# Delta Function Well and Uncertainty Principle

1. Nov 9, 2007

### dwintz02

[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$$^{2}$$> for the Delta Function Well.
The answer given is:
< p$$^{2}$$> = ($$m\alpha$$/$$\hbar$$)^2

The wave function given by the book is:

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

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 -$$\infty$$ to 0 and the other from 0 to +$$\infty$$. Is this not enough? If not, how do I calculate the jump in the 2nd derivative at 0? Thanks in advance!

2. Nov 9, 2007

### 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.

3. Nov 9, 2007

### 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!

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