Question

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.

Short answer

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

Step-by-step solution

Significance of Schrodinger’s equation

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$\hat{H}={\hat{H}}^0+\lambda \hat{H}$

Determination of the eigenvalues

Write equation 6.17.

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

Write equation 6.22.

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

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.