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Homework Help: An infinite square well problem

  1. Oct 28, 2013 #1
    1. The problem statement, all variables and given/known data

    Particle in well:

    V(x)=0 for |x|<[itex]\frac{L}{2}[/itex]
    V(x)=∞ for |x|>[itex]\frac{L}{2}[/itex]

    initial wave function [itex]\Psi[/itex](x,0)=[itex]\frac{1}{√L}[/itex][cos[itex]\frac{\pi*x}{L}[/itex]+ i*sin[itex]\frac{2*\pi*x}{L}[/itex]]

    a) calc P(p,t) (momentum prob density)

    2. Relevant equations

    Anything from Griffiths QM

    3. The attempt at a solution

    I'm getting tripped out from the initial wave function. It is perfectly clear to me the process in which to solve for ψ(x,t) , given the initial wave function ψ(x,0); however, I'm not sure what to do in this case.

    I know that, given ψ(x,0), we must do the following:

    1)normalize [itex]\Psi(x,0)[/itex]
    2)compute the expansion coefficients (aka, c[itex]_{n}[/itex])
    3)compute E[itex]_{n}[/itex] and plug into the time dependent solution
    4)plug in c[itex]_{n}[/itex], Normalized "A" value, and E[itex]_{n}[/itex] into ψ(x,t)
    Last edited: Oct 28, 2013
  2. jcsd
  3. Oct 28, 2013 #2

    king vitamin

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    Gold Member

    If the initial wave function had been

    $$ \Psi(x,0) = \sqrt{\frac{2}{L}}sin(\pi x/L) $$

    do you know what [itex]\Psi(x,t)[/itex] would be?
  4. Oct 28, 2013 #3
    If you're trying to see that I know how to do the most basic infinite square well problem , yes, and i've done it for various situations (shifted wells and whatnot)


    where, in ψ(x,t), h = h-bar (didn't know how to find h-bar in latex)

    was the plan to aid in clarification of the problem at hand once i prove that I'm not just looking for "answers" ?
  5. Oct 28, 2013 #4


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    Staff: Mentor

    It's \hbar. :biggrin:
  6. Oct 29, 2013 #5

    king vitamin

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    I didn't have a "plan," I'm just trying to help based on info given to me. Given that you have no problem attaining [itex]\Psi(x,t)[/itex], do you know the equation you use to find the momentum probability density? You mention expansion coefficients in your OP, but I'm not sure what expansion coefficients you're referring to (are they for momentum?).
  7. Oct 29, 2013 #6


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    Staff Emeritus
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    So what's stopping you from doing these steps? It would help if you were a little more specific than saying you're "getting tripped out."
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