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Boundary value problem

  1. Jan 24, 2007 #1
    Schodinger's equation for one-dimensional motion of a particle whose potential energy is zero is
    [tex]\frac{d^2}{dx^2}\psi +(2mE/h^2)^\frac{1}{2}\psi = 0 [/tex]
    where [tex] \psi [/tex] is the wave function, m the mass of the particle, E its kinetic energy and h is Planck's constant. Show that

    [tex]\psi = Asin(kx) + Bcos(kx)[/tex] ( where A and B are constants) and [tex] k =(2mE/h^2)^\frac{1}{2}[/tex] is a solution of the equation.
    Using the boundary conditions [tex]\psi=0[/tex] when x=0 and when x=a, show that
    (i) the kinetic energy [tex]E=h^2n^2/8ma^2[/tex]
    (ii) the wave function [tex]\psi = A sin(n\pi\times x/a)[/tex] where n is any integer. (Note if [tex] sin(\theta) = 0 then \theta=n\pi[/tex])

    My attempt:

    A*sin(0) + B*cos(0) = 0, => 0 +B =0 => B = 0.
    Therefore
    A*sin(k*a)=0, Therefore [tex] (2mE/h^2)^\frac{1}{2}a = n\pi => E=n^2\pi^2h^2/2ma^2[/tex]
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Jan 24, 2007
  2. jcsd
  3. Jan 24, 2007 #2
    (ii) should read A*sin(n*pi*x/a), I'm just notgood at boundary value problems.
     
  4. Jan 24, 2007 #3

    HallsofIvy

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    Well, first, you haven't shown that that [itex]\psi[/itex] does, in fact, satisfy the differential equation. After that, yes, B= 0. Now, assuming A is not 0, that is that [itex]\psi[/itex] is not itself identically 0, then yes, we must have sin(ka)= 0 so that [itex]ka= (2mE/h^2)^{\frac{1}{2}}= n\pi[/itex]. E follows eactly as you say.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution[/QUOTE]
     
  5. Jan 25, 2007 #4
    I was able to show that [tex]\psi[/tex] is a solution of the equation, it is (i) and (ii) that I had trouble doing espically (ii) i.e., [tex]\psi = A sin(n\pi\times x/a)[/tex]. Where did he get the argument "[tex]n\pi\times x/a[/tex]", or more importantly how does he expect me to get that argument of the sine.
    Also remember in (i) the answer he has for E [tex]=h^2n^2/8ma^2[/tex], not what I got for E. Thanks for the help.
     
  6. Jan 26, 2007 #5
    I hope someone can tell me how [tex]\psi[/tex] can go from [tex] Asin((2mE/h^2)^\frac{1}{2}x) to Asin(n\pi\times x/a)[/tex]. Thanks for the help.
     
  7. Jan 26, 2007 #6

    HallsofIvy

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    Well, when x= a, the argument is [itex]n\pi[/itex] what is [itex]sin(n\pi)[/tex]? Remember that you were told that [itex]\psi(0)= 0[/itex] and [itex]\psi(a)= 0[/itex]. Knowing that cos(0)= 1 tells us that the second constant, B, must be 0. That leaves Asin(kx). We must have Asin(ka)= 0 and we don't want A= 0 (that would mean our function is always 0) so we must have sin(ka)= 0. For what x is sin(x)= 0? Multiples of [itex]\pi[/itex] of course:
    [itex]ka= n\pi[/itex]. For that to be true, k must be equal to [itex]n\pi/a[/itex]

    You were also told that [itex]k= \sqrt{2ME/h^2}[/itex] and you now know [itex]k= n\pi/a[\itex] so [itex]n\pi/a= \sqrt{2ME/h^2}[/itex]. Solve that for E.
     
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