For the infinite square-well potential, schrodinger eqation

AI Thread Summary
The discussion focuses on calculating the probability of a particle in the fourth excited state of an infinite square-well potential being located in specific segments of the box. The initial calculations were incorrect due to a misunderstanding of the quantum number "n," which for the fourth excited state is actually 5, not 4. Participants emphasized the importance of correctly incorporating "n" into the sine function of the probability equation. After correcting this error, the calculations yielded the expected results. The final formula for determining the probability was confirmed to be accurate with the correct value of "n."
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Homework Statement



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)

Homework Equations


∫ψ^2= Probability

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?
 
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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?
 
n should be 4, maybe I'm missing a n in the sine function
 
dawozel said:
n should be 4

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

maybe I'm missing a n in the sine function

Yes, check that.
 
still not working, might be missing something else
 
What value are you getting for the probability of the particle to be in the range 0 < x < L/3 ?
 
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
 
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})##?

Is your calculator in radian mode when you make the calculations?

Can you show the formula that you used with the numbers plugged in for the case of n = 3?
 
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 = ?
 
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  • #10
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
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
dawozel said:
Omg how could i be so blind, thank you! the 4th exited state is actually n=5! thanks the answers coming out correct now

OK. Good work.
 
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