Probability of finding a particle in a 1D Box

Click For Summary
SUMMARY

The discussion focuses on calculating the probability of finding a particle in a one-dimensional box of size L in specific regions of size 0.01L while in its ground state. The relevant equations include the wave function \(\Psi(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n \pi x}{L}\right)\) and the probability density \(P = |\Psi(x,t)|^2\). The correct approach involves integrating the probability density over specified intervals, leading to the final probabilities of 0%, 1%, 2%, 1%, and 0% at the locations x = 0, 0.25L, 0.5L, 0.75L, and L, respectively.

PREREQUISITES
  • Quantum mechanics fundamentals
  • Understanding of wave functions and probability densities
  • Integration techniques in calculus
  • Familiarity with the concept of boundary conditions in quantum systems
NEXT STEPS
  • Study the derivation of the wave function for a particle in a box
  • Learn about normalization of wave functions in quantum mechanics
  • Explore the concept of probability density and its applications
  • Investigate the implications of boundary conditions on quantum states
USEFUL FOR

Students and educators in quantum mechanics, physicists working on particle behavior in confined systems, and anyone interested in the mathematical foundations of quantum probability.

scottnoplot
Messages
3
Reaction score
0

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
 
Physics news on Phys.org
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
 

Similar threads

Replies
28
Views
2K
Replies
8
Views
1K
  • · Replies 30 ·
2
Replies
30
Views
8K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K