Can changes in spring constant affect the eigenstates of a harmonic oscillator?

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Discussion Overview

The discussion centers on the effects of changing the spring constant in a harmonic oscillator on its eigenstates, particularly examining the implications of Hamiltonians that may or may not commute. Participants explore the mathematical relationships and physical interpretations involved in this scenario.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant asserts that if two Hamiltonians commute, they share the same eigenstates, but questions this in the context of changing the spring constant from k1 to k2.
  • Another participant challenges the assertion that the Hamiltonians commute, providing a calculation that shows they do not commute due to the difference in spring constants affecting the commutation relation.
  • A third participant agrees with the previous point about non-commutation and elaborates that the eigenvectors for two commuting observables pertain to a single system, while two Hamiltonians represent different systems, suggesting this is more of an academic question.
  • Another participant mentions the sudden approximation, indicating that if the new eigenvectors are the same as the old ones, one can determine transition probabilities accurately.
  • One participant acknowledges that there may be specific applications where the previous points hold, but emphasizes that generally, a system is described by a unique Hamiltonian and a set of observables.

Areas of Agreement / Disagreement

Participants express disagreement regarding the commutation of the Hamiltonians and the implications for eigenstates. There is no consensus on whether changing the spring constant affects the eigenstates, as differing views are presented on the nature of the Hamiltonians involved.

Contextual Notes

Participants discuss the implications of changing the spring constant on the eigenstates without resolving the mathematical details of the commutation relations or the physical interpretations of the Hamiltonians.

hokhani
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We now that if [A,B]=0, they have the same eigenstates. But consider a harmonic oscillator with the spring constant k1. If we change k1 to k2, then [H1,H2]=0 and the above expression implies that the eigenstates should not change while they really change!
Could you please tell me if i am wrong?
 
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I don't think [H1,H2] is 0, [H1,H2]=[p^2+k_1*x^2, p^2+k_2*x^2]=k_1[x^2,p^2]+k_2[p^2,x^2]=(k_1-k_2)[x^2,p^2], neither of the two factors is 0.
 
1) kof9595995 is right, the two Hamiltonian do not commute b/c H2 = H1 + (k2-k1)x2 and therefore [H1,H2] = [H1,H1 + (k2-k1)x2] = [H1,H1 + (k2-k1)x2] = (k2-k1)/2m [p2,x2]

2) the eigenvectors for two commuting observables are relevant for two observables in one system; but two Hamiltonians describe two different systems, so this is an academic question
 
1) agree

2) In the sudden approximation you apply a new Hamiltonian to an "old" set of wave functions to find the transition probabilities due to the change. If the "new" Eigenvectors are the same as the old ones, you know exactly where you end up.
 
regarding 2) there may indeed be some applications where this makes sense; but in general you are intersted in a system described by a unique H and by a set of observables {H, O1, O2, ...}.
 

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