Given a matrix $H$ we can find it eigen value on a given basis set $|\phi_i\rangle$. Suppose the eigen vector is

$|\psi \rangle = \sum_i \alpha_i |\phi_i\rangle$

Put in the eigen equation

$H|\psi\rangle = \epsilon |\psi\rangle \implies \sum_i H|\phi_i\rangle \alpha_i = \epsilon \sum_i |\phi_i\rangle \alpha_i$

We can act $\langle \phi_k|$ on the left, but in general, the basis set are not orthogonal.

$H_{ij} \alpha_j = \epsilon S_{ij} \alpha_j \to H\cdot \alpha = \epsilon S\cdot \alpha$

$H_{ij} = \langle \phi_i|H|\phi_j\rangle , S_{ij} = \langle \phi_i|\phi_j\rangle$

This is the General Eigen Value Problem.

One way to solve the problem, is the “reconfigure” the basis so that it is orthogonal. However, in computation, non-orthogonal basis could give supreme advantage. So, the other way is split the problem. First solve the $S\cdot v = \sigma v$,

The eigen system of $S$ is

$S = P^T \cdot \sigma \cdot P, P\cdot P^T = P^T \cdot P = I$

Here, $\sigma$ is a diagonal matrix of eigen value. Now, we define a new non-unitary matrix

$\displaystyle R_{ij} = \frac{P_{ij}}{\sqrt{\sigma_j}}$

Notices that $R^T \neq R^{-1}$

Thus,

$R^T \cdot S\cdot R = I \implies S = R^{-T} \cdot R^{-1}$

We know that, the form $R^T \cdot Q \cdot R$ is a transform that from one basis to another basis, i.e.

$|\hat{\phi}_i \rangle = R|\phi_i\rangle$

$|\psi \rangle = \sum_i \alpha_i |\phi_i\rangle = \sum_i \alpha R^{-1} |\hat{\phi}_i\rangle = \sum_i \beta_i |\hat{\phi}_i \rangle \implies \alpha = R\cdot \beta$

and for any operator,

$\hat{Q} = R^T\cdot Q\cdot R$

We put this back to the general problem

$H\cdot\alpha = \epsilon S\cdot \alpha = \epsilon R^{-T} \cdot R^{-1} \alpha$

$\implies R^T\cdot H \cdot \alpha = \epsilon R^{-1} \alpha$

$\implies \hat{H}\cdot \beta = R^T\cdot H \cdot R \cdot \beta = \epsilon \beta$

Thus, we can solve the $\hat{H}$, get the eigen system, then use $R\cdot \beta = \alpha$

For example,

$\displaystyle \phi_1 = (1,0), \phi_2 = \frac{1}{\sqrt{2}}(1,1)$

$S = \begin{pmatrix} 1 & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & 1 \end{pmatrix}$

The eigen system is

$P = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}$

$\sigma = \begin{pmatrix} 1 + \frac{1}{\sqrt{2}} & 0 \\ 0 & 1 - \frac{1}{\sqrt{2}} \end{pmatrix}$

The matrix $R$ is

$R = \begin{pmatrix} \frac{1}{\sqrt{2+\sqrt{2}}} & -\frac{1}{\sqrt{2-\sqrt{2}}} \\ \frac{1}{\sqrt{2+\sqrt{2}}} & \frac{1}{\sqrt{2-\sqrt{2}}} \end{pmatrix}$

We can verify that

$R^T \cdot S \cdot R = I$

$R^{-T} \cdot R^{-1} = S$

$R^T \cdot R = \begin{pmatrix} \frac{1}{2}(3-\sqrt{5}) & 0 \\ 0 & \frac{1}{2}(3+\sqrt{5}) \end{pmatrix}$

$R \cdot R^T = \begin{pmatrix} 2 & -1 \\ -1 & 1 \end{pmatrix}$

Let a Hamilton is

$H = \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}$

The Hamiltonian in the basis is

$H_S = [H_S]_{ij} = \phi_i \cdot H\cdot \phi_j = \begin{pmatrix} 2 & 2 \\ 2 & 3 \end{pmatrix}$

The eigen values in the S frame are

$\displaystyle \sigma_s = \frac{1}{2}(5 \pm \sqrt{17})$

which is not the correct eigen values.

Now, we transform the Hamiltonian into orthogonal basis

$\displaystyle H_R = R^T \cdot H_S \cdot R = \begin{pmatrix} \frac{15+\sqrt{5}}{10} & -\frac{1}{\sqrt{5}} \\ -\frac{1}{\sqrt{5}} & \frac{3}{2} - \frac{1}{2\sqrt{5}} \end{pmatrix}$

The eigen values are

$\sigma_\pm = 2, 1$

where are the correct pair. The un-normalized eigen vector are

$\displaystyle \beta_{\pm} = \left(\frac{1}{2}(-1 \pm \sqrt{5}), 1\right)$

We can verify that

$\alpha_{\pm} = R\cdot \beta_{\pm}$

$H_S \cdot \alpha_\pm = \sigma_\pm S\cdot \alpha_\pm$