Misunderstanding about measuring the spin of the electron

  1. micromass

    micromass 19,224
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    Hello,

    I have been researching a bit of quantum physics, and I have encountered a conceptual problem.

    Basically, we have an electron whose spin is aligned along some axis, and we wish to measure the spin along the z-axis. So what we do, is that we put the electron between two magnets. This automatically aligns the spin of the electron so that the spin is pointing up. If the spin was originally pointing up, then no photon is emitted. If the spin was originally pointing down, then a photon is emitted. If the spin was pointing in an arbitrary direction, then a photon can or can not be emitted depending on a probability. Is this correct so far?

    However, after the measurement, the electron is thrown into one of the eigenstates. So if I measure the spin being up, then the electron is thrown into the state where the spin is up. And if I measure the spin being down, then the electron is thrown into the case where the spin is down. OK?

    Here is my problem: suppose that the original alignment of the electron is such that the spin is down. Then measuring the spin will emit a photon (with probability 1). And this causes the spin to be realigned in the up-direction.
    However, after the measurement, they say that the electron is thrown into the state where the spin is down. Aren't these two statements contradictory?? How does the electron leave the experiment?

    Thanks for your time :smile:
     
  2. jcsd
  3. I like Serena

    I like Serena 6,194
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    Hmm, I'd say the electron is thrown into the state where the spin is down before the flip.
    The electron leaves the experiment spin-up.

    Note that a probability function also has time as a parameter.
    The measurement collapses the probability function.
     
  4. micromass

    micromass 19,224
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    Thank you ILS!! This is a possibility that occured to me too. But wikipedia says:

    Doesn't this section indicate that if we enter the measurement as spin-down, that we leave the measurement spin-down??

    I mean, if the experiment changes my spin-down to spin-up, why do the subsequent experiments also measure spin-down??

    I've read about the Starn-Gerlach experiment and I understand that perfectly. But somehow I think my explanation:

    must be wrong.

    For the record, I've got this from the Susskind lectures. Here is a short summary: http://www.lecture-notes.co.uk/susskind/quantum-entanglements/lecture-2/electron-spin/
     
  5. micromass

    micromass 19,224
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    Or is the thing that the wikipedia article is talking about a different experiment than mine??
    That is: my experiment should change spin-down to spin-up. But the wikipedia article might use something like Starn-Gerlach which keeps the spin-down.

    That would be a resolution, since then ILS's explanation that it is thrown into the state before the flip is most likely valid.
     
  6. I like Serena

    I like Serena 6,194
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    As far as I can tell, the wikipedia section you refer to says nothing about flipping.
    It only says that up has a certain probability and down has a certain probability and it shows what the bra-ket notation for this is.
    Furthermore it says that once you measure it, the probability collapses.
    It doesn't say to which state it collapses nor from which state it came from.
    Nor does it refer to the Stern-Gerlach experiment.
     
    Last edited: Nov 11, 2011
  7. micromass

    micromass 19,224
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    Doesn't the wikipedia article state that the particle will collapse to the corresponding eigenstate AFTER the measurement. And that all following measurements will yield the same value??

    Or am I reading too much into this?

    I am refering to Stern-Gerlach, because the wikipedia article doesn't mention how the measurement is made. Their measurement method might be different from what I mentioned in my OP.
     
  8. I like Serena

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    It holds true for *any* measurement, or more accurately, for any possibility of a measurement.

    When you measure the electron in your experiment, you effectively always measure it as up.
    Or you could say you measured it as down since it emitted a photon, after which it was flipped up, after which the probability function is still 100% up.
     
  9. I like Serena

    I like Serena 6,194
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    Note that if you reuse your electrons after making them all spin-up, and you put them in a reverse magnetic field, *every* electron will emit a photon.

    After that all electrons will be 100% spin-down.
     
  10. micromass

    micromass 19,224
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    OK, it makes sense now. Thanks a lot!!
     
  11. kith

    kith 916
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    That would be no measurement of the initial state, but a preparation of the |up> state. This isn't what's happening in the SG experiment. There, in order to measure the ratio of |up> and |down> of your intial state, you introduce a spatial separation.

    To do this, you use a magnetic field gradient. The potential energy of a electron in a magnetic field is V = -µ B, where µ is the dipole moment associated with the electron's spin. Since force is F = -grad V, the direction of the force your particles are expierencing, depends on their spin orientation which leads to the spatial separation.
     
  12. micromass

    micromass 19,224
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    Yes, it isn't the SG experiment. But it's another type of experiment outlined in the link I've put above. It may very well be that this type of experiment isn't valid or won't work, but I honestly don't know.
     
  13. kith

    kith 916
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    This is a question of what you want to do.

