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Homework Help: Planar wave solution to zero potential Schrödinger equation

  1. Feb 8, 2016 #1
    1. The problem statement, all variables and given/known data
    Schrödinger.PNG

    2. Relevant equations
    \begin{align}
    \begin{split}
    \psi(x, t) = e^{(ikx- i \omega t)}
    \\
    V(x) = 0
    \end{split}
    \end{align}

    3. The attempt at a solution
    For a free particle, the Schrödinger equation can be put in the form of ##\psi(x, t) = e^{(ikx- i \omega t)}##. With constant potentials for all x, ##\forall x : V = 0##, or put equally succinctly: ##V(x) = 0## (the function of which is time independent for all x in the domain)

    \begin{align}
    \begin{split}
    i \hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} + V \psi
    \end{split}
    \end{align}

    Make the substitution ##\psi(x, t) = e^{i(kx-\omega t)}##

    \begin{align}
    \begin{split}
    i \hbar \frac{\partial}{\partial t} e^{i(kx- \omega t)} = \frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} e^{i(kx- \omega t)}
    \end{split}
    \end{align}

    Now, take the partial derivative with respect to time of the left hand side of the equation
    \begin{align}
    \begin{split}
    i \hbar \frac{\partial}{\partial t} e^{i(kx- \omega t)} = -i^2 \hbar \omega e^{(ikx-\omega t)}
    \end{split}
    \end{align}

    And the second order partial derivative with respect to x, of the right hand side of the equation

    \begin{align}
    \begin{split}
    \frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} e^{i(kx- \omega t)} = \frac{\hbar^2}{2m} (i^2k^2) e^{i(kx- \omega t)}
    \end{split}
    \end{align}

    Moderator's note: post edited to fix the LaTeX. Use double # for inline equations, not $.
     
    Last edited by a moderator: Feb 8, 2016
  2. jcsd
  3. Feb 8, 2016 #2

    DrClaude

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    Staff: Mentor

    You should get rid of the ##i^2##. Otherwise, do you have a question?
     
  4. Feb 8, 2016 #3

    BvU

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    And (2) ##\Rightarrow ## (4) + (5) = 0 gives you ?
     
  5. Feb 8, 2016 #4
    @BvU: A homogenous partial differential equation

    @DrClaude: I imagined that I would simplify a little later
     
  6. Feb 8, 2016 #5

    BvU

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    $$ a = b\\c = d $$ gives a pde ## a=c ##, I agree :wink:.
     
  7. Feb 8, 2016 #6
    \begin{align}
    \begin{split}
    \frac{\hbar^2}{2m} (k^2) e^{i(kx- \omega t)} - \hbar \omega e^{(ikx-\omega t)} = 0
    \\
    \frac{\hbar}{2m} (k^2) e^{i(kx- \omega t)} = \omega e^{(ikx-\omega t)}
    \end{split}
    \end{align}

    And cancelling an $\hbar$ right away
     
  8. Feb 8, 2016 #7

    BvU

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    Bingo !

    -- kudos for using the \split goody. Didn't know it. But it seems to confuse the \align thingy, perhaps because of the label rhs appearing twice ...​
     
  9. Feb 8, 2016 #8
    I see no reason not to cancel the following term

    \begin{align}
    \begin{split}
    e^{i(kx- \omega t)}
    \end{split}
    \end{align}

    but I've been wrong before :P
     
  10. Feb 8, 2016 #9

    BvU

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    (6) has to be true for all t and all x, independently.
    What is happening here is separation of the variables x and t.

    Oh, and ##e^{iy} \ne 0 ## always, so why not ?
     
  11. Feb 8, 2016 #10
    So for a free particle (in a box) we have

    \begin{align}
    \begin{split}
    k^2 = \frac{p^2}{\hbar^2}
    \end{split}
    \end{align}

    But the wavenumber

    \begin{align}
    \begin{split}
    k = \frac{\pi}{\lambda} \\
    \left( \frac{\pi}{\lambda} \right)^2 = \frac{p^2}{\hbar^2} \to \left( \frac{\pi}{\lambda} \right) = \frac{p}{\hbar}
    \end{split}
    \end{align}
     
  12. Feb 8, 2016 #11

    DrClaude

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    Work with the equations ou already have. Don't introduce p here (which is an operator, since you are working in position space).
     
  13. Feb 8, 2016 #12

    BvU

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    And wasn't ##k = {2\pi\over \lambda } ## :wink: ?
    (But I agree with Claudius)
     
  14. Feb 8, 2016 #13
    So I can use the following identity to separate variables

    \begin{align}
    \begin{split}
    e^{iy} = \cos(y)+ i \sin(y)
    \end{split}
    \end{align}

    And use the fact that

    \begin{align}
    \begin{split}
    e^{i(kx- \omega t)} = e^{ikx} \times e^{- i \omega t}
    \end{split}
    \end{align}
     
  15. Feb 8, 2016 #14
    Since e^(iy) \neq 0 always, I could go ahead and cancel the term on both sides of the equation?
     
  16. Feb 8, 2016 #15

    BvU

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    All in favor !
     
  17. Feb 8, 2016 #16
    Therefore k and omega need to be related in the following way

    \begin{align}
    \begin{split}
    \frac{\hbar}{2m} k^2 = \omega(k)
    \end{split}
    \end{align}
     
  18. Feb 8, 2016 #17

    BvU

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    Isn't it nice ? And if we now take a sneak preview: with ##E = \hbar\omega\ \ \& \ \ \vec p = \hbar \vec k\ \ ## it looks like ##E = {p^2\over 2 m}## :smile: !
     
  19. Feb 8, 2016 #18
    It's a ridiculously satisfying result :))

    Should I let the angular velocity depend on k?
     
  20. Feb 8, 2016 #19

    BvU

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  21. Feb 8, 2016 #20
    Thanks :biggrin:
     

    Attached Files:

    Last edited: Feb 8, 2016
  22. Feb 8, 2016 #21
    schroedinger-b-png.95531.png
    Which function do you think the author of exercise b asks me to use?
     
  23. Feb 8, 2016 #22

    BvU

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    The one in (1).
    You were already embarking on dissecting it in
    (and you remember that coefficients are complex numbers in QM)
     
  24. Feb 8, 2016 #23
    That yields the following

    \begin{align}
    \begin{split}
    e^{i(kx-\omega t)} = \cos{(i(kx-\omega t))} + i \sin{(i(kx-\omega t))} \\
    Closest: \ cosh(k x-t \omega)-sinh(k x-t \omega)
    \end{split}
    \end{align}
     
  25. Feb 8, 2016 #24

    BvU

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    Re:
    This ##\omega## is a totally different angular velocity.
    I have become quite enthousiastic about these Feynman lectures (the link came from Simon Bridge) -- Richard Feynman talks about photons, but later on you understand that this is true for all and everything (except radioactivity and gravity :smile: ).

    Re:
    I seem to remember something like ##e^{ix} = \cos x + i\sin x## :smile:
     
  26. Feb 8, 2016 #25
    The linear combination resulting is then

    \begin{align}
    \begin{split}
    e^{i(kx-\omega t)} = \cos{(kx-\omega t)} + i \sin{(kx-\omega t)}
    \end{split}
    \end{align}

    I'm very fond of watching Feynman's lectures as a way to cover more ground and to develop insight in physics
     
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