Today I Learned
Interesting Facts on Eigenstates
The eigenstates of a matrix
\[\begin{pmatrix} E_1 & 0 \\ 0 & E_1 \end{pmatrix}\]are
\[\begin{pmatrix} 1 \\ 0 \end{pmatrix}, \qquad \begin{pmatrix} 1 \\ 0 \end{pmatrix}.\]However if we put in two real off-diagonal elements
\[\begin{pmatrix} E_1 & \epsilon \\ \epsilon & E_1 \end{pmatrix},\]the eigenstates becomes
\[\begin{pmatrix} 1 \\ -1 \end{pmatrix}, \qquad \begin{pmatrix} 1 \\ 1 \end{pmatrix}.\]If we are talking about real matrix,
\[\begin{pmatrix} E_1 & 0 \\ 0 & E_2 \end{pmatrix}\]even though it is true that the eigenstates are
\[\begin{pmatrix} 1 \\ 0 \end{pmatrix}, \qquad \begin{pmatrix} 1 \\ 0 \end{pmatrix},\]the eigenstates are not that simple if we add in off-diagonal elements that make sure the matrix is still Hermitian.
In fact the eigenstates for the eigenequation
\[\begin{pmatrix} E_1 & \epsilon \\ \epsilon^* & E_2 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \lambda \begin{pmatrix} a \\ b \end{pmatrix},\]is given by
\[(E_1-E_2) a b + \epsilon b^2 - \epsilon^* a^2 = 0,\]where $b^2 = b \cdot b$ is not the modulus squared.