Raising and lowering operators

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SUMMARY

The discussion clarifies that any term formed by four raising and lowering operators, with a lowering operator positioned at the rightmost end (e.g., A-A+A+A-), results in a zero expectation value in the ground state of a harmonic oscillator. This is due to the behavior of the lowering operator, which, when applied to the ground state wave function ψ₀, yields zero, indicating that no further operations can be performed. The participants confirm that the lowering operator acting on the first excited state ψ₁ returns the ground state ψ₀, but applying it again results in zero, reinforcing the conclusion.

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  • Understanding of quantum mechanics concepts, specifically harmonic oscillators.
  • Familiarity with raising (A+) and lowering (A-) operators in quantum mechanics.
  • Knowledge of wave functions, particularly the ground state (ψ₀) and first excited state (ψ₁).
  • Basic proficiency in calculating expectation values in quantum systems.
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Students and professionals in quantum mechanics, particularly those studying harmonic oscillators and the application of raising and lowering operators in quantum systems.

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Homework Statement



Can anyone please explain why any term which is a product of 4 raising and lowering operators with a lowering operator on the extreme right (eg. A-A+A+A-) has zero expectation value in the ground state of a harmonic oscillator?



Homework Equations





The Attempt at a Solution

 
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What happens if you let the lowering operator work on [itex]\psi_1[/itex]? What do you get when you let the lowering operator work on [itex]\psi_0[/itex]?

Calculate [itex]a \psi_1[/itex] and [itex]a \psi_0[/itex]. Does [itex]a \psi_0[/itex] exist?
 
I get it now... thank you. When the lowering operator operates on v1, I get v0. When it operates on v0, I get zero. A- v0 doesn't exist so the rest of the operators in the product have nothing to operate on.

Cheers.
 

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