Stern-Gerlach Spin-1 spin state transition tutorial

Discussion in 'Physics & Math' started by geistkiesel, May 30, 2005.

  1. geistkiesel Valued Senior Member

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    The tutorial begins with learning the transition rules for Stern-Gerlach transitions of spin-1 partricles. Soin-1 simply means the particle can be poilarized into one of three possible states +S, +-S, or -S (or any approprioate segment designator other than S , +T, +-T or -T for example) . The magentic orientation of the spin vectors will be discussed later.

    First the transisition rules. By +S I mean the particle is polarized to take the + or upper channel when passing through a Stern-Gerlach S segment magentic field/gradient volume. By convention particle coordinate line S is imposed parallel with the lab z-axis (floor to ceiling). +, +- or - indicates the direction of motion of the particle wrt the z-axis of the segment: up, horizontal or down.

    The following schematic shows the sideview of four transition experiments. From the top row there is a +S state particle entering an S segment with the middle and lower channels obstructed. The +S particle always takes the upper channel and all +S particles will pass through this S segment. This is a domestic-to-domestic (DTD) transition.

    The +S particle then enters an SG T segment, identical to the first S segment but rotated around the direction of motion of the particle, the y-axis. Here the T segment also has the middle and lower channels obstructed. The particle takes one of the three channels crashing into the middle or lower obstructions or passing through in the + channel. Hence the "1/3" fraction of the beam that survives the transition. Notice the particle is now in the +T state, where all previous +S information is zeroed, nada. These four experiments describe everything you need to know about transition rules, the physics on top of this is another matter. However, "We will not serve any theory before its time."

    Finally the +T state particle now enters another S segment identical to the initial S segment. The +T particle is polarized to one of the three possible S states and where statitically 1/3 of the input beam survive. This is an alien-to-domestic (ATD) transition as was the +S to +T transition.

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    The second experiment is physically the same as the first with different channels blocked. Here the output of the T segnment is a +-T (historically referred to as the '0T' state) that then enters the final S segment with the predictable results. !/3 of the beam survive in the +-S state in this ATD transition. BTW +S is not related to +T, nor is +-T related to +-S, the states are different , period. No S state can be identical to any T state, ever.

    Notice in ATD transitions the input particle state is always changed to one imposed by the host SG segment.

    The last two experimental arrangements are identical to the first with the exception that the obstructions are completeley removed. The T segments are "wide open". As before in ATD transitions the +S particle is polarized to one of the three possible T states, and upon exit the particle reforms into the original input state, +S, as if the segment were not present. The +S state then enters an identical S segment from which it originated. Look at all the input states in the experiments. All +S particles using the upper channel and surviving. So too the particle here.

    The last experiment shows the predicted of what has been learned so far. The particle enters the S segment, that has the upper and lower channels obstructed, and uses the upper channel and strikes the obstruction. Zero particles survive the transition.

    The only question remaining is why the permamnent change of particle state after polarization when the particle transitions through obstructed segments. The particle trajectory is unaffected, but the obstructions are crucial to the causitve chain of events leading to the permanent change of state.

    It has to be some action at a distance, and absent the manifest existence of forces applied spatially between the obstructions and the particle the assignement of nonlocal status to the force carrying channels is appropriate.

    In the presence of he magnetic field/gradient the particle exhibits the property of extending some otherwise nonloclocal probes to the vicinity of the obstructions, which are, the channels the particle would have taken had it been polarized as such. There uis an internal particle state generating system crucially dependent on the existence of nonlocal elements of the observed state +S.

    More as the turtorial devlops. You can check the language attached to the figure with the currently typed material.
    Geistkiesel
     
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  3. James R Just this guy, you know? Staff Member

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    It should be noted that the description of these experiments given by Feynman, and by quantum mechanics, is very different to the one given by geistkiesel.

    In particular, quantum mechanics does not require "immediate polarisation into one of the three possible states" when a +S particle enters the T segment, as geistkiesel requires. Nor does quantum mechanics require the existence of any mysterious "non-local forces" of the kind which geistkiesel introduces.

    Interested readers are advised to read Feynman's lectures on physics. Compare Feynman's explanation of quantum physics with geistkiesel's and make up your own mind.
     
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  5. geistkiesel Valued Senior Member

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    Not so. James R is mistating Feynman ('Lectures on Physicvs' Vol. III chapter 5).
    The transition rules are identical to what feynman wrote. In fact he spends most of the chapter developing the transition rules used here.​

    Whatever QM requires, the fact of polarization is certain and certian to occur as the particle enters the Stern-Gerlach magnetic field and gradient volume. This is seen by the immediate change of motion of the particle as it enters the field and becomes polarized to one of the three allowed trajectories., i.e. directions of motion.​
    So far there has been no departure from Feynman or quantum mechanics by the description above. The fact is and recognized by Feynman, the insertion of the obstructions are causally related to the permanent change of spin state of the particle that is transitioning through an obstructed segment as described. Remove the obstructions in the T segment and the +S state particle will always reform to the +S state [it was in before entering the T segment]. Add the obstructions and the particle is permanently changed when exiting the T segment to whatever the particle was polarized to when entering the segment. OK the obstructions are not directly affecting the particle, after all it survives the transition, but if the obstructions are miles away, relatively speaking, the effect of the obstructions must be action at a distance, by definition. There is no other explanation. What is occuring is the obstructions acting through nonlocal force exchange channels, in the presence of the mnagnetic field/gradient volume, effectively set the permanent state switch of the particle to the current polarized state. Without the switch set the particle can reform to the remembered state prior to entering the Tsegment as if the unobstructed or , "wide open" segment, "were not present" (quotes by Feynman).

    Those that continue here will see a difference in QM and the tutotial, but not contradictions as James R will have you believe. The differences are primarily in James R's misundertstanding of what QM teaches as outlined by Feynman in the referenced 'Lectures'. You should get the book, vol III chaptrer 5.
    Geistkiesel ​
     
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  7. James R Just this guy, you know? Staff Member

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    Once again, I advise readers to read the original, and not geistkiesel's version of it. I have already spent considerable time in other threads explaining to geist exactly where he is wrong, but it seems that in the past month he has forgotten those discussions. Interested readers can search the Physics forum for "Feynman" and "Stern-Gerlach" to find the relevant discussions, if they are interested.

    The amplitude expressions are correct. Geistkiesel's stipulation that transitions happen instantly on entering the apparatus is false, and not supported by Feynman, as I said.

    In fact, this is false. But in any case there is no way to know from your description, since you don't measure the state of polarisation inside the apparatus, but only after exiting the apparatus.

    That is incorrect. Bell showed that "hidden variable" theories are incompatible with quantum mechanics. He showed no need for non-local forces. Some quantum "communication" between particles appears to involve "spooky action at a distance", or non-local effects, but the term "force" is misapplied by geist in that context.

    Read my previous post for a few of the departures.

    What was that you were saying about Bell above? What you are saying sounds very much like hidden variables to me. A "remembered" state would need a hidden variable for the memory, and Bell showed that such variables are incompatible with quantum physics.

    James R vouches that he has studied much more quantum mechanics than geistkiesel has. He invites readers to judge for themselves who has a better understanding of the theory. We could play "My ego is bigger than your ego" for a long time, but it wouldn't get us very far, would it?
     
  8. geistkiesel Valued Senior Member

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    AS I remember the exchange it was I spending the consderable time educating JJames R, of well....
    James R speaks as if he had accomplished something in a previous exchange. This is not so. The particles must be polarized as the alien particle enters the domestic segment as in +S -> -T for instance. If the particle were polarized later then the magnetic field and gradient of the host segment would be applying force on the magbnetic spin vector oruineted in a +S direction and would veer off the normal trajectory seen to be taken consistently by yher particles such that the trajectory u would be erratic and probably the particle would not be detectecd exiting the segment.


    Feynman doesn't state specifically that the particles are not polarized immediately. He probably didn't consider the question. If the particle retained the prepolarized orientation of its magnetic spin vector rigidly the particle would not take a defined trajectory and would crash into the walls of the segment. James R refuses to give any references to the claim that the polarization is not immediate or near immediate upon the particle entering the segment.


    Feynman makes no claims regarding the time it takes a praticle to become polarized, James R keeps saying the polarization isn't immediate. or sudden after entering the field gradient region of a Stern-Gerlach segment. AllRF's drawings , however, imply an immediate polarization, bnut like I said, RF probably didn't consider thje question here in chapter 5.



