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Angular Momentum

  1. Dec 10, 2009 #1
    Well I've been looking at angular momentum...

    I have an idea why a function f cannot be an eigenfunction of 2 different non-commutating operators, but has anyone a nicely precise reason??

    If the angular momentum is resolved into its components, and we look at one, say L_z, then:

    L_z.f = h.m.f

    I am letting h be h-bar, m the quantum number, f the wave function, and L_z is the operator/

    Is the total angular momentum quantized in m then also?? How do we find the total angular momentum from just this equation for a COMPONENT?

    is it possible to select a direction so that the total angular momentum is L_z?
     
  2. jcsd
  3. Dec 10, 2009 #2

    diazona

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    Let's see if I can remember this: say that we have two operators A and B. If there is a simultaneous eigenfunction of both, call it f. Then
    [tex]Af = af[/tex]
    [tex]Bf = bf[/tex]
    I can use those to calculate
    [tex]BAf = Baf = aBf = abf[/tex]
    and
    [tex]ABf = Abf = bAf = baf = abf[/tex]
    (I've used the fact that real numbers commute with each other and with all linear operators) Now putting the last two together,
    [tex][A,B]f = ABf - BAf = abf - abf = 0[/tex]
    So if there is a simultaneous eigenfunction f, the operators commute, at least when applied to the function f. Thus if they don't commute (for any function), there can't be any simultaneous eigenstates.


    Total angular momentum is quantized, but it has a different quantum number, [itex]l[/itex]. Well... what we actually calculate is the total angular momentum squared, [itex]L^2 = L_x^2 + L_y^2 + L_z^2[/itex]. (Since angular momentum is a vector, we need to square it to get a "total".) You can't find the total angular momentum just by knowing one component.

    No, because of the uncertainty principle. When you have two noncommuting operators, like [itex]L_z[/itex] and [itex]L_y[/itex], you can't know both of them precisely because there are no simultaneous eigenstates. Any eigenstate of [itex]L_z[/itex] is a mixture of several eigenstates of [itex]L_y[/itex], so the total angular momentum could correspondingly have several different values.

    If you have access to David Griffiths' book on introductory quantum mechanics, take a look at chapter 4 where he offers a nice diagram regarding exactly that point.
     
  4. Dec 10, 2009 #3
    Diazona explained this beautifully but I just want to add this:

    The total angular momentum is a vector that has three non-commuting components so it's usually impossible to measure it with a single measurement. But in spin-half systems the eigenvalue of L^2 is always equal to the average value you obtain by different measurements:

    [tex]
    L^2 = L_x^2 + L_y^2 + L_z^2 = 3 \hbar^2 / 4
    [/tex]

    because you either get hbar/2 or -hbar/2 the squares of which are always h^2/4. And remember that the eigenvalues of L^2 are given by l * (l+1) , l is 1/2 here.

    so without carrying out separate measurements, you can know the eigenvalue. Of course this is a very special property and immediately fails for spin-one systems for instance. Because you could get all kinds of results ( 0, hbar, 3 hbar, 2 hbar from different measurements of L^2) and they may not always be 2hbar -- which is the eigenvalue of L^2 for l=1.


    Edit: The fact taht L^2 is a combination of non-commuting angular momentum components doesn't mean that the eigenvalues of L^2 are not measurable -- they may be inferred from other experiments, such as the energy spectra of molecules - which yield direct information about the number l, hence the eigenvalue of L^2.

    I hope I didn't make it worse for you.
     
    Last edited: Dec 10, 2009
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