- #1

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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.

Help please :)

##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.

Help please :)

Last edited: