What Is the Probability of Finding a Quantum Particle in a Section of a 1D Box?

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SUMMARY

The probability of locating a quantum particle in a one-dimensional infinite potential well of length L, specifically in the second excited state (n=3), is calculated using the wave function ψ = sqrt(2/L)sin(nπx/L). The integral of the squared wave function from 0 to L/3 yields a probability of 1/3. The confusion regarding the limits of integration arises from the relationship between the excited state and the defined length L, which clarifies why the limits are set from 0 to π instead of π/3.

PREREQUISITES
  • Understanding of quantum mechanics principles, particularly wave functions.
  • Familiarity with integration techniques in calculus.
  • Knowledge of the concept of probability density in quantum mechanics.
  • Experience with solving problems involving infinite potential wells.
NEXT STEPS
  • Study the derivation of wave functions for different quantum states in infinite potential wells.
  • Learn about the normalization of wave functions in quantum mechanics.
  • Explore the implications of quantum state transitions on probability distributions.
  • Investigate the differences in probability calculations for various excited states (n=1, n=2, n=3, etc.).
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Students and educators in quantum mechanics, physicists working with quantum systems, and anyone interested in the mathematical foundations of quantum probability distributions.

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Homework Statement


A quantum particle is free to move in a infinite 1d potential well of length L excited on the second excited state n=3.What is the probability to locate the particle between 0 and L/3.


Homework Equations


∫ψ^2=1
ψ=sqrt(2/L)sin(n∏x/L)

The Attempt at a Solution


∫ψ^2dx=∫(2/L)sin^2(3∏x/L) dx limits of integration from 0 to L/3. then I set y = 3∏x/L then dx = Ldy/3∏. So that gives (2/3∏)∫(1/2)-(1/2)cos(2y) dy then the limits change to 0 to ∏
which gives (2/3∏)((y/2)-(1/4)sin(2y)) now use the limits 0 to pi which gives a probability of 1/3. What I don't understand is why use 0 to pi why not ∏/3 confused there . what I assume is that it has to do with the excited state the particle is in. So for instance if it was n=1 it would be 0 to ∏/3 am I correct.
 
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NVM I was dumb I figured it out it has to do with the length being defined as L it helps to read that being said I need help with a conceptual problem.
 
Uggg ok neverming maybe my point wass valid I just did a problem that was similar but in the ground state and it used limits from 0 to npi/3 help again.
 

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