# For the infinite square-well potential, schrodinger eqation

1. Dec 8, 2013

### dawozel

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

For the infinite square-well potential, find the probability that a particle in its fourth excited state is in each third of the one-dimensional box:
(0 to L/3)
(L/3 to 2L/3)
and (2L/3 to L)

2. Relevant equations
∫ψ^2= Probability

3. The attempt at a solution
So from ∫ψ^2 for the first third of the problem i got that my probability equation should be
1/3-(1/((2πn))(sin(2π/3)) - 0 where n is that excited state but I am getting a wrong answer of about .10 off , is there something wrong with my equation?

Last edited: Dec 8, 2013
2. Dec 8, 2013

### TSny

Hello, dawozel.

Shouldn't there be an "n" in the argument of the sine function?

Just to check: what is the value of "n" for the 4th excited state?

3. Dec 8, 2013

### dawozel

n should be 4, maybe I'm missing a n in the sine function

4. Dec 8, 2013

### TSny

What is "n" for the ground state? What is "n" for the first excited state?

Yes, check that.

5. Dec 8, 2013

### dawozel

still not working, might be missing something else

6. Dec 8, 2013

### TSny

What value are you getting for the probability of the particle to be in the range 0 < x < L/3 ?

7. Dec 8, 2013

### dawozel

Well I've been testing this out with an example I Know the answer to (where N=3) but I'm getting an answer of .306 when it should be .299

8. Dec 8, 2013

### TSny

Hmm. The probability for the particle to be between x = 0 and x = L/3 when n = 3 is exactly 1/3.

Did you decide whether or not there should be a factor of n in the argument of the sine function in your result of

$\frac{1}{3} - \frac{1}{2 \pi n} \sin(\frac{2 \pi}{3})$?

Can you show the formula that you used with the numbers plugged in for the case of n = 3?

9. Dec 8, 2013

### TSny

I think there is still a misunderstanding of what the value of n should be for the fourth excited state.

Can you specify the value of n for each of the following?

(1) Ground state. n = ?
(2) First excited state. n = ?
(3) Second excited state. n = ?

10. Dec 9, 2013

### dawozel

Omg how could i be so blind, thank you! the 4th exited state is actually n=5! thanks the answers coming out correct now

11. Dec 9, 2013

### dawozel

For reference the necessary function to calculate this is

1/A - 1/(2πN) x Sin(2πN/3)

Where N is the state
A is the amount of the box you want
Just plug in your limits and solve

12. Dec 9, 2013

### TSny

OK. Good work.