Infinite Square Well (Quantum Mechanics)

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SUMMARY

The discussion focuses on calculating the probability of finding an electron in an infinite square well of width 0.300 nm. For the ground state, the probability of locating the electron within 0.100 nm of the left wall is determined using the wave function derived from the time-independent Schrödinger equation. The calculations for the 99th excited state (n=100) follow similar principles, emphasizing the correspondence principle's consistency. The wave function is expressed as Ψ(x,t) = Asin(kx)e^{-iωt}, where A is the normalization constant to be determined.

PREREQUISITES
  • Understanding of quantum mechanics concepts, specifically infinite potential wells.
  • Familiarity with the time-independent Schrödinger equation.
  • Knowledge of wave functions and normalization in quantum mechanics.
  • Basic grasp of eigenfunctions and their properties in quantum systems.
NEXT STEPS
  • Calculate the normalization constant A for the wave function Ψ(x,t).
  • Explore the implications of the correspondence principle in quantum mechanics.
  • Investigate the probability density function for different quantum states in infinite potential wells.
  • Learn about the energy levels and their derivation for an infinite square well.
USEFUL FOR

Students and educators in quantum mechanics, particularly those studying wave functions and probability distributions in infinite potential wells. This discussion is beneficial for anyone tackling problems related to quantum states and their properties.

ElijahRockers
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Homework Statement



An electron is trapped in an infinitely deep potential well 0.300nm in width. (a) If the electron is in its ground state, what is the probability of finding it within 0.100nm of the left-hand wall? (b) Repeat (a) for an electron in the 99th excited state (n=100). (c) Are your answers consistent with the correspondence principal?

(This question comes from Serway / Moses / Moyer Modern Physics Third edition)

Homework Equations



P(x,t)dx = \Psi^*\Psi dx

\Psi (x,t) = \psi (x) \Phi (t)

\Phi (t) = e^{-i\omega t}, \omega = \frac{E}{\hbar}

The Attempt at a Solution



Since the potential does not depend on time, I can use the time independent Schrödinger equation to solve for psi (x).

psi(x) = Asin(kx) + Bcos(kx)

E_n = \frac{n^2 \pi^2 \hbar^2}{2m(.3*10^{-9})^2}

k=\frac{\sqrt{2mE}}{\hbar} = \frac{n\pi}{L} , L=.3nmIn the regions associated with infinite potential, the eigenfunction psi = 0. Since it must be continuous along all regions, psi(0) = 0 = B.

So,

\Psi(x,t) = Asin(kx)e^{-i\omega t}

And this should be the wave function, right?

Now I could use this to get information about the particle, but I need A first. How do I find A?
 
Last edited:
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normalization I suppose.
 

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