Calculating Probability of Particle in a Box

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The discussion focuses on calculating the probability of finding a particle in a box of width L within the region [L/4, 3L/4] for both the ground state and the first excited state. The probability density function used is (2/L)sin²(nπx/L) for even n and (2/L)cos²(nπx/L) for odd n. The user initially made an error in their integration and calculations, resulting in a probability value greater than 1. After clarification, they corrected their boundaries to -L/2 and L/2 and identified mistakes in adding fractions post-integration. The final probabilities for n=1 and n=2 were confirmed as (2+π)/(2π) and 1/2, respectively.
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Homework Statement



A particle is in a box of width L. Calculate the probability to find the particle in the region [L/4, 3L/4] when the particle is a) in the ground state b) in the first excited state.

Homework Equations



(2/L)sin(n*π*x/L)^2 dx is the probability in [x, x+dx]

The Attempt at a Solution



Integrating that gives me 2/L[x/2-[L/(4π)]sin(2n*π*x/L)], boundaries being L/4 and 3L/4. For a) n=1 and b) n=2, right? After I plug in the values, I get value greater than 1. Where have I gone wrong?

Hopefully this is readable, no LaTeX. :cry:
 
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The probability density is (2/L)sin²(nπx/L) for even n and (2/L)cos²(nπx/L) for odd n.

EDIT: Sorry, this is only if you take the boundaries of the box to be -L/2 and L/2.
 
Last edited:
No, it's the same formula for all n!
After integration i got 2/L[x/2-[L/(4π n)]sin(2n*π*x/L)]
For n = 1 -> (2+π)/(2π)
for n = 2 -> 1/2
 
Yes sorry, I took the boundaries to be -L/2 and L/2.
 
Thanks for the replies, guys! I forgot the one n in my first post. I found out my error was in the easy stuff after the integration, I'd done a mistake in adding fractions. :redface:
 
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