
#91
Jan1212, 05:17 PM

P: 661

http://www.amazon.co.uk/reader/03064...TF8&query=moon (The relevant section starts on p 273 called 'When Can We Ignore Symmetrization and AntiSymmetrization?') The point is that the the type of effect Cox tried to popularize, is in fact completely negligible in practice, even if quantum mechanics, as we currently formulate it, is exactly theoretically correct. But he did link to lecture notes where this point was made explicit to ~50 decimal places in his first post on the thread (several weeks ago) 



#92
Jan1212, 07:19 PM

P: 27

Wikipedia can be a good starting point, right? From there you can check all the references to see if the authors are from a university, research facility, or published in a reputable journal. Peer reviewed is more reliable and clearly, arXiv is not peer reviewed. It can contain some dubious eprints but most of the authors care about what they write. If the website ends with .gov or .edu it’s probably a good source, right? Can you think of anything thing else to add? List of Scientific Journals How the Scientific Peer Review Process works Misconceptions about science What is Science? P.S. If you’re such a stickler, here’s a suggestion for your next write up. Why does a photon slow down in a medium? There are tons of explanations out there. Here is ZapperZ’s explanation from in here and another from yahoo. Is either of these explanations accurate? If not, then perhaps you could provide a better one on your blog. http://www.physicsforums.com/showpos...93&postcount=4 http://answers.yahoo.com/question/in...8084206AALZBC5 



#93
Jan1212, 11:42 PM

Sci Advisor
P: 8,002

Let me add that although the topic was first introduced in scenario where the effect is so small as to be practically negligible  the antisymmetry of fermionic wavefunctions that Cox talks about is very important. Matter would not be stable without it, nor neutron stars exist.
http://rmp.aps.org/abstract/RMP/v48/i4/p553_1 http://www.astro.umd.edu/~miller/tea.../lecture17.pdf 



#94
Jan1312, 09:12 AM

P: 3

I agree that Wikipedia is a good starting point. Contrary to popular opinion, Wikipedia has a very high fidelity, in physics at least. I hear from specialists in other fields, such as art history, that the pages do not generate enough interest from editors to be reliable. However, in physics there seems to be a good supply of specialist contributors. The only disadvantage I have found is that for a nonspecialist, the pages can be difficult to understand. But Wikipedia is a reference source, not an educational program. I agree with what you say about the other sources, but would always read them with a skeptical mind. As I mentioned in the article, I think the best source for basic physics comes from Walter Lewin's MIT course. As for the photon question, that's a pretty difficult one to answer, and I can't claim to fully comprehend all the details of modern theory! I think the explanation you linked was right to avoid single atom explanations, but did not address the faulty assumptions in the question. As the Double Slit Experiment aims to elucidate, we are not able to measure what happens between a photons emission and its arrival without changing the conditions sufficiently to alter the experiment. And the double slit experiment summarized the very counterintuitive results concerning detection of photons. They arrive as particles, but do not seem to behave as particles on their journeys. Encapsulated in the Copenhagen Interpretation of QM is a policy of not trying to speculate about 'where the photon goes' from source to detector. We might have some mathematical equipment to calculate the probabilities of where the photon might end up, but we don't (or can't) know which path it took. Indeed, QED calculations assumed you need to consider every permissible path to determine the probabilities. So we can't appeal to the mathematical calculations for a satisfactory answer. Thus, to as 'why' and expect a deterministic 'then the photon does this...' type of narrative asks too much of quantum mechanics. But, the question could be answered by describing why the extra calculations for the material seems to delay the probability of a photon's arrival, compared with it traveling through empty space. I don't have sufficient quantum mechanical answer for this! 



#95
Jan1312, 10:07 AM

PF Gold
P: 3,069

What's more, you are overlooking the fact that there may be a reason that Dr. Cox is getting this hour (and a blackboard!), and neither you nor I are he has proven the ability to entertain and energize his viewers. Personally I think I could put together something that would be entertaining and enlightening also, which you might find less occasion to criticize if we share similar educational values, but I'm not going to get the opportunity to reach such a huge audience. I'm just not, the issue is moot. So I can see value in a certain tradeoff there yes, perhaps there is an overemphasis on what is titillating rather than what is good basic science, but it's not such a bad exchange to get these ideas out there to people, to help them see that scientists are not just in ivory towers discovering arcane looking equations that somehow helps us build better iPads. Instead, we are getting glimpses deep into the workings of our reality, and getting quite amazed in the process, and we are inclined to want to share some of that experience with a larger audience. 



