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Partial Differentiation in free Schrodinger Equation

  1. Mar 5, 2009 #1
    Hello again! This time I have another calculus question for you, coming straight out of my study of the free Schrodinger equation, since I am not that experienced with that kind of derivative.

    It all starts with a given wavefunction (which I think is 2-dimensional,correct me if wrong)
    [itex]\psi(x,t)=e^{i(px-Et)}/\hbar[/itex] (1)

    My book says|: We derive 3.4 according to t and x so we have:
    [itex]\frac{\partial \psi}{\partial t}=-\frac{iE}{\hbar} \psi[/itex]
    and [itex]\frac{\partial \psi}{\partial x}=\frac{ip}{\hbar}\psi[/itex].

    What I don't understand is how this partial deriving works. What I can figure out is that given the (1) equation, you derive the [itex]i(px)/\hbar[/itex] and [itex]i(-Et)/\hbar[/itex] parts differently, one by one not together. Is that it? So you basically remove the e (2.71)(this must be the way to derive from exponential equations, I don't remember clearly though) and derive the power...partially?

    Alas, I am confused.. Any understandable link or advice is appreciated!
  2. jcsd
  3. Mar 5, 2009 #2
    i have no idea what you said but



    then you compute the partial derivatives by taking regular derivatives while considering the other variables as constants. hence what the second and third equation say is that

    [tex]\frac{\partial}{\partial t}\psi(x,t) = \frac{-iE}{\hbar}e^\frac{i(px-Et)}{\hbar}=\frac{-iE}{\hbar}\psi(x,t)[/tex]

    this means [itex]\psi(x,t)[/itex] is an eigenfunction of the [itex]\frac{\partial}{\partial t}[/itex] operator with eigenvalue [itex] \frac{-iE}{\hbar}[/itex].
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