    Quite the opposite, again James R,. Bell showed the absolute necessity of hidden variables being absolutely necessary to completely explain QM activity. Bell proved that no local model was suffiient for completeness in QM modeling purposes.

    No, force is not implied erroneously. Once one has accepted the reality of nonlocal activity as affecting the state of the particle, there must necessarily be a nonlocal/local interface such that the nonlocal activity is realized on the observed level. Nothing else makes any sense. James R is embarrassing himslelf. There is nothing I can do about that except suggest he look at the literature and determine the reality of QM as it is.


    Are you suggesting that the schematic I drew does not represent the true activity of particles transitioning through SG segments as described in the schematic?


    Bell is directly in opposition to what James R said here.

    JS Bell is quoted:
    But if qm were embeddable in a local causal theory (16) would apply, with a -> 0, b -> phi, and c, the implicit specification of the production mechansim, held fixed. The right hand side of (22) should then be negative. So qm is not embeddable in a locally causal theory as formultated above".​
    See Page 99 'the theory of local beables' in "Speakable and unspeakabkle in QM, a collection of Bell's papers.

    In Bell's chapter 8 he says on page 64,
    " . . .we allow fully for the effect of measuring equipment by allowing A and B to depend not only on the initial values lambda of the hidden parameters but also on the parameters a and b specifying the measuring device."​
    Bell has chapter 13 titled:

    "Atomic -cascade and quantum mechanical nonlocality"​

    You rememberered correctly. Feynman in chapter five states that the information regarding the previous state is lost by the "blocking masks" . When transitioning through unblocked T segments like the 3rd and 4th experiment in the original postn the prevuious state is reformed, remembered.

    Quoting Feynman in Ch 5:
    "The infiormation about the original (+S) state is retained -- it is as though the T apparatus were not there at all."​

    This is what I mean about memory of the previous state.

    Again Feynman states on page 5-9 that an atom in the +S state transitioning through a blocked T+ channel and blocked T- channel segments will exit the T segment with a loss of 2/3 of the beam particles] as :

    "The atoms which come out of T are in the base state of 0T and have no memory that they once were in the state of (+S)."​

    I use +-T instead of 0T for a reason to be shown later.


    No it wouldn't get us very far. I started this thread with the purpose in mind of showing the results that Feynman arrived at as far as the rules of transitions are pertinent. When describing the whys anad wherefors RF and myself will diverge considerably, but this is for your benefit to make up your own minds what is occuring. But we haven't diverged yet Rf and myself. It is James R that has diverged from the basics of QM as described by Feynman in spin-1 transition experiements through SG segments as described in ch 5..

    The readers, if any, are getting screwed if they rely on James R's rememberance of what constitutes proper Stern-Gerlach trasnsition acttivity. All I can urge is look at the experiments, described in the opening post,and subsequent posts to come. Look for consistency of the rules, or contradictons, there are none I claim. But look for them just the same.

    The four experiments described in the opening post are excised from chapter 5. I placed my own notation scheme on the matter, but the results are left intact as they must be. Once one sees the rules as incorporated fully in the four experiments the rules become second nature.

    What is polarization of the spin-1 state?

    When a particle enters the blocked S segment, say an unknown state enters, but certainly not an S particle (except by the rarest of conditions assuming a virgin spin-1 particle source) the magnetic spin vector gets reoriented to the demands of the host SG S segment. The S coordinate for the particle aligns with the z-axis and consequently, the magnetic field and gradient direction of each segment. The +, +- and - symbol indicate the direction of motion of the particle along S , or the z-axis of the segment, which is the observed state of the transitioning particle. This is what it means to say which state the aprticle is in: the direction of motion along the z-axis after polarization.

    With the exception of the domstic to doemstic transitions, S to S, T to T, the state of the particle during transition is not known unless a measuremnt is taken by placing a detector in two of the three channels of the segment under scrutiny. Two channels blocked, by detectors, will stop 2/3 of the beam and the 3rd unblocked channel, by default tells us the partricle state transitioning through the unblocked channel.

    In an unblocked segment, only measuring the particle state after the particle exits from the T segment can the reformed input state be verified. Remember, Feynman tells us that the +S particle transitioning through the unblocked T segment is, "as if the segment were not there". This means that a +S polarized state produced exiting the S segment (blocked in the +- and - channels), will resuilt in the reformation of the +S state after exiting the "wide open" an unobstructed Tsegment. However, the particle must necessarily have been temporarily polarized into one of the three T states allowed while transitioning through the segment. As RF says, "the particle has nowhere else to go".

    Anyone can play at this game, so don't be shy or afraid to ask questions. Or don't be afraid to stump me and show me in error!!.

    Geistkiesel​
     
  9. James R Just this guy, you know? Staff Member

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    geistkiesel:

    As I explained to you in another thread, Feynman is working in the "sudden approximation", where the polarisation does not adiabatically follow the changing magnetic field direction. Any decent book on intermediate to advanced quantum mechanics will explain this concept for you.

    Actually, what I am saying is that the physical direction of polarisation does not change at all. What changes is the basis in which that polarisation is expressed, which is defined by the orientation of the magnetic field in the apparatus. The basis change is assumed to happen "suddenly" as the particle enters the apparatus. Thus, a polarisation which is an eigenstate in one segment of the apparatus can become a superposition of eigenstates in the next segment.

    You appear to be confusing "hidden variables" and non-locality. I specifically talked about the former, not the latter. I agree that quantum mechanics is non-local. However, Bell's theorem rules out hidden variables under reasonable conditions.

    Already done that. Have you?

    Below, you quote Bell almost at random. Quotes this far out of context aren't worth much, but I'm happy to comment on them for what they're worth.

    i.e. quantum mechanics is non-local. No argument from me there.

    Presumably, Bell then goes on to show that hidden parameters are incompatible with quantum theory. Did you read beyond this point in the book?

    My explanation for the unblocked segment is that the particle state is not "remembered" via some kind of hidden variable, but is simply unchanged by the presence of the T segment from what it originally was.

    That's right.

    Right again. No hidden variables required.

    I am not relying on some vague memory here. This stuff is very current for me.

    Just so we're clear, my position is that this is completely incorrect. I say the particle's spin vector never changes direction until a measurement is made (either by blocking the particle or detecting its state after exiting the apparatus).
     
  10. geistkiesel Valued Senior Member

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    Your books are very interesting, I am sure. My dog-eared copy of David Bohm "Quantum Theory", was and remains very interesting to me to this day. I found references in Bohm’s book that verifies your statement.

    I agree you see? The SG transitions are sudden, not adiabatically stretched out, where the polarization force, here the magnetic field/gradient volume of the SG segment is entered infinitely slowly.

    I cannot understand the meaning of James R’s statement above. This is what I have been saying that the particles are immediately polarized when entering the SG segment. {Yet later below you insist the particles do not change their state when entering an ATD transition, alien-to-domestic transition, or at least this is how I read your comments. Straighten me out here JR.]


    Here is what the particles see, the inhomogerneous field on the left, essentiaklly flat and not qadiabatic.

    The particles are whizzing through the say 20 cm long Stern-Gerlach segments. The deflection of the particles is perhaps on the order 1mm in the distance traveled. If the particle were not instantaneously polarized when entering the intense magnetic field and gradient volume, the particle could pass through in a straight line contrary to the experimental results guaranteeing the 1/3 statistical distribution of states in a pure unpublicized source gas of particles.

    There have been no physical data demonstrating that which you claim, has there? Perhaps I am wrong, then if I could call upon your relative vast excess in information regarding the QM reality Would truly appreciate the effort to provide the reference.

    At first considered the issue of the polarization rate to be distracting to the thread. Do you understand what I am saying? But I see it differently now. On page 450 of David Bohm’s ‘Quantum Theory’, advanced undergraduate, graduate level, maybe? Bohm discusses the adiabatic case of the polarization as occurring if the particle is subject to the infinitely slow exposure to the forced oscillation term that is neglected when the event is “sudden”. In the Stern-Gerlach, SG, transitions the transition from field free to maximum intensity is very sudden. The field-gradient extends –Z (the floor of the segment to +Z direction of the low–to-high magnetic field density. The inverted roof concentrates the uniform field to a narrow band oriented parallel to the Z-axis. I am remembering here but increases in magnetic field/gradients reach typically 20k times the linear field density wrt the un-concentrated field. The particles enter the segments abruptly where the no-field/positive-field interface is basically a flat wall of magnetic field/gradient the particle enters. This is far from the “theoretical “ adiabatic case James R could have been referring.

