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Insights Why the Quantum | A Response to Wheeler's 1986 Paper - Comments

  1. Aug 4, 2018 #1

    RUTA

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  2. jcsd
  3. Aug 4, 2018 #2

    stevendaryl

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    So the quantum weirdness in an EPR-type experiment is due to a combination of two things, neither of which is weird in itself:
    1. Conservation laws (conservation of angular momentum)
    2. Discreteness of measurement results (always getting ##\pm \frac{\hbar}{2}## for the spin measurement in any direction)
    But it seems that there is something else going on in EPR, which is a collapse-like assumption: When you measure a fermion's spin along some axis ##\vec{a}##, then it is as if, afterward, it is definitely in that direction. That's different from an imagined classical measurement that is somehow constrained to give a discrete result. You could imagine (this is Bell's toy model) that the electron as an associated spin vector, ##\vec{s}##, and measuring spin with respect to an axis ##\vec{a}## would return ##+1/2## if the angle between ##\vec{s}## and ##\vec{a}## is less than 90o, and ##-1/2## otherwise. This would give a discrete result, but the result would not be the actual spin vector of the electron.
     
  4. Aug 4, 2018 #3

    RUTA

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    Thnx for your comments, stevendaryl. 1 and 2 are spot on, but the collapse of some definite vector in that fashion doesn’t reproduce the quantum correlations (see the example in the Dehlinger paper referenced therein). The quantum correlations assume +1 or -1 is the “magnitude” and either Alice or Bob can claim they are always measuring the magnitude in each trial, it’s the other person who is getting the average, no collapse necessary. How can you not be impressed with such perspectival invariance?
     
  5. Aug 12, 2018 at 1:47 AM #4

    bhobba

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    Very nice post/paper

    My thoughts on the matter have tended to be more along the lines of generalized probability models and QM being the simplest one after ordinary probability theory that allows continuous transformations between pure states. But it is quite abstract - yours is much more physical.

    Very thought provoking.

    Thanks
    Bill
     
    Last edited: Aug 12, 2018 at 1:58 AM
  6. Aug 12, 2018 at 7:37 AM #5

    stevendaryl

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    I'm trying to get an intuitive understanding of the way that EPR probabilities (for anti-correlated spin-1/2 particles) are in some sense the closest we can get to the criteria:
    1. The sum of the spins is zero.
    2. Spin measurement always gives ##\pm 1/2##
    If Alice measures her particle's spin along axis ##A## and Bob measures his particle's spin along axis ##B##, then it is impossible to satisfy both criteria, because unless ##A## and ##B## are aligned, none of the following combinations adds up to zero:
    1. ##\frac{1}{2} (+\vec{A} + \vec{B})##
    2. ##\frac{1}{2} (+\vec{A} - \vec{B})##
    3. ##\frac{1}{2} (-\vec{A} + \vec{B})##
    4. ##\frac{1}{2} (-\vec{A} - \vec{B})##
    What the quantum probabilities do instead is the following:
    • Filter only those events in which Alice gets +1/2. (That includes possibilities 1&2 above)
    • Compute the vectorial average of the spin sums: This will be given by: ##\frac{1}{2} (P_1 (\vec{A} + \vec{B}) + P_2 (\vec{A} - \vec{B}))## (where ##P_1## is the probability of possibility 1 above, and ##P_2## is the probability of possibility 2).
    • This average is still not zero, but its projection onto ##\vec{B}## is zero.
    This uniquely determines the probabilities ##P_1## and ##P_2##:
    1. ##P_1 + P_2 = 1##
    2. ##(\frac{1}{2} (P_1 (\vec{A} + \vec{B}) + P_2 (\vec{A} - \vec{B}))) \cdot \vec{B} = 0##
    The latter equation becomes:

    ##\frac{1}{2} (P_1 (cos(\theta) + 1) + P_2 (cos(\theta) - 1)) = 0## (where ##\theta## is the angle between ##A## and ##B##)

    These equations have the unique solution: ##P_1 = \frac{1}{2} (1-cos(\theta)) = sin^2(\frac{\theta}{2})##, ##P_2 = \frac{1}{2} (1+cos(\theta)) = cos^2(\frac{\theta}{2})##

    Those are the quantum probabilities for anti-correlated spin-1/2 particles.

