Correction to the Eigenvalue -- doubly degenerate case

[Imperatore]

Determine the corerctions to the eigenvalue in the first approximation and the correct functions in the zeroth approximation, for a doubly degenerate level.

The solution:
Equation $$\left| V_{nn'}-E^{(1)}\delta_{nn'}\right|=0$$ has here the form

$$\left|\begin{array}{ccc}V_{11}-E^{(1)}&V_{21}\\V_{12}&V_{22}-E^{(1)}\end{array}\right|=0$$

Solving, we find:

$$E^{(1)}=\frac{1}{2}\left[ V_{11}+V_{22} \pm h\omega^{(1)}\right]$$

Solving also equation $$\sum_{n'}(V-{nn'}-E^{(1)}\delta_{nn'})c_{n'}^{(0)}$$ with these values of $$E^{(1)}$$, we obtain for the coefficients in the correct normalized function in the zeroth approximation , $$\psi^{(0)}=c_{1}^{(0)}\psi_{1}^{(0)}+c_{2}^{(0)}\psi_{2)^{(0)}$$$$c_{1}^{(0)}=\left\{ \frac{V _{12} }{2\left| V _{12} \right| } \left[ 1 \pm \frac{V _{11}-V _{22} }{h\omega ^{(1)} } \right] \right\} ^{ \frac{1}{2} }$$

$$c_{2}^{(0)}= \pm \left\{ \frac{V _{21} }{2\left| V _{12} \right| } \left[ 1 \mp \frac{V _{11}-V _{22} }{h\omega ^{(1)} } \right] \right\} ^{ \frac{1}{2} }$$

The first equation I know, the task is just to solve the problem $$(V _{11}-E ^{(1)})(V _{22}-E ^{(1)})-V _{21}V _{12}=0$$ ut why there is$$\left| V _{12} \right| ^{2}$$ factor in the final formula ? What it represents ? And the most important, how to derive the formulas for these coefficients $$c_{1}^{(0)}$$$$c_{2}^{(0)}$$ ?

 

[eljose79]

The factor \left| V _{12} \right| ^{2} represents the magnitude of the off-diagonal element in the Hamiltonian matrix, which is important in determining the coefficients c_{1}^{(0)} and c_{2}^{(0)}. This is because the off-diagonal element represents the coupling between the two degenerate states, and its magnitude will affect the relative contribution of each state to the overall wavefunction.

To derive the formulas for c_{1}^{(0)} and c_{2}^{(0)}, we can use the orthogonality condition of the eigenfunctions:

\int\psi_{1}^{(0)*}\psi_{2}^{(0)}dx=0

Using the definition of the coefficients c_{1}^{(0)} and c_{2}^{(0)}, we can rewrite the above equation as:

c_{1}^{(0)*}\int\psi_{1}^{(0)*}\psi_{1}^{(0)}dx+c_{2}^{(0)*}\int\psi_{1}^{(0)*}\psi_{2}^{(0)}dx=0

Since the eigenfunctions are normalized, the integrals in the above equation will be equal to 1, so we can simplify it to:

c_{1}^{(0)*}+c_{2}^{(0)*}\int\psi_{1}^{(0)*}\psi_{2}^{(0)}dx=0

Substituting the expression for \psi^{(0)} from the given equation, we get:

c_{1}^{(0)*}+c_{2}^{(0)*}\int c_{1}^{(0)*}\psi_{1}^{(0)*}\psi_{2}^{(0)}dx+c_{2}^{(0)*}\int c_{2}^{(0)*}\psi_{2}^{(0)*}\psi_{2}^{(0)}dx=0

Using the orthonormality condition again, we can simplify this to:

c_{1}^{(0)*}+c_{2}^{(0)*}\int c_{1}^{(0)*}\psi_{1}^{(0)*}\psi_{2}^{(0)}dx+c_{2}^{(0)*}=0

Solving for c_{1