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Can we rewrite Schrodinger equation using observable variable?

  1. Aug 18, 2013 #1
    We know that in Schrodinger equation, Ψ is called wave function, which is not observable, while Ψ·Ψ* is the probability, which is observable.
    can we rewirte the Schrodinger equation to a form without Ψ but only Ψ·Ψ*?

    because I think, in this way can I figure out all conservations in the equation. Although I can make it with present Schrodinger equation, it's obvious that the Schrodinger equation will change if I make t→-t transformation.

  2. jcsd
  3. Aug 18, 2013 #2


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    Write down the equation of motion for [itex]\psi^*[/itex] given that [itex]\psi[/itex] fulfills the usual Schrödinger equation. Then change [itex]t \rightarrow -t[/itex]!
  4. Aug 18, 2013 #3
    you mean, this equation set(containing two equation, Ψ and Ψ*) does not change?
    by the way, could you tell me, can Schrodinger equation be rewrited to Ψ·Ψ* mathemetically?
  5. Aug 18, 2013 #4


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    No, we can't.


    ##\psi = R \, e^{iS}##

    with two real variables R and S.

    Then introduce

    ##\rho = \psi^\ast \psi = R^2##

    Now we see that an equation in R or ρ is an equation in one single real variable, whereas the original equation was an equation in two independent real variables R and S.
    Last edited: Aug 18, 2013
  6. Aug 18, 2013 #5
    That's essentially the idea behind density functional theory (DFT).
  7. Aug 18, 2013 #6


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    But that's an approximation.
  8. Aug 18, 2013 #7
    If you mean that the DF theory itself is an approximation - no, it's exact. Practical implementations are approximations, however.
  9. Aug 19, 2013 #8


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    But DFT works only for the ground state, whereas the SG works for all states including all bound and scattering states.
  10. Aug 19, 2013 #9
    DFT works for the lowest states of a given symmetry, of which there might be several. The calculation of excited state properties (e.g. absorption spectra) can be carried out using time-dependent DFT. More information here: http://en.wikipedia.org/wiki/Time-dependent_density_functional_theory.
  11. Aug 19, 2013 #10


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    I still think that this answer is missleading in our context. Otherwise you would have to prove that DFT is fully equivalent to the SG including the complete set of states (bound plus scattering states) in Hilbert space plus all derived phenomena like superpositions, interference, entanglement etc.
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