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Stationary states for even parity potential

  1. Apr 1, 2010 #1
    Hi I know that stationary states in a system with an even potential energy function have to be either even or odd.

    Why does the ground state have to be even, and not odd???? This is asserted in Griffiths, page 298.
  2. jcsd
  3. Apr 2, 2010 #2


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    Basically, in order for the state to be odd, it has to pass through zero. In order to do this, it needs to change more quickly, which means higher energy. So a state with fewer nodes (places where the wave funtcion goes to zero) always has lower energy than a state with more nodes. Typically the ground state has no nodes, so it can't be odd.
  4. Apr 3, 2010 #3
    I suppose it makes sense that stationary states with higher energy have a larger (magnitude of) second spatial derivative on average given the form of the time independent Schrödinger equation.

    But that would be true in general. Why would the ground state have an even wavefunction given an even potential?
  5. Apr 3, 2010 #4


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    As phyzguy said ... the energy of a state is proportional to the "curvature" of the wavefunction (the true mathematical description of curvature is a bit more complicated that just the 2nd derivative, but that simpler version will suffice for this discussion). Thus, states with fewer nodes have lower energies. So you just have to ask youself, what is the smallest number of nodes a wavefunction can have? It's zero, isn't it? So, zero nodes means a wavefunction that doesn't pass through zero, and thus must have even symmetry.
  6. Apr 3, 2010 #5
    Errr nevermind let me think about this
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