Energy State Probability Particle in a Box

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SUMMARY

The discussion focuses on calculating the probability of obtaining energy states for a particle in a box, specifically using the wave function \(\Psi(x) = \sqrt{\frac{30}{L^{5}}}(x)(L-x)\) for \(0 < x < L\). The probability is derived from the expression \(|c_n|^2 = \frac{240}{n^6\pi^6}[1-(-1)^n]^2\), where \(c_n\) is calculated using the integral \(c_n = \int \psi^*_n \Psi(x) dx\). Participants confirm that while the integral can become complex, it is manageable and follows a standard form involving polynomials and sine functions.

PREREQUISITES
  • Understanding of quantum mechanics principles, specifically wave functions.
  • Familiarity with the concept of probability amplitudes in quantum systems.
  • Knowledge of integration techniques, particularly integration by parts.
  • Experience with Fourier series and orthogonality of functions.
NEXT STEPS
  • Study the derivation of energy eigenstates for a particle in a box using wave functions.
  • Learn advanced integration techniques, focusing on integrals involving polynomials and trigonometric functions.
  • Explore the implications of probability amplitudes in quantum mechanics.
  • Review the properties of orthogonal functions and their applications in quantum mechanics.
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Students and educators in quantum mechanics, physicists working on wave function analysis, and anyone interested in the mathematical foundations of quantum probability distributions.

GrantB
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Homework Statement



Show that the probability of obtaining En for a particle in a box with wave function

[itex]\Psi[/itex](x) = [itex]\sqrt{\frac{30}{L^{5}}}[/itex](x)(L-x) for 0 < x < L

and [itex]\Psi[/itex](x) = 0 for everywhere else

is given by |cn|2 = 240/(n6[itex]\pi[/itex]6)[1-(-1)n]2

Homework Equations



cn = [itex]\int[/itex][itex]\psi[/itex][itex]^{*}_{n}[/itex][itex]\Psi[/itex](x)dx

The probability is cn squared.

Shouldn't have to use eigenvalues and eigenfunctions.

The Attempt at a Solution



I used the integral from (2) and used the given uppercase Psi and used the sqrt(2/L)sin(n*pi*x/L) lowercase psi (conjugate), from 0 to L.

The integral quickly turned messy with integration by parts and such.

I would like to know if I am on the right track here... If I am, I will just work through the integral until I get the right answer.

I'm hoping there is a much easier way to do this.

Thanks!

P.S. Sorry this is a repost from another section. I wasn't getting any responses from the Intro Physics section so thought I would try here.
 
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Yep, you're on the right track. Yes, the integral takes a little bit of work, but it is of the form
[tex]\int(\text{polynomial in }u)\sin u\;\mathrm{d}u[/tex]
and those integrals are doable.
 

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