Question: In the text I asserted that the first-order corrections to an n-fold degenerate energy are the eigen values of the Wmatrix, and I justified this claim as the "natural" generalization of the case n = 2.

Prove it, by reproducing the steps in Section 6.2.1, starting with

${\psi }^0=\sum _{j=1}^n{\alpha }_j{\psi }_j^0$

(generalizing Equation 6.17), and ending by showing that the analog to Equation6.22 can be interpreted as the eigen value equation for the matrix W.

The Schrodinger wave equation is a linear partial differential equation that governs the wave function of the Quantum Mechanical System.

The following can be interpreted from the equation.

$\hat{H}\psi =E\psi$and ${\hat{H}}^0{\psi }^0=E^0{\psi }^0$and$\hat{H}={\hat{H}}^0+\lambda \hat{H}$

Write equation 6.17.

${\psi }^0=\alpha {\psi }_a^0+\beta {\psi }_b^0$

$\alpha W_{aa}+\beta W_{ab}=\alpha E^1$

It is known that $E_n=E_n^6+\lambda E_n^1+{\lambda }^2{E}_n^2+...$and $\psi ={\psi }^0+\lambda {\psi }^1+{\lambda }^2{\psi }^2+...$and ${\psi }^0{=}_{j=1}^n{\alpha }_f{\psi }_j^0$

Apply Schrodinger's equation.

$$\begin{gathered} \left({\hat{H}}^0+\lambda \hat{H}\right)=\left[{\psi }^0+\lambda {\psi }^1+{\lambda }^2{\psi }^2+...\right]=(E_n^0+\lambda E_n^1+{\lambda }^2{E}_n^2+...)\left[{\psi }^0+\lambda {\psi }^1+{\lambda }^2{\psi }^2+...\right] \\ ={\hat{H}}^0{\psi }^0+\lambda \left[{\hat{H}}^0{\psi }^1+{\hat{H}}^{\text{'}}{\psi }^0\right]+{\lambda }^2\left[{\hat{H}}^0{\psi }^2+{\hat{H}}^{\text{'}}{\psi }^1+...\right] \\ =E_n^0{\psi }^0+\lambda \left[E_n^0{\psi }^0+E_n^1{\psi }^0\right]+{\lambda }^2\left[E_n^0{\psi }^2+E_n^1{\psi }^1+E_n^2{\psi }^0\right]+... \end{gathered}$$

Obtain the first-order correction for the energy eigen values by equating the coefficients of the same order.

$=H^0{\psi }^1+{\hat{H}}^{\text{'}}{\psi }^0=E_n^0{\psi }^1+E^1{\psi }^0$

Multiply both sides by $\psi \_j^{0*}$.

It is known that${\psi }^0=\sum _{j=1}^n{\alpha }_f{\psi }_j^0$ and$\left({\psi }^0\overline{){\hat{H}}^0=E_n^{0*}}\right({\psi }^0\overline{),{\hat{H}}^{\text{'}}={\hat{H}}^{0*}}$ , . Then the

role="math" localid="1658211098621" $$\begin{gathered} E_n^0=E_n^0 \\ \left({\psi }^0\overline{){\hat{H}}^0=E_n^{0*}}\right({\psi }^0\overline{)} \end{gathered}$$

Write the generalized form of the equation.

Thus, the analog function can be interpreted as the eigenvalue equation for the matrix W.