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Simultaneous measurement on 1 photon

  1. Apr 3, 2005 #1
    suppose two photons are created, they have spin 0, because before their creation the total spin was zero.
    Now two simultaneous spin measurements on the same photon take place. (within the same minimum time inteval, one Planck time). Amazing equipment.
    What can be said of the outcome of these two spin measurements. Can / must they be the same?
  2. jcsd
  3. Apr 3, 2005 #2


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    No. One has spin +1 around some axis and the other has spin -1 (the opposite direction), so they preserve the original 0.

    If they are made along the same spin axis, they will be tha same. If along two different axes, different. Spins around different axes don't commute, and so the first spin measured does not contribute to the second measurement. This is a famous physics experiment, often used, as by Feynmann, to illustrate the non-intuitive reality of quantum physics.
  4. Apr 3, 2005 #3
    I would appreciate a few words more. Specifically related to "quantum entanglement." Let me tell what I think, but am not sure of:

    I assume that all we know is that the sum of the two spins is zero, but nothing about the orientation of either until we measure one. Then, even though the second is measured with very little delay after the first (so little delay that not even light could span their separation before the second measurement is made). I am not sure the following actually makes any sense, but it will help make my amazement more clear: suppose that the first to be measured has "spin up" in the local vertical direction (prior to measurement) but I don't know that and have set my "spin direction detector quantizer" to observe either spins aligned +50 or +130 degrees form the local vertical. I think it probable, the indicated (observed) direction will be +50, but there is some probability that +130 will be indicated. In any case the second measurement on the other photon will give the opposite indication, assuming the other "spin direction detector quantizer" was aligned parallel to the first.

    I think this is pretty standard (but very strange) quantum theory. Now I want to assume that my friend making the second measurement according to our pre-arranged schedule with accurately synchronized clocks, set his "spin direction detector quantizer" to +5 and +185, (because the "0" of my "50" did not get printed on his copy of the protocol) Thus his "spin direction detector quantizer" axis is actually rotated 45 degrees from mine. (Both of us did tilt the top of the "spin direction detector quantizer" due East as planned, so only the angle from vertical is "wrong.")

    Frankly I am confused as to what to expect. If my observation is +50, then I think that prior to his slightly delayed measurement the spin of his photon that will very soon arrive at his "spin direction detector quantizer" has become +130 but will be forced into either +5 or +185 by this second observation. with probabilities that could be calculated from assumption that the "pre observation spin is in fact +130 (not sure this makes any sense, but hope you are still with my concerns) -All this seems some what reasonable to me, but quantum entanglement is very strange.

    Now suppose that instead of placing his "spin direction detector quantizer" half a meter farther from the atom that emitted the entangled photons, he placed it half a meter closer - no problem. His is the first measurement/ observation and we just reverse "roles." The probability of my +50 and +130 results can be calculated under that assumption that his prior measurement has forced my approaching photon to be 180 from his measurement.

    Now is where I get completely lost: Suppose he thought i was the one who was to be half a meter more distant so he set his "spin direction detector quantizer" at the exact same distant as mine. Neither of us make a measurement before the other. How would we predict the probabilities? Is it a 50/50 mixture of the prior results? or something else? If the two "spin direction detector quantizers" are almost parallel, say his is at +51 and +131, I assume that Almost all the time when I observe +50 he observes +131, but perhaps there are just enough +50 and +51 pares of observation to make on the over all average his photon exactly 180 degrees from mine?

    Quantum entangement had not been observed when I ceased to be very interested in physics (more interested in how the mind functions for last 15+ years) so I have not read many paper on it. If you know of a good one and don't think there is any thing you can say to help me here, just refer me to it, but I bet there are many here, who although they may not have initially had my concern, do now and would like to know any light you can shed on this puzzle. Thanks in advance.
  5. Apr 4, 2005 #4


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    Interestingly, the results do not in any way change - as best as can be determined experimentally - according to which is observed first. Assuming we call the polarizer settings A and B: there are only 2 possibilities:

    1. One is A, the other is A+90 degrees
    2. One is B, the other is B+90 degrees

    Either of the above are conpatible with experimental results, regardless of which is observed first. The results are NOT compatible with the idea that either photon polarization had any other value than one of the two above.
  6. Apr 4, 2005 #5
    We were talking of electron spins, up or down (180degrees) but I have the same problem when two entangled photons are measured at the same time by polarizers that are not aligned 90 degrees from each other, or by:

    Assume that near the source of the two entangled photons they pass thur exactly 90 different orientation polarizers then at exactly the same distance of travel later they encounter a pair also with exactly 90 degree relative to each other, but 30 degrees for the first "filter pare."

