Delta Function Well and Uncertainty Principle

[dwintz02]

[SOLVED] Delta Function Well and Uncertainty Principle

Homework Statement

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])

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!

 

[Avodyne]

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

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

You need to use these when computing derivatives of the wave function.

 

[dwintz02]

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!