#96
Jan1312, 12:14 PM

P: 27

Sorry, but I couldn’t resist. However, I’ll refrain from linking the video. You’re young, handsome, and your accent makes you sound intelligent, but here’s some womanly advice. Critics should cover their own butt and stick to the bare necessities, don’t cha think? What’s up with the banana? Thanks again. Cheers! 



#97
Jan1312, 02:02 PM

P: 661

There are several science programs on bbc tv and radio, some more populist than others. Brian Cox's are more at the entertainment end of the scale, but I for one quite enjoyed the four episodes in The Wonders of The Universe series, for example (even with the ott music in the first series of broadcasts).
The target audience is certainly not elitist types, and you should probably avoid these programs if you have 'a stick up your bottom' attitude to such populist science. There're always the online lectures of Susskind for example if you want a dry Diracesque introduction to QM. Feynman's style can be seen in the Messenger Lectures http://www.microsoft.com/education/f...&c1=enus&c2=0 (requires silverlight  microsoft compatible only) , I personally doubt his doubleslit lecture (lecture 6) will enlighten the uninitiated any more than Cox's attempts. 



#98
Jan3012, 04:35 PM

P: 1

I am still surprised by what was said about the consequences for electrons throughout the Universe of warming a diamond in one's hand. For a start, diamond is an electrical insulator with a large energy gap of more than 5 electron volts whereas the average thermal energy of an electron at room temperature (3/2 kT) is only 0.04 eV. Increasing this by at most 5% falls far short of the minimum needed to cause any electrons to jump into higher energy levels (assuming the "box of carbon atoms" contains no impurities); it will just cause the atomic lattice to vibrate a bit more.
Ignoring anomalies (if any?) caused by relativistic effects such as electron creation and annihilation or the lack of any FTL signals, the Pauli Exclusion Principle does of course hold for all electrons everywhere, regardless of whether they are pictured as bound to nuclei, zipping along on their own at almost the speed of light or just drifting about in a plasma. The doublewell example is fine as far as it goes, but only bound states corresponding to fixed separations of the wells are considered. In a gas, unless two nuclei are part of the same molecule, they will not usually remain a fixed distance apart and therefore will not give rise to a set of stationary states with exact electron energy levels. I think I'm right in saying that at present, the conventional view of astronomers is that a good 90% of ordinary (baryonic) matter (nearly all H) is in the plasma state. If this is correct, then around 90% of all electrons are not bound to any nuclei at all! When two of these "free" electrons are in relative motion, there could always be some inertial observers for whom their energies are equal alongside others for whom they are unequal. Therefore, I do not see how it is possible in general to substitute rules about electron energies for the basic requirement of antisymmetry of the electron component of the total wavefunction, a property which is both observerindependent and permanent. I agree of course that quantum mechanics does imply that "everything is connected to everything else" through entanglement, but I don't think the scenarios chosen to illustrate this amazing idea were at all convincing. 



#99
Feb212, 08:30 AM

P: 1





#100
Mar112, 08:06 AM

P: 4





#101
Mar112, 08:54 AM

PF Gold
P: 3,069

I'm sure Dr. Cox understands conservation of energy. His viewpoint is simply that if there is a probability that an electron will be in an energy state, this affects the accessibility of the state, so if I remove energy from an electron such that it would have a higher probability of moving into some state, and there is already some probability of an electron being in that state, the fact that all electrons are entangled (by their indistinguishability) implies that they are all "affected" in some sense. I think the real problem here is that Dr. Cox's words are being overinterpreted the key point is that electrons are identical, and thus entangled. Hence, any counterargument that first pretends the electrons have separate identities is already missing the point. Perhaps he was not careful to make this distinction it is crucial that all language like "this electron" or "that electron" be avoided when one is discussing Pauli exclusion.




#102
Mar112, 10:34 AM

P: 4

"I think the real problem here is that Dr. Cox's words are being overinterpreted."
I agree; the real problem is to try to find the right words to describe the situation in terms of a layman's frame of reference while minimizing the possibility of misleading them. 



