Particle in a box, boundary co-ordinate change

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Discussion Overview

The discussion revolves around the wavefunction of a particle in a one-dimensional box with varying boundary conditions, specifically comparing a box defined from 0 to L with one defined from -L/2 to L/2. The focus is on the implications of these boundary conditions on the wavefunction and its physical significance.

Discussion Character

  • Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant states that the normalized wavefunction for a particle in a box with boundaries at 0 and L is given by ## \psi (x) = \sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})##.
  • Another participant questions whether the wavefunction would instead be ##\sqrt{\frac{2}{L}}\cos(\frac{n\pi x}{L})## for a box defined from -L/2 to L/2.
  • A third participant agrees that the wavefunction would change to the cosine form, but notes that this change has no physical significance as the problem remains a square well of width ##L##.
  • A later reply emphasizes that the choice of boundary conditions is flexible and can be made based on convenience for calculations.

Areas of Agreement / Disagreement

Participants generally agree on the mathematical forms of the wavefunctions for the different boundary conditions, but there is a discussion about the physical significance of these changes, indicating a nuanced view rather than a consensus.

Contextual Notes

The discussion does not resolve the implications of changing boundary conditions on the physical interpretation of the wavefunctions, nor does it clarify any assumptions made in the mathematical derivations.

I_laff
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If you have a particle in a 1-d box with a finite potential when ##0 < x < L ## and an infinite potential outside this region, then the normalised wavefunction used to describe said particle is ## \psi (x) = \sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})##.
However, if you had say instead a finite potential when ##\frac{-L}{2} < x < \frac{L}{2}## and an infinite potential outside this region, then wouldn't the wavefunction now change?
 
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Actually would the wavefunction be ##\sqrt{\frac{2}{L}}\cos(\frac{n\pi x}{L})## instead?
 
I_laff said:
Actually would the wavefunction be ##\sqrt{\frac{2}{L}}\cos(\frac{n\pi x}{L})## instead?
Assuming you've done the math right (I haven't checked), yes. However, this change has no physical significance because the problem is still the same square well of width ##L##. It's analogous to what happens if we decide to draw the prime meridian through something other than the Greenwich observatory - everything that is a function of longitude changes, but the surface of the Earth does not.

You can set the problem up either way, and in practice you will pick whichever way makes the algebra easiest for you.
 

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