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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

    if

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

    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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