    That's sort of interesting, but my understanding of the motivation is a little muddled. I understand that you can't have perfect cancellation if the axes ##\vec{A}## and ##\vec{B}## are not aligned. But why ask for cancellation (on the average) along axis ##\vec{B}##?
     
  7. Aug 12, 2018 at 10:20 AM #6

    RUTA

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    You've arrived at the heart of the spin singlet state (uniquely producing the max deviation from the CHSH-Bell inequality, i.e., the Tsirelson bound). There is nothing special about axis ##\vec{B}##, indeed you could have done the analysis looking at Bob's results along axis ##\vec{B}## and required projection along axis ##\vec{A}## to be zero in producing the correlation ##\langle a,b \rangle## . So, in which direction is angular momentum for the quantum exchange of momentum actually being conserved? It's being conserved on average from either Bob or Alice's perspective, i.e., along either ##\vec{B}## or ##\vec{A}##. In classical physics there is a definite direction for angular momentum ##\vec{L}##, so with Alice and Bob measuring along random directions in the classical case we would expect neither ##\vec{A}## nor ##\vec{B}## to align with ##\vec{L}##. Consequently, in the classical case, Alice and Bob should always (essentially) be measuring something less than the magnitude L of the conserved quantity ##\vec{L}## (as shown in the picture of the SG experiment in the Insight). But, in the quantum case, it's as if there is no ##\vec{L}## independent of Alice and Bob's measurements. That is, you can't account for the quantum correlation using a hidden variable or Mermin "instruction sets" on a trial-by-trial basis (giving classical correlations satisfying the Bell inequality). No, the bottom line is that the quantum correlation satisfies a truly frame-independent conservation principle.

    As I said in the Insight, this is reminiscent of another frame-independent principle, the light postulate of SR. That postulate also led to "weird consequences," e.g., length contraction, time dilation, and relativity of simultaneity, and it was also opposed because it was something postulated not explained. Making this reference to the light postulate was motivated by quotes from Hardy and other reconstructionists in QIT. You read in many places in the QIT literature things like this Hardy quote
    So, in our paper we point out that this explanation of the Tsirelson bound should satisfy the desideratum of QIT.

    We're still waiting for Bub's response, he was the one who asked us to bring our adynamical approach to bear on his question "Why the Tsirelson bound?" when we gave a talk on our book at his QIT seminar last April. He wrote a nice blurb for that book, so we're hoping he now sees the relevance of adynamical/constraint-based explanation for QIT.

    Any suggestions for where to submit the paper? I was thinking PRA, since they do QIT stuff.
     
  8. Aug 12, 2018 at 3:05 PM #7

    RUTA

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    To fill in the blanks (not for you, I know you get it, but for others who might not have followed our exchange), rewrite ##\frac{1}{2} (P_1 (\vec{A} + \vec{B}) + P_2 (\vec{A} - \vec{B}))\cdot\vec{B}## as ##\frac{1}{2} ((P_1 + P_2)\vec{A} + (P_1 - P_2)\vec{B}))\cdot\vec{B} = \frac{1}{2} (\vec{A} + (P_1 - P_2)\vec{B}))\cdot\vec{B}## where ##\frac{1}{2}## is the magnitude of Alice's measurement along ##\vec{A}## (note that both ##\vec{A}## and ##\vec{B}## are unit vectors). Now we're only considering those outcomes for which Alice measured ##+\frac{1}{2}## (first bullet point), so the average value Alice would expect to measure along ##\vec{B}## for her ##+\frac{1}{2}## outcomes along ##\vec{A}## is ##+\frac{1}{2}\vec{A}\cdot\vec{B} = +\frac{1}{2}cos(\theta)##. Since we need ##\vec{L}## conserved to zero on average, we need Bob's average result along ##\vec{B}## to cancel this ##+\frac{1}{2}cos(\theta)##. His average is ##+\frac{1}{2}P_1 + -\frac{1}{2}P_2 = (+\frac{1}{2}P_1\vec{B} + -\frac{1}{2}P_2\vec{B})\cdot\vec{B}##. Thus, we need ##+\frac{1}{2}cos(\theta) + \frac{1}{2}P_1 + -\frac{1}{2}P_2 = \frac{1}{2} (\vec{A} + (P_1 - P_2)\vec{B}))\cdot\vec{B} = 0##. Again, you can divide up the results the same way for Bob and demand that Alice's average outcomes cancel his ##+\frac{1}{2}cos(\theta)## to derive the same quantum correlations.
     
