# Probability of finding a particle in a 1D Box

1. Feb 1, 2010

### scottnoplot

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

Determine the probability of finding a particle in a 1-D box of size L in a region of size 0.01L at the locations x = 0, 0.25L, 0.5L, 0.75L and L when it is in its ground state. As percentages

2. Relevant equations

$$\Psi$$(x)=$$\sqrt{\frac{2}{L}}$$sin($$\frac{nxPI}{L}$$)
P=$$\Psi$$(x,t)^2

3. The attempt at a solution

I'm not even sure if i've got the right equations there, i've tried loads of different ways of doing this and cannot get the right answers, which are 0%,1%,2%,1%,0%

Cheers

2. Feb 1, 2010

### Mulder

ok, finding probabilities means integrating the probability density - P=|$$\Psi$$(x,t)^2| - over parts of the box

try integrating over the whole box first, what does this give you?...

3. Feb 1, 2010

### scottnoplot

$$\int$$|$$\Psi$$(x,t)^2| = |$$\Psi$$(x,t)^3| /3

I think that's right.

Last edited: Feb 1, 2010
4. Feb 1, 2010

### cpt_carrot

Your answer cannot be a function of x, since you are integrating over x. Try substituting your expression for $\Psi(x)$ from your first post before you integrate.

5. Feb 1, 2010

### Zhivago

You need to integrate |psi^2| from x=0 to x=0.01L , from x=0.245L to 0.255L, etc

Try to plot psi, |psi^2| and the integral of |psi^2| from 0 to x (remember that the integral is the probabilty of finding the particle in the limits of integration...)

6. Feb 1, 2010

### scottnoplot

Thanks guys, got a bit of extra help and the penny has dropped now. cheers