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Schrodinger's equation

  1. Feb 22, 2010 #1
    The general solution of Schrodinger's equation is givrn by --------
    \Psi= A e^{kx-wt}.
    And this satisfies the equation .

    But the general solution of 3-D sinosidal wave is given by
    Psi= A Sin(kx-wt)
    And this also satisfies the schrodinger's equation.

    Schrodinger is credited to find the solution as complex phase factor (to signify matter waves) .

    Now the question is what is the need of depicting matter waves as complex phase factor ?
  2. jcsd
  3. Feb 22, 2010 #2
    I think it has to do with keeping the total energy linear while taking a second derivative. For example: (d2/dt2)(exp[iEt]) = -E^2exp[iEt]. Where as i(d/dt)(exp[iEt]) = -E(exp[iEt]).

    In a complex wave equation (i) acts as a derivative because it changes the phase by the same amount (90 degrees) while preserving linear total energy in the solution.
  4. Feb 22, 2010 #3


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    You're missing an "i" in that argument of the exponential.

    No it doesn't. Not without some funky potential.
  5. Feb 22, 2010 #4
  6. Feb 22, 2010 #5


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    No. That's the time independent solution. zodas is claiming that the full time dependent wave function \Psi(x,t) = A sin(kx-wt) satisfies the SE. Just work it out...it's not difficult.
  7. Feb 22, 2010 #6


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    Where is the time dependence of the wavefunction? It is not covered in that link. The point is that for an eigenstate of a quantum system, the time-dependent phase is always complex. So it is the "omega-t" term in the sine function mentioned by the OP that makes it not a solution of the TDSE. Try plugging that sine function into the TDSE and see what you get ... you will find that, as bapowell said, it requires a "funky" potential.
  8. Feb 22, 2010 #7
  9. Feb 22, 2010 #8
    Thanks guys !
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