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The ways of Erwin

  1. Sep 27, 2009 #1
    I see how one can arrive at the Schrodinger equation by starting with a complex plane wave

    [tex]

    \psi = \psi _0 e ^{i(k \cdot x - \omega t)}

    [/tex]

    taking its first partial derivative with respect to time, second partial derivative with respect to space, making the quantum substitutions

    [tex]
    k=p/ \hbar \hspace{10 mm} \omega = E / \hbar
    [/tex]

    as well as the classical one

    [tex]

    E=p^2/2m + V(x,y,z)

    [/tex]

    and putting it all together.

    But why does this work when the [itex] \psi [/itex]'s that one finds in quantum mechanics are typically not plane waves? Luck? I've read that one should accept the Schrodinger equation as an axiom and not worry about "deriving" it, and yet...
     
  2. jcsd
  3. Sep 28, 2009 #2
    Just plug it back in, and you will see your solution is not valid because of the potential V.

    Namely, the first partial derivative with respect to x is:

    [tex]\frac{\partial}{\partial x} e^{i (kx - (\frac{p^2}{2m}+V(x)))} = (ik - iV'(x))e^{i (kx - (\frac{p^2}{2m}+V(x)))}[/tex]

    The second one will be even more complicated. So your solution doesnt satisfy the schroedinger equation at all!

    A second reason why you can see this: you're trying to find the energy eigenfunctions. The energy eigenvalues that you find are not allowed to depend on x. They should be constants. But in your case E clearly depends on [itex]V(x)[/itex].
     
  4. Sep 28, 2009 #3
    Actually, since the total energy of a particle in a conservative field is assumed to be constant even if it moves around the field,

    [tex]

    \frac {p^2}{2m} + V(x)

    [/tex]

    is a constant and

    [tex]

    \frac {\partial}{\partial x} ( \frac {p^2}{2m} + V(x) ) = 0.

    [/tex]
     
  5. Sep 30, 2009 #4
    Does anyone else out there have any thoughts about this? It looks to me like one can in fact derive the Schrodinger equation by assuming that [itex] \psi [/itex] is a complex plane wave, even though the solutions to it are for the most part not plane waves. Does that make sense? Was the equation a kind of educated guess that happened to work with spectacular success?
     
  6. Sep 30, 2009 #5

    jtbell

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    Staff: Mentor

    Schrödinger didn't use this particular "derivation," but his own "derivation" was indeed an "educated guess." See this post for a summary of what Schrödinger actually did.
     
  7. Oct 1, 2009 #6
    Thanks, jtbell; that looks very interesting. Maybe I can find the original paper translated into English at my local physics library.

    Aside: To those for whom such things matter, I would like to apologize for the rather silly title I've given this thread. "The ways of Erwin" was the dummy title I was using while I tested the LaTeX. I was going to change it to something more specific like, "on the origin of the Schrodinger equation", but before I did I accidentally hit something or other on the computer I was using (not mine) and posted the question. I know one can edit a post, but if there's a way to change a thread title I never learned about it.
     
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