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B A quick question about entanglement

  1. Dec 17, 2016 #1

    rede96

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    I'm just an interested layman with no back ground in physics so please excuse my ignorance.

    As I understand it, due to the uncertainty principle it isn't possible to know both the X and Z spin component for example, of an electron simultaneously.

    So if I took pair of entangled electrons and measured one in the X axis and got an 'Up' result, what would I observe if I measured the second in say the Z axis?
     
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  3. Dec 17, 2016 #2

    Nugatory

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    You would have a 50% chance of measuring up on the z-axis, 50% chance of measuring down.

    That's because the angle between the x and z axes is 90 degrees. In general, if you measure "up" on the x-axis, your probability of getting "up" on any other axis will be ##\sin^2\frac{\theta}{2}## where ##\theta## is the angle between them. (That's assuming the most commonly case of spin-entangled electrons, where the spins are always opposite on the same axis).

    The uncertainty principle is at work here because when you measured the pair and found that it is in the state "one is spin-up on the x-axis, the other is spin-down on the x-axis" you became uncertain about the result of the z-axis measurement.
     
    Last edited: Dec 17, 2016
  4. Dec 17, 2016 #3

    vanhees71

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    The uncertainty principle doesn't simply tell you that you cannot know the values of the spin of the electron along the ##x## and ##z## direction but that indeed these two observables cannot take both determined values, i.e., you cannot prepare an electron such that both of these spin components have a determined value. The mathematical formalism of quantum theory tells you that you can only know the spin component in one direction and the total magnitude of the spin, which is ##s(s+1) \hbar^2##. For an electron ##s=1/2## (one says that the electron is a spin-1/2 particle). The possible values of the spin component in an arbitrary direction (usually chosen as the ##z## direction of a Cartesian coordinate system) are ##\sigma_z \in \{\hbar/2,-\hbar/2 \}##.

    However, as Nugatory said, you know the probability for measuring ##\sigma_x=\hbar/2## if the electron is in a state of definite ##\sigma_z##. That's the content of quantum states: It gives the probabilities for measuring values of any observable.
     
  5. Dec 17, 2016 #4

    rede96

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    Thanks for the reply. I thought that might be the case but what I didn't quite get is that if I measure a pair of entangled particles and get say 'Up' in the X axis and 'Up' again in the Z axis of the other particle then I know both components of Spin, which I thought was allowed?

    So it it just as simple that if I measure one component then the other result I get can only be assumed to be a random result and not representative of both components of spin? If that is the case, how can we prove that it is just random? If that makes sense.
     
  6. Dec 17, 2016 #5

    Nugatory

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    No, you know one component of spin (the one that you actually measured) for each electron. The unmeasured components do not have any value (not "we don't know what the value is because we haven't looked", but "no value, the same way I don't have a lap when I'm standing up"). Quantum mechanics tells us what the probability is of getting this result or that result if we measure, but says nothing about things that aren't measured.

    We take a very large number of electron pairs, one member of which has been measured spin-up on the x-axis, and measure the z-axis spin of the other. When half the z-axis measurements come out spin-up and the other half come out spin-down, we have that 50/50 random. (That ##\sin^2## rule I mentioned above is experimentally confirmed in the same way).
     
  7. Dec 17, 2016 #6

    DrChinese

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    Good question. You can ask it either way: how can you prove you DO know the spin of both?

    If you take a particle with X axis UP and measure again in the X directions, you will get UP again. And again and again. If that "other" electron was Z=DOWN, and you measured this one's Z, you will have a 50-50 chance of getting Z=DOWN. That is because it does not have simultaneously well defined X and Z spin components.

    Bell's Theorem starts with the idea that an electron DOES have simultaneously well defined X and Z spin components (various 2 non-commuting components actually) and derives a contradiction with the predictions of Quantum Mechanics.
     
  8. Dec 18, 2016 #7

    rede96

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    I think the point I was making is that we can physically measure the different components of spin of a pair of entangled electrons. (X on one and Z on the other) Hence why I asked the question. And I assumed as you pointed out, that if I measure the X component of one of the pairs and get 'UP' then I will always get 'UP' and if I measure the Z component of the other one of the pairs and get 'Down' then I will always get 'Down' is that right?

    I know that if I measure a different component of spin on the same particle then then the outcome is a 50/50 probability, (at 90 degrees to the original angle) but that's the same of any particle, entangled or not.

    So where I am getting confused is that if I did the Bell test (Sorry not sure what its actually called) with just random particles that were not entangled then I would get the same result. So it doesn't really tell me anything about how entanglement works. If that makes sense?
     
