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What does space quantization of angular momentum actually signify?

  1. Nov 7, 2009 #1
    I have just come to learn (Physics, with modern physics, Richard Wolfson, J M. Pasachoff, second edition) that not only angular momentum's magnitude is quantized, but also its direction.
    Its given that, Cos[tex]\theta[/tex]min= l / [tex]\sqrt{l(l+1)}[/tex]
    Telling that, [tex]\theta[/tex]min is the minimum angle between any orbital angular momentum (l) and any arbitrary chosen axis.

    How can the orbital angular momentum always make certain minimum angle [tex]\theta[/tex]min with any arbitrarily chosen axis???
    What if my chosen axis happens (by chance) to be the same axis of the orbital angular momentum?
    Last edited: Nov 7, 2009
  2. jcsd
  3. Nov 7, 2009 #2
    It seems that my question was wrong. (no reply for so long)
    What I am basically asking is --> What do direction quantization (of angular momentum) signify??? Does it signifiy that
    1. there should be only discrete angles between different angular momenta
    2. As I said in OP, there should be only discrete angles (greater than [tex]\theta[/tex]min) between any angular momenta and any chosen arbitrary axis
    Last edited: Nov 7, 2009
  4. Nov 8, 2009 #3

    Hi there,

    One of the reasons you're not getting an answer is that in the orthodox interpretation of QM, the answer is 'the question is meaningless since we can never know the reality of a quantum system'.

    In other interpretations such as many-worlds people only ever concern themselves with eternal verities and speaking in logically consistent sentences rather than concerning themselves with ontology or 'what really might be going on'. So they're not going to answer you either.

    So conventionally one may speak of 'angular momentum' but you have to do 'air quotes' with your fingers as if to say 'but not really'.

    In the de Broglie-Bohm interpretation however, the answer is as usual, straightforward. Particles have trajectories, pushed around by an objectively existing wave field represented mathematically by the wave function. And orbital angular momentum really is the angular momentum of particles 'orbiting' the nucleus. Of course this is a bit difficult if you believe that particles only exist when you look at them (as in the orthodox view).

    Why is it quantized? You have to remember that in de Broglie-Bohm theory quantities are well-defined and continuously variable for all quantum states - values for the subset of wave functions which are eigenstates of some operator have no fundamental physical significance. So this characteristic features of QM - the existence of discrete energy levels - is due to the restriction of a basically continuous theory to motion associated with a subclass of eigenfunctions.

    Now what happens with angular momentum? Let's say we prepare the system with a wave function which is an eigenfunction of the [tex]{\hat L}_z[/tex] operator (where the direction of the z-axis is arbitrary). Then the z component of the angular momentum will coincide with the eigenvalue of [tex]{\hat L}_z[/tex], and the total orbital angular momentum with the eigenvalue of [tex]{\hat L}^2[/tex]. Conventionally one would say that the x- and y- components are 'undefined'. However, in de Broglie-Bohm they are perfectly well-defined. In fact you have:

    [tex]L_x= -m \hbar \cot \theta \cos \theta[/tex]

    [tex]L_y = m\hbar \cot \theta \sin \theta[/tex]

    [tex]L_z = m\hbar [/tex]

    So the difference is that along the trajectory [tex]L_z[/tex] and the total angular momentum are conserved, but [tex]L_x[/tex] and [tex]L_y[/tex] are not. Thus, although the classical force is central the motion is asymmetrical, which reflects the fact that a particular direction in space has been singled out as the quantization axis.

    Hope this helps, though I suspect it won't for some reason.
  5. Nov 8, 2009 #4
    Quantization does not mean that there may not be intermediary values.

    Take a superposition of different wave functions and you will obtain an average value between the discrete values. Normally the wave functions are superpositions rather than the eigenstates.
    Last edited: Nov 8, 2009
  6. Nov 8, 2009 #5
    First of all, thank you for answering.
    To make matter clear: I am not confused to accept that Energy and angular momentum is conserved. I can readily accept it by saying that, its what mathematically follows, if you solve Schrödinger's equation for electron bound to a nucleus and apply boundary conditions for the wave function.
    What I am confused about is, direction quantization of Angular Momentum; from which follows a absurd sounding statement that --> whichever arbitrary axis you chose in space, the angular momentum isn't aligned to that axis!!!(it always makes some minimum angle with my arbitary axis)?

    I hope you understand what I meant.
  7. Nov 8, 2009 #6
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