Hamiltonian matrix - Eigenvectors

  • #1
87
1

Main Question or Discussion Point

Hello everybody,

From a complete set of orthogonal basis vector ##|i\rangle## ##\in## Hilbert space (##i## = ##1## to ##n##), I construct and obtain a nondiagonal Hamiltonian matrix
$$
\left( \begin{array}{cccccc}
\langle1|H|1\rangle & \langle1|H|2\rangle & . &. &.& \langle1|H|n\rangle \\
\langle2|H|1\rangle & . & . &. &.&. \\
. & . & . &. &.&. \\
. & . & . &. &.&. \\
. & . & . &. &.&. \\
. & . &. &. &.& \langle n|H|n\rangle \end{array} \right)
$$

After diagonalization, I obtain diagonal ##n\times n## matrix that represent the eigenvalues of the Hamiltonian, and another ##n\times n## matrix ##V## composed of scalars that represent the eigenvectors of the same Hamiltonian.

My question is, what is the link between the scalar matrix ##V## and the complete set of orthogonal basis vector ##|i\rangle## that I choose in the beginning ?

Thank you very much everybody.

Konte
 

Answers and Replies

  • #2
BvU
Science Advisor
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As far as I know the ##\left | {i} \right \rangle ## are the base vectors for ##V##.

so ##V_{ni}## are the coefficients ##\left\langle e_n \middle | i \right \rangle## ; in other words:

Column n of ##V## are the coordinates of eigenvector ##e_n## on the basis ##\left | {i} \right \rangle ##
 
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Likes Konte and extranjero
  • #3
87
1
Column n of ##V## are the coordinates of eigenvector ##e_n## on the basis ##\left | {i} \right \rangle ##
Thanks.

Konte
 

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