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}


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