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Quick QM question

  1. Sep 29, 2006 #1
    The schrodinger eq is 1st order in t. Why does that matter to the probability interpretation?

    Should be an easy question, but I can't seem to get it.
     
  2. jcsd
  3. Sep 29, 2006 #2

    quasar987

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    Here's a suggestion you might want to investigate. When solving the SE by the method of separation of variables, we find that the time dependant part of the solution is [itex]\exp{iEt/\hbar}[/itex], and the position dependant part satisfies the time-independant SE. Denote [itex]\psi(x)[/itex] the solution to the time independant SE for a given potential. Then the general solution to the SE is [itex]\Psi(x,t)=\psi(x)e^{iEt/\hbar}[/itex], and according to the Born interpretation, [itex]\Psi \Psi^*[/itex] is a probability density function for the position of the particle. But [itex]\Psi \Psi^* = \psi\psi^*[/itex]. I.e. the probability density is is time independant!

    So the question is, would the time dependant part of the [itex]\Psi[/itex] still be such that the probability is time independant if the t "dependance" of the SE was not of first order?
     
  4. Sep 29, 2006 #3
    Interestingly, Schrödinger originally started with a relativistic equation, but didn't know what to do with the negative probability densities that resulted, so came up with the final non-relativistic equation instead.
     
  5. Sep 30, 2006 #4
    Ok I understand what you said there (I think), but I can't make the jump to what would happen if it wasn't 1st order in t. I know how to derive the equation from dealing with wave packets, I know why it's 1st order & how to get the time independent SE eigenvalue equation.

    Call it a lack of imagination. I can't see how it could be different. I know the difference between the SE & the familiar wave equation, but I can't see how to get probabilities out of the latter.
     
  6. Oct 1, 2006 #5

    quasar987

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    It should have to do with what Daverz said. I.e. that a different order in t will give us negative probabilities... but how could that be? Whatever the solution [itex]\Psi(x,t)[/itex] to a modified SE, the complex conjugate of [itex]\Psi[/itex] is still it's norm squared, which is still positive no matter what. :confused:
     
  7. Oct 17, 2007 #6
    Bump. I was going through old posts & needed some closure here. Thanks to whoever moved this.
     
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