    If the one of the later two observations were made before the other, then your answer is correct, clearly. But I am not sure what to think if the final observations are made at the same time. I.e which one forces the other to be 90 degrees from it? Or even if this is true. Was the "forcing observtion the one that was 30 or 60 degrees misaligned with the "prepared polarization" state? The probability of the different possible observations will differ, depending on how you think about this.
    Last edited: Apr 4, 2005
  7. Apr 4, 2005 #6


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    After the particles pass through the first measurement device, the entangled state ends. Subsequent measurements will not reveal additional information about the other member of the pair.
  8. Apr 4, 2005 #7
    I guess it depends on what if any thing happens there (at first polarizers). Perhaps they can be eliminagted and a magnetic field align the orbitals of the radiating atom so that it emits entangled pair of photons, wirh some degreee of knowledge about their polarization (without any measurement of it), if that would make you happier. I did not call it a "measurement", but a "filter." I admit that very few photons would pass thru randomly oriented polarizer with no interaction. Clearly it is not important wether or not some human makes an observation/ measurement. I am not sure that all photons passing thru "filters" would be "untangled" but will not argue strongly with you on this. (I am too ignorant.) I don't know how they preserved "entanglement" and made the observation a few years back in two different towns about 10 km apart. I suspect that distance record is much greater now. In short, I am not sure you are correct that there is no way to "prepare them" without destroying "entanglement"

    Honestly before you switched me to photons form electron spins, I had in the back of my mind someting like the Stern Gerlock setup, with never any particular location for the "filter". Perhaps having the source of photons inside a weakly optically active solution would be the approach for me to suggest. I don't know enough to pose the problem well. What bothers me is that the results of two measurements of entangled quantum objects are easily predicted (but hard to understand why) if one measurement forces a mixed state wavefunction into only one Eigen state - I.e. the results of the other (second) measurement is fixed by the quantum mechanical "choice" made during the first measurement. My problem is what if the two measurements are made at the same time.

    In some ways this puzzle is easier to see/ think about/ with entangled photons. Start with Detector "A" more remote than "B" and slowly decrease the distance of "A" from the source. If we can arrange entangled photons so that we know (perhaps by the magnetic field applied to the source) the polarization of one is "vertical" and the other is "horizontal" only we don't know which is which prior to any measurement, then half will pass thru "perfect polarizer A" if it is also "vertical" and half will be absorbed by it. (Also true even if we don't "prepare" the emission and use a random polarization source.)

    If we have prepared the orientation emitted to be vertical and horizontal, then when A is closer to the source and inclined to the vertical by 30 degrees, only 25% of those that happen to be vertical will pass thru and less than 25% of the "horizontal ones" that strike polarizer A at 60 degrees (from the "pass axis") will pass thru. That is, the total fraction passing thru will drop from the 50% that would if "A" were vertical.

    We both agree that what subsequently happens at more distant "B" is determined by what happened at "A" if they are still "entangled" (definition of "entangled")

    My quandry is what happens as the distance to from source to "A" slowly decreases. When it is less than "B" to source distance, then the events at "B" are "calling the results" at "A." Who is "in charge" when the distances are equal? Does it make any difference if (1) A and B are exactly orthogonal or (2) if they are not exactly orthogonal. (In the net fraction of photons passing thru?)

    I think "who is in charge" surely does in case (2) because the selection/ "choice" at the "master polarizer" can help the photon pass thru the "slave" polarizer. For example:
    if two polarizers are orthogonal and perfect, no photons pass thru both, but if a third is placed between the two at 45 dgrees to both, some do pass thru all three. That is the "forcing into pure Eigen state" at the polarizer "in charge", (the "master") will change the probability of passing thru the slave, just as the forcing at the 45 degree polarizer did when insterted between the two orthogonal ones.

