Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Step 1

Let me formulate the problem to convey my notation. I have a matrix $A$ which is hermitian - and is diagonalisable by a transformation $$ U_A A\,\,U_A^{-1} = A_{diag}$$

Now the matrix is changed, using the small parameter $\lambda$. Therefore, $$ A \rightarrow A-\lambda B $$ where $B$ can be not Hermitian, in general.

I tried to calculate the first order changes in eigenvalues and eigenvectors the following way -

Step 2

The matrix $A-\lambda B $ has to be rotated by $U^\prime = e^{i\lambda \alpha}U_A$ to be diagonalized, $\alpha$ being a generator of rotation - and it makes sense that the "extra" rotation has to be proportional to $\lambda$.

$$ A^\prime_{diag} = A_{diag} + C_1\lambda + C_2\lambda^2 + \mathcal{O}(\lambda^3) $$

I wrote this down using the BCH formula and set the linear coefficient, $C_1= 0$ and got the constraint $U_AB\;U_A^{-1} = \left[i\alpha,A_{diag}\right]$.

The final form I obtained for $A^\prime_{diag}$ was $$ A^\prime_{diag} = A_{diag} -\frac{1}{2}\left[i\alpha,U_AB\;U_A^{-1}\right]\lambda^2 + \mathcal{O}(\lambda^3)$$

Question

Can anyone tell me if -

  1. I am on the right track - and how to proceed from here to solve for $\alpha$ in terms of the known matrices.

  2. I need to approach this differently.

  3. it would help if $B$ were Hermitian?

EDIT

This is exactly the problem of doing perturbation theory using the SW transformation. Can anybody cite a reference?

share|improve this question
add comment

1 Answer 1

The Schriffer Wolff transformation was developed to remove the effect of the effect of the perturbation term to the first order by performing a similarity transformation on the hamiltonian. $\tilde{H}=SHS^{-1}$. I am giving you a paper where it was first developed...

http://link.aps.org/doi/10.1103/PhysRev.149.491

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.