Probability of finding a particle in a 1D Box

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scottnoplot
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Homework Statement



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

Homework Equations



[tex]\Psi[/tex](x)=[tex]\sqrt{\frac{2}{L}}[/tex]sin([tex]\frac{nxPI}{L}[/tex])
P=[tex]\Psi[/tex](x,t)^2

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

try integrating over the whole box first, what does this give you?...
 
Mulder said:
ok, finding probabilities means integrating the probability density - P=|[tex]\Psi[/tex](x,t)^2| - over parts of the box

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

[tex]\int[/tex]|[tex]\Psi[/tex](x,t)^2| = |[tex]\Psi[/tex](x,t)^3| /3

I think that's right.
 
Last edited:
scottnoplot said:
[tex]\int[/tex]|[tex]\Psi[/tex](x,t)^2| = |[tex]\Psi[/tex](x,t)^3| /3

I think that's right.

Your answer cannot be a function of x, since you are integrating over x. Try substituting your expression for [itex]\Psi(x)[/itex] from your first post before you integrate.
 
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 probability of finding the particle in the limits of integration...)
 
Thanks guys, got a bit of extra help and the penny has dropped now. cheers