    Here is what the partocl;e sees, a sudden imposition of the force of the magnetic field. The eleft figure indicates an inhomogemeous magnetic field/gradient volume.

    This is what the particle sees as it enters the segment: The figure on the left, with the indicated field/gradient that is effectively flat on the “front door”. Moving through the magnetic field/gradient there is the effect of the particle experiencing an electric field component due to the time varying magnetic field generated by the particle motion. The forces acting on the particle are sudden, as opposed to the infinitely slow adiabatic changes of state.
    I understand this to mean that the particle is polarized instantly to the magnetic orientation of the host field/gradient segment. We are apparently discussing different aspects of similar processes.

    The magnetic polarization vector always orients to the direction of the host magnetic field/gradient, otherwise it would not take the proper physical trajectory, or channel, and would behave as if unpolarized to the current segment field direction and would move as if polarized to the +S direction. If S and T z-axes are 36 degrees apart the particle would be pulled to the S segment z-axis pulling the particle into the x-y frame of the T segment. This is not the observed case.

    So whatever the superposition scheme James R mentioned it is most likely something other than mentioned by Feynman in Ch 5, and Geistkiesel above.


    Here is what I just copied from the post you are referring to regarding JS Bell:
    We aren't playing a word game here I assume. What can "nonlocality" be if not drenched in "hidden variables"?
    OK, I will take you on face value. Perhaps I spoke in haste here. I only state then that what I am saying is definitely not ‘quantum mechanics’. I am merely discussing chapter 5 of Feynman's Text book "Lectures on Physics", Vol.III, Ch 5. I would like to stay on this as a guide and limiting information on the direction for the thread to take.

    I am referring to your statement that Bell rejected the concept of "hidden variables". I have heard the rumor that Bell wanted to legally change his name to "HiddenVariables", but his wife talked him out of it.

    Did you see the quote where Bell rejected the claim of any "local model" as properly describing the events?


    Good


    What do mean by "presumably"? Does he or doesn't he?

    I have read the book from cover to cover, and intensely studied various chapters and paragraphs through out. I know the book. The book is a testament to QM nonlocality of which the absence of "hidden variables" would be a contradiction in terms. If there is nonlocality at work that affects the observed results then there must be a hidden variable or two, or three that effectively interacts with the local, the ‘ what we observe’ attributes. There is no other way to look at the matter.

    If you really want to stick me with “no hidden variable” rule you should refer to the Bohm book I mentioned, “Quantum Theory”, published 1952, where on page 611 (out of 628 pages) Bohms offers a proof supporting the von Neumann “proof” that “hidden variables” are impossible. Within a year of publication Bohm immediately changed his tune and adopted the classic model patterned after de Broglie’s “pilot wave“ theory. He later, one year after publishing his “quantum Theory”, in 1953, disproved his own proof, as did JS Bell later.

    Nonlocality isn't a theoretical phrase hiding some mathematical concept. Nonlocality is meant in a physical sense-- as if widely separated particles could share the same, but negative end of the other's spin vector, for instance, or similarly share in the + and – ends of the instantaneous angular momentum direction [vector].

    Here is the first paragraph of Bells opening paper entitled, “On the problem of hidden variables in QM”. He says
    “To know the quantum mechanical states of a system implies, in general, only statistical restrictions on the results of measurements. It seems interesting to ask if this statistical element be thought of as arising, as in classical statistical mechanics, because the states in question are averages over better defined states for which individually the results would be quite determined. These hypothetical ‘dispersion free’ states would be specifies not only by the quantum mechanical state vector but also by additional ‘hidden variables’—- ‘hidden ‘ because if states with prescribed values of these variables could actually be prepared quantum mechanics would be observably inadequate.”​

    After a long discussion of von Neumann’s ‘proof’, which Bell proved erroneous, he says, when discussing non-commutating operators,
    ”Nevertheless, they give logically consistent and precise predictions for the results of all possible measurements, which when averaged over lambda, [the hidden parameter] are fully equivalent to the quantum mechanical predictions. In fact, for this trivial example, the hidden variable question as posed informally by von Neumann in his book is answered in the affirmative.”​
    Is this horse dead yet? I understand you, you understand me, OK?


    If the particle's magnetic spin vector orientation is not changed you are saying there is no polarization. The particle is always slaved to the orientation of the particle spin vector to the host segment magnetic field/gradient direction. This is major point Feynman makes. Do you have a copy of "Lectures"?

    If the forces aren’t hidden the forces aren’t nonlocal. So where are the forces directed at maintaining the previous +S state in the superposition scheme James R mentioned?

    So if the +S particle goes through the "wide open " T segment, whose magnetic field gradient direction is not parallel with the orientation of the polarized magnetic spin vector of the entering +S particle and exits from the segment in a state of +S it is as if the segment were not present?
    I paraphrased the question slightly different from the original but the same answer would result I am sure.

    The particles are polarized to the +-T state (historically termed the ‘0T’ state) for example, when they enter the T segment; otherwise, they crash into the walls. When exiting they return to the +S state.

    If there are obstructions in the +-T and -T channels, then 2/3 of the input beam crashes into the obstructions. The surviving +T exits the T segment in a permanent state of +T with all +S previous state memory erased. The particles must have been polarized to the T state, some T state.

    Repeat the same transition but remove the obstructions and the particle exits in the +S state. The +T particle earlier was our +S before the particle entered the T segment. If I understand you, the presence of the obstructions have what effect on the particles? Where does the superposition scheme intervene here?

    The particles that use the lower two channels crash into the obstructions, hence 2/3 of the beam lost. How could the particles get to the middle and lower channels if they weren’t polarized to be driven there by the force of the magnetic field/gradient acting on the particle in the +-T or the – T state?

    If the particle scrutinized, the surviving particle didn’t have a nonlocality interaction scheme going here, how do you explain the effect of the obstructions? The particle exits in a +T state if the T segment is obstructed, and the input +S state exits as +S state if the T segment is unobstructed.?

    Where is the causal chain that keeps it all local, unhidden, observed?[

    What do the obstructions perturb that isn’t perturbed in the unobstructed case?


    In the unobstructed case the +S -> +T -> +S transition for all +S particles, 100% survive the unobstructed segment. The elements that guarantee the reformation of the +S state are not seen in the expression, that is because they aren’t observed and physical expressions generally are concerned only with the observable, or the observed. To keep the score properly adjusted. The +S term takes on a new look. In trivial functional notation,
    Y(+S) = Y(1 00[+S])

    where “1“ refers to “+”, which says in words the state function +S is composed of the observed +S state that is a function of existent critical elements (excess), the nonlocal 00[+S] guaranteeing the +S state reformation.

    The ‘00’ is an arbitrary assignment of “two nonlocal spaces”. No physical attributes of these nonlocal elements are inferred directly.

    A physical polarization (P) operator, acting on the Y(+S) is (P)Y(+S) = (P) Y(+S)00[+S])) -> Y(_ 00[+S] _ ) -> Y(1 00[+S] 00[+T])
    The 1 is understood to be a + wrt the T segment, the 00[+T] the nonlocal elements of the temporary T state. The underscore mean the process is not instantaneous and is used for instructional purposes though they must part of the be the intermediate steps in the process of polarization.

    When exiting the segment the particle reforms to the +S state in the inverse process (dP)Y -> Y(_ 00[+S] _) -> Y(1 00[+S]).
    The nonlocal elements were not perturbed in the unobstructed transition. When obstructed the 00[S+] elements are perturbed as are the 00[+T], but only the 00{+T] are compatible with the current polarized +T state.

    In the unobstructed case the instability of the hybrid T state Y(1 00[+S] 00[+T^]) the reformation of +S favors the ‘stable and unperturbed nonlocal elements 00[+S]’ , which are necessary and sufficient for the reformation of the +S state.

    The particles are whizzing through pretty fast, 60k km/hour or so?

    The particle has to be in one of the three states as exhibited by their motion when entering the T segment.

    If there is someone who can provide some arbitration in the form of some literature to solve this difference, I would be eternally grateful?

    I am not taking a poll.

