- #1

Haorong Wu

- 417

- 90

- 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!

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!