Quantum Mechanics - Infinite Potential Well

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SUMMARY

The discussion focuses on a particle trapped in an infinite potential well, specifically analyzing its wavefunction and energy states. The wavefunction at time t=0 is given as ψ = 1/√(2a) for -a < x < a, and the energy levels are defined by the equation E_n = n²ħ²π²/(8ma). The probability of the particle having an energy of 9ħ²π²/(8ma²) is derived by projecting the initial wavefunction onto the appropriate cosine function, with the sine solution being invalid due to the even parity of the well.

PREREQUISITES
  • Understanding of quantum mechanics concepts, specifically infinite potential wells.
  • Familiarity with wavefunctions and their normalization.
  • Knowledge of energy quantization in quantum systems.
  • Proficiency in mathematical techniques for projecting functions and calculating probabilities.
NEXT STEPS
  • Study the normalization of wavefunctions in quantum mechanics.
  • Learn about the implications of parity in quantum systems.
  • Explore the mathematical techniques for calculating projections of wavefunctions.
  • Investigate the differences between sine and cosine solutions in infinite potential wells.
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Students and professionals in physics, particularly those studying quantum mechanics, as well as educators seeking to clarify concepts related to infinite potential wells and wavefunction analysis.

TheBaker
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Homework Statement



A particle is trapped in an infinite potential well, with the infinite walls at ±a. At time t=0, the wavefunction of the particle is

\psi = \frac{1}{\sqrt{2a}}

between -a and a, and 0 otherwise.

Find the probability that the Energy of the particle is \frac{9 \bar{h}^2 \pi^2}{8ma^2}

Homework Equations



E_n = \frac{n^2\bar{h}^2\pi^2}{8ma}

\psi = A \cos{\frac{(2r+1) \pi x}{2a}} for |x| < a
\psi = 0 otherwise

The Attempt at a Solution



I've calculated the above equations, but I'm unsure how to get from them to the probability of the particle having a certain energy. This could be really simple and it's me just having a brain dead moment, but any help would be very much appreciated.
 
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btw there is also a sin solution with an argument (in your notation) 2r(pi)x/a

personally I prefer the notation n(pi)x/a n even

however the cos solution you wrote is the one you want with r=1, that 2r+1 thing is just a way of writing n so that n is always odd.

so just take the projection of psi at t=0 on your cos function and square the answer
 
The sin solution isn't valid because this well has even parity (i.e. it's symmetric).

How do I find A? Presumably I need to use the initial condition of Psi, but I found when doing that that A is x dependent, when it should be a constant.
 

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