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Why the even Eigen states have less energy than odd ones?

  1. Dec 24, 2011 #1
    In the symmetrical potentials the Eigenstates are either even or odd. why the even states have
    less energy than odd states?
  2. jcsd
  3. Dec 24, 2011 #2

    Vanadium 50

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    What makes you think this is true?

    Consider a 1D harmonic oscillator. The n=1 state has less energy than n = 2. The n = 3 state has less energy than n = 4. The n = 5 state has less energy than n = 6.
  4. Dec 26, 2011 #3
    I meant, why the ground state in symmetric potentials is always an even state?
  5. Dec 26, 2011 #4
    For one dimensional problems the eigenstates are ordered by their number of nodes (the points in which the wavefunctions is 0).
    So an odd wavefunction can not be the ground state.
  6. Dec 26, 2011 #5
    All right, thanks
    But, what is the analytic justification here?
  7. Dec 27, 2011 #6
    I found a general demonstration in Courant & Hilbert "Methods of Mathematical Physics" Vol.I p 452 (you can read it from google books).

    But for one dimension it can be intuitively understood by elementary methods.

    Consider a real function which solves the equation for a generic energy E, but obeys the border condition only on one side, say the left one, so E is not an eigenvalue.
    For border conditions take as usual that the function vanishes at infinity.
    When E is very low (with respect to the potential) you can prove that it has no nodes (because it has the same sign on the whole domain), and that it does not go to 0 on the right side (for example something like e^x).
    Then let E grow until it obeys the border condition on the right side: this is the fundamental level and it still has no nodes.
    Then for bigger E, looking at the signs of f(x) and f''(x) you can prove that the number of nodes continue increasing.

    Does this make any sense to you?
  8. Dec 28, 2011 #7
    Yes; to some extent.
    But i am trying to explain it by using the concept of curvature of the wave function.
    Last edited: Dec 28, 2011
  9. Dec 28, 2011 #8

    Vanadium 50

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    It's probably a good idea for you to stop and to reformulated your question carefully. This is twice now where your response to an answer was along the lines of "I didn't really mean...".

    It's better for you to be careful than to hope we come upon your meaning by guessing.
  10. Dec 28, 2011 #9

    Ken G

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    Do you understand how curvature of the wave function connects to its kinetic energy? That's a key point, but it's not the whole story because there are two very different flavors of bound states. One is generally framed in terms of a positive potential energy, and it often extends to infinity (or is even infinite at infinity), as for the particle in a box or the simple harmonic oscillator. This first type generally uses the reference potential to be zero at the center of the spatial domain, and then the lowest number of nodes will have the smallest kinetic energy and the least (positive) total energy. But there is a second type, where the potential energy is usually regarded as negative, and zero at infinity (this is often made necessary by the potential energy going to negative infinity at the center of the spatial domain of interest, as for the Coulomb potential). With this second type, the lower number of nodes actually correspond to higher kinetic energies, because the scale of the wavefunction is so much smaller that it has higher curvature even though it has fewer nodes. But because the potential energy is negative and large at the center, the more concentrated wavefunctions, with fewer nodes and higher kinetic energies, end up having the smallest total energy because the total energy has the largest magnitude but is negative.
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