    Here is the issue:

    Specific Question: Does the polarization of the input particle state in a +S state, for example, enter a T segment and is the +S state transformed into either a +T, + -T or –T, as described by Feynman in ‘Lectures on Physics’ Vol III ch 5?”

    What I mean by the +S state is, the particle will take the upper + channel when entering an S segment and will be polarized to one of three allowed T states when entering a T segment identical to the S segment, but rotated around the y-axis of particle travel.

    Passing the plane of the obstructions sets the lock that keeps the particle state permanently oriented in the now, +T direction. The particle will move up the z-axis of any T segment. The S and T segment designators mean only that S and T z-axes are not parallel.


    OK , I will agree and I will also direct you to the QM reality that placing the particle in a magnetic field is a measurement.

    Bohm said
    ”Thus the magnetic field of the earth can be regarded as part of a mass spectrograph that that separates cosmic–ray particles according to their energies and charges.”​
    Bohm was commenting that the observing apparatus need not be constructed by man and located in a lab.

    Why measure a particle that was produced in an S segment as a +S particle that transitioned through a wide-open T segment? We know what state the particle is in, a +S state, because the unobstructed segment has no permanent affect on the particle state. It is as if the segment wasn’t there, but it was there, and the particle wouldn’t make it through if the particle weren’t polarized into a T state during the transition through the segment.

    I read James R as saying the magnetic field/gradient has no affect on the state of the particles.

    Again, putting the particle in a magnetic field is equivalent to what qm considers a measurement. We know from huge numbers of transitions experiments what the predictable out come is under any generalized Stern-Gerlach conditions, even to the conditions of particle collisions.

    The particle's direction is altered when entering the magnetic volume. I read James R to mean that if 100% of 1 million particles survive transition through an S segment that produces +S particles, that one will never know what the actual state of the particles are that exit from the +S channel until they are measured, how? If the particle exits from the upper + channel of an SG segment it is in the permanent state as defined by the trajectory used by the particle in the obstructed host segment.

    If the particle makes it through it is a +S particle. The test is over. The particle state is determined long before exiting the segment. If one million particles in succession all pass through the S segment where only the upper trajectory is open is it fair to say that their input state into the S segment was also all +S?

    We need not collapse the wave function because we know the state of the particle. If all particles exiting a +S state producing S segment enters the wide open T segment and exits the T segment in a +S state then filtering through the final S segment also with the + channel open only, is redundant. We know what the state is, +S.

    To simplify the matter, if a +S particle just produced in an S segment with the middle and lower trajectories blocked (the + - S and the -S channels) then immediately passes through an identical S segment, 100% of the beam survives the second segment transition using the upper + channel of the second segment.

    The diagram tells everything you need to know regarding spin-1 SG transitions. Alien-to-domestic are S--> or T --> S, domestic-to-domestic transitions are S -> S or T -> T.

    Notice how always the ATD transition results in polarization of the incoming alien particle to one of three states in the host domestic segment.

    In all DTD transitions the particle takes the trajectory that it was polarized to before entering the second S segment.

    In wide open T segments in ATD transitions the final state is the polarized state that survived the transition. The obstructions merely seal the fate of the one time hybrid mixed state of the particle.

    here is a summary of all the transition rules, even some duplications.
    remember the number on the upper right corner is the fraction surviving that transition, The “+S” to the right of the segment is the output state of the preceding transition and the input state to the next transition coming up. The S and T below the segment define the segment type, where the T z-axis is rotated wrt the S segment Z—axis.

    The +S to T transition in the 3rd to fourth segment is the described by the same physics as the +-T to +-S in the 8th to 9th transition, both ATD transitions.
    Geistkiesel.

     
  11. James R Just this guy, you know? Staff Member

    Messages:
    39,397
    geistkiesel:

    You've given me a lot to respond to. I apologise in advance that I won't address everything you've written.

    Here's an analogy for you. Think of the +S,-S and oS states as being like the x,y and z axes of a coordinate system. The general polarisation vector of a particle is like a vector which points in a certain direction in the (x,y,z) space. We can express the general vector as some combination of the basis vectors x, y and z. For example, we might have:

    P = 3x + 4y + 2z.

    Now, suppose we take our coordinate system and rotate it in some way, while leaving the vector P to point in exactly the same direction as before. Then, P will be expressed as some other combination in terms of x,y and z, because the x, y and z directions have changed, even though P has not changed.

    This is exactly the kind of thing which happens in Feynman's SG example. The S segment has its x,y and z directions set up in a particular way. The T segment is rotated so that x,y and z are different. When a particle passes through, the physical direction of P never changes, but its decomposition in terms of x, y and z can change.

    If you block one or more of the x,y and z channels, then the measurement you make of P will vary depending on which channels you blocked. For example, with the P given above, if we "blocked" the x and y directions, we'd measure a final P as being:

    P = 2z

    So, we'd "measure" P as being in the "z state" after the apparatus. Then, if we rotated the apparatus and put this "z state" through, the state after the passage would be, again, some new combination of x, y and z.

    The point is, without any blocking, the actual, physical direction of P doesn't change. All that changes is the basis (x,y,z) as we rotate the apparatus.

    That's all correct.

    Agreed. And this is exactly why the actual direction of the polarisation vector doesn't change. It just doesn't have time since it hits the "wall" so suddenly. The wall corresponds to a sudden change of basis, and that is all.

    There is no time for the particle to be physically rotated by the field. That's the whole point. Instead, it becomes a superposition of polarisation states, just as P is expressed as a new combination of x,y and z when it enters a new segment in the state 2z.

    To the contrary, Bell shows that the apparent nonlocality CANNOT be explained away as being due to hidden variables.

    That's precisely the view which Bell demolishes.

    Right. That's the starting point. We ask: could quantum behaviour be due to hidden variables?

    Yes. At first glance, the hidden variable theory seems like a viable explanation of apparent nonlocality. But this is still early in the book, isn't it? Bell goes on to show why hidden variables of the von Neumann type are not viable.

    The obstructions effectively "remove" one or more of the components of the polarisation vector, like removing x and y in my example above.

    It is like this, in my analogy: Each part (x,y,z part) of P follows a different path through the apparatus. If, at any stage, a measurement is made, the state of P is observed as being in one of the three states (x,y,z), with probability in proportion to its component in that "direction". A single particle can exist, inside the apparatus, in all three states at once, or more accurately in a superposition of the three states.

    No. The particle can be in all three states at once, just as P is in all three states x,y and z at once.

    Without a measurement, my answer is unequivocally NO. There is no physical tranformation - just a change of expression to a different basis.

    Correct. The only effect of the gradient is to spatially separate the state components, until such time as a measurement is made, at which time the state collapses to one of the basis states.

    No. The measurement only occurs when you detect the particle, perhaps by projecting it on a screen, or allowing it to collide with a particle detector.

    More accurately, each STATE's direction is altered. The particle can be in a superposition of states, each of which follows a different path.

    We can know the state exactly, yet still be unable to avoid collapse. In my example, we know P = 3x + 4y + 2z, but when we measure the state, it MUST be found to be either x,y or z. We can't MEASURE a superposition, even if we know it exists. This is the basis of the probabilistic interpretation of the quantum wavefunction.
     
  12. geistkiesel Valued Senior Member

    Messages:
    2,471
    I understand everything you just said operfectly. The problem is not near as complex from my side of the segment. The +S, what I mean by _S is thagt this is thje pbserbved sgtate of the particle, observed or observable. Th eparticle will always in teh up or '+' direction parallel to the S line which is parallel to any S segment z-axis. KI told you I wasn't exponding on QM and I meant it. here is no real contradiction here , but an interpretationn barrier.

    If you block one or more of the x,y and z channels, then the measurement you make of P will vary depending on which channels you blocked. For example, with the P given above, if we "blocked" the x and y directions, we'd measure a final P as being:

    P = 2z

    So, we'd "measure" P as being in the "z state" after the apparatus. Then, if we rotated the apparatus and put this "z state" through, the state after the passage would be, again, some new combination of x, y and z.
    [/quote]
    If this is QM then we dio have a problem. IN all "normal" ATD gtransitions, (alien to donmestic transition) as in S - > T where the T segment is blocked in the middle and lower channels, the particle that survives is in the +T state with all memory of the previous S sgtate erased, gone zero, nada.
    Emopahtically no. James R you are mising the reality of the forces at work here, the magnetic field gtadient of the T segment acting on the magnetic properties of the particle. Maybe not you, but the math is missing the forces.

