Second-order correction to energy in degenerate purturbation

[Haorong Wu]

Homework Statement

If $H^{'}$ is diagonalised, could I treat the problem with non-degenerate purturbation theory?



Here is the problem.



$H_0$ has two eigenvalues of $E_1^0$ and $E_2^0$. The eigenstate of $E_1^0$ is $\psi _1$, and the two eigenstates of $E_2^0$ are $\psi_2$ and $\psi_3$. In the space of $H_0$, $H^{'}=\begin{pmatrix} 0 & b & c \\ b & -a & 0 \\ c & 0 &a \end{pmatrix}$. What is the second-order corrections to the energies?

Relevant Equations

none

Let's focus on the degenerate case.

Since in the subspace of the denegerate eigenvectors, $H^{'}=\begin{pmatrix} -a & 0 \\ 0 &a \end{pmatrix}$ is diagonal, then I treat $\psi _2$ and $\psi _3$ with the undegenerate perturbation theory. And I got
$E_2=E_2^0+\left < 2 \right | H^{'} \left |2 \right > +\frac {{\left | H^{'}_{12} \right | }^2} {E_2^0 - E_1^0} = E_2^0 -a +\frac {b^2} {E_2^0 - E_1^0}$

$E_3=E_3^0+\left < 3 \right | H^{'} \left |3 \right > +\frac {{\left | H^{'}_{13} \right | }^2} {E_3^0 - E_1^0} = E_3^0 +a +\frac {c^2} {E_2^0 - E_1^0}$

Or should I still appy the second-order correction equations for degenerate perturbation?

Such that, I got
$\begin{pmatrix} G_{22} - E_2^2 & G_{23} \\ G_{32} & G_{33} - E_2^2 \end{pmatrix} \begin{pmatrix} c_1 \\ c_2 \end{pmatrix} =0$
where $G_{22}=\frac {{\left | H^{'}_{21} \right | }^2} {E_2^0 - E_1^0}=\frac {b^2} {E_2^0-E_1^0}$, $G_{33}=\frac {{\left | H^{'}_{31} \right | }^2} {E_2^0 - E_1^0}=\frac {c^2} {E_2^0-E_1^0}$, $G_{23}=\frac { H^{'}_{21} H^{'}_{13} } {E_2^0 - E_1^0}=\frac {bc} {E_2^0-E_1^0}$, $G_{32}=\frac { H^{'}_{31} H^{'}_{12} } {E_2^0 - E_1^0}=\frac {bc} {E_2^0-E_1^0}$

Solving the equation then $E_2^2=\frac {b^2+c^2} {E_2^0-E_1^0}$ or $0$.

So $E_2 = E_2^0 -a +\frac {b^2+c^2} {E_2^0-E_1^0}$
$E_3 = E_2^0 +a +\frac {b^2+c^2} {E_2^0-E_1^0}$

I am no sure which one is correct. A textbook reads, if the degenerate eigenvectors are splited with the first-order correction, then I should treat the corrected good states with undegenerate perturbation. Could this apply to this problem?

Thanks!

 

[Dr Transport]

H&#039; is not block diagonal, so you just can't take out a piece of the Hamiltonian and treat it separately. You have to treat the entire problem.

 

[Haorong Wu]

H&#039; is not block diagonal, so you just can't take out a piece of the Hamiltonian and treat it separately. You have to treat the entire problem.

Thanks, Dr. Transport.

I have another problem. In the second solution, I got $E_2^2=\frac {b^2+c^2} {E_2^0-E_1^0}$ or $0$; then how should I decide which one goes to $E_2 = E_2^0 -a + E_2^2$ and the other one goes to $E_3 = E_3^0 +a + E_3^2$? I think that $E_2^2=\frac {b^2+c^2} {E_2^0-E_1^0}$ goes to both $E_2$ and $E_3$ is incorrect.

 

[Dr Transport]

The Hamiltonian contains 3 states, you'll get three solutions, they could be degenerate. I haven't had time to work this problem independently so I can't say anything more.