# Time independent perturbation (Quantum Mechanics)

1. Sep 11, 2013

### Telemachus

Hi there. I have to find the energy corrections through the perturbation method, and then give the exact result for the hamiltonian:

$H= \begin{pmatrix} E_A & \epsilon & \epsilon & \epsilon \\ \epsilon & E_B & 0 & 0 \\ \epsilon & 0 & E_B & 0\\ \epsilon & 0 & 0 & E_B \\ \end{pmatrix}$

So I've got:

$H=H_0+W= \begin{pmatrix} E_A & 0 & 0 & 0 \\ 0 & E_B& 0 & 0 \\ 0 & 0 & E_B & 0\\ 0 & 0 & 0 & E_B \\ \end{pmatrix} + \begin{pmatrix}0 & \epsilon & \epsilon & \epsilon \\ \epsilon & 0 & 0 & 0 \\ \epsilon & 0 & 0 & 0\\ \epsilon & 0 & 0 & 0 \\ \end{pmatrix}$

So, perturbation theory gives for the non degenerate energy:
$E_A(\epsilon)=E_A+ \left < A \right | W \left | A \right > + \sum_{i=1}^3 \displaystyle \frac{| \left < B_i \right | W \left | A \right > |^2}{E_A-E_B}=E_A+\frac{3\epsilon^2}{E_A-E_B}$

Now, the matrix $\hat W^n$ corresponding to $W$ inside the eigensubspace generated by the B states gives me zero:
$\hat W^n=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0& 0 \\ 0 & 0 & 0 \\ \end{pmatrix}$

So I think there'll be no corrections in the $E_B$ energies. I'm not sure if this is right.

On the other side, I've tried to diagonalize the Hamiltonian, to get the exact values for the energies. But I've found some trouble. Basically I didn't get the energy that I've found before, $E_A(\epsilon)$ as one of the roots for the determinant of the Hamiltonian. I don't know if I've made a mistake at any step, or if I'm not getting this right.

This is what I have:

$H= \begin{pmatrix} E_A & \epsilon & \epsilon & \epsilon \\ \epsilon & E_B & 0 & 0 \\ \epsilon & 0 & E_B & 0\\ \epsilon & 0 & 0 & E_B \\ \end{pmatrix}$

So when I try to diagonalize this Hamiltonian, I get to this secular equation:

$-\epsilon^2(E_B-E_n(\epsilon))^2+(E_B-E_n(\epsilon))[(E_A-E_n(\epsilon))(E_B-E_n(\epsilon))^2-2\epsilon^2(E_B-E_n(\epsilon))]= \\ (E_B-E_n(\epsilon))^2[3\epsilon^2+(E_B-E_n(\epsilon))(E_A-E_n(\epsilon))]=0$

So I have a double root $E_{1,2}(\epsilon)=E_B$, and then I solve the quadratic equation:

$3\epsilon^2+(E_B-E_n(\epsilon))(E_A-E_n(\epsilon))=3\epsilon^2+E_BE_A-En(\epsilon)(E_B+E_A)+E(\epsilon)_n^2=0 \: \: \displaystyle \left ( 1 \right )$

Then I get: $E_{3,4}(\epsilon)=\displaystyle\frac{E_A+E_B±\sqrt{E_B^2+E_A^2-2E_AE_B -12\epsilon^2} }{2}=E_{3,4}(\epsilon)=\displaystyle\frac{E_A+E_B±\sqrt{(E_B-E_A)^2-12\epsilon^2} }{2}$

And this doesn't look like what I've found when using time independent perturbation theory.

Last edited: Sep 11, 2013
2. Sep 11, 2013

### Telemachus

After thinking of this for a while I came to thought that perhaps I should try expanding the square root on $E_{3,4}$ in Taylor series. I'll try with that.

Any help, or suggestion will be appreciated.

Alright, I've found $E_A$ through the taylor expansion. And the correction for one of the $E_B$ which tells me that what I did before was wrong, when assuming there were no corrections on any $E_B$, now I need to know how to get this right through the perturbation method. I usually get this through the diagonalization of the submatrix corresponding to the eigensubspace spanned by the degenerate states. But in this case in which that matrix is zero I'm not sure what I have to do. I'll see what I can do.

Ok, it's done. It was pretty obvious, I had to use perturbation method for degenerate case to second order in the energies.

From $\displaystyle \left ( 1 \right )$ its clear: $E_{1,2}=E_B$

Also:
$If \: \epsilon<<E_A: \sqrt{(E_B-E_A)^2-12\epsilon^2}\approx E_B-E_A-\frac{6\epsilon^2}{E_B-E_A}$

Then
$E_{3}(\epsilon)=E_B-\frac{3\epsilon^2}{E_B-E_A}, \: \: E_{4}(\epsilon)=E_A+\frac{3\epsilon^2}{E_A-E_B}$

Last edited: Sep 11, 2013