  9. Dec 18, 2016 #8

    rede96

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    Ah ok, so measuring a different component of spin of an entangled particle is no different than measuring a different component of spin of any particle? In that it is probabilistic? But we still get a positive outcome from the measurement right? E.g. it is still either Up or Down? rather than a no result?
     
  10. Dec 19, 2016 #9

    zonde

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    If you measure different spin components of entangled particles you get uncorrelated results. Say if you measure X component of one of the pair and get 'Up' then measuring Z component of the other one of the pair will give you 'Up' or 'Down' with equal probability.
     
  11. Dec 19, 2016 #10

    DrChinese

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    I wouldn't say it was "no different". They are entangled, after all. Entangled state statistics are different than product state statistics, although the particular example you gave (x vs z axes) yields similar predictions.

    Check out Bell's Theorem to get more information about the significance of my comment.
     
  12. Dec 19, 2016 #11

    rede96

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    Hi and thanks for the reply. I think all I was asking is if I measure the X component of particle 'A' and get UP, then measure the Z component of the particle 'B' and get DOWN, then when ever I measure these same components again and again, will I always get the same result 100% of the time?

    Yes sorry. I was probably thinking of the example I gave.... honest!

    I've tried on a number of occasions to understand Bell's Theorem. I sort of get the math and the principle of it, but I can't grasp just how it relates to entanglement and disproving that the spin properties aren't set at the time of pair production. But let me go though it again and see if I can ask some more constructive questions around it.
     
  13. Dec 19, 2016 #12

    zonde

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    Yes, if you measure the same spin component of the same particle you will get the same result.
    As I see this example shows most clearly that the spin properties (or photon polarizations in this case) aren't set at the time of pair production:
    https://www.physicsforums.com/threads/a-simple-proof-of-bells-theorem.417173/#post-2817138
     
  14. Dec 19, 2016 #13

    DrChinese

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    This is the basic question that Bell's Theorem addresses. I have a web page that explains the idea using entangled photons rather than electrons, but the principle is the same. It starts with the assumption than properties are set at the time of pair production, and then derives a contradiction with the predictions of QM. Experiment, as you might expect, agrees with QM. Check out:

    Bell's Theorem with Easy Math

    Another way to describe the issue: other than at a very few specific angle settings (90 degrees for example), there are no preset values for entangled pairs that match the predictions of QM (unless you know the angle settings in advance).
     
    Last edited: Dec 19, 2016
  15. Dec 20, 2016 #14

    rede96

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    Thank you for the information. I have read through Bell's theorem a number of times, even with the easy math. But either I still haven't had that Eureka moment or it just hasn't convinced me! Sorry.

    If you don't mind me dumping my thoughts I think there are a few things I struggle with. Firstly, the theorem could apply to any pair of random particles, entangled or not the prediction would be the same in so much is that the theorem lists all possible outcomes that can be achieved classically. (By the way, if I did the same actual experiment with non entangled particles, what result would I get? Still 0.25? Or something different?)

    I know that is one of the key points, there is no test we can do classically that will get the QM result, but just because we can't find a classical equivalent doesn't mean the properties of the entangled particles aren't set at pair production. It just means they don't behave in a classical way doesn't it?

    Also, I sort of thought we could use the same rationale used to eliminate the hidden variable theory, to eliminate action at distance? For example couldn't I assume that their properties are only set when I measure one of the particles, and if so, what results would I expect to get if I measured the other... In my mind it would be the same Math as the other way around. Which I suppose is a bit like the God argument. Just because I can't prove God exits, doesn't mean there is no God and just because I can't prove God doesn't exist doesn't mean there is a God.

    I know I am probably missing something, but I have tried a lot with this and like I said, just not had that Eureka moment. But I really appreciate the responses! It's such a fascinating topic.
     
  16. Dec 20, 2016 #15

    PeterDonis

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    What does it mean for "the properties of the particles to be set at pair production"? This is one of the key questions (to my mind the key question) that Bell's analysis illuminated. He gave a very simple mathematical definition of what this means: it means that the expression for the probability of a given pair of measurement results (one for each particle) must factorize.

    In other words, if we have two particles ##a## and ##b##, and two measuring devices ##A## and ##B##, then the most general expression for the probability of a given pair of results ##a, b## is (using sloppy notation, but hopefully the meaning is clear)

    $$
    P(a, b) = F(A, B, \lambda)
    $$

    where the ##F## on the RHS is some function, and its arguments are: ##A##, the settings of measuring device ##A##; ##B##, the settings of measuring device ##B##; and ##\lambda##, which is just a general expression for any "hidden variables" that might affect the result. Any "properties of particles set at pair production" would be included in ##\lambda##.

