# Bogoliubov transformation with a slight twist

Given a Hamiltonian of the form

$$H=\sum_k \begin{pmatrix}a_k^\dagger & b_k^\dagger \end{pmatrix} \begin{pmatrix}\omega_0 & \Omega f_k \\ \Omega f_k^* & \omega_0\end{pmatrix} \begin{pmatrix}a_k \\ b_k\end{pmatrix},$$

where $a_k$ and $b_k$ are bosonic annihilation operators, $\omega_0$ and $\Omega$ are real constants and $f_k$ is a complex constant.

How does one diagonalise this with a Bogoliubov transformation? I've seen an excellent answer to a similar Phys.SE question here, but I'm not quite sure how it translates to this example. Any hints or pointers much appreciated.

• Dose similarly assume that $c_k=u_c a_k+v_c b_k$ and $d_k=u_d a_k+v_d b_k$? Mar 11 '14 at 10:57

This is an eigenvalue problem.

Let's assume your Bogoliubov transformation is of the form: $(a_k,b_k)^T=X(c_k,d_k)^T$. What this transformation do is let your Hamiltonian become: $H_k=w_1c_k^\dagger c_k+w_2 d_k^\dagger d_k$, with the anti-commute relation holds for new field operators $c_k$ and $d_k$.

Now you can check that $X$ is just the matrix where its columns are just the Normalized Eigenvectors of your original matrix.

I would just like to point out that the given Hamiltonian does not require a Bogoliubov transformation to be diagonalized, since it is of the form of a single-particle operator (nevertheless in second quantization) i.e. does not contain 'off-diagonal' terms of the form $a a$,...

You can simply diagonalize it by diagonalizing the coupling matrix.

@leongz: although this matrix is also Hermitian for the true Bogoliubov case, you will generally get the wrong answer for the eigenenergies and modes if you diagonalize it. The resulting modes would not be bosonic, i.e. it would not be a canonical transformation. You can obtain the right answer (which is much more powerful than the typical ansatz for the Bogoliubov operators) by diagonalizing $\Sigma H$, where $\Sigma$ is the pseudonorm on the sympletic space you're working on. Note however, that this matrix is not always Hermitian (and not always diagonlaizable - but this is physical: one bosonic mode is missing for each Goldstone mode).

Hamiltonian is already diagonalized by momentum. You need to define new Bose-operators
$c_k = u_k a_k + v_k b_k \\ d_k = w_k a_k+x_k b_k$
This is general form, with some complex constants $u_k, v_k, w_k, x_k$ for each $k$ independently. There are also $c^+_k$ and $d^+_k$, conjugated with previous one. Now you need $c_k$ and $d_k$ correspond to some quasi-particles, so
$[c_k, c_k^+] = 1 \\ [d_k, d_k^+] = 1 \\ \text{(all other commute to zero)}$
This equation give you some constraint on constants $u_k, v_k, w_k, x_k$. But to find them definitivly, you must substitute them to hamiltonian. After that, you must obtain
$H = \sum_k C_1 c^+_k c_k + C_2 d^+_k d_k + C_3 c^+_k d_k + C_4 d^+_k d_k.$
Constants $C_1, C_2, C_3, C_4$ derived from $\omega_0, \Omega, f_k$ and $u_k, v_k, w_k, x_k$. You must then solve $C_3 = 0, C4 = 0$ equations to obtain $u_k, v_k, w_k, x_k$. Then you'll get
$H = \sum_k C_1 c^+_k c_k + C_2 d^+_k d_k,$
with found $C_1, C_2$. That completes diagonalization.

• I would add that the diagonalization is always possible because the matrix is Hermitian. Mar 11 '14 at 21:09

The Hamiltonian can be written as

$\sum_k \psi^\dagger M \psi$

where $\psi=\begin{pmatrix}a_k \\ b_k\end{pmatrix}$ and $M=\begin{pmatrix}\omega_0 & \Omega f_k^* \\ \Omega f_k & \omega_0\end{pmatrix}$.

We introduce a new set of operators $\phi=\begin{pmatrix}c_k \\ d_k\end{pmatrix}$, via $\psi=U \phi$ where $U$ is neccesarily a 2x2 matrix. This gives us

$\psi^\dagger M \psi = \phi^\dagger N \phi$

where $N = U^\dagger M U$. We wish for this new form of the Hamiltonian to be diagonal. aka we wish for the matrix $N$ to be diagonal. As per the standard process of diagonalising a matrix, a matrix $M$ is diagonalised by $M \rightarrow U^\dagger M U$ where $U$ is the matrix with the eigenvectors of $M$ as its columns.

Therefore, first we find the eigenvectors of $M$, substitute those as columns into a 2x2 matrix $U$, diagonalise $M$ so that $N=U^\dagger M U$, then our diagonalised Hamiltonian is

$H=\sum_k\phi^\dagger N \phi$

where $\phi=U^{-1} \psi$.

Thanks to @luming and @Vladimir for the pointers.