Probability of finding a particle in a 1D Box

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Homework Help Overview

The discussion revolves around calculating the probability of finding a particle in a one-dimensional box of size L, specifically in a region of size 0.01L at various locations when the particle is in its ground state.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to integrate the probability density over specified regions of the box. There are attempts to clarify the correct equations and integration methods, with some questioning the validity of initial equations and expressions.

Discussion Status

Some participants have provided guidance on integrating the probability density and suggested plotting functions to visualize the probability. There appears to be a productive exchange of ideas, with one participant indicating a breakthrough in understanding.

Contextual Notes

There is mention of specific locations for calculating probabilities and a focus on the ground state of the particle. The original poster expresses uncertainty about the equations used and the expected results.

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



\Psi(x)=\sqrt{\frac{2}{L}}sin(\frac{nxPI}{L})
P=\Psi(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=|\Psi(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=|\Psi(x,t)^2| - over parts of the box

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

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

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

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