    The fiorst figure below is an S =T => S transition where the segment is is not obstructed. The magnetic spion vector is incribed in the aprticle anad ios more than juswt for sjow I mean it as a literal statement. IN any event you do see the logis of the figure and the results are consistent with experiment. that is the particle reforem to the input S state.

    The "magnetic spin vector ientration is stated for a purpose. If the MSV
    did not reorient to the demands of the fiels of the domestic segment the MSV would remain whjere you see it and the field wowuild then operate on this vector in qan irregular way. The particle would be driven off of a straight line trajectorhy, Actually, straight-line wrt the y-axis of travel and into the z-y plane.
    //ourworld.cs.com/Sandgeist/forum_tutorial/segment.GIF

    Notice the orientation along the T segment z-axis, until the particle exits the segment, then it reorients to the inpuit state, +S. I read your post as if the mathematics left the playing field of observational physics.
    In any event the results are consistent with experiment correct? The particle exits the segmjent in the +S state?
    Now lets add top obstructions. Tjere is nothing new for this particle description until the particle reaches the place containing the obstructions. The polarization is the same, the particle takes the +T channel and exits the segment in a +T base state. This state is produced identicallhy with e production of the +S state. Can yoju see this?

    Now though the orientaion does not vary from the +T orientation after passing through the plane of the obstructions. The obstructions very definitely and unambiguously are causally related to the permanent change of state.of the particle as were the obstructions in theoriginal S segment that produced the +SD state of the particle.

    //ourworld.cs.com/Sandgeist/forum_tutorial/TS+Transition.GIF

    That's all correct.


    James the particle dioen's get through the segment if the magnetic spin vecotr doesn't chanbge. This is a physical reality. If we are in some agreement regarding the results of the transitionsgthen you must integrate the reality of the forces acting on the oparticle. into your mathematical miodel, other wise you are being led astray.If you paid good money to learn what yiou just stated you have a substantial refund due in the form of returned tuition fees.

    James R, the original SG experiments in th e20-30s brouight out the error in assuming the state vector was randomly oriented and rigidly attached to the particle. Were this model proper the output particles would produce a continuousi band of impacts along the z-axis of teh target screen. This was not he case there were only two points [once they worked the bugs out of the experiment.. I spojke hastily when I said the particles would be dragged into the x-y plane. The particles would not exhibit quantum mechanical attributes. Your understanding of the pysics is not only contrary to mine but to Feynman and the general body QM theorists.

    1. What is a 2x state?
    2, What authority do you have saying there isn't enough time to change the state?
    3.You realize do you that Feynman agrees with me and disagrees with you on this?
    4. You realize there are no measurements showing any x and y components of the magnetic spin vector that is polarized as in +S or +T state, for example don't you? don't you? If thios ios copntrary to This is contray to


    Show nme a Bell sattement that unambiguously says that nonlocality canniot be exoplained awahy due to ghidden variable?



    That's precisely the view which Bell demolishes.



    Right. That's the starting point. We ask: could quantum behaviour be due to hidden variables?



    Yes. At first glance, the hidden variable theory seems like a viable explanation of apparent nonlocality. But this is still early in the book, isn't it? Bell goes on to show why hidden variables of the von Neumann type are not viable.



    The obstructions effectively "remove" one or more of the components of the polarisation vector, like removing x and y in my example above.



    It is like this, in my analogy: Each part (x,y,z part) of P follows a different path through the apparatus. If, at any stage, a measurement is made, the state of P is observed as being in one of the three states (x,y,z), with probability in proportion to its component in that "direction". A single particle can exist, inside the apparatus, in all three states at once, or more accurately in a superposition of the three states.



    No. The particle can be in all three states at once, just as P is in all three states x,y and z at once.



    Without a measurement, my answer is unequivocally NO. There is no physical tranformation - just a change of expression to a different basis.



    Correct. The only effect of the gradient is to spatially separate the state components, until such time as a measurement is made, at which time the state collapses to one of the basis states.



    No. The measurement only occurs when you detect the particle, perhaps by projecting it on a screen, or allowing it to collide with a particle detector.



    More accurately, each STATE's direction is altered. The particle can be in a superposition of states, each of which follows a different path.



    We can know the state exactly, yet still be unable to avoid collapse. In my example, we know P = 3x + 4y + 2z, but when we measure the state, it MUST be found to be either x,y or z. We can't MEASURE a superposition, even if we know it exists. This is the basis of the probabilistic interpretation of the quantum wavefunction.[/QUOTE]
     
  13. geistkiesel Valued Senior Member

    Messages:
    2,471
    Yes it was a long post. I will make the most serious effort to streamline, abbreviate, to single sylableize the next post to the maximum extent!!!.

    Are you familiar with the term the "excess" elements of the particle state?
    Next post is on that very subject.. In the S -> T -> S transition the elements that guarantee the reformation of the S staste are not expressed in the transition statement. Therfore, the elements are nonlocal, unobserved, hidden. The elements are also the absence of gthe elements means nio observed state; existence critical elements are the "excess".
    The problem with your analogy James R is that +S has a physical meaning. It says the particle will move up along the z-axis when transitioning through an S segment.And this is basically all it says.

    Where is Feynman's Feynman's example that you just mentioned.
    1.are you saying the direction of P is always the same, for eternity?
    2.where how are the components of P established,
    3.are the components, mass? EM? Photon" Fundamental particle, if so what kind?
    4,have the components wver been measured directly, if so where?


    James R. went I through the Feynman references, the Bell references and the Bohmm references. You are totally off base. I have never heard antything approach your model which is totally foreign I understand everything you just said perfectly. The problem is not near as complex from my side of the segment. The +S, what I mean by +S. as well is that this is the observed state of the particle, observed or observable. The particle will always move in the the up or '+' direction parallel to the S line which is parallel to any S segment z-axis. I told you I wasn't expounding on QM and I meant it. There is a real contradiction here because we aren't talking the same languag e when we point to the same object we are descrbingt. This is deeper than a simple interpretation barrier.

    Here is what RF said on page 5-5:
    ' as careful as we have been to make sure we have the atoms in a definite condition, the fact of the matter is that if it goes through an apparatus which is tilted at a different angle it has , so to speak to "reorient" itself--which it does, don't forget, by luck. We can put only one particle thtrough at a time, and then we can only ask the question: What is the probhability that it gets through? Some of the atoms that have gone through S will end in a (+T) state, some of them will end in a (0T) state, and some in a (-T) state. - all with different odds.​


    On 5-6
    "Supposse we had atoms filtered into a the (+S} state, then we put them through a second filter, say into a (0T) state, and then through another +S filter. (We'll call that last filter S' just so we can distinguiish it from the first S-filter).Do the atoms remember they were once in a (+S) state? In other words we have the following experimet​

    Please Register or Log in to view the hidden image!



    But the direction of the vector would not change? is that it?

    What is the physical nature of P you describe? I am trying to sort it all out.
    If this is QM then we do have a problem.However, you are confusing the issue here james R. drastically. In all "normal" ATD transitions, (alien to donmestic transition) as in S - > T, where the T segment is blocked in the middle and lower channels, the particle that survives is in the +T state with all memory of the previous S state erased, gone, zero, nada.
    Emphtically no. James R you are missing the reality of the forces at work here, the magnetic field gradient of the T segment acting on the magnetic properties of the particle. Maybe not you, but your math is missing the forces.

    The first figure below is an S =T => S transition where the segment is not obstructed. The magnetic spin vector is inscribed in the particle and is more than just for show I mean it as a literal statement. In any event you do see the logic of the figure and, the results are consistent with experiment. that is the particle reforms to the input S state.

    The "magnetic spin vector" orientration is stated for a purpose. If the MSV
    did not reorient to the demands of the fields of the domestic segment the MSV would remain where you see it and the field would then operate on this vector in an irregular way. The particle would be driven off of the allowed trajectory. The particle would not behave as observed if your description were applied.

    Please Register or Log in to view the hidden image!



    Notice the orientation along the T segment z-axis, until the particle exits the segment, then it reorients to the inpit state, +S. I read your post as if the mathematics left the playing field of observational physics.
    In any event the results are consistent with experiment correct? The particle exits the segment in the +S state?

