about that! After I have measured the qubit and the result was down, I can say that it is (now) in the down state and that it is not in the up state. Right? I can also say that before the measurement it couldn't have been in the up state. Right? Just to be clear, I am not saying anything about values of observables. I am saying that whatever the state was before the measurement it wasn't |up>. Values of observables don't exist prior to measurements, it is meaningless to talk about them, so stop repeating that. But surely, you will agree that the system is in some state at any time, and it is meaningful to ask questions about the state of a system. NOT values of observables.

ReplyDeleteNo, no, no. First of all, they do talk - and you did talk - about objects' being in some state in the classical sense even though the corresponding observable obviously and demonstrably was *not* measured. By that, they violate the basic rules of quantum mechanics.

ReplyDeleteAt the beginning of your latest comment, you pretend that you are not making this mistake, but at the end of the comment, you are explicitly doing it again. You ask whether I agree that the object is in a specific state at any moment. I surely agree that it is.

Holy crap, I surely don't. This is what I just wrote 5,000 times. Objects are not in any specific state prior to the measurement. It is absolutely wrong to make any classical assumption about the objects' state before the measurement. How can one say this elementary things more clearly so that even retarded people start to get it?

Ok, this is your blog and your rules, so I will stop this discussion as it seems to lead nowhere. But I want one more clarification. You say that objects are not in a specific state prior measurement. I don't know what to make out of this! At least not in the context of the example above about the qubit. The state is a vector(well a line) in the corresponding Hilbert space. What does it mean for the qubit not to be in a specific state!? It doesn't have a wave-function!

ReplyDeleteDear BBB, you previously meant a specific state in the classical sense - assigning sharp values of observables. It is a basic feature of quantum mechanics that physical systems don't have well-defined values of observables.

ReplyDeleteOne may describe the *knowledge* about a physical system by a state vector, an element of the Hilbert space, but one must realize that 1) this vector is in no way an objective fact about the world that different observers would have to agree about, 2) perhaps even more importantly, such a state vector inevitably, unavoidably, and ünäwöitábly, changes in the process of the measurement, so it is definitely wrong to think that the sharp eigenstate prepared by the measurement is the same state that the physical system had before the measurement.

The measuments inevitably modify the measured system - they "collapse" it in a way, if I use the misleading language of popular books - have you really never heard of that? It is a basic feature of any measurement in quantum mechanics in general and every quantum mechanical theory in particular.

Ok, then what is wrong with the example. I have a quibit, which I measure and obtain result 'down'. Now I can say that the state vector of the qubit is |down>. Before the measurement the state vector could have been |down> or it could have been something else, any of a continuum possibilities. But I can also say that before the measurement the state vector of the qubit was not |\up>, otherwise I couldn't get 'down' from the measurement. Now going back to my first comment, I may misunderstand the paper (or your post) but the way I understood the paper their statement is similar to that of the qubit example. After the measurement if the result is 'i' for all three, the state vector of the system is |i>|i>|i> and before the measurement the state vector couldn't have been |two are in the same box>.

ReplyDeleteI have answered this idiocy of yours about 6,000 times already. You're such an immense waste of time, such a persistent moron, that I have placed you on the blacklist.

ReplyDeleteYou can't just that the spin wasn't "up" before the measurement just because you measured it to be "down" during the measurement. The result "down" is partly a result of the measurement and its unavoidable influence on the measured system. It may occur whenever the probability for "down" was nonzero before the measurement, but that doesn't mean that the probability was zero to have "up".

If one describes the system by a generic superposition of "up" and "down", it's just wrong to interpret this state by saying that the "state is not up". It *is* up with some probability, and it is down with some probability.

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ReplyDeleteYes, the very idea that the roles of men and women in these matters are symmetric is flawed. The text which you linked highlights this although I would regard it more a matter of common sense and experience than of science.

ReplyDeleteCheck this paper out.

ReplyDeleteObservation of a quantum Cheshire Cat in a matter wave

interferometer experiment

http://arxiv.org/pdf/1312.3775v1.pdf

Sorry for being late to the party, but I just came across the pigeon-hole paper and am new to this blog. My QM knowledge is pretty basic (though greater than zero), and would like to ask Lubos what he would expect the actual experimental result to be, following the three electron experiment, including post-selection at the detector. The authors at http://arxiv.org/pdf/1407.3194v1.pdf show Fig 2(a). From what I gather, you would expect the right hand configuration to be observed? And the left-hand configuration to be wrong? Or maybe even some other configuration? Thank you!

ReplyDeleteChrist, thank you for blacklisting him. I started to believe he was strictly trolling.

ReplyDeleteThanks for this article discussing the now-famous "quantum pigeon-hole effect" which is clearly plastered on New Scientist's front page for days now. It's also on ArsTechnica which I frequent.

