# Homework Help: Probability for finding particle outside potential well

1. Sep 16, 2010

### Eric_meyers

1. The problem statement, all variables and given/known data
A particle in the harmonic oscillator potential is in the first excited state. What is the probability of finding this particle in the classically forbidden region?

2. Relevant equations

probability of finding particle = integral of abs[psi squared] [a,b]

3. The attempt at a solution

So, I'm using mathematica to find the intersection point of my x probability distribution with the oscillator potential function.

Solve[ ((m*w)/(\[Pi]*h))^(1/4)*(2*m*w/h)^(1/2)*
Exp[-((m*w)/(2*h))*x^2] - (1/2) (w^2)*m*x == 0, x]

{{x -> -(Sqrt[h] Sqrt[
ProductLog[(8 Sqrt[(m w)/h])/(h^2 Sqrt[\[Pi]] w^2)]])/(
Sqrt[m] Sqrt[w])}, {x -> (
Sqrt[h] Sqrt[
ProductLog[(8 Sqrt[(m w)/h])/(h^2 Sqrt[\[Pi]] w^2)]])/(
Sqrt[m] Sqrt[w])}}

and then when I use this in my integral I get the answer
1/Sqrt[\[Pi]]2 ((m w)/h)^(3/2) If[Re[1/h] Re[m w] > 0

which obviously isn't a number as expected and seems to contain imaginary numbers. I don't understand what I'm doing wrong but this problem doesn't seem like it should be so complicated.

2. Sep 16, 2010

### Dick

It doesn't make any sense at all to intersect the probability distribution with the potential. They don't even have the same units. The classically allowed region of x is the region where E>V(x), where E is the energy of the first excited state.

3. Sep 16, 2010

### Eric_meyers

Right, I've corrected that part using the energy of the first excited state and finding those x intersection points. However the integral I get still involves an error function, and I'm wanting to get a numerical approximation.

4. Sep 16, 2010

### Dick

Try doing a change of variables so the exponential part becomes exp(-u^2). What's u in terms of x? If you do that the other constants in the integral should drop out too.