    Thus without going thru a lot of calcualtional "WHAT IFs" it seems to me there is a problem here, but I can not define it very clearly. Hope you can understand what I am trying to state.
  9. Apr 5, 2005 #8


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    What you are describing are the famous EPR experiments ! They were first thought of by Einstein, Podolsky and Rosen in the 30ies, and the first experimental tests were done (famously) by Alain Aspect in the beginning of the 80ies, but now these experiments have been improved a lot.

    Normally, you can find stuff here:
    (but the link seems to be down right now).

    Anton Zeilinger is one of the leading figures in the Austrian research specializing in these photon entanglement situations.

    Last edited: Apr 5, 2005
  10. Apr 5, 2005 #9


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    You touched on a lot of points, so I will try to address a few...

    1. Whether it is an electron and a Stern-Gerlach device (which can be oriented at any angle) or a photon and a polarizer, the results for entangled systems will be essentially the same in EPR setups. The issue is whether there is simultaneous local reality of non-commuting observables.

    2. Considering that the spin orientation is always random, you never see any obvious difference when you measure one before the other - or if it is simultaneous. Reason: Time itself is not a variable in the results.

    3. In the long distance versions of the setup - one or more kilometers - the photons pass through fiber optics and their entangled state is not disturbed by this process. See: http://arxiv.org/PS_cache/quant-ph/pdf/9810/9810080.pdf [Broken].

    FYI: I might also point you to a couple of urls which provide some additional background:

    a. My own site: [URL [Broken] Bell & Aspect: The Original References (in PDF Format)
    [/url] which has the key papers on the subject.

    b. Wikipedia: EPR Paradox will take you to plenty of other spots to pick up general information.
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  11. Apr 5, 2005 #10
    I was just wondering whether "simultaneous" measurements were even possible?
    Would not the two measuring photons be measuring the effects of each other on what they were measuring?
    peace and love,
    love and peace,
    (kirk) kirk gregory czuhai
    http://www.altelco.net/~churches/BlueRoses.htm [Broken]
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  12. Apr 5, 2005 #11
    I have not yet visited your sites, but as you told me that even what I called a "filter" would destroy the entanglement, I wonder if you can (people will laugh at me now) say a few "feel good words" about what type of interactions do and which do not "destroy entanglement." Surely a photon being guided by optical fiber for kilometers is having a great deal of interaction with the glass during it flight thru the fiber.

    I will read you reference later and probably get back to you with more questions as it is obvious you know more and may be able to help me understand EPR a little, but I doubt I will ever "feel good" about it.
  13. Apr 5, 2005 #12


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    Not sure how you would ever tell if it was actually simultaneous, since you wouldn't expect to see any specific effect. The only real thing that happens is that the entanglement ends for both upon the measurement of an attribute of one.
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  14. Apr 5, 2005 #13


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    It is beyond my scope to try to explain the nature of the quantum measurement process and the uncertainty relations.

    It does seem strange that the light passes through a long cable without a measurement or other "decoherence" events taking place. However: when you picture the photon as Feynman's path integral notation, you realize that any photon anywhere takes all possible paths from place to place anyway. The paths of some of them would be expected to interact with something and yet the average of the ensemble does not. That is what is happening when a photon moves through a coiled fiber optic cable. The photon path is not a straight line as we know it; but the average effect is "no measurement" before arriving at the polarizer.
  15. Apr 5, 2005 #14
    Thanks. Site is back up now. In brief search attempt, I got no hits on "destruction of entanglement"

    As DrChinese has stated that passing a "filter" polarizer would destroy entangled photons (even those that were by chance aligned with it) and we both agree that photons can travel long distances thru optical fibers without having the entanglement destroyed, I have become interested in having some idea what type of interactions with matter (definitely not "observations" / "measurements" even if no human ever looked at the results) are permitted without destruction of "entanglement."

