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Commit 2f9ad78f authored by Maciej Topyla's avatar Maciej Topyla
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Update src/6_eigenvectors_QM.md

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|\psi_2\rangle = \begin{pmatrix} \psi_{2,1} \\ \psi_{2,2} \\ \psi_{2,3} \\ \vdots \end{pmatrix} \,, \quad \ldots \, , |\psi_n\rangle = \begin{pmatrix} \psi_{n,1} \\ \psi_{n,2} \\ \psi_{n,3} \\ \vdots \end{pmatrix} \, .
$$
An important property of eigenvalue equations is that if you have two eigenvectors
$ |\psi_i\rangle$ and $ |\psi_j\rangle$ that have associated *different* eigenvalues,
$\lambda_{\psi_i} \ne \lambda_{\psi_j} $, then these two eigenvectors are orthogonal to each
other, that is
$$
\langle \psi_j | \psi_i\rangle =0 \, \quad {\rm for} \quad {i \ne j} \, .
$$
This property is extremely important, since it suggest that we could use the eigenvectors
of an eigenvalue equation as a *set of basis elements* for this Hilbert space.
!!! tip "Orthogonality of eigenvectors"
An important property of eigenvalue equations is that if you have two eigenvectors
$ |\psi_i\rangle$ and $ |\psi_j\rangle$ that have associated *different* eigenvalues,
$\lambda_{\psi_i} \ne \lambda_{\psi_j} $, then these two eigenvectors are orthogonal to each
other, that is
$$
\langle \psi_j | \psi_i\rangle =0 \, \quad {\rm for} \quad {i \ne j} \, .
$$
This property is extremely important, since it suggest that we could use the eigenvectors
of an eigenvalue equation as a *set of basis elements* for this Hilbert space.
Recall from the discussions of eigenvalue equations in linear algebra that
the eigenvectors $|\psi_i\rangle$ are defined *up to an overall normalisation constant*. Clearly, if $|\psi_i\rangle$ is a solution of $\hat{A}|\psi_i\rangle = \lambda_{\psi_i}|\psi_i\rangle$
......
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