    If you want to measure the spin in z-direction, your setup certainly doesn't work, because regardless of the initial state, you always end up in the |up> state. This is what I meant by calling it a "preparation". The SG experiment is able to tell you the ratio of |up> and |down> of an arbitrary initial state, so it really measures the spin in the z-direction.
     
  14. micromass

    micromass 19,224
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    So Susskind was wrong??
     
  15. kith

    kith 916
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    No, sorry for adding confusion!

    I didn't read your posts carefully enough and thought that the final state of the electron was the measurement signal, while actually it is the detected photon. So it's true that you always prepare the electron in the |up> state, but it's not true that you can't measure the spin with the setup.

    Susskinds experiment is easier to understand for laymen, but I would say it is less straightforward if you aim for the mathematical concepts, because it is not the final states of the particle itself that carry the information.
     
  16. micromass

    micromass 19,224
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    OK, that makes sense. Thank you a lot!!!!
     
  17. Fredrik

    Fredrik 10,474
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    Sorry for being late to the party. I didn't see this thread before. Not sure if everything has been sorted out already. I'll add a few comments anyway.

    I would say that the setup that Susskind describes is not a spin component measurement, because the result is never "down". (It's either "up", or there's no result at all). This doesn't make Susskind wrong, because he only talked about "wanting" to measure the spin in a section titled "classical detection" that ends with the words "this does not happen". He's not saying that what he's describing is a measurement.

    If a photon is detected, the electron is now prepared in the "up" state. If no photon is detected, the electron's spin state hasn't changed since it was prepared by another pair of magnets earlier.

    If you run this experiment over and over on electrons that were all prepared in the same spin state, then you can figure out how the first pair of magnets were aligned. This isn't what one would normally consider a "measurement" in QM. A measurement is an interaction between the system and the measuring device that puts a component of the measuring device, that I'll call "the indicator component" here, into one of many possible final states labeled by numbers. The indicator component must appear as a classical object to a human observer, and its possible final states must be distinguishable. Otherwise, it wouldn't be of any use as an indicator. The number corresponding to the final state is considered the "result" of the measurement.

    According to what I just said, the setup defines a measuring device with only one possible final state of the indicator component. I can't consider it a spin component measurement, since it doesn't have two possible final states. I also can't disqualify it from being considered a measurement just because we don't get a result every time we put a particle in its vicinity, because then an ordinary particle detector that simply goes "click" when it detects a particle wouldn't be considered a measuring device either. (A detection is considered a position measurement)
     
  18. micromass

    micromass 19,224
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    Thanks a lot, Fredrik!! It does make sense.

    This is interesting. So you say that if an electron comes in with a state like [itex]\frac{1}{\sqrt{2}}|\text{up}> + \frac{1}{\sqrt{2}} |\text{down}>[/itex] and if it doesn't emit a photon, then it remains in this state?? I thought it would change to [itex]|\text{up}>[/itex] regardless on if it emitted a photon or not.

    I do see why this wouldn't be called an actual measurement.
     
  19. Fredrik

    Fredrik 10,474
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    One last quick reply before I go to bed...

    Yes, I believe that's accurate. This is what I'm thinking: A superposition like that is an eigenvector of [itex]\vec n\cdot\vec S[/itex], where [itex]\vec S=(S_x,S_y,S_z)[/itex] and [itex]\vec n[/itex] is a unit vector. I haven't calculated the direction of [itex]\vec n[/itex] in this example, but it should be in the xy plane. An electron has a magnetic moment [itex]\vec m[/itex] in the same (or is it opposite?) direction as [itex]\vec n[/itex]. So it's also in the xy plane. Classically, a magnetic moment m in a magnetic field B has potential energy [itex]U=-\vec m\cdot\vec B[/itex]. To rotate the magnetic moment from the xy plane to the z (or -z) direction will (at least in the classical theory) change the energy by an amount [itex]|\vec m|\,|\vec B|[/itex]. I expect the energy of the photon to correspond to this energy. So either the electron goes through unchanged, or it emits a photon. With perfect detection abilities, no photon detected should mean that the spin state hasn't been changed.
     
  20. micromass

    micromass 19,224
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    OK, thanks a lot Fredrik!! This was very helpful!!

    I'm studying a little bit of QM because I'm teaching a course on functional analysis soon. And I think I owe it to the students to at least give them a hint on why functional analysis is an important field. And perhaps I'll get them interested enough in the applications to actually get them to study it themselves :biggrin:

    Thanks a lot!!
     
  21. dextercioby

    dextercioby 12,314
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    There's very little functional analysis associated with spin, because the spin states space is finite dimensional, isomorphic to C2s+1, where s is the spin value. Linear operators acting on finite dimensional spaces are not that interesting, not to me at least.
     
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