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Central Potential

  1. Mar 27, 2007 #1
    When we solve the Schrodinger equation for a central potential, such as the coulomb potential for the hydrogen atom, we get eigenstates which are not, in general, spherically symmetric. So they depend on the choice of coordinates.

    If we can get two distinct solutions, just by choosing different axes, how do we know which solution is the correct one?

    My prof says that they are the same solution, no matter which way they are oriented, and I disagree with him. If we solve the angular dependence using [itex] \phi [/itex] and [itex]\theta[/itex] in the usual definition, we get solutions (of |Psi|^2) with cylindrical symmetry about the z-axis, but if we use angles [itex]\phi ' [/itex] and [itex]\theta ' [/itex] such that [itex]\theta ' [/itex] is the polar angle from the y-axis and [itex] \phi ' [/itex] is the azimuthal angle around from the z-axis toward the x-axis, we get solutions with cylindrical symmetry around the y-axis. And in general [itex] Y_l^m(\theta, \phi) \neq Y_l^m(\theta ', \phi ') [/itex]

    The only solution I can think of is if you can make the eigenstates that you get using one set of coordinates, from linear combinations of the eigenstates you get using any other set of coordinates.

    Anyone know the answer? Are they in fact the same solution, and I'm confusing myself by adding orientation to the problem, when it doesn't matter? Or can you make the eigenstates you find from one set of coordinates from the eigenstates from another set? Or is there some other solution that I can't see.
    Last edited: Mar 27, 2007
  2. jcsd
  3. Mar 27, 2007 #2
    Yes, the eigenstates are oriented differently, and have a different appearances, as you found. The critical similarity, however, is they are energetically degenerate.
    For example, 2px, 2py, and 2pz all give the same energy for the hydrogen atom.
    Moreover, since they are energetically degenerate, the px, py, and pz can mix
    to produce solutions having the same energy but not clearly identified as a pure state.
    In fact, some of the solutions are imaginary, but linear combinations may be taken to give all real solutions.
    Finally, any completely filled shell (p,d, f, etc.) is spherically symmetric, although the separate components are not. So, neon, etc. are spherically symmetric and possess a completely filled p shell.
  4. Mar 27, 2007 #3


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    That's the essential point ! In fact, the set of solutions, as a set (a vector space), is a representation of the rotation group. That means exactly what you write. Now, the 3 "basis solutions" (for p orbitals) that you prefer, is dependent on the coordinate system in which you work them out, but the set spanned by them, not. In other words, the basis solutions found in another coordinate system are linear combinations of those in the first - as you point out. In fact, their linear combinations are given by the same (only in 3D, a bit more sophisticated but similar in nD) rotation matrices as those that rotate the axes! This is why it is a representation of the group.
  5. Mar 28, 2007 #4
    It's a nice point about the symmetry by vanesch. To which I would add that when you start looking at molecules, a similar principle applies. Any symmetry - not just spherical symmetry - which has a degenerate representation may have degenerate occupied orbitals. In such cases, you are free to take linear combinations of the orbitals in a degenerate set. The highest occupied pi orbitals in benzene are an example.
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