How do I diagonalize the following BCS (Bardeen-Cooper-Schrieffer) Hubbard Hamiltonian: \begin{equation} H= \sum\limits_{k \in [-\frac{π}{2}, +\frac{π}{2}[} \begin{bmatrix}c^\dagger_k & c^\dagger_{k+\pi} \end{bmatrix}\hat{\cal H}\begin{bmatrix}c_k \\ c_{k+\pi} \end{bmatrix} \space\space\space\space\space with\space\space\space\space\space \hat{\cal H}= \begin{bmatrix} \epsilon_k & v \\ v & -\epsilon_k \end{bmatrix} \end{equation} using the following Bogoliubov transformation: \begin{equation} a^\dagger_k=u_k c^\dagger_k + v_k c^\dagger_{k+\pi}\\ b^\dagger_k=v_k c^\dagger_k - u_k c^\dagger_{k+\pi}\\ with \space\space\space\space u^2+v^2=1 \end{equation} The result should be \begin{equation} H= \sum\limits_{k \in [-\frac{π}{2}, +\frac{π}{2}[} E_k (b^\dagger_{k}b_{k}-a^\dagger_{k}a_{k}) \space\space\space with \space\space\space E_k=\sqrt{\epsilon_k^2+v^2}\end{equation}

I tried to reverse the Bogoliubov transformations to find the transformation for $c_k$, $c^\dagger_k$,$c_{k+\pi}$, $c^\dagger_{k+\pi}$ and insert it, but this gives only a sum of 16 terms involving linear combinations of $a^\dagger_ka_k,b^\dagger_kb_k,a^\dagger_kb_k,b^\dagger_ka_k$. And the terms $\epsilon_k$ and v only show up linearly, so I don't see how a term like $\sqrt{\epsilon_k^2+v^2}$ should show up.

Thank you in advance


1 Answer 1


At first, you should recognize that the eigenvalues of Hamiltonian is nothing but $\sqrt{\epsilon_k^2+\nu^2}$, according to $|\lambda I-H|=0$.

Then, you can find eigenvectors expressed by $c_k$ and $c_{k+\pi}$ also $c_k^\dagger$ and $c_{k+\pi}^\dagger$, which is nothing but the Bogoliubov transformation.

I believe that above procedure helps you to understand Bogoliubov transformation.

Usually in textbook, people write down Bogoliubov transformation without any explanation, just some mathematical expression. So you don't know why you should mix $c_k$ and $c_{k+\pi}$ also $c_k^\dagger$ and $c_{k+\pi}^\dagger$, and what will do next step.

For your case, you have expressed Hamiltonian by $a$ and $b$, then you should find proper values of $u$ and $v$ to make Hamiltonian satisfy the diagonal form, in which equations are not difficult to solve.

  • $\begingroup$ Can you be more explicit. I have problems diagonalizing an operator in second quantization formalism. (Finding the eigenvalues is clear to me, but the rest not) $\endgroup$
    – QMphysics
    Commented Apr 16, 2018 at 17:12
  • $\begingroup$ @QMphysics You can find detailed calculation in some lecture note: portal.ifi.unicamp.br/images/files/graduacao/aulas-on-line/… or lassp.cornell.edu/clh/Book-sample/7.3.pdf. I don't know why that physics people like finding proper value to make Hamiltonian satisfy some diagonal form, instead of diagonalize it via linear algebra technique. Your question is indeed based on linear algebra, from which you learned how to find eigenvalues and eigenvectors of a matrix. You should review it. $\endgroup$
    – qfzklm
    Commented Apr 17, 2018 at 3:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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