  9. Aug 13, 2018 at 8:58 AM #8

    bhobba

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    Have ordered a copy from Amazon.

    Looking forward to reading it.

    Interesting Hardy is the one that got me into the probabilistic view of QM foundations.

    Do you know if he has moved away from that?

    Thanks
    Bill
     
  10. Aug 13, 2018 at 1:44 PM #9

    RUTA

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    Hardy revised his original (2001) set of axioms "replacing the simplicity axiom with more a compelling axiom" in 2011 (https://arxiv.org/abs/1104.2066). Per Hardy, "We show that classical probability theory and quantum theory are the only two theories consistent with the following set of postulates." His new postulates are Sharpness, Information Locality, Tomographic Locality, Permutability, and Sturdiness, which follow from two simple axioms:

    Axiom 1 Operations correspond to operators.
    Axiom 2 Every complete set of physical operators corresponds to a complete set of operations.

    In the original version of our paper (as presented in the IJQF workshop last month), we advocated "quantum-classical contextuality," where physical reality isn't "quantum rather than classical, but fundamentally both." Thus, we made explicit reference to Hardy's 2011 statement and postulates. We nixed that when we decided to submit the paper to a physics journal.

    Be forewarned about our book -- as a mathematician, you'll want to avoid the philosophical threads. The main thread is probably already too philosophical for you :-)
     
  11. Aug 14, 2018 at 3:07 AM #10

    vanhees71

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    I'm not sure, whether I understand your problem. This is an example for the fact that a single-particle quantity (like single-particle spin) in a many-body system can be determined for the system as a whole (here the total angular momentum) while the single-particle quantities are indetermined. That's an implication of entanglement.

    In the here discussed case you have a total angular momentum 0 state of a two-particle system of spin-1/2 particles, i.e., the two-particle spin state is
    $$|\Psi \rangle=\frac{1}{\sqrt{2}} (|1/2,-1/2 \rangle - |-1/2,1/2 \rangle).$$
    That's obviously a simultaneous eigenstate of ##|\hat{\vec{S}}^2 \rangle## and ##|\hat{S}_z \rangle## to the eigenvalues ##S=0##, ##\sigma_z=0##. Here
    $$\hat{\vec{S}}=\hat{\vec{s}} \otimes \hat{1} + \hat{1} \otimes \hat{\vec{s}}.$$
    Note that the ##S=0## state is very special, because in this case all three components of ##\vec{S}## are determined although these operators do not commute.

    Nevertheless the single-particle spins are maximally undetermined, i.e., there probabilities are given by the Statistical operator
    $$\hat{\rho}_A=\mathrm{Tr}_B |\Psi \rangle \langle \Psi|=\frac{1}{2} \hat{1}$$
    and
    $$\hat{\rho}_B=\mathrm{Tr}_A |\Psi \rangle \langle \Psi|=\frac{1}{2} \hat{1}.$$
    Now measuring the angular momentum component at particles A and B in different directions you get the probabilities you quote, and that's all you know about the outcome of measurements of the single-particle angular momenta. Of course, the measured outcomes do not add up to 0. Why should they? Even in classical physics it doesn't make too much sense to add components of vectors with respect to basis vectors in different directions. Of course, if you measure the angular-momentum components for both particles wrt. the same direction, then they must add up to 0 due to the preparation of the two-body system in the ##S=0## state. As explained above here you have the special case of a preparation of all three angular-momentum components to have the determined value 0. This is special, because that's possible only for the ##S=0## state and is due to the complete rotational symmetry (isotrophy) of this one special state. So you can have sometimes common eigenvectors for incompatible observables, and that's the most common example for this fact.

    I don't see any further specialty in this example, despite the fact that it's the most simple example to explain entanglement, Bell's inequality and all such unusual quantum correlations without a classical counterpart. It's only a problem, if you don't accept the quite abstract mathematical formulation of quantum theory and its minimal probabilistic interpretation in terms of Born's rule. Due to our persistent intuition from our experience with classically behaving (quantum) objects (aka many-many-many-...-body systems) we sometimes think we have to "explain" something more with quantum theory than there is contained in it, but that's pretty misleading.
     
