How Can Dirac Notation Be Used to Determine Eigenvalues and Eigenfunctions?

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SUMMARY

Dirac notation is utilized to determine eigenvalues and eigenfunctions through the eigenvalue equation Au = pu, where u represents the eigenstate and p denotes the eigenvalue. In the discussion, the eigenvalues identified are +1 and -1, with corresponding states expressed as (φ + Bφ) and (φ - Bφ). The participant seeks a more algebraic method to derive eigenvalues and eigenfunctions rather than relying solely on substitution. Additionally, the concept of projection operators is explored, particularly in relation to eigenstates with a zero eigenvalue.

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  • Understanding of Dirac notation and its application in quantum mechanics
  • Familiarity with eigenvalue equations and their significance
  • Knowledge of projection operators and their properties
  • Basic algebraic manipulation skills in the context of linear algebra
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  • Research the algebraic methods for finding eigenvalues and eigenfunctions in quantum mechanics
  • Study the properties and applications of projection operators in linear algebra
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  • Learn about the implications of zero eigenvalues in quantum systems
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Students and professionals in quantum mechanics, physicists working with linear algebra, and anyone interested in the mathematical foundations of eigenvalue problems.

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Homework Statement


I have the following question (see below)

Homework Equations


The eigenvalue equation is Au = pu where u denotes the eigenstate and p denotes the eigenvalue

The Attempt at a Solution


I think that the eigenvalues are +1 and - 1, and the states are (phi + Bphi) and (phi-Bphi)
however I got this by just substituting these in from the symmetry of the operator.

Is there are neat algebraic way to work out the eigenvalues and eigenfunctions as opposed to just substitution?

I am stuck on working out how to express the 0 eigenvalue eigenstates in terms of the projection operator as well ...

Thank you very much
 

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About the projection operator , I'm not very well versed in them but if |x> is an eigenvector of it's projection operator with an eigenvalue of 1 ,
|x> <x| |x> = |x>
and any vector orthogonal to the |x> shown above is an eigenvector with eigenvalue zero. Also if
|x> <x| |A> = |x> <|A>
then if A is our zero eigenstate it seems a little meaningless.
Not sure how this fits in with a 0 value eigenstate though.
Sorry if I'm wrong.
 

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