    I think some thought about this might be useful in trying to understand the boundary between the classical and QM worlds - I.e. what is an "observation."

    I note that optical fibers often come rolled up, 500m long pieces and as they are expensive and the "space like" separation used in EPR experiments is often much less, the excess optical fiber is just left rolled up on the drum. That is, the photon goes round and round the drum, clearly interacting with the glass, without becoming "disentangled." Yet DrChinese want me to think that it could not pass thru any sort of "filter" with becoming disentangled.

    I am nearly sure he is right. He knows much more than I do about this than I do. I have now visited the sites he listed for me (and many of their links) Unfortunately, I have yet to find anything discussing what type of interactions with matter can occur without producing "disentanglement."

    Because the photon can whip around a drum inside an optical fiber, I suspect it could also pass thru a weak optically active crystal or solution - a very strong one is effectively a type of polarizer, so that will destroy the entanglement, I think. I have a sense something important is tied up in my question about what is the nature of "disentangling interactions" with matter. I bet "adiabatic" has something to do with the answer.

    Please point me to any links that touch on this question. DrChinese' page is terrific. - It has joined the few I have in my favorites on physics, at least till I lose interest in this.
    Last edited: Apr 5, 2005
  16. Apr 5, 2005 #15
    Thanks for reference. I looked at most and like your approach to Bell's "negative probabilities" - I will return and read more slowing.

    See my adjacent post.

    How sure are you that it is impossible to "prepare" entangled photons with some non random direction of polarizations, using magnetic fields to align orbitals of the source atoms, and/or split degenerate energy levels by quantum number "l"? What do you think happens when source of entangled pair is inside weakly optically active solution - i.e. is weak "adiabatic" transform of the polarization direction emitted going to destroy the entanglement? - I guess not, but you know much more than me about this. If it does not and we make the optical active paths different, i.e. let only one pass thru a glass window an enter into an evacuated pipe, its plane of polarization will continue to rotate to stray orthogonal to the one still in the OA solution! QM is so strange - this would not surprize me.

    As stated in the other adjacent post, I sense there is something important to understand here. Thanks again and congratulations on well written and interesting site.

    PS by "edit" - At one time I could easily have derived the uncertainity principle equation for you - All I can do now it tell you it falls out of the fact that certain pair of operators do not commute under the Hamiltonian. I never have been very confortable with the Copenhagen school's requirement of classical "observations" as something outside of the QM theory. That is why I am interested to know more about what does and what does not destroy entanglement.
    Last edited: Apr 5, 2005
  17. Apr 6, 2005 #16


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    Strange as it may seem, you will not "disentangle" photons. What will eventually happen is that you ENTANGLE them, separately, with other stuff, and it is THAT what puts an end to observation of interference phenomena, such as EPR. Don't confuse "destruction of entanglement" with "destruction of coherence". In fact, they work in the opposite way: entanglement destroys coherence!

    Look at it this way, in symbolic ket notations:

    We have an outside system S1 which will interact with photon 1, and we have photon 2. In the beginning, system S1 is in a state, independent of our two photons which are in an entangled state ("independent" means: a tensor product state ; the opposite of "entangled").

    So our overall state is:
    |S0> (x) ( |1+>|2-> - |1->|2+> )

    As long as system S0 does NOT interact "irreversibly" with photon 1, we can neglect this |S0> (x) part of the state, and work with the entangled photon state, to do funny things such as EPR measurements.
    But imagine now that S0 interacts with photon 1. It will interact differently according to whether photon 1 is in a + state or in a - state, so our system will then end up in two different states |S+> in the first case, and |S-> in the second. Because of the linearity of the time evolution operator, we obtain:

    |S+>|1+>|2-> - |S-> |1-> |2+>

    So we now have MORE entanglement, not less. Well, this destroyed the coherence between the photon states, if S is a big system. Because if S is a big system, S+ and S- will evolve differently, but we will always in a good approximation have that <S+ | S-> = 0. And this means that all possible "interference terms" (which are the indication of "entanglement") will drop out any expectation value, and the system will behave as if you had a statistical mixture of 1/2 of the systems in the |1+>|2-> state, and 1/2 of the systems in the |1->|2+> state (if you repeat the experiment often) and NOT as if all the systems were in the original, entangled mode.