  12. Aug 14, 2018 at 5:26 AM #11

    stevendaryl

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    The issue was not to derive the quantum probabilities from quantum mechanics, but to see if those probabilities can be derived from the assumptions that:
    1. The measured angular momenta of the two particles separately yields a discrete answer: ##\pm \frac{\vec{A}}{2}## for the first measurement and ##\pm \frac{\vec{B}}{2}## for the second measurement.
    2. The sum of the spins must add up to zero (in some average sense).
     
  13. Aug 14, 2018 at 7:02 AM #12

    bhobba

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    Don't worry - I sort of figured that out with its emphasis on the Blockworld which I am not a fan of. But as I often say - my views mean jack shite - I am sure it will contain interesting insights. Every interpretation of QM I have read, transnational, MW, BM etc etc have helped me in understanding the formalism better. My personal views are pretty well known, and are virtually identical to Vanhees, but to hold any view you must subject it to what other views say.

    Thanks
    Bill
     
  14. Aug 14, 2018 at 7:15 AM #13

    bhobba

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    This whole approach is very new to me. I am very interested in QM foundations, but have been taking a back seat and listening, rather than participating until I feel more comfortable commenting.

    Thanks
    Bill
     
  15. Aug 14, 2018 at 9:28 AM #14

    vanhees71

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    That's a different question, which is not describing the statistics you expect from quantum theory as detailed above!
     
  16. Aug 14, 2018 at 9:31 AM #15

    vanhees71

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    Thanks for the warning. In general, philosophy on quantum theory rather confuses the reader than to help him or her. Maybe your book is an exception. Nevertheless the word philosophy in connection with quantum theory (or even physics in general) should be read as a caveat sign ;-)).
     
  17. Aug 14, 2018 at 9:56 AM #16

    stevendaryl

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    I'm not sure what you mean. Both the heuristic argument (not original with me; I was paraphrasing the Insights article) and the quantum theory make the same predictions: If Alice measures her particle's spin along axis ##\vec{A}## and Bob measures his particle's spin along axis ##\vec{B}##, then the conditional probabilities are:

    • ##P_1## = the probability that Bob will measure spin-up given that Alice measures spin-up = ##sin^2(\frac{\theta}{2})##
    • ##P_2## = the probability that Bob will measure spin-down given that Alice measures spin-up = ##cos^2(\frac{\theta}{2})##
    (where ##\theta## is the angle between ##\vec{A}## and ##\vec{B}##).
     
  18. Aug 14, 2018 at 9:57 AM #17

    vanhees71

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    Yes, but the spin components in non-collinear directions need not to cancel each other. Why should they?
     
  19. Aug 14, 2018 at 9:58 AM #18

    stevendaryl

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    That was what I asking RUTA for clarification about.

    The sense in which there is cancellation on the average is this:

    Among those events where Alice measures spin-up along direction ##\vec{A}##, the expectation for
    ##(\vec{S}_A + \vec{S}_B) \cdot \vec{B}## = 0

    (where ##\vec{S}_A = + \frac{1}{2} \vec{A}## and ##\vec{S}_B = \pm \frac{1}{2} \vec{B}##, depending on whether Bob gets spin-up or spin-down)
     
    Last edited: Aug 14, 2018 at 10:15 AM
  20. Aug 14, 2018 at 10:27 AM #19

    vanhees71

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    It's again not clear to me what you want to calculate. The joint probability that A finds ##\sigma_A=\pm 1/2## and B find ##\sigma_B=\pm 1/2## is, of course
    $$P(\sigma_A,\sigma_B) =|\langle \sigma_A | \otimes \langle \sigma_B |\Psi \rangle|^2.$$
    Here ##|\sigma_A \rangle## and ##|\sigma_B \rangle## are the eigenvalues of the operators
    $$\hat{\sigma}_{A}=\vec{A} \cdot \hat{\vec{\sigma}}, \quad \hat{\sigma}_{A}=\vec{A} \cdot \hat{\vec{\sigma}}.$$
    I'm to lazy to explicitly figure this out, but I still don't see the point of the exercise :-(.
     
  21. Aug 14, 2018 at 10:42 AM #20

    RUTA

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    There are philosophers who are interested in foundations of physics (FoP) and our book was written for them as well as physicists interested in FoP. As a typical physicist, I tend to make unarticulated assumptions and the philosophers are good at identifying those. My interest in FoP is based on my desire for a model of objective reality for all of physics. See Becker's book for the value in this.
     
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