    Now lets add two obstructions. There is nothing new for this particle description until the particle reaches the place containing the obstructions. The polarization is the same, the particle takes the +T channel and exits the segment in a +T base state. This state is produced identically as the production of the +S state. Can you see this?

    Now though the orientation of the MSV does not vary from the +T orientation after passing through the plane of the obstructions. The obstructions very definitely and unambiguously are causally related to the permanent change of state.of the particle as were the obstructions in the original S segment that produced the +S state of the particle.

    Please Register or Log in to view the hidden image!



    all correct.The figure speaks for itself in a way. This shows the unambuguous affect of the obstructions, but notice the obstructions do not affect the particel directly it is safely above the obstructions.


    What physical aspect or attribute of the particle does the filed/gradiemt affect.?
    James the particle doesn't get through the segment if the magnetic spin vecotr doesn't change. This is a physical reality. If we are in some agreement regarding the results of the transitionsg then you must integrate the reality of the forces acting on the particle into your mathematical miodel, other wise you are being led astray. If you paid good money to learn what yiou just stated you have a substantial refund due in the form of returned tuition fees.

    James R, the original SG experiments in th e20-30s brought out the error in assuming the state vector was randomly oriented and rigidly attached to the particle. Were this model proper the output particles would produce a continuousi band of impacts along the z-axis of teh target screen. This was not he case there were only two points [once they worked the bugs out of the experiment].. I spoke hastily when I said the particles would be dragged into the x-y plane. The particles would not exhibit quantum mechanical attributes. Your understanding of the physics is not only contrary to mine but to Feynman and the general body QM theorists.

    1. What is a 2x state? please describe the state, even if hypothetical.
    2, What authority do you have saying there isn't enough time to change the state?
    3.You realize do you that Feynman agrees with me and disagrees with you on this?
    4. You realize there are no measurements showing any x and y components of the magnetic spin vector that is polarized as in +S or +T state, for example don't you? don't you? If thios ios copntrary to This is contrary to


    No one has ever tried to "expalin away" nonlocalioty by "hidden bariables".
    Show me a Bell statement that unambiguously says that nonlocality cannit be explained away due to hidden variable? James show me just one unambiguous reference that Bell stated that nonlocality could not be explained away by hidden variables. I hope this will also explain "explained away".

    You just confused the issue, and issues don't get confused , issuers get confused. What does the statement mean that hidden varioable cannot explain away nonlocality. Bell never tried to "explaain away" nonlocality. James R, I mean this in the most bland and objective observation I can muster. It sounds like you are making all this stuff up as you go along. You aren't making sense. .

    Wrong agaqin James R, this is the arguament that Bell made, the argument he created. YYou are makingh this up I know. You cannot hide the fact you just don't kniow what you are talking about. James R I cannot sit here and let you do this without comment. I just cannot understand your motive. WHat are you embarrassed that you have no konlwedge one way or the other of what you say?

    Wrong again Jamesz R. Bell was talking about if the so-called hidden variables could be prepared, then they wouldn't be 'hidden 'and QM would be observably inadequate. Obserbvably james R. Read what the man said.
    WEll that's why I am here I say to myself, as I quit complaining.



    NOnono. The only wqay you can make sufcha sttement is if you are completely ignorant of the Bell's book, or your lying. Either waymakes no difference because whaty you say is totally fabricated. Bell argues with Eisnstein, Podolsky and Rosen who argues that QM theory should be supplemented with additional variables. These variables were to restore to the theory causality and locality, say EPR, , Bell says to this in the EPR chaopter,
    "These additional variables were to restore the to the theory causality and locality. In this note that idea will be formulated mathematically and shown to be incompatible with the statistical precdicitons, of quantum mechanics.It is the requirement of locality or more precisley that the result of a measuremnt on one system be unaffected by the operation on a distant with which theit has interacted in the apst , that creates the essential difficulty . "[Bell discusses attempts by others to show no 'idden variables' interpretation is possible, then Bell contionues] These attempts have fouind to be wanting, Moreover, a hidden variable interpretation of elementary QM theory has been explicitly constructed [by David Bohm].That particlular interpretation has indeed a grossly non-local stricture. . . . .there is no difficulty in giving a hidden variable account of spin measuremnts on a single particles, . . .the statistical features of QM arise because of the value of this variable is unknown in individual instances. ​

    James R the Bell book is a collection of published papers over a period opf 20 years.( 64 to 84) . Your characterization of the book leaves something to be desired in how you formulate your arguements how you construct your physical models. It is all too apparent here that you need some catching up to do. Certainlhy your library has access to Feynman's Vol III ch 5 and Bells speakabe and unspeakable QM.

    I can understand that, so what is the physical nature of the x and y components of you vector? Or in real world, other than your hypothetical, what are the physical nature of the components of your spin vector and what exactly is the pjysical interdiction of the obstructions
    n the particle state.
    1.are foprces applied.
    2. What is th nature of the force, mass, EM, other?
    3, what brings out the component such that the obstruction in the unused channels are the difference. If the obstructions were in the x axis the obstruction would have no affect on the particle state.


    Absolutely not.


    NO NO A thousand times no.
    Whatever you mean three states at once the particle takes one of three trajectorie. On page 5-3 and 5-4 he discusses the taking of a single path as basic to SG transitions. This is why I suggested you were confused, perhaps thinking about some other aspect of QM?

    THis is a mathematical function the change to a differentr basis?



    This sounds reaonably coherent except for the statement that the only purpose of teh gradient is to separate the state components. The gradient is what gives the particles z component motion, +, - or +-(0).. Havee you noticeds there is never any X component motion, only z and y? The y-axis because that is the direction the partuicle beam is generated from out side the segment. The z-axis due to the QM attributes of the particle, but X axis? I see none, show me..



    Or subjecting it to an ingtense magnetic field and gradient.


    What are the forces directing the partuicle? What is a state "direction"?



    [/quote=James R] We can know the state exactly, yet still be unable to avoid collapse. In my example, we know P = 3x + 4y + 2z, but when we measure the state, it MUST be found to be either x,y or z. We can't MEASURE a superposition, even if we know it exists. This is the basis of the probabilistic interpretation of the quantum wavefunction.[/QUOTE]
    Howe can you "kniow it exists" if you canno measure it?
    Are you saying the particle does not have a particular definitive state ever?

    How do you know the three states are there if you can consistently measure only one of the states? I know the rationale, or can deduce it, but how do you prove the three states with a single state. How is a state expressed?..how is the state measured.?
    What is so convincing to you that the three states exiasts simultaneoyuslty?

    Geistkiesel ​
     
  14. James R Just this guy, you know? Staff Member

    Messages:
    39,397
    geistkiesel:

    In my analogy, you can take x,y and z to correspond to +S,oS and -S.

    For this experiment, P is the same until a channel is blocked, or a measurement is made (i.e. the state of the particle is observed by detecting it).

    They are established at the start of the experiment. One way is to take a group of particles with random P directions, and put it through an SG apparatus with two channels blocked, after which all particles will be measured to have the same P.

    Components of the polarisation? The polarisation direction of a particle can be taken to be the direction its spin axis points in space, if you like.

    Put the particle through a SG apparatus, and you can measure the component of P in one direction (the z direction of the apparatus).

    That may be true, but Feynman certainly claims to be expounding on QM.

    It is just a vector which always points in the same spatial direction, regardless of how you rotate the x,y and z axes of the coordinate system in which P is expressed.

    The forces are not the point. The forces only let us distinguish the states at the end of the experiment. They do not affect what the state is. The only thing which alters the state in these experiments is which channels are blocked or unblocked.

    No attributes. The gradient exerts a force on the particle, but does not change its polarisation.

    If only that were true. My PhD thesis would cause quite a stir if it was really saying something fundamentally new about this.

    It is analogous to a +S state. The "2" is just a measure of "how much" +S there is, compared to oS or -S, which would be y or z states in the analogy.

    The polarisation won't change if the Larmor precession frequency of the particle is much greater than the perceived rate of change (by the travelling particle) of direction of the magnetic field as the particle transitions through the apparatus. That is the physically required condition for the state not to change, though I don't think Feynman goes into that kind of detail in his lectures. He just assumes implicitly that the condition is met.

    No.

    My discussion of x,y and z, as I said, is an analogy. x,y and z do not refer to directions within the SG apparatus, which are an entirely separate matter.

    You have Bell's book. Look it up yourself. If you think he didn't say that, find me an unambiguous quote which says so. I'm not going to do your leg work for you.

