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I Symmetry of Hamiltonian and eigenstates

  1. Mar 29, 2016 #1
    Suppose we have an electron in a hydrogen atom that satisfies the time-independent Schrodinger equation:
    $$-\frac{\hbar ^{2}}{2m}\nabla ^{2}\psi - \frac{e^{2}}{4\pi \epsilon_{0}r}\psi = E\psi$$

    How can it be that the Hamiltonian is spherically-symmetric when the energy eigenstate isn't? I was thinking along the lines of rotations with angular momentum operators but I'm not sure I can come up with a nice explanation. Can someone help me see why this is the case?
     
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  3. Mar 29, 2016 #2

    A. Neumaier

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    It just means that the eigenspaces are invariant under the symmetry group. Since symmetry groups of interest rarely have a nontrivial 1-dimensional representation, this means in practice that most energy eigenstates states are degenerate. One can easily verify this in the hydrogen atom, or in a pair of equal harmonic oscillators.
     
  4. Mar 29, 2016 #3

    Strilanc

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    Solving a differential equation tends to exponentiate things. Consider the complex circle ##2 e^{i \theta}##. What happens when we exponentiate it? Make a parametric plot of ##e^{2 e^{i \theta}}## and you get this:

    Screenshot from 2016-03-29 14:29:34.png

    Which is not circularly symmetric like its input was.

    That's not a full explanation, but I hope it gives an intuition for why the spherical symmetry could be lost.

    (Actually, because the Hamiltonian is scaled by ##i##, exponentiating tends to introduce cycling. But the cycles are in the phases of amplitudes, not in physical space.)
     
  5. Mar 29, 2016 #4
    Thank you for this - it's a nice thought. Is there any way you could apply this mode of thinking to the situation where ##\psi = R(r)Y(\theta, \phi) = R(r)cos(\theta)## say?
     
  6. Mar 29, 2016 #5

    Strilanc

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    I actually don't know enough about it to do that, unfortunately. Actually I'm half-expecting the next poster to complain that it's a really misleading way to think about a Hamiltonian, so we'll see.
     
  7. Mar 29, 2016 #6
    To see this intuitively consider Newtonian orbital dynamics. The hamiltonian is very similar, with a central 1/r^2 force and a kinetic energy term, mv^2/2 - or, if written as in Schroedinger's, p^2/2m. So ask the same question about planetary orbits: the Hamiltonian is spherically-symmetric but orbits don't have to be, why?

    The reason is, the kinetic energy term is formally (as written) symmetric but not when we plug in actual values for a planet's initial position and velocity. If those initial conditions are just right we get a spherically-symmetric orbit (circle) but much more likely it will be an ellipse (or hyperbola, but that's irrelevant here). True the angular momentum is always symmetric, meaning: at any angle it's the same. But position and velocity (or, momentum) are not.

    The hydrogen atom case is, roughly, the same idea, but it's more difficult because the electron's a wave function not a solid body like the Earth. You get complicated standing wave patterns instead of simple ellipses. But it's still true that the position and momentum (NOT angular) operators will give non-spherically-symmetric values in a given eigenstate (except s-orbitals).

    I may be wrong on some detail or terminology, but I hope this intuitive picture helps
     
  8. Mar 29, 2016 #7

    Strilanc

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    Nit: a circular orbit also isn't spherically-symmetric. Good example.
     
  9. Mar 29, 2016 #8
    Good catch! But actually I did cover that objection:
     
    Last edited: Mar 29, 2016
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