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Can someone verify my work

  1. Nov 17, 2009 #1
    1. The problem statement, all variables and given/known data

    I have an infinite square well and I am asked to show why E = 0 and E < 0 does not satisfy the Schrodinger's equation. I must do this by applying the boundary conditions.

    For E = 0:

    I argued that the second derivative of the wave function is zero.

    So,

    [tex]\Psi(x) = A + Bx[/tex]

    By imposing the boundary conditions [tex]\Psi (0) = \Psi (a) = 0[/tex] I get:

    [tex]\Psi(x) = Bx[/tex]

    and

    [tex]\Psi (a) = Ba = 0[/tex]

    Therefore I concluded that:

    (1) [tex]B[/tex] cannot be zero, or else we get [tex]\Psi(x) = 0[/tex] which is physically unacceptable
    (2) [tex]a\neq0[/tex] since [tex]a[/tex] is the upper bound.

    Perhaps before presenting my second solution to E < 0 I should make sure all the above is correct.

    Is it valid?

    2. Relevant equations

    ...



    3. The attempt at a solution

    ...
     
  2. jcsd
  3. Nov 18, 2009 #2

    Redbelly98

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    Looks good so far.
     
  4. Nov 18, 2009 #3
    E < 0:

    The second derivative of the wave function I set equal to [tex]k^{2}\Psi(x)[/tex]

    The solution is [tex]\Psi(x) = Ae^{-kx} + Be^{kx}[/tex], where the first term on the right cancels since it blows up at infinitely large values of [tex]x[/tex].

    Thus, [tex]\Psi(0) = Be^{k0} = 0[/tex] (after imposing the boundary conditions), which does not give us a satisfying solution for [tex]B[/tex] or [tex]e^{0}[/tex]. Neither does it work for [tex]\Psi(a) = 0[/tex].

    So, the boundary conditions don't work.
     
  5. Nov 18, 2009 #4

    Redbelly98

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    Note that the solution and coefficients A and B are only valid within 0<x<a. So it is okay if this Ψ(x) expression blows up at infinity, since the actual Ψ(x) will be a different expression for x<0 or x>a.

    Use both boundary conditions (at x=0 and a), and the expression you wrote for Ψ(x).
     
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