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Stern Gerlach Experiment

  1. Aug 25, 2009 #1
    How do they make sure the "measurement" process in conventional Stern-Gerlach experiments are clean quantum measurements?

    Spins dynamically precess in magnetic fields (uniform or non-uniform) and Stern-Gerlach (especially sequential SG setups) make precise predictions regarding the resultant beam.

    You could easily end up rotating a z-beam making it an x-beam accidentally.

    How is this difficulty avoided?
     
    Last edited: Aug 25, 2009
  2. jcsd
  3. Aug 26, 2009 #2
    Nobody has an idea... Everybody is an expert in the interpretation though...
     
  4. Aug 26, 2009 #3

    Doc Al

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    How would that happen "by accident"? You'd have to rotate your magnet--not something one does by accident.

    :confused:
     
  5. Aug 26, 2009 #4

    Vanadium 50

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    Based on not getting a response for two whole hours?

    Doc Al is right - how do you rotate your magnet without noticing?
     
  6. Aug 26, 2009 #5
    Rotating the magnet?

    Spins, themselves WILL be rotating in the magnetic field because of the magnetic field)

    how do you make sure there's no precession while the electron is travelling WITHIN the magnetic field?
     
  7. Aug 26, 2009 #6
    Based on not getting a response for a whole day. Did I say anything about rotating a magnet?
    I am talking about the rotation of the individual SPINS while they are travelling inside the magnetic field.
     
  8. Aug 26, 2009 #7

    Doc Al

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    Again, I'm not quite sure what you're talking about. You had said:

    If the inhomogeneous magnetic field is aligned along the z-axis, then the beams will select for the z-component of spin. I don't see how a "z-beam" can accidentally turn into an "x-beam". (In a z-beam, the x-component of spin is random.)
     
  9. Aug 26, 2009 #8
    Let me try again: Electron spin responds to an external magnetic field, by making precessional movements when they are inside the field. I suppose you very well know this. torque ~ u x B (u= mag. moment, if there's no damping, such as in LLG equation)

    If you write down the Hamiltonian for a z-beam, obviously you need to include the magnetic field, and independent of the force that's exerted on the electron that separates the up and down components, electrons would be precessing about the principal axis of the non-uniform magnetic field.

    So when you try to measure the x-component of a z-beam, how do you know that the spins are not precessing around the x-axis?? Because that's exactly what would happen, a z-directed spin would precess about the x-axis, and if you carefully arrange the parameters (B-field and the magnet thicknesses etc..) you could end up with a y-beam just by rotation... or a minus z-beam.. or somewhere in between

    Now the question is HOW is that taken into account?

    Check this if you still don't see my point:
    http://en.wikipedia.org/wiki/Larmor_precession
     
    Last edited: Aug 26, 2009
  10. Aug 26, 2009 #9

    Vanadium 50

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    Oops...it was 14 hours, not 2.

    But again, like Doc Al, I don't see how the beam can turn into an x-beam. Remember, once the beam splits in two, I don't have to preserve the spins, because the selection has already occurred.
     
  11. Aug 27, 2009 #10
    still very confusing, no convincing answers in two days.
    I think details of this is important!
     
  12. Aug 27, 2009 #11
    The components transverse to the magnetic field do oscillate at the Larmor frequency, while the longitudinal component can be taken constant. The Larmor frequency in a typical Stern-Gerlach is so large that it averages to zero. Still, the point of the experiment is that in the classical theory, any longitudinal component would be possible and the atoms would be deflected continuously according to their arbitrary component. However, in the quantum theory we obtain only J "jets" of width in agreement with Heisenberg.

    See for instance Feynman vol2 35.2
     
    Last edited: Aug 27, 2009
  13. Aug 27, 2009 #12
    How come "any arbitrary component" could NOT be zero if they are precessing wildly (in classical theory)? Because, if what you are writing here is correct, even in the classical theory, any non-longitudinal component of spin would precess with a large frequency (therefore average out to zero, as you say) while there will only be TWO and ONLY TWO components present that do not precess because they are collinear with the non-homogeneous (but unidirectional) magnetic field : + longitudinal and - longitudinal.

    You just proved that even classically, only two discrete spots are expected. (Because all other directions precess and average out to zero)
     
    Last edited: Aug 27, 2009
  14. Aug 27, 2009 #13

    Cthugha

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    I am puzzled about what you mean exactly. Classically you have the projection of spin on the longitudinal axis, which does not precess and the remaining spin components, which precess wildly and average to zero. In a "classical description of spin" only the total spin is given and the projection on the longitudinal axis can have any magnitude between - and + total spin and therefore there should be a line instead of two spots classically. Or did I miss your point somewhere?
     
  15. Aug 27, 2009 #14
    That's your understanding, but that's not what "I proved", that's not what's in any elementary introduction book to quantum mechanics, and that's most importantly not what this experiment tells us.
     
