QM - Ladder Operator QHO - factorization

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SUMMARY

The discussion centers on the factorization of the quantum harmonic oscillator (QHO) operators, specifically the lowering operator A and the raising operator A†. The equation discussed is (A A† - 1 + 1/2) ħω [Aψ] = A (A† A - 1 + 1/2) ħω ψ. The confusion arises from the ordering of the operators when factoring out A. Jane clarifies that another A exists in [Aψ], which necessitates pulling this A into the bracket, resulting in (AA†A - A + 1/2A) before factoring A out to the left.

PREREQUISITES
  • Understanding of quantum mechanics principles, particularly the quantum harmonic oscillator (QHO).
  • Familiarity with operator algebra in quantum mechanics.
  • Knowledge of the significance of lowering and raising operators (A and A†).
  • Basic grasp of the notation and operations involving ħ (reduced Planck's constant) and wave functions (ψ).
NEXT STEPS
  • Study the mathematical properties of quantum harmonic oscillator operators, focusing on their commutation relations.
  • Learn about the implications of operator ordering in quantum mechanics.
  • Explore the derivation of energy eigenstates for the quantum harmonic oscillator.
  • Investigate the role of the reduced Planck's constant (ħ) in quantum mechanics and its applications in operator equations.
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This discussion is beneficial for physics students, quantum mechanics researchers, and anyone studying the mathematical framework of quantum harmonic oscillators and operator theory.

JaneHall89
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Hi, quick question with A being the lowering operator and A the raising operator for a QHO

(A A - 1 + 1/2) ħω [Aψ] = A (A A - 1 + 1/2) ħω ψ

By taking out a factor of A. Why has the ordering of A A swapped around? I would have thought taking out a factor of A would leave it as

A (A - 1 + 1/2) ħω ψ

Jane
 
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There is another A in [Aψ]. Pull this A into the bracket from the right to get (AAA - A + 1/2A). Now pull out A out to the left of the bracket.
 
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