(adsbygoogle = window.adsbygoogle || []).push({}); [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!

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Delta Function Well and Uncertainty Principle

**Physics Forums | Science Articles, Homework Help, Discussion**