  16. Aug 27, 2009 #15
    I am trying to figure out which axis you are using for your magnetic field? Conventionally we talk about z but in this quote you appear to be taking it about y. Later you talk about z rotating to y, which would happen if the field is along x. What is your intended orientation?
     
  17. Aug 27, 2009 #16
    Why do you dodge your "own" remarks and your OWN explanation? Why do you have to senselessly attack instead of focusing on the physics?

    YOU said that: ANY non-longitudinal component of spin WILL precess about the magnetic field direction. Rotation can be resolved in three coordinates clasically, so ANYTHING that does not coincide with the magnetic field axis will average out to zero, GIVING TWO DISTINCT directions.

    This is what you say. And I am pretty sure, you understand this kindergarden logic, but now you are confused, you don't want to admit it. You don't know it any better than I do, you are just in a state of denial about it.
     
  18. Aug 27, 2009 #17
    No. Rotating Z to Y could happen in a Y-directed field as well, because of the precession...

    A bicycle wheel hanging on a z-directed rope would make a 2*pi rotation ABOUT the z-axis, passing through both y-axis and the x-axis.

    Depends on the rotation (if there's only precessional motion)

    But since the field is non-uniform , I guess the motion is more complicated, still unknown to me.
    Humanino's answer by the way, has to be wrong, he apparently solved the mystery.
     
    Last edited: Aug 27, 2009
  19. Aug 27, 2009 #18
    Sokrates, you are the one constantly attacking. You began with your nasty remark "everybody has an interpretation, but nobody to answer my question".

    I am not confused at all. As I am trying to explain calmly to you, classical theory does not predict only J possible projections along the magnetic field direction, but any possible projection. The deflection will be proportional to the component of the moment along the magnetic field direction. So in the classical setting, you get a continuum of possible deflections, including zero.

    In the quantum setting, whatever the initial polarization can be written in a basis of proper states along any arbitrary direction. But whatever the choice you make in the direction, only J projection along this direction are possible. In particular, you can see if you take an electron and write it down as a superposition of two possible polarization projections along the magnetic field direction, you get only two possible discrete deflection, not a continuum. You will get different interpretation of what's going on if you decide to use another direction to write the state, but you will compute the same output possible BTW.
     
  20. Aug 27, 2009 #19
    You are right. "Magnitude" quantization is the key here. IT cannot be understood classically, but I wonder why all this complicated spin dynamics is avoided in discussing SG experiments?
     
  21. Aug 27, 2009 #20
    The point of the Stern Gerlach experiment is that it is probing a quantum spin, not a bicycle wheel. If the angular momentum is not large compared to hbar, then you can clearly distinguish the possible projections along the quantization axis. For a bicycle wheel, the number of projections is so large that it appears continuous.

    Seriously Sokrates, a more civilized tone would be appropriate.
     
  22. Aug 27, 2009 #21
    This is clear to me and has nothing to do with what I asked. I know the significance of the experiment, I have been talking about the spin dynamics, which involve aligning with the magnetic field, precession, etc....
     
  23. Aug 27, 2009 #22
    What are you talking about? I am civil to everyone who politely engages in my discussions, and let me remind you, you are the one who is constantly stalking my posts pointlessly attacking me. I don't have a problem with you, just discussing physics is my objective here.

    You don't even understand my points before you post, the precession of an electron (IF YOU ASSUME IT HAS INTRINSIC ANGULAR MOMENTUM) could just as well be demonstrated by a bicycle wheel, it has nothing to do with the quantization.... Same equations! and same interpretation

    I won't conclude by saying: See for example, Feynman I, precession.

    as you do.
     
  24. Aug 27, 2009 #23
    If this is so clear to you, then what is wrong with this simple answer : write down the wavefunction as a combination of proper states along the magnetic field and your precession is gone. It does not matter what the original direction of the spin is, I can always choose to write it down in the basis along this convenient direction.
     
  25. Aug 27, 2009 #24
    You already managed to have a mentor a scientific advisor not interested in this discussion anymore.
     
  26. Aug 27, 2009 #25
    Look... You are simplifying things. What you say here is in every textbook. So I can just as well go read Sakurai, which I have already done.

    Precession "averaging out" to zero is added here, WE KNOW that it averages to zero, I don't think it comes out of the SINGLE-PARTICLE Schrodinger equation.

    If you want to write down the states, you have to include the magnetic field through the vector potential representation to the Hamiltonian, then solve this Hamiltonian assuming that you are only interested in the magnetic field region and then two distinct states, and electrons going up, etc... etc.. should come out. BUT:

    I don't think precession averaging to zero comes out of such a naive picture.

    But instead, if you say, I KNOW my states, and they have only TWO possible outcomes, of course the problem becomes much simpler and probably they yield the same results,

    but it's just NOT that trivial, if you think about the details, and that's my whole point.
     
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