#103
Mar112, 10:36 AM

PF Gold
P: 3,069

Exactly. I'm sympathetic of that problem we might not all agree with how Dr. Cox negotiates it, but we're all in glass houses on that score. If one person thinks Cox is doing more harm than good by stressing the more mystical elements, another can say he is doing more good than harm by simply getting people interested in some of the more fascinating new elements of what we have discovered. The fact is it might take centuries before we really understand what all this means, remember Feynman's wonderful words about quantum mechanics:
"We have always had a great deal of difficulty understanding the world view that quantum mechanics represents. At least I do, because I'm an old enough man that I haven't got to the point that this stuff is obvious to me. Okay, I still get nervous with it.... You know how it always is, every new idea, it takes a generation or two until it becomes obvious that there's no real problem. I cannot define the real problem, therefore I suspect there's no real problem, but I'm not sure there's no real problem." 



#104
Mar112, 12:22 PM

P: 661

When a supernova explodes it undoubtedly has a significant effect on the state vector of the universe, but it ought to be consistent with unitary evolution according to the Schrödinger Eqn. Of course, this isn't an issue if you don't believe in macroscopic wavefunctions, especially one describing the entire universe, but in that case you need corrections to the current standard formulation of QM. The nocommunication theorem says a measurement in one place cannot change the probability distribution of any observable outside the future lightcone of the first measurement. But science has no consensus on the nature of freewill, and such theorems may not apply. However, if freewill does break unitarity in a deterministic way then we may also need a reformulation of relativity since we would otherwise have the possibility of causal paradoxes. 



#105
Mar112, 12:35 PM

PF Gold
P: 3,069





#106
Mar112, 01:21 PM

P: 661





#107
Mar712, 10:11 AM

P: 27

Disclaimer: Precoffee
I thought that quantum entanglement had to be created by direct interactions between subatomic particles, but this guy says that the entire universe is in this entangled state. I don’t know but I don't like it. Was Brian Cox Wrong? Bell Himself Explaining the Implications of his Inequality Does it prove that the entire universe is in an entangled state simply because there are methods of creating entanglement? Is quantum nonlocality equivalent to entanglement? Aren’t there limits to quantum nonlocality, e.g. Tsirelson's bound? BTW, doesn’t he look like Johnny Depp as Willy Wonka? “Oh, you should never, never doubt what nobody is sure about.”~ Willy Wonka 



