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Spherical symmetry implies zero angular momentum?

  1. Jan 19, 2009 #1
    why does the spherical symmetry of the ground state orbital of hydrogen necessitate zero angular momentum?

    I really don't have any intuition for "quantum angular momentum"
  2. jcsd
  3. Jan 19, 2009 #2
    Even in classical physics, angular momentum is a vector quantity (it has an axis), and hence contradicts spherical symmetry. (That's not to say that every spherically symmetric potential necessarily has no angular momentum in its particular ground states.)
  4. Jan 19, 2009 #3
    mm, that's what I thought - if there exists angular momentum, it has a direction - if the reference frame is rotated, the angular momentum vector changes direction, violating the spherical symmetry of the system

    nevertheless, if the electron is not actually "in orbit" how can we think of zero vs nonzero angular momentum in terms of its motion?
  5. Jan 19, 2009 #4
    It has been argued that the (even intrinsic) angular momentum corresponds to actual rotating currents of the 'probability cloud', but then it may be more fruitful not to try visualising it at all.
  6. Jan 20, 2009 #5
    it still has spin though. spin can also be though of as actual rotating currents of the 'probability cloud',

    http://modelingnts.la.asu.edu/pdf/MysteriesofDirac.pdf [Broken]
    Last edited by a moderator: May 3, 2017
  7. Jan 20, 2009 #6
    I'm not sure that I do either. The way I think about ordinary momentum is that the gradient of a function is a vector that points in the direction in which the function increases fastest; so the momentum operator picks out the direction in space in which you're most likely to find the particle. Angular momentum (like any observable) in QM is defined by replacing a classical dynamical variable with an operator, in this case r x i(hbar)grad. In that sense, it's obvious that it comes out of the maths that the gradient of a spherically symmetric function evaluated at a point will be parallel to the position vector of that point, so the cross product (and hence the angular momentum) will be zero. But trying to extend the way I think about linear momentum to angular momentum is a little trickier. The nearest I can get is that it's sort of like the wavefunction is dragging the particle around to lie on some axis of greatest probability.

    There's two caveats I should point out about my way of thinking, however. The first is to emphasise that there's no time derivative in the gradient operator; so any conception of the particle really moving is entirely in my head. It's perhaps nearer the idea of virtual work in classical mechanics. The second is to point out that probability is determined by the square modulus of the wavefunction, so a decrease in its value can lead to an increase in the probability. I really need to think through the complex analysis in a little more detail to work out what implications that has on my picture.
  8. Jan 20, 2009 #7
    To clarify this: are the "rotating currents" roughly contained within static orbitals (by which I mean spatial regions of high likelihood of finding the particle), or is there a suggestion of these orbitals rotating in time? The former would perhaps make more sense to me than the latter.
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