Probability density function problem

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SUMMARY

The discussion centers on the properties of probability density functions (PDFs) in quantum mechanics, specifically regarding the action of the lowering operator \(\hat{a}\) on eigenstates \(|n\rangle\). It is established that the operation \(\langle n|\hat{a}|n\rangle\) results in \(\langle n|\sqrt{n}|n-1\rangle = 0\), indicating that the lowering operator produces functions orthogonal to the original eigenstate. This conclusion is supported by the theorem stating that eigenfunctions of a Hermitian operator corresponding to different eigenvalues are mutually orthogonal, confirming the utility of the lowering operator in generating distinct eigenfunctions of the Hamiltonian.

PREREQUISITES
  • Understanding of quantum mechanics principles, particularly eigenstates and operators.
  • Familiarity with Hermitian operators and their properties.
  • Knowledge of the mathematical formulation of quantum mechanics, including wavefunctions.
  • Basic grasp of linear algebra concepts related to orthogonality.
NEXT STEPS
  • Study the properties of Hermitian operators in quantum mechanics.
  • Learn about the implications of eigenvalues and eigenfunctions in quantum systems.
  • Explore the role of raising and lowering operators in quantum harmonic oscillators.
  • Investigate the mathematical proofs of orthogonality for eigenfunctions of Hermitian operators.
USEFUL FOR

This discussion is beneficial for physics students, quantum mechanics researchers, and anyone interested in the mathematical foundations of quantum theory, particularly those studying the behavior of quantum systems and operators.

Chronos000
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Homework Statement



Is the PDF of something between two different bases or wavefunctions always 0?

For instance, if you have the lowering operator [tex]\hat{}a[/tex] -

<n|[tex]\hat{}a[/tex]|n>

that changes to <n|[tex]\sqrt{}n[/tex]|n-1> =0

I'm not sure I understand the physical scenario if this is true however.
 
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Does the lowering operator produce a function orthogonal to the original? If it's guaranteed to produce an eigenfunction of the Hamiltonian with a different eigenvalue (it should, or else it's not a very useful lowering operator), then all functions produced by the operator have to be orthogonal. This is due to a more general theorem which says that all eigenfunctions of a Hermitian operator with different eigenvalues are mutually orthogonal.
 
ok, I think I get this thanks
 

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