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Angular dependence in QM?

  1. Mar 16, 2008 #1
    When calculating eigenstates in the hydrogen atom one finds plenty of eigenstates with angular dependence. The s orbitals are spherically symmetric, but an orbital like 2p is not, there is some angular dependence through the spherical harmonics. But why is there angular dependence? It is a totally spherically symmetric problem so how can there be? Certainly, starting from the nucleus and going out in one direction should not be any different from going out in any other direction. How is the angular dependence to be interpreted?
     
  2. jcsd
  3. Mar 16, 2008 #2

    reilly

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    Recall that most planets have elliptical orbits while undergoing a central force, the sun's gravity.
    Regards,
    Reilly Atkinson
     
  4. Mar 16, 2008 #3

    kdv

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    Consider a completely classical situation
    Even if the force is spherically symmetric, the initial conditions of the motion don't have to be!!

    If the force is a central force, the angular momentum will be conserved. But the motion is not necessarily spherically symmetric since you can choose whatever initial conditions you want!

    Consider tossing a comet at some arbitary direction in the solar system (and assume it doe snot have enough energy to escape the potential). It will move following an ellipse which is obviosuly not spherically symmetric.

    Again the key point is that you must consider both the symmetry of the force and the symmetry of the initial conditions.
     
  5. Mar 18, 2008 #4

    olgranpappy

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    rephrasing the last post in "quantum terms", even thought the potential which appears in the schorodinger equation is spherically symmetric the solutions to the schrodinger equation are not necessarily spherically symmetric.
     
  6. Oct 28, 2011 #5

    Great! The solutions of an aquation may have less symmetry than the equation.
     
  7. Oct 28, 2011 #6

    alxm

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    Welcome to PF, Jiadong. Just a heads-up, though - you might want to check the date (upper-left corner of the post) before responding; you've answered a few that are a bit old.
     
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