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Variational Method

  1. Oct 16, 2009 #1
    I recently saw the Rayleigh Ritz variational approach used in spectral graph theory, so I was curious to look it up again in the quantum mechanics context. Anyway, there was a real sticking point quite quickly...

    When we pick our trial wave function, because we want our overlap integrals Sij=Sji, we pick real valued basis functions. Moreover, because f(z) = z conjugate is not holomorphic, we choose our constants to be real as well. So our trial wave functions is the sum of real constants and real functions.

    So my question is: is it true that we can approximate well complex valued wave functions with pure real quantities? It seems really sketchy, and my gut feeling is that this wouldn't work well at all.
     
  2. jcsd
  3. Oct 16, 2009 #2

    alxm

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    Why do you think the wave function is complex? Typically you'd be using the variational method to determine the time-independent ground state wave function, which (in the absence of external fields) is real-valued.
     
  4. Oct 16, 2009 #3
    Interesting, I had no idea that the ground state wavefunction is real...can you provide some intuition/justification/proof?
     
    Last edited: Oct 17, 2009
  5. Oct 17, 2009 #4

    alxm

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    The wave function under time-reversal is its complex conjugate (this is easily shown):
    [tex]\psi(x,-t) = \psi^*(x,t)[/tex]
    So if the wave function is time-independent, you have:
    [tex]\psi(x,t) = \psi(x,-t) = \psi^*(x,t)[/tex]
    So the wave function must be real for a stationary state.
     
  6. Oct 17, 2009 #5
    Hmm, it doesn't look like you've used the hypothesis that we're in the ground state -- that's a little worrying as otherwise it would be true for all states. Do you use the hypothesis under the time reversal => complex conjugate part?

    Also, clearly the stationary state is time independent, but is it really true [tex] \psi (x,t) = \psi (x,-t) [/tex] for all t. Surely that complex exponential [tex]e^{-iE_n t/h} [/tex] isn't always 1 for the ground state.
     
  7. Oct 17, 2009 #6
    Ugh, this is trivial -- the real and complex components of the time independent Schrodinger equation are independent of each other if the potential is purely real. So the real component is a solution on its own with the desired eigenvalue.
     
  8. Oct 18, 2009 #7

    alxm

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    The ground state is a stationary state, or at least the 'most stationary' of states. An excited state OTOH is obviously not stationary; it can decay to a lower state over time.
     
  9. Oct 19, 2009 #8
    Stationary state is what it means: eigenstates of the time-independent H operator. Excited states are just as stationary.

    The reason real excited state would decay is of course the system is coupled to some time-dependent potential. If you do have an external time dependent potential, none of the states can stay stationary, ground or excited states. Sure, spontaneous emission makes going down more likely than going up. I would consider in this case none of the states are stationary.

    By "real" I suppose one means the WF only has a global, no spacial dependent phase factor.
    If a WF have a spacial dependence phase factor, then it tends to have a node somewhere. Then it is possible to construct another WF very similar to the original one except with no nodes: just take the absolute value of the original one and "smooth" out the kinks where the nodes were. The new wave function will have similar potential energy but smaller kinetic energy and therefore smaller total energy. Therefore the original WF cannot be the ground state. This Feynman's No Node Theorem applies for a single particle WF or manybody bosons WF and fails for manybody fermions.
     
    Last edited: Oct 19, 2009
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