# 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.

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