Particle trapped in an infinite well (1d) - find probability

Click For Summary
SUMMARY

The discussion focuses on calculating the probability of a particle trapped in a one-dimensional infinite potential well using the Schrödinger equation. The key equation derived is the wavefunction ## \psi(x) = \sqrt{\frac{2}{L}} \sin(\frac{n \pi x}{L})##, which is essential for determining the probability density. Participants express confusion over why probabilities in different regions of the well do not simply divide evenly and emphasize the importance of integrating the wavefunction to find probabilities in specified intervals. The integral method is highlighted as necessary for accurate probability calculations.

PREREQUISITES
  • Understanding of quantum mechanics concepts, particularly wavefunctions
  • Familiarity with Schrödinger's equation and its applications
  • Knowledge of probability density functions in quantum systems
  • Basic calculus skills for performing integrals
NEXT STEPS
  • Study the derivation of the wavefunction for a particle in a box
  • Learn how to perform integrals of wavefunctions to find probabilities
  • Explore the concept of probability density and its implications in quantum mechanics
  • Investigate the differences in probability distributions for various quantum states
USEFUL FOR

Students of quantum mechanics, physics educators, and anyone seeking to deepen their understanding of probability calculations in quantum systems.

Brianrofl
Messages
21
Reaction score
0

Homework Statement


http://puu.sh/bTtVx/ba89b717b8.png

Homework Equations


I've tried using the integral method of Schrödinger's eq, getting:

(X/L - (1/4pi)sin(4xpi/L) from x1 to x2.

The Attempt at a Solution



I've tried plugging in the values of x given in the problem to the above equation and got reasonable answers, but they weren't right. I've tried multiple things, and I'd love to just keep trying to figure this problem out but unfortunately if I try any more I'll just miss out on the points and not bother learning it anyway (I really hate online HW systems for this reason. Puts cheating above learning the material).

Also, if you have a width L, and you're finding the probability of it being in either the first third of the width, the second third of the width, or the last third of the width, why is the probability of finding it in each region not just 100/3 or 33.33%?

Overall, I don't understand these probability questions much. Why can I use the generic (2/L)sin^2(pix/L) for small regions but have to do the integral of P(x) for others? Explanation would really be appreciated.
 
Last edited by a moderator:
Physics news on Phys.org
Also, I'm not even sure where to begin on this question: http://puu.sh/bTxr5/7109166e04.png .

I'm really not understanding this stuff, his lectures were super confusing and the textbook isn't helping either.
 
Last edited by a moderator:
1. First derive the wavefunction ## \psi(x) = \sqrt{\frac{2}{L}} sin(\frac{n \pi x}{L})##. Make sure you know how this is derived, trust me - it's important for you to understand this.
2. To find probability of the particle being in region ##x_0## to ##x_1##, take the overlap of wavefunction: ## \int_{x_0}^{x_1} (\psi)(\psi^*) dx##

Brianrofl said:
Also, if you have a width L, and you're finding the probability of it being in either the first third of the width, the second third of the width, or the last third of the width, why is the probability of finding it in each region not just 100/3 or 33.33%?
Hint: It is not equally likely to find the electron anywhere (why?).
 
Last edited:
Also, I'm not even sure where to begin on this question
Well, you could make a table for nx = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 .. ny = 0,1,2,3,4,5,6,7,8,9 ..

Oh, and : try to avoid intertwining two threads. Start a new one when you have a new exercise.
 

Similar threads

Replies
28
Views
2K
Replies
14
Views
2K
Replies
10
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
18
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 1 ·
Replies
1
Views
4K