Stationary states for even parity potential

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Discussion Overview

The discussion revolves around the characteristics of stationary states in quantum mechanics, specifically in systems with an even potential energy function. Participants explore why the ground state must be even rather than odd, referencing concepts from quantum mechanics and the implications of wavefunction symmetry.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant asserts that stationary states in a system with an even potential must be either even or odd.
  • Another participant explains that an odd state must pass through zero, which requires a faster change and thus higher energy, suggesting that the ground state, having no nodes, cannot be odd.
  • A different participant notes that higher energy states have a larger average second spatial derivative, questioning why the ground state specifically must have an even wavefunction.
  • One participant elaborates that the energy of a state is related to the curvature of the wavefunction, indicating that states with fewer nodes have lower energies, leading to the conclusion that the ground state must have zero nodes and thus even symmetry.
  • A participant expresses uncertainty and decides to reconsider their thoughts on the matter.

Areas of Agreement / Disagreement

Participants present multiple viewpoints regarding the relationship between wavefunction symmetry and energy states, indicating that the discussion remains unresolved with competing explanations and interpretations.

Contextual Notes

Participants reference the relationship between nodes and energy without fully resolving the implications of wavefunction curvature or the mathematical details involved.

Nick R
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Hi I know that stationary states in a system with an even potential energy function have to be either even or odd.

Why does the ground state have to be even, and not odd? This is asserted in Griffiths, page 298.
 
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Basically, in order for the state to be odd, it has to pass through zero. In order to do this, it needs to change more quickly, which means higher energy. So a state with fewer nodes (places where the wave funtcion goes to zero) always has lower energy than a state with more nodes. Typically the ground state has no nodes, so it can't be odd.
 
I suppose it makes sense that stationary states with higher energy have a larger (magnitude of) second spatial derivative on average given the form of the time independent Schrödinger equation.

But that would be true in general. Why would the ground state have an even wavefunction given an even potential?
 
As phyzguy said ... the energy of a state is proportional to the "curvature" of the wavefunction (the true mathematical description of curvature is a bit more complicated that just the 2nd derivative, but that simpler version will suffice for this discussion). Thus, states with fewer nodes have lower energies. So you just have to ask youself, what is the smallest number of nodes a wavefunction can have? It's zero, isn't it? So, zero nodes means a wavefunction that doesn't pass through zero, and thus must have even symmetry.
 
Errr nevermind let me think about this
 

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