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Delta Function Well and Uncertainty Principle

  1. Nov 9, 2007 #1
    [SOLVED] Delta Function Well and Uncertainty Principle

    1. The problem statement, all variables and given/known data
    Griffiths Problem 2.25.

    I need to calculate < p[tex]^{2}[/tex]> for the Delta Function Well.
    The answer given is:
    < p[tex]^{2}[/tex]> = ([tex]m\alpha[/tex]/[tex]\hbar[/tex])^2

    The wave function given by the book is:

    [tex]\Psi(x,0)[/tex]=[tex]\frac{\sqrt{}m\alpha}{\hbar}[/tex] EXP([tex]^{-m\alpha|x|/\hbar^{2}}[/tex])



    3. The attempt at a solution
    The result I get from just calculating the expectation value is -(ma/hbar)^2. The negative from the operator is still hanging around. However, the hint given in the book specifies that the derivative of Psi has a discontinuity at zero and this affects the value of p^2 because it puts a 2nd derivative on Psi. The book says to use the result from the previous problem which is:

    If a function is defined piecewise such that:

    "F=1 for x>0
    F=0 for x<0
    And for the rare case that it matters F(0)=1/2
    The result is that dF/dx equals the delta function." I don't know how to incorporate this.

    What I am currently doing is plugging in the operator, breaking the integral into 2 parts, one from -[tex]\infty[/tex] to 0 and the other from 0 to +[tex]\infty[/tex]. Is this not enough? If not, how do I calculate the jump in the 2nd derivative at 0? Thanks in advance!
     
  2. jcsd
  3. Nov 9, 2007 #2

    Avodyne

    User Avatar
    Science Advisor

    (d/dx)|x| = sign(x)

    (d/dx)sign(x) = 2 delta(x)

    You need to use these when computing derivatives of the wave function.
     
  4. Nov 9, 2007 #3
    Awesome, that works out--the extra term from including those in the derivatives makes up the difference from the answer I got and the correct answer. Thanks tons!
     
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