    I don't disagree with that.

    Yes. Bell argues that hidden variables are in fact inconsistent with quantum mechanics.

    Yes.

    Right. That's what I've been saying all along. Hidden variables are incompatible with quantum mechanics. Bell agrees with me. I agree with Bell.

    Here, you are talking about directions within the SG apparatus (a completely different topic to my x,y and z components of P). I have no argument with anything you've written here.

    No. Subjecting it the gradient, by itself, does not constitute a measurement. No observation is made while the particle is subjected to the gradient. It is only observed after it exits the apparatus.

    It's theoretical. And you can deduce part of what the state must have been from the measurements you make.

    The whole point of quantum mechanics is to describe how states of a system evolve.

    No. A particle has a definite state all the time. This is often a combination of measureable eigenstates. A measurement on a quantum particle causes the quantum state to "collapse" to one of the eigenstates, with probability given according to the complete the quantum state. AFTER a measurement is made, the complete state is actually altered so that it "starts again" as a "pure" eigenstate.

    In my analogy, suppose we have a particle with P = 2x + 3y + 4z. This means the probability of the particle being measured as being in state x, as compared to y or z, is in the ratio 2:3:4.

    If we make a measurement, suppose we find that the particle is in eigenstate y. Then, after the measurement, that particle will have P=3y. The state has "collapsed" from a superposition of the three eigenstates to the single eigenstate y.

    That's what Feynman is trying to demonstrate. What you are doing is assuming that only one eigenstate is present at all times - that there are no superposition states. However, it can be shown that that approach does not work. It doesn't reproduce real-world results.

    The fact that quantum mechanics works.
     
  15. geistkiesel Valued Senior Member

    Messages:
    2,471
    Fine, but why do you need an analogy? When I say +S I recognize that +S is incomplete description of the particle state because this is all it says. The +S says nothing of those unobserved states that guarantee the existence of the +S state.

    If a channel is blocked and the +S state particle with spin vector P enters the segment when do the obstruction affect the particle?

    Where do the obstructions affect the particle P direction?

    What is the force that affect the particle P direction?

    This sounds like we agree here. The particles are all transformed into +S or +-S or -S when passing through a blocked S segment. Only the +S particles survive. Is thios correct?

    I am asking what are the components of the polarization vector P. Do you know?
    WE ahve agreed that the polarization vector is P, or +S did we not? This is what I am ghetting at here: What is the physical nature of the components of the vector, the physical vector?

    That is the vector isn't it? I mean the +S particle moves up along the Z-axis, so is this what youi are refering to as the components of the spin vector? Pardon me, but it sounds like we've gone into a circle. Straighten this out OK?

    So if the +S polarized particle enters a T segment the +S state remains pointed at the same direction it was pointed at when passing through the obstructed S segment that produced the +S state, correct?
    What is the source of the physical forces that change the state of the particle?

    If the magnetic field and gradient only let us distinguish the state of the particle after it leaves the segment then are you saying that any P direction of any +A +D +H +K +Q +R +S +U ...+Z state particle will all be pointing at different directions when entering and exiting a T segment?
    I am asking what attributes of the particle are affected by the field and gradient of the segment. Does the field exert forces on the quarks, the mesons , the neutrons of the particle, what attribnute of the particle does the field affect?
    [
    What does Feynamn mean then when he says when entering a T segment the +S spin vector is reoriented to the direction of the T segment z-axis?

    Are you saying the same thing as 'how many' +S state particles pass through a doubly obstruicted S segment as a proportion of the total number of particles in the beam that enter the S segment?

    So, if we assume the Larmor precession frequency is below some cutoff frequency in all the +S particles that pass through the T segment then all the particles' polarization directions will change to that imposed by the magnetic field of the T segment. In this case we can easily determine the state of the particles in the segment before the exit the segment. At least we know the particle will be in one of theallowed T states while passing through the T segment, correct?
    Are you aware of any measuremnts of other than Z -axis components of a +S state particle passing through another S segment?

    First you deny the concept of hidden variables,and you add that Bell said nonlocality cannot be explained away by "hidden variable"s At least this is what I understand you to say.
    Bell said none of this. Bell emphasized the reality of nonlocal fiorces and the real affect of hidden variables, The terms nonlocal and hidden variables are equivalent. Your statement makes no sense.

    No, Bell says Quantum mechanics is incomplete without hidden variables.And that QM statistical predistive poweer would inconsistent with additional local variables.
    Bell of course was talking about others who had hoped tha additional vriables would restore causality and locality to qm.
    The quote was taken from the opening paragraph of the EPR paradox paper. The words are describing the belief by some that aditional local variable would restore causlity and locality to quantum mechanics. Bell is saying that these additional variables are incompatible with the statistical predicitions of QM . If we are to say that measuruing particle A affects a distant particle B that no additional local variables will work as these are inconmpatible with the statistical predicitons opf quantum mechanics.

    James R, Bell was criticising what others had thought that additional local variables would restore causality and locaclity as being inconsistent with the statistical predicitve power of QM

    In the same paragraph Bell is saying that "others" have tried to show that no 'hidden variable' theory is possible. Bell replies
    "These attempsts have been examined and found wanting. Moreover a hidden variable interpretation of elemntary quantum theory has been explicitly constructed."​

    James R you have to get out of that rut where you think Bell has discounted "hidden variables", Bell is Mr, Hidden Variable personified. Get it in your head man. You make me think that you discount "Hidden Variables " yourself. Do you discount the viability of hidden variables in quantum mechanical models?

    OK, we make progress in understanding. You are saying then that the polarization vector, the magnetic spin vector can change direction inside the SG segment, but outside in field free region the vector is pointing in a different direction, is this it?

    If one polarizes one million particle to the +S state in a suitable SG S segment and measures the first 100000 particles to be in th +S state by checking the state by passing them through another S segment all 100000 are found to be in the +S state, then are you saying that the state of the next 900000 particle passing through the second S segment is unknown until measured after passing through the second segment?
    Then are you saying that QM state predicitons cannot be explicitly verified by experiment and that the dedeutions of part of the states is equivalent to measuring the state exlicitly?
    I never said that the particle had one eigenstate at all times, but yes as you use the term "superposition of states", I do disagree. If the +S particle is moving up in an S segment it has a definte state of +S. There are components of that state though the components are nonlocal, unobserved. QM claims that a definte +S state will result in the indefinite +-1 state along the x -axis, though they cannot be measured at the same time. Therefore I conclude that qm predicts x components for the observed Z-axis motion of a +S particle moving in a SG segment.

    While I have promised not todiscuss QM Imust point out a trivial flaw rfegardng thesuperposition of states that has gone a longa way gtolead QN dioen an incomplete and dead end path.

    When you perform the superposition of states don't blindly follow the mathematical models that work in observed realms of realiy and asume that an observed Z-axis state will be complemented with additional +-x state components. Rotate the x components onto the z-axis where all the observed action is seen to take place. Actually rotate the x-axis and and shift such that the components are all positioned on the z-axis.

    Another roration scheme which seems more symetrical is to simply rotate the x-axis ontothe z-axis from the get go. and procede on the assuption that the state generation and componetrs of the state are all =z-axis dioscoverable.

    Finally, consider the 0S, or 0U state to be reflecting the +-U state, which is a partial picture of the entire particle state gnerating structure which produces all the sattes on a continuous basis until polarized and the current observed state is designated the default polarixzed state still axtive genrating alternate + and - states as + - + - + - + -+ etc as a y(+,-) state, which is, if you are perceptive, a modified spin-1/2 particle state generator. This leads to a better understanding of two hole diffraction where the spin-1/2 particle, the observed electrons, transitions through one of two holes where the + and - states are generated in the accompanying hole. This reduces the mathematical complexity and removes the necesity of constructing an ad hoc "particle /wave" duality and remioves the riddle of
    one hole-two hole mystery where theoerists have struggled to find a way to get one electron through two holes. Electrons do not easily transform into wispy aethereal matter and coperatively use both holes in the transition. Have electrons ever shown that they easily and forcelessly separate into two mass parts and recombine with no observable reaction one way or the other of separation and recombination?
    The answer toall this is no..

