Shrodinger equation/Particle in a box

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SUMMARY

The discussion centers on the application of the Schrödinger equation to a particle in a box scenario, specifically analyzing the term U(x)ψ(x) within the interval 0 < x < L. Participants noted that U(x) is set to zero, leading to the second derivative of ψ(x) being expressed as -(nπ/L)²Csin(nπx/L). A miscalculation was identified when attempting to derive the energy expression, resulting in (ħnπ)²/2mL² * C * sin(nπx/L), which was deemed incorrect. Clarity in problem presentation and the use of LaTeX for mathematical expressions were emphasized as critical for effective communication.

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  • Review the derivation of the Schrödinger equation for a particle in a box
  • Study the implications of setting potential energy U(x) to zero
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Linus Pauling
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1. What is U(x)psin(x) in the interval 0 < x < L? (less than or equal to)



2. In previous problems I had found the following:

Second deriv. of psi(x) = -(n*pi/L)2*C*sin(n*pi*x/L)

U(x)*psin(x) = 0





3. From the above and the Shrodinger equation, I added U(x) = 0 to the second derivative of psi given above multiplied by -h-bar/2m, obtaining the incorrect answer of:

(h-bar*n*pi)2/2mL2 * C * sin(n*pi*x/L)
 
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I don't think you will receive much help if you don't clearly explain your problem. For example, what is U(x)? Also, you might want to use latex or people will be put off trying to decipher what you wrote.
 

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