    Entanglement with large systems (thermal environment) destroys coherence: transforms effectively a single, pure, entangled state into a statistical mixture ; at least if you keep your eye on the photons only.
    This is the essential message of decoherence theory.

    So you should not be surprised that you do not find a lot of references to "destruction of entanglement" :-)

  18. Apr 6, 2005 #17
    Again thank you. I was making this mistake.
    It has been many years since I worked with bra|kets, so be patient with me. I think your "(x)" is just indicating the product of the independent system state with the antisymmetric entangled state in the parentis following it and the + and - are generalized ways to state binary possibilities, like up or down for spins and horizontal/ vertical for polarizations. (Right?) My questions may also help others to follow.

    Is there only the antisymmetric state or does the now more complex (tied together) System S1 and 2 photon state also exist with + replacing the - above?

    What you are telling me is that Bohm may have a stronger point than I realized before in his book "the Undivided Universe" - but I still don't think he is correct in his basic idea of "guiding waves" going thru all paths, but particles only going thru one. (My objection centers on fact single electron waves can interfere with themselves and when many are present, each must "recognize" it own "guiding wave" and /or the guiding wave must recognize them- that requires all electrons to have a "tag" or "unique name" i.e. not be identical. - This problem is especially clear in a two path interferometer with different path lengths so that the guiding wave associated with electron that enters interferometer slightly before a second one has its guiding wave traveling the longer path overlap with the second electron going thru the shorter path.- First electron's guiding wave should not affect second electron's subsequent path, but would if it can not recognize it as not "its own" electron.)

    Hope you are familiar with Bohn's views - I gave this counter argument because I know DrChinese has read some of Bohm's works, and at least he should be interested in the above counter argument.

    I have spent a little time googling on entanglement. Unfortunately quantum secure exchange of code keys is such a hot topic it is hard to find simple things about the set up and destruction of coherence of the photons. I have learned that UV incident upon nonlinear crystal can produce pair of very well entangled visible photons. (I already knew one could "down shift" to get two, and even that one must try to use wavelengths and crystal orientations where there is little dispersion between f and f/2 to give longer interaction distance within the crystal. - That they are entangled is new information for me.)

    Because I think I understand the mechanism by which the non linear terms in crystal's response to UV photon's E field makes the two new photons, I would think that both have their E fields orientated parallel to the original UV's E field. DrChinese has been telling me I can't prepare entangled photons with a known polarization direction, if I have understood him correctly, but this seem to be easy to do, if I am correct about the mechanism by which they usually are made. i.e. Just pass the UV photon thru a polarizer before it enters the down splitting crystal. - Any comments by either of you?)
    Last edited: Apr 6, 2005
  19. Apr 7, 2005 #18


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    It was just an example, where system S interacts with photon 1, and ends up the way I wrote down. As I didn't specify exactly WHAT I meant with S+ and S-, you can take it as their definition. For instance, if somehow you can make a thing which, upon the passage of a photon in the state +, sets some water to boil and upon the passage of a photon in the state -, freezes the water, then S+ is the state "boiling water" and S- is the state "ice cubes".
    What is important, however, is that the interaction took place only with photon 1.

    Yes, there have been some discussions on this forum about Bohm which induced me to learn more about it. I have to say that Bohm wins in getting known: the more you learn about it, the nicer it seems. However, I'm not a Bohmian :-) I think it is a great model, illustrating a lot of misconceptions about quantum theory (such as that the probabilities are not "classical probabilities" ; the very existence of Bohms model shows that this idea is wrong). My main difficulty is that it screws (in its guiding equation) all the nice symmetries that led us to the correct quantum theory, the most important one being Lorentz invariance. But for non-relativistic quantum theory, it is really a nice "replacement".

    As far as I understand it (I'm no expert) there are 2 types of "parametric down conversion": I and II (hehe). The difference between the two is the orientation of the spins.

  20. Apr 9, 2005 #19
    Thnx for the contributions. It helps me to understand the strange phenomena of QM
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