    The claim that the the +S particle has indefinte local x components is patently false as should be obvious to you and the rest of the observing world. You alluded to this previously above where you stated that only the obstructions, or not, affect state changes. To change a state means a change of the components of the state, and as the obstructions are located along the z-axis only, no affect on any x-axis component is possible and that the components of the z-axis motion all lie along the z-axis. Now the problem arises from the fact that the particle that survives the tranisiton in the unobstructed channel is physically untouched yet the obstructions are nevrtheless crucial conditions for the change of state that are observed in the particle that is outwardly unaffected by the obstructions..

    If we follow your logic as stated that the +S particle passing through a wide open T segment does not reorient the magnetic spin vector, to a T segment magnetic field imposed direction , yet adding obstructions results in the change of state of the particle and orientation then you are arguing for an exchange of force between the particle and nonlocal force channels that is explicitly manifest at the point where the trajectory of the particle in the obstructed channels would have been had the particle been polarized to one of the obstructed channel states. This is not so dreadfullly complicated that it cannot be sorted out. I believe you can follow what is said here.

    There is no other explanation for the mechanism of the obstructions being a crucal role in the permanent change of particle states.

    You introduce an unnecessary variable in the matter by insisting the particle retains the +S state in the wide open channels, but ceded the state change to obstruted segments. This then implies the state change occured not only by the obstructions, but added is the orientation is also changed by the obstructions all of which the particle does not contact. This is so as the first half of the transition trajectory is obstruction free, identical to the obstructed segment, which leaves only the position of the obstructions which can account for the permanent vector state change, which is what I have been saying.

    However, at this time it seems that James R is in the painted corner because now you must explain the physical mechanism that resulted in the exhange of state and thepernmanency of the change and direction for the spin vector relating to the obstructions. This is adaunting task for a model that is purely statistical in nature and comes very close to denying any underlyinh reality to the quantum world.

    My scenario doesn't have the problems assigned to QM as I have accounted for the change of spin vector direction at the entrance to the magnetic field and gradient due to the construction of a hybrid mixed element T state. The mixed elements are nonlocal elements of the +S and now observed and polarized, T state.This solves another problem for me that you have also rejected when you say the spin vector doesn't change direction but that eigenstates change when the particle is directed through a domestic SG segment. AS I read you the nonlocal states change but the local observed sates remain invariant [when they do that is].

    Infact you have expressly state this in aniother thread (I will look it up if you like).

    At the present you are stuck to the particle passing though the plane of the obstructions as the key point, and only point actually, where the particle state and spin vector direction occur. However you will not be able to explain this as an effectively complete descriptive model of the state change without first including the nonlocal force channels and the local/nonlocal force channel interface which is screaming in your face from the obstructions in the segment.

    I truly hope you see the necessity of including nonlocality and hidden variables in the causal chain describing the spin vector reorientaion in the obstructed segment.

    If you accept the direction change, at the intsance of polarization entering the T segment, to a temporary and unstable hybrid state of +T as we have been discussing then you need not ask the obstructions to change either vector direction and state by their presence. You only ask the obstruction to make permanent the temporary hybrid +T state as it is observed and known to be whenpassing through the plane of the segment obstructions. When the obstructions are removed you need only to recognize that the instability of the mixed hybrid T state is resolved in the favor of the +S state upon exiting the field of the segment. Remember the force of the field provides the stability of the hybrid T state platform which cannot be mantained in the field free region.The obstructions perturb the mixed and nonlocal hidden elements of the +S and +T . In the wide open segment the nonlocal elements of the hybrid state are not perturbed and the competition for the survival of the 00[+S] nonlcal elements prevail over the 00[+T] nonlocal elements.

    The commin compass needle requires the force of the earth's magnetic field to reorient froma perturbed direction. The polarize particle ahs its "north" built in which is thedirection of the segment that produced the permanent particle spin sate such as the +S state and +T sates we have been discussing.

    One way to look at this is when the 00[+S] and 00[+T] are perturbed , wldly, neither of those nonlocal elements will survive without someplace to go. The +T is available as an existing observed state for the 00[+T] elements at the instant the particle passes through plane of obstructions. The 00[+S] elements have nothing observed to latch onto are not heard from again, but the total state component count is preserved perfectly.

    In the wide open case, the observed +S state, latches onto the nonlocal 00[+S] and there is no +T state to latch onto the 00[+T] elements as the +T disappears when the field disappears which occurs when the particle enters the field free region. hence the nonlocal oo[+T] disappear into the void and again the conservation of component count is preserved.


    So did Ptolemy's prediciton of the motion of stellar objects work with his unique circles within circles model of the solar syatem. Yes QM 'works' yet it is grossly incomplete, as Bell tells us, and as experimental results patiently tells us every time we run one of these experiments, grossly incomplete when the QM model is exlicitly void of hidden variables or sometimes referred to as nonlocal force structures.


    Geistkiesel​
     
    Last edited: Jun 4, 2005
  16. James R Just this guy, you know? Staff Member

    Messages:
    39,397
    geistkiesel:

    I don't need an analogy. I already understand quantum mechanics. I introduced the analogy in the hope that by explaining things to you in a different way we might make some progress. But now I think I just caused more confusion.

    To save myself some effort, I am not going to discuss my (x,y,z) analogy of polarisation any further in this thread. Let's just stick to +S, -S etc.

    When the particle hits the obstruction. Of course, if the particle is in a superposed state, only some parts of the particle wavefunction will hit the obstructions.

    The forces involved in hitting the obstructions, if any.

    Yes, in a segment with two channels blocked.


    I just told you. The physical nature of the polarisation of a particle is the direction its spin axis points in space.

    Yes.

    The only forces which change the state of the particle are forces exerted by whatever blocks the particle beam inside the apparatus.

    Yes, provided the T segment is unblocked.

    The force is a magnetic interaction between the external field of the SG apparatus itself and the intrinsic magnetic dipole moment of the particle passing through. If that particle is an atom, the magnetic dipole moment is due to the orbital angular momentum and spin of the electrons in the atom, as well as possibly a contribution from the intrinsic dipole moment of the nucleus. I don't see why all this is relevant.

    I assume he means that the basis changes, but I'd have to see the exact quote to be sure.

    Yes. In that case, all the particles would smoothly rotate from the +S alignment to the +T alignment, and all would exit the T segment in the +T state.

    Finally, from what you've written below, I understand why we have some confusion about Bell. There are two potential types of "hidden variables" - local hidden variables and non-local hidden variables.

    Can we agree that Bell showed that LOCAL hidden variables are incompatible with quantum mechanics, but that NONLOCAL hidden variables may exist?

    If so, then we're on the same page. I have been talking all along about local hidden variables. It seems you have been talking about nonlocal hidden variables.

    I agree. That clears that up, then.

    Yes.

    No.

    No. When a measurement is made on a superposed state, you can't tell from the measurement what the state was immediately before the measurement, though you know exactly what the state is immediately after the measurement. Measuring changes the state - it collapses the state to one of the eigenstates.

    Yes. But the same particle can be partially in the state +S and partially in states -S and oS. Only the +S part moves up in the S segment. The other parts follow different paths.

    This paragraph obvious comes from a mixing of my (x,y,z) analogy with the actual SG experiment, and the resultant misunderstanding. I don't think I need to address it further.

    I don't really follow what you're proposing here.

    If a car collides with a wall, is the car changed? Then why should a particle not be affected when it collides with a physical barrier?

    Let me ask you. Your view is that a +S particle entering a T apparatus instantly changes to one of the three allowed T states upon entering the apparatus. What force causes that change? And how is the particular T state selected? Why do some particles end up in the +T state, and others in the -T or oT state? What's your theory?

    Nonlocality is not required in these examples. This has nothing to do with EPR experiments, entanglement, or any of those kinds of issues. Nor are hidden variables of any kind necessary to explain the behaviour of quantum particles in these (SG) experiments - local or nonlocal.

    Right. That's an adiabatic motion, not a sudden transition such as the ones we're discussing here.

    Ptolemy's model didn't work to 1 part in 10<sup>16</sup>, as some quantum results have been established to.
     
  17. geistkiesel Valued Senior Member

    Messages:
    2,471
    James R I thought I would try this and then let youpick it apart. The second one is the brioefer of the two, the 1st should have been entitled overkill.
    Geistkiesel
    http://ourworld.cs.com/sandgeist/EQTP_Home_Page.html


    http://ourworld.cs.com/Sandgeist/index.html

    Your discalimer the SG transitions are strictly a local effect, well I question that,.nut How did you come to such a conclusion.
     

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