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Schrodinger equation, lowest potential energy

  1. Oct 25, 2012 #1
    Classically, particles seek configurations of least potential energy. Something like this happens in QM: the wavefunction will usually be densest in those areas the potential energy is smallest. But looking at the Schrodinger equation itself, I can't see intuitively why this should be.
     
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  3. Oct 25, 2012 #2

    Bill_K

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    But this is not true, either in classical or quantum mechanics. As an example look at the one-dimensional harmonic oscillator. The particle does not spend most of its time at the bottom of the well, rather it likes to be at the turning points. Same holds in QM - the wavefunctions have their greatest amplitude at the turning points. See page 8 here, for example.
     
  4. Oct 26, 2012 #3

    tom.stoer

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    It seems to be true for the ground state, however I don't know if there's a rigorous proof; look at the qm harmonic oscillator, the hydrogen atom etc.

    This idea is used directly in the variational approach to find approximations to the ground state energy and wave function. One starts with an Hamiltonian H and a trial wave function ψα depending on some parameters α. Instead of solving the Schrödinger equation one minimizes the energy expectation value <E> wr.t. α

    [tex]\text{min}_\alpha\, E(\alpha) = \text{min}_\alpha\,\langle\psi_\alpha | H | \psi_\alpha \rangle \;\to\, \nabla_\alpha\,E(\alpha) = 0[/tex]

    For the potential term in H only it is obvious that this could be achieved by maximizing the wave function in a region where the potential has its minimum. But due to the kinetic term ~p² it will not be possible to strictly localize the wave function; it will spread out around the minimum of the potential.

    It can be shown that for this approximation the relation

    [tex]\langle\psi_\alpha | H | \psi_\alpha \rangle \ge \langle\phi_0|H|\phi_0\rangle[/tex]

    holds, i.e. the trial wave function does never underestimate the ground state energy.
     
  5. Oct 26, 2012 #4

    DrDu

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