Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I About the premises behind the Schrödinger equations

  1. Dec 25, 2017 #1
    Long ago I read on these forums that one cannot derive the Schrödinger equations because they're fundamental scientific laws. But I have noticed that I can generate them by making the following physical assumptions and then doing a trivial amount of substitution and differentiation:
    (1) [tex] \frac {\partial ^2 \psi }{\partial x^2} + k^2 \psi = 0 [/tex]
    (2) [itex] \lambda = h/p [/itex]
    (3) Total energy = PE + [itex] p^2/2m [/itex]
    and (4) [itex] E=h \nu [/itex]
    where (4) is only needed for the time dependent form.

    What bothers me is that (1) assumes a [itex] \psi [/itex] with a definite wavelength — that is — a momentum eigenfunction, and (4) was arrived at for photons, and of course the S.E.'s are used to deal with wavefunctions that are not necessarily momentum eigenfunctions and for particles that are not photons.

    Thoughts?
     
    Last edited: Dec 25, 2017
  2. jcsd
  3. Dec 26, 2017 #2

    mfb

    User Avatar
    2017 Award

    Staff: Mentor

    You derived the Schrödinger equation limited to free particles with fixed momentum. It turns out that the equation is more general than that.
     
  4. Dec 26, 2017 #3
    I did not derive it at all here, I'm saying that putting (1),(2) and (3) together it is easy to arrive at
    [tex]
    \frac {\partial ^2 \phi} {\partial x ^2} + \frac {8 \pi ^2 m }{h^2}(E - PE) = 0
    [/tex]
    and then adding (4) we can get
    [tex] i \hbar \frac {\partial \phi} {\partial t } = E \phi [/tex]
    and these forms are not limited to momentum eigenfunctions, nor are they for photons, yet (1) and (4) are connected to those physical premises.

    In case you're wondering, I'll show you explicitly what I mean in an upcoming post.
     
    Last edited: Dec 26, 2017
  5. Dec 26, 2017 #4
    What I had in mind was . . if we start with
    [tex]
    \phi = \phi_0 e^{i(kx-\omega t)}

    [/tex]
    and take the derivative with respect to x twice and rearrange terms, we get (1) in the original post.
    Then if we plug in [itex] k=2 \pi / \lambda [/itex] , [itex] \lambda = h/p [/itex], [itex] KE = p^2 /2m [/itex] and E = KE + PE, we get the time independent form I cited in entry #3 above.

    If instead we start with
    [tex]
    \phi = \phi_0 e^{i(kx-\omega t)}

    [/tex]
    and take the time derivative once, then substitute [itex] \omega = 2 \pi f [/itex], f=E/h, [itex] \hbar = h/ 2 \pi [/itex] and rearrange terms, we get the time dependent form I cited in entry #3 above.

    Thus we start with a [itex] \psi [/itex] with a definite wavelength and definite speed and yet end up with the general (one dimensional) Schrödinger equations.
     
    Last edited: Dec 26, 2017
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: About the premises behind the Schrödinger equations
  1. Schrödinger equation (Replies: 11)

Loading...