ReplyDeleteI couldn't wrap my head around the extremely vague comments in the pop-sci sites so I went to the original paper. Admittedly I do not understand the math well - I'm an electrical engineer so my QM background is very limited. But after I understood the premise of their proposed experiment, I was thinking that it just wasn't making sense (QM sense). Math aside, their claims seem unintuitive (QM-wise). I will have to mull over what you've posted here a few more times but it sounds a whole lot more valid.

Interesting that this is the author of the 'weak-measurement' stuff... also made pop-sci headlines... also seemed impossible and illogical to me at the time it was making news. I'm going to have to remember Aharonov's name so I can spot garbage before wasting much thought on it.

Lubos, why do you have to be such an unfailing asshole? You make some valid criticisms of the paper, but you then write yourself out of the dialogue with all these vicious ad hominem attacks. You're the one who ends up sounding like a crackpot. So why do you do it?

ReplyDeleteProvided we have a spin 1/2 particle, so |up> is then orthogonal to |down>, BBB's last comment is correct in the following sense: You can exclude the possibility that someone else set/measured the system to be in the state |up> just prior to you measuring "into" |down>. I suppose this is what BBB meant in the last comment.

ReplyDeleteDoes their description of the electron MZI make sense to you? I've got only a rudimentary knowledge of QM, but I can't follow this claim:

ReplyDelete|+> = 1/sqrt2 (|L> + |R>)

This is the state after BS1. Then they define:

|± i> = 1/sqrt2( |L> ± |R>)

What is the "i" doing in there? The RHS appears to simply be |±>. This differs from the definition on page 1, where:

|+i> = 1/sqrt2( |L> + i|R>)

Anyway, they then say:

"If the state before PS is |+i> the electron ends, with certainty at D1 while if it is |−i> it ends with certainty at D2."

How could the state before PS be anything but |+i> (=|+>)?

Sorry for the poor notation (and for the novice question).

I think you've confused yourself. I was in this |confused> state for a few days too. The problem is their _wording_ in certain critical parts of the paper is totally misleading. But what they do is really simple and can be summarised thus: imagine a system in a state |S>. Assume this system could collapse with nonzero probabilities to either of the two _orthogonal_ states |A> and |B>. Now perform S. Let's say the answer is "yes", so the system collapses to |S>. Now ask about A. Again, suppose the answer is "yes", so the system collapses further to |A>. Now according to QM the following is true: if you ask now about B, you are guaranteed to get the answer "no" (since |A> and |B> are orthogonal). For example, using a single spin-1/2 particle consider |S> = |spin up>, |A> = |spin left> and |B> = |spin right>. So this is all they do, with their |S> equal to |Psi>, their |A> equal to |Phi>, and their |B> equal to |Psi'>. The reason for the initial preparation state |Psi> is to weed out the possibility of their |B> being NOT orthogonal to their |A> (this non-orthogonality would happen if, for example, the second coefficient in front of |R>|R> in their formula (6) was "i" instead of "+1". QED.

ReplyDeleteWhat you write is just not true - it's an elementary misunderstanding of quantum mechanics. What's wrong is the heart of the content, not just the wording.

ReplyDeleteOnce you measure S, you inevitably change the measured object, so even though you measured a different AvsB variable later to be A, it does *not* mean that the variable wasn't B before you measured S.

It's the paper's abysmal wording that's tripping you up. The point is that you first perform S, then perform A. If the answers to both S and A were "yes", then if B were to be performed NEXT (i.e. AFTER A), it would definitely say "no". What's confusing in the paper (it's probably the result of too many authors in the kitchen) is that they silently switch A and B around when they discuss the experimental setup (such reversal is allowed for what they are after). And then on top of that, some of that "reversed" terminology has creeped to the first part (probably some cutting and pasting gone awry), suggesting (incorrectly) that B was being merely _inferred_ to exist before A. Most explanations of this paper in pop-sci magazines contain the same mistake.

ReplyDelete(such reversal is allowed for what they are after)

ReplyDeleteBy the way, very recently, Em. Prof Stephen Parrott released a criticism of the pigeon paper

ReplyDeletehttp://www.math.umb.edu/~sp/pigeonco.pdf

that is very similar to mine except that his is much wittier. I will post it as a guest blog if he agrees and sends me the source.

Forget about the reversal then, it only applies to the experimental setup. Let's just pretend the paper is only about theory (pages 1 and 2). So the point is there is nothing wrong with their main claim on those pages, it's just that the paper is VERY BADLY written, to the point of being 100% misleading, mainly through the wrong use of the word "intermediate" in the left column on p. 2 of the paper. You have to besically reverse engineer the argument yourself and ignore their description.

ReplyDeleteI don't see Motl's reply about this claim. What say you, Lubos? Here is my response to it:

ReplyDeletehttp://arxiv.org/abs/1410.1522