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Discretization of Lz

  1. Jun 8, 2009 #1
    I just learned about L and Lz, I can accept the fact that L is discreted in quantum world, but it does not make sense at all for me that Lz too is bound in term of m multiple. I mean, if we choose an axis that is tilted just a little bit, then our value for Lz changes immideately and m is not an integer any more right?
     
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  3. Jun 8, 2009 #2

    diazona

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    The thing is, you can't choose one axis, measure the angular momentum, then choose a different axis and measure the same angular momentum again. When you go to make the second measurement, the angular momentum will have changed, and you'll still get a multiple of [itex]\hbar[/itex].

    It is a very tricky concept to get used to. The angular momentum just doesn't have a definite direction in 3D space. It only has a value along whatever axis you measure it on.
     
  4. Jun 19, 2009 #3
    put it this way

    I can think of a scalar, having a discrete value, like energy for instances.

    But yet I cannot think of a vector having a discrete value, not only in magnitude, but also in direction. From my point of view, the symmetry just collapse. Its like having a symmetrical system with asymmetric behavior.
     
  5. Jun 20, 2009 #4
    You're right, the direction isn't really quantized. That wouldn't make sense, for the reasons you've said.

    What happens when you tilt the axis is that an electron which had an "exact" spin in the z direction now has a mixed spin in the z' coordinates of, say, 80% up and 20% down. By the particular algebra of spinors, this is exactly the same as the original. In fact, whatever axis you pick, the actual spin of the electron is going to be in a superposition of spin states. Just like an electormagnetic wave is always going to be in a superposition of states "x-polarized" and "y-polarized" no matter what you choose as your axes. Except with an electromagnetic wave, you add the components as vectors to get the true polarization direction, but with electrons you have to know how to add the components as "spinors".

    One last point: with electromagnetic waves, you'd theoretically have to add THREE polaraization directions to get an arbitray result; but in practise, there is never any polarization along the direction of propagation (conventionally "z"). So you add two vectors because your polarization is restricted to a plane. But with spinors, you actually get any arbitray direction in 3-space by adding only two spinors together. If it seems like you don't have enough degrees of freedom to accomplish this, remember that the complex phase also carries information...
     
  6. Jun 20, 2009 #5
    You can only measure L along the axis you are measuring. If you measure Lz, then Lx and Ly do not have definitive values. They are 'smeared out'. This is basically the HUP; you cannot know Lx and Lz at the same time for example.
     
  7. Jun 26, 2009 #6
    and so, Nick89, does it means that if I tilt my Lz value into Lz', then i simply not know any of all 3 values, Lx Ly and Lz, but only Lz'?
     
  8. Jun 27, 2009 #7

    Fredrik

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    No, it's worse than that. It's not about knowing the values. The system won't have values for those variables. This can be proved by considering a pair of spin 1/2 particles prepared in a state such that if you measure Lz of one of them and get the result +1/2, you know with certainty that a measurement of the other will yield the result -1/2. See e.g. this.

    If your system is in an eigenstate of Lz, and you rotate your coordinate system a bit, then the system is in an eigenstate of [tex]\hat n\cdot\vec L[/tex], where [itex]\hat n[/itex] is a unit vector determined by the rotation. The eigenstates of this operator can be expressed as

    [tex]|\hat n\pm\rangle=\frac{1}{\sqrt 2}\Big(|+\rangle+\frac{n_1+n_2}{\sqrt{n_1^2+n_2^2}}\mbox{ sign}(n_3\pm 1)|-\rangle\Big)[/tex]

    where |+> and |-> are the eigenstates of Lz. (I had this already written down, because I calculated this as an excercise some time ago, and made a note. I think I verified that it's correct, but I'm not 100% sure).
     
  9. Jun 28, 2009 #8
    Sorry Fredik but I do not understand anything past your second sentence. I think I must have slacked too much in my physics class.
     
  10. Jun 29, 2009 #9
    Let's try it in the Bohm interpretation then (where the particles do have continuously existing properties, their trajectories are guided by the wave function, and the angular momentum components really are angular momenta of particles moving around an axis). Let's do it with no equations.

    If you happen to prepare the system with a wave function which is an eigenstate of the Lz operator, then the angular momentum of the particle around the z-axis is conserved. It is a constant of the motion. The angular momentum around the x- and y-axes is not conserved (i.e. it varies with time, and one can write an explicit formula for the time dependence).

    In the traditional interpretation one would say that the Lx and Ly components are 'undefined' in this situation. This is fair enough - it is probably the best you can do if you refuse to believe that particles exist if you're not looking at them.

    Note that if you then change the z-axis by changing the external potential in some appropriate way, then the wave function will no longer be an eigenfunction of L_z and the angular momentum about the new axis will no longer be constant.

    You probably won't have heard of this way of looking at things, because it's largely banned from textbooks, but it works and is completely observationally equivalent to the traditional interpretation.
     
  11. Jun 29, 2009 #10
    wow yea thx so much zenith8.

    I understand very well why such explanation is banned

    and now I think i have a much better idea about this. and the whole idea of switching or tilting axis.


    and now I want to make sure I do have the correct understanding regarding the even more basic principle. So if I have a hydrogen atom, I will automatically know that the atom has an quantized angular momentum for just exactly one axis just by solving the schrodinger equation alone. am I right this far? which that one axis is completely random and it has equal probability to point in any direction, like tossing a dice, and for simplicity we usually call that direction Lz. do i miss something? or my understanding is correct in the strict textbook sense?
     
  12. Jun 29, 2009 #11

    diazona

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    I wouldn't say the axis is "random" really... it's "chosen" by what kind of interaction the atom undergoes with the rest of the universe. So if you have a magnetic field, for example, the direction of the magnetic field is going to be the axis.
     
  13. Jun 29, 2009 #12
    if this happen to be an isolated system? just the atom and nothing else? will it be random?'

    actually if there is a magnetic field, the atom would precess am i right?
     
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