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Linearity of Schrodinger Equation

  1. Oct 20, 2006 #1
    Could someone please address this question?
    How do you algebraically demonstrate the superposition principle revealed by the Schrodinger equation (ie. If Psi1(x,t) and Psi2(x,t) are both solutions then Psi(x,t)= Psi1(x,t)+Psi2(x,t) is also a solution.)?
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  3. Oct 20, 2006 #2

    George Jones

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    What do you get when you substitute Psi(x,t) into the left side of the Schrodinger equation?
  4. Oct 20, 2006 #3
    I am not quite sure what you are asking me to do...Psi(x,t) is already in there?
  5. Oct 20, 2006 #4
    No, [tex]\psi(x,t)[/tex] is not in there. However, you do know that [tex]\psi_1(x,t)[/tex] is a solution -

    [tex]-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi_1(x,t)+V(x)\psi_1(x,t)=i \hbar \frac{d}{dt}\psi_1(x,t)[/tex]

    And [tex]\psi_2(x,t)[/tex] is also in there:

    [tex]-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi_2(x,t)+V(x)\psi_2(x,t)=i \hbar \frac{d}{dt}\psi_2(x,t)[/tex]

    However, you have to show that [tex]\psi=\psi_1+\psi_2[/tex] also satisfies this dynamic equation.

    And then you can extend your result to [tex]a\psi_1 + b\psi_2[/tex] for any coefficents a,b.
    Last edited: Oct 20, 2006
  6. Oct 20, 2006 #5

    George Jones

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    You have to show that

    -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} \Psi(x,t) + V(x) \Psi(x,t) = i \hbar \frac{\partial}{\partial t} \Psi (x,t).

    In the left, substitute [itex]\Psi = \Psi_1 + \Psi_2[/itex], and, using that [itex]\Psi_1[/itex] and [itex]\Psi_2[/itex] both statisfy Schrodinger's equation, work your way to the right side of the equation you must show true.
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