How to Calculate Eigenvectors of the Unperturbed Hamiltonian?

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SUMMARY

The discussion focuses on calculating the eigenvectors of the unperturbed Hamiltonian represented by the matrix H = Vo * diag(1, 1, 2) when e = 0. The eigenvalues identified are Vo, Vo, and 2Vo. The confusion arises regarding the calculation of eigenvectors, particularly the misconception that they could be zero. The standard procedure for finding eigenvectors involves solving the equation (H - λI)v = 0, where λ represents the eigenvalue and I is the identity matrix.

PREREQUISITES
  • Understanding of linear algebra concepts, specifically eigenvalues and eigenvectors.
  • Familiarity with matrix operations and diagonalization.
  • Knowledge of Hamiltonian mechanics in quantum mechanics.
  • Basic proficiency in solving linear equations.
NEXT STEPS
  • Study the process of diagonalization of matrices in linear algebra.
  • Learn about the application of eigenvalues and eigenvectors in quantum mechanics.
  • Explore the method of solving (H - λI)v = 0 for eigenvectors.
  • Review the implications of the trivial solution in the context of eigenvector calculations.
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Students of quantum mechanics, particularly those studying linear algebra and its applications in physics, as well as educators looking to clarify concepts related to Hamiltonians and eigenvector calculations.

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Homework Statement


The Hamiltonian of a system has the matrix representation

H=Vo*(1-e , 0 , 0
0 , 1 , e
0 , e , 2)

Write down the eigenvalues and eigenvectors of the unperturbed Hamiltonian (e=0)

Homework Equations


when unperturbed the Hamiltonian will reduce to Vo* the 3x3 matrix with 1,1,2 along the diagonal. the eigenvalues are therefore Vo,Vo,2Vo (right??)

I am a bit confused about how to calculate the eigenvectors. I have tried looking this up but still get confused. Would they not all be zero since if you sub the eigenvalue Vo back into matrix you would get for the first row

Vo(1-Vo,0,0) * (x,y,z) = (0,0,0) where (x,y,z) is a vertical matrix?
 
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What is the problem then? You should know that the 0 vector is the trivial solution, so it is not count for the eigenvector cos it cannot span any solution space.
 
So how would you calculate the other eigenvectors. Sorry I am still confused.
 
for e=0,
H=
(Vo , 0 , 0
0 , Vo , 0
0 , 0 , 2Vo)
eigenvalue is Vo,Vo,2Vo as you said. So what is the standard procedure to find the eigenvector? I assume that you should take at least one linear algebra before you take QM.
 

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