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Homework Help: Wavefunction Evolution

  1. Jun 23, 2008 #1
    Problem
    Describe the evolution in time of [tex]\phi_1 = A \sin \omega t \cos k(x+ct)[/tex].

    Attempt at Solution
    We have that

    [tex]\partial^2 \phi_1 / \partial x^2 = -Ak^2 \sin \omega t \cos k(x+ct)[/tex]
    [tex]\partial \phi_1 / \partial t = A \sin \omega t (-kc \cdot \sin k(x+ct)) + A \omega \cos \omega t \cos k(x+ct)[/tex]

    Now, by Schroedinger's Equation,

    [tex]-h^2/2m \cdot \partial^2\phi_1 / \partial x^2 = i \hbar \cdot \partial \phi_1 / \partial [/tex]

    So, substituting, we have

    [tex]\hbar^2 / 2m \cdot Ak^2 \sin \omega t \cos k(x+ct) = ihA \sin \omega t \cdot kc \sin k(x+ct) - i\hbar A \omega \cos \omega t \cos k(x+ct)[/tex]

    [tex]\iff \hbar^2/2m \cdot k^2 \cdot \sin \omega t = i \hbar \sin \omega t \tan k(x+ct) - i \hbar \omega \cos \omega t[/tex]

    [tex]\iff \hbar^2/2m \cdot k^2 = i\hbar \tan k(x+ct) - i \hbar \omega \cot \omega t[/tex]

    [tex]\iff \hbar k^2 = i \cdot 2m (\tan \k(x+ct) - \omega \cot \omega t)[/tex]

    Is this it? What am I to do now?
     
  2. jcsd
  3. Jun 25, 2008 #2
    What is [itex]\phi[/itex]?

    (Sidenote: Looking at the function, the first term is a time dependent term whereas the second term is basically a "traveling wave" (for want of a better term...its actually the wavefunction of a free particle).)

    I don't know what [itex]\phi[/itex] is, but if its the total wavefunction (we usually use [itex]\psi(x,t)[/itex] for it..lol :tongue2:) and if your algebra is correct, the last equation doesn't make sense: the left hand side is a constant, whereas the right hand side is a function of space and time...for it to hold, [itex]x+ct[/itex] and [itex]\omega t[/itex] both should be constants (by a simple argument).

    You could play around a bit by writing the whole thing as a bunch of complex exponentials, using Euler's Theorem...

    [tex]\phi(x,t) = A\left(\frac{e^{i\omega t}-e^{-i\omega t}}{2i}\right)\left(\frac{e^{i(kx+kct)}+e^{-i(kx+kct)}}{2}\right)[/tex]

    See if that helps...
     
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