#108
Mar712, 03:45 PM

P: 203

As far as I know, the discussions on this issue are still ongoing. I thought I'd describe the situation from the viewpoint of my armchair.
Regardless of the discussions regarding whether Brian Cox should perhaps have said “quantum state”, rather than “energy level” in the TV show, this whole discussion has made me try to understand the applicability of the concept of entanglement to a situation such as this. Certainly Cox and Forshaw in their book did have entanglement in mind in connection with this issue, since they state: The model Cox and Forshaw are using is the double rectangular potential well. This model is described here. The energy eigenstates of a single rectangular well are split into pairs of energy eigenstates with very closely spaced energy eigenvalues. One member of a pair is a wavefunction with odd reflection symmetry about the origin and the other has even reflection symmetry. We now populate the double well system with a pair of fermions. For simplicity, they could be spinless electrons, which would have to be in different states to respect their fermionic nature. As an example, they could be in each of the two lowest energy eigenstates, so the system state would be [tex]\Psi \rangle={1\over{\sqrt{2}}}(E_1 \rangle E_2 \rangleE_2 \rangle E_1 \rangle) \ \ \ (0)[/tex] The sort of question we would like to ask is whether or not there is entanglement between quantities measured in the left hand well and quantities measured in the right hand well? Conventionally, entanglement questions would be treated by decomposing the full Hilbert space in the form [tex]{\mathcal{H}=\mathcal{H_{L}}\otimes\mathcal{H_{R}}}[/tex] For example, in the "classic" EPR entanglement scenario, this sort of decomposition is clear  [itex]\mathcal{H_{L}}[/itex] is the two dimensional Hilbert space of spin states of a LHtravelling spin 1/2 decay product of a spin 0 singlet state, and [itex]\mathcal{H_{R}}[/itex] the RHtravelling equivalent. For any pure state [itex]\Psi\rangle\in\mathcal{H}[/itex]I can choose an orthonormal basis [itex]\{\Psi^L_{i}\rangle\}[/itex]for [itex]\mathcal{H_{L}}[/itex] and [itex]\{\Psi^R_{i}\rangle\}[/itex]for [itex]\mathcal{H_{R}}[/itex] such that [tex]\Psi\rangle=\sum\limits_{i}\alpha_{i}\Psi^L_{i} \rangle \otimes\Psi^R_{i}\rangle \ \ \ (1)[/tex] here [itex]\alpha_{i}[/itex] are a bunch of coefficients (which can be chosen to be real and positive). This is the Schmidt decomposition. Given this, a good measure of entanglement  namely the entanglement entropy  can be defined as [tex]S_{A}=\sum\limits_{i} \alpha_{i}^2log \alpha_{i}^2[/tex] The higher the entropy of a state, the more entangled it is. Now trying to apply this to the double well scenario, we immediately run into trouble, because it is not clear how to perform the decomposition [itex]\mathcal{H}=\mathcal{H_{L}}\otimes\mathcal{H_{R}}[/itex]. If we want to ask the question "is there any entanglement in the double well model?" a key problem is that the two electrons in the system are indistinguishable fermions, so when one tries to construct a two particle state, it must be antisymmetric in the two electron identities. For example, ignoring spins, the position wavefunction representation of a two particle state might be constructed from single particle wavefunctions as: [tex]\Psi(x_1,x_2)={1\over{\sqrt{2}}}(\psi(x_1)\phi(x_2)\psi(x_2)\phi(x_1)) \ \ \ (2)[/tex] An nparticle state would be the same, except it would be a normalised sum over all the even permutations of [itex]x_1,x_2,...x_n[/itex] minus all the odd permutations. Such states/wavefuctions are sometimes called Slater determinants. Now, there is a fairly large body of literature around which discusses entanglement in multifermion systems. However, much of it is concerned with treating entanglement in systems appropriate to quantum computing  for example entanglement between quantum dots. In these cases, the mere fact that you cannot express a two particle state as a product state, but rather a difference of such, like in (2), is deemed *not* to constitute entanglement. For example Shi defines entanglement in a multifermion system to be the inability to express the state (by choosing a suitable single particle basis) as a single Slater determinant (like (2) for the case of 2 particles). In other words, a state is *not* entangled if you *can* express it as a single Slater determinant. Adopting this definition would immediately rule out the double well energy eigenstate (0) as being entangled – it's a single Slater determinant. But is this criterion really appropriate for the double well discussions? As far as I can tell, the reasoning behind considering (2) as unentangled has immediately made an assumption regarding remote exchange correlations, namely that they can be ignored due to the large separation. Schliemann, whilst arguing the case for using Slater rank as the entanglement criterion states ( where I've substitued the wavefunctions in (2) for his notation) states: There are other approaches to entanglement of fermions, such as the one discussed by Zanardi et al(http://arxiv.org/abs/quantph/0308043). They state that it is meaningless to discuss entanglement of a state This approach avoids the need to perform the decomposition (1) and instead focuses on the properties of various observables on the state being checked for entanglement. The criterion of Zanardi et al seems quite complex, but its essence is captured in a simpler formulation described in a reference by Kaplan, to which I was referred by PF user Morberticus. Basically the question of whether or not a state is entangled is asked *with respect to a pair of observables* [itex]A[/itex], and [itex]B[/itex]. A state [itex]\Psi[/itex] is deemed entangled with respect to [itex]A[/itex], and [itex]B[/itex] if the covariance function [tex]C_{AB}\equiv \langle\PsiAB\Psi\rangle\langle\PsiA\Psi\rangle\langle\PsiB\Psi\rangle \ \ \ (3)[/tex] is non zero. However, to apply this to our double well system, we need to be able to define the operators A and B appropriate to "making an energy measurement in the LH well" and "making an energy measurement in the RH well". The only energy operator I can think of that would be consistent in the twofermion system would be the total energy operator [itex]E_1+E_2[/itex]. This is symmetric in permutation of the electron identities 1 and 2 as it should be. However, to evaluate (3) to check for entanglement, I'm still left with the job of defining a "left hand well energy operator" [itex]E^{A}_1+E^{A}_2[/itex] and a "right hand well energy operator" [itex]E^{B}_1+E^{B}_2[/itex]. I've no idea how to do such a thing, and I'm inclined to agree with the conclusion of Arnold Neumaier in his answer to my question on physics stackexchange (http://physics.stackexchange.com/que...lewellsystem), namely that there is no simple way to progress this discussion ! 


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