    Now, what Bell proposed was that, if the properties of the particles are "set at pair production", then the expression for ##P(a, b)## should factorize into two functions, as follows:

    $$
    P(a, b) = F_a(A, \lambda) F_b(B, \lambda)
    $$

    This seems natural enough: the measurement of particle ##a## should not "care" what the settings of measurement device ##B## are, and vice versa. But Bell proved that no function of this form can reproduce the predictions of quantum mechanics. In other words, any function that can reproduce the predictions of QM must be of the first form above, not the second, factorized form.

    The beauty of this is that it replaces a vague set of ordinary language words that are hard to interpret (your expression was "the properties of the particles are set at pair production", but other words like "locality" are also used) with a precise mathematical criterion that is easy to check. So perhaps it will help if you focus on the question: what are the implications of the mathematical statements I made above? What is QM telling us when it tells us we can't make the right predictions if we try to factorize the probability ##P(a, b)## for a given set of joint measurement results?
     
  17. Dec 21, 2016 #16

    zonde

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    Look, it's very basic reasoning and have nothing much to do with "classical" or "non classical" behavior.
    If you propose local model to explain entanglement what are the basic properties of this model to consider it "local"? It can be considered "local" when measurement of particle with freely chosen measurement angle have no effect on remote measurement of other particle and vice versa i.e. two measurements are independent. So within such model you can independently speak about "what result Alice will register if she measures photon polarization with angle 'a' " and "what result Bob will register if he measures photon polarization with angle 'b' ". And it turns out you can't get predicted correlations even with handpicked outcomes for different measurement angles.
    So it's not like we can't think of reasonable model how measurement results are determined but instead that measurement results themselves don't stick together if you consider them as independently determined.

    Say source produces some number of entangled pairs.
    Alice can set her measurement angle to ##\alpha_1## or ##\alpha_2##. For ##\alpha_1## setting Alice gets set of measurements A1 and for ##\alpha_2## - A2.
    And Bob can set his measurement angle to ##\beta_1## or ##\beta_2##. For ##\beta_1## setting Bob gets set of measurements B1 and for ##\beta_2## - B2.
    And there are no such sets of measurements A1, A2, B1, B2 that give predicted correlations for any combination of A#, B#.
     
  18. Dec 21, 2016 #17

    DrChinese

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    Adding to what PeterDonis already said about this point: You CAN prove that there aren't any theories in which the spin values are set at pair production that also agree with the predictions of QM (this is Bell's Theorem). You can't even make it up by hand to agree!

    There is perfect correlation at all angles for entangled pair spin, correct? At 0 degrees, 1 degree, 2 degrees, etc to 360 degrees. So whatever your hypothetical preset values are, they must be predetermined across the board IF they can be measured independently and still yield the perfect correlations. If that is the case, then what happens if you measure at different angles? That's when Bell comes in. He showed there was no way to come up with values at particular angle pairs that would be consistent with QM. As I say, you can hand pick the values for certain sets of angles (I use 0, 120 and 240 degrees in my photon examples, following Mermin). If you attempt to do so, you will quickly see that your results will not match either QM or observation. QED.

    If you were to try it - no one ever bothers to do this step - you would immediately see that your hypothesis is untenable.
     
  19. Dec 23, 2016 #18

    rede96

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    I think this might be where I am confusing myself. I didn't think of the paired particles as exactly correlated,in fact not even exactly identical. But they did have similar properties that would mean if one was detected at a certain angle then the other would be too. So 100% correlation in those cases but that might not follow for other angles.

    For example, I thought that maybe each paired particle had a different 'value' like a dice. Maybe one particle had of value '4' and the other of value '5'. And I imagined the binary property we measure would be like saying Up, at a given angle, would detect all particles with values 4,5 and 6 and Down would detect all values 1,2 and 3. But Up and Down at a different angle it might be all odd versus even numbers for example. So in that case the outcome would be probabilistic.

    So while we would expect perfect correlation at certain angles of measurement, other angles of measurement may not be perfectly correlated. This would represent what we see in the QM results and why Bell's inequality is violated at certain angles, but it didn't necessarily mean that there wasn't some sort of hidden variable.

    Of course I honestly don't expect it to be that simple or that I am right! I fully accept Bell's test and understand it a lot better now. I'm just enjoying playing around with it all as I find it so fascinating.

    Again, really appreciate everyone's responses.
     
  20. Dec 23, 2016 #19

    rubi

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    Last edited by a moderator: May 8, 2017
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