Finite Square Well: Bond States and Asymmetric Potential Wells

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Discussion Overview

The discussion revolves around the existence of bound states in asymmetric finite square wells compared to symmetric ones, focusing on the conditions under which bound states can be found in these potential wells. The scope includes theoretical considerations and mathematical reasoning related to quantum mechanics.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants express uncertainty about the existence of bound states in asymmetric potential wells when the energy is less than the potential barriers, questioning why this differs from the symmetric case where solutions always exist.
  • One participant suggests that the asymmetric finite square well does indeed have bound state solutions, indicating a potential misunderstanding of the question.
  • Another participant emphasizes that bound state solutions are not guaranteed in all cases for asymmetric wells, contrasting this with the symmetric square well, which always has a bound state solution.
  • A later reply mentions that an explicit equation for the energies of bound states in asymmetric wells can be derived, suggesting that at least one bound state exists under certain conditions.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the existence of bound states in asymmetric potential wells, with multiple competing views presented regarding the conditions under which bound states may or may not exist.

Contextual Notes

The discussion highlights potential misunderstandings regarding the conditions for bound states in asymmetric versus symmetric wells, but does not resolve the underlying mathematical or conceptual uncertainties.

LagrangeEuler
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I am not sure how is it possible that asymetric potential well does not have bond states if ##E<U_1<U_2##. In symmetric case solution always exists. Why this is a case?
 
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LagrangeEuler said:
I am not sure how is it possible that asymetric potential well does not have bond states if ##E<U_1<U_2##. In symmetric case solution always exists. Why this is a case?
I may be misunderstanding the question here... The asymmetric finite square well does have bound state solutions.
 
But not always. Right? Symmetric square well always has bound state solution.
 
For square (asymmetric) potential well an explicit equation with respect to energies of bound states is not difficult to derive. The existence of at least one bound state follows from that equation almost immediately.
 

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