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About ground state

  1. Dec 29, 2008 #1


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    If I know the explicit form of potential, the energy and a specific eigenstate, but I don't know the general form of eigenstates and eigenvalues, can I tell if the state is ground state or not?
  2. jcsd
  3. Dec 29, 2008 #2


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    No i dont think so, why are you asking?
  4. Dec 29, 2008 #3


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    A ground state wave function usually has no nodes.
  5. Dec 29, 2008 #4

    I think you can prove this statement by noting that the ground state wave function minimizes
    [tex]\left < \psi \right| H \left | \psi \right>[/tex]
  6. Dec 30, 2008 #5
    The Variational Principle guarantees that

    E_g \leq \langle \psi|H|\psi \rangle \equiv \langle H \rangle
    Last edited: Dec 30, 2008
  7. Dec 30, 2008 #6
    In doing some QM problems a few weeks ago I noticed that, for the PARTICULAR potential I was working with, the product of the standard deviation of the position (sx) and the standard deviation of the momentum (sp) was exactly hbar/2 in the ground state (sx*sp=hbar/2), and the product grew larger for higher energy states. So the uncertaintly principle was closest to getting violated in the lowest energy state. I assume that's NOT true for other potentials, but maybe something you could look into depending upon what exactly you are trying to do.
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