3
$\begingroup$

I was able to diagonalise $H=\hbar\omega a^{\dagger}a+\hbar\omega b^{\dagger} b+\hbar(u a^{\dagger}b+u b^{\dagger} a)$ using the Bogoliubov transformation \begin{equation} a=\cos{\alpha}c_1+\sin{\alpha}c_2,\qquad b=-\sin{\alpha}c_1+\cos{\alpha}c_2 \end{equation}

However, I can't seem to work it out for the case of different frequencies, i.e. for

$$ H=\hbar\Omega a^{\dagger}a+\hbar\omega b^{\dagger} b+\hbar(u^{*} a^{\dagger}b+u b^{\dagger} a) $$

I have tried to follow a similar procedure by using the Bogoliubov transformation \begin{equation} c_i=u_ia+v_i b, \qquad c_i^{\dagger}=u_i^{*}a^{\dagger}+v_i^{*}b^{\dagger} \end{equation} for $i=1,2$ with $[c_i,c_j^{\dagger}]=\delta_{ij},[c_i,c_j]=0$. Inverting: $$ \begin{aligned} a &= (v_2 c_1 - v_1 c_2)\frac{1}{u_1 v_2 - u_2 v_1}=(v_2 c_1 - v_1 c_2)\frac{1}{detUV} \\ b &= (-u_2 c_1 + u_1 c_2)\frac{1}{u_1 v_2 - u_2 v_1}=(-u_2 c_1 + u_1 c_2)\frac{1}{detUV} \end{aligned} $$

Then I eventually arrive at

Now substitute these expressions into the Hamiltonian: \begin{equation} \begin{aligned} H/\hbar=\Omega a^{\dagger}a+\omega b^{\dagger} b+u^{*}a^{\dagger}b+u b^{\dagger} a\\ =\frac{1}{|detUV|^2}\\ \Omega(|v_2|^2 c_1^{\dagger}c_1-v_1v_2^{*}c_1^{\dagger}c_2-v_1^{*}v_2c_2^{\dagger}c_1+|v_1|^2c_2^{\dagger}c_2)\\ +\omega(|u_2|^2c_1^{\dagger}c_1-u_1u_2^{*}c_1^{\dagger}c_2-u_1^{*}u_2c_2^{\dagger}c_1+|u_1|^2c_2^{\dagger}c_2)\\ +u^{*}(-u_2v_2^{*}c_1^{\dagger}c_1+u_1v_2^{*}c_1^{\dagger}c_2+u_2v_1^{*}c_2^{\dagger}c_1-u_1v_1^{*}c_2^{\dagger}c_2)+u(-u_2^{*}v_2c_1^{\dagger}c_1+u_2^{*}v_1c_1^{\dagger}c_2+u_1^{*}v_2c_2^{\dagger}c_1-u_1^{*}v_1c_2^{\dagger}c_2)\\ =\frac{1}{|detUV|^2}\\ (\Omega|v_2|^2+\omega|u_2|^2-u^{*}u_2v_2^{*}-uu_2^{*}v_2)c_1^{\dagger}c_1\\ +(\Omega|v_1|^2+\omega|u_1|^2-u^{*}u_1v_1^{*}-uu_1^{*}v_1)c_2^{\dagger}c_2\\ +(-\Omega v_1v_2^{*}-\omega u_1u_2^{*}+u^{*}u_1v_2^{*}+uu_2^{*}v_1)c_1^{\dagger}c_2\\ +(-\Omega v_1^{*}v_2-\omega u_1^{*}u_2+u^{*}u_2v_1^{*}+uu_1^{*}v_2)c_2^{\dagger}c_1 \end{aligned} \end{equation}

Assuming this is correct, I don't know how to kill off the off-diagonal terms.

$\endgroup$

1 Answer 1

6
$\begingroup$

A good start could be to re-write the Hamiltonian as $$ H=\hbar\Omega a^{\dagger}a+\hbar\omega b^{\dagger} b+\hbar(u^{*} a^{\dagger}b+u b^{\dagger} a)= \begin{pmatrix}a^\dagger&b^\dagger\end{pmatrix} \begin{pmatrix}\hbar\Omega&\hbar u^*\\\hbar u&\hbar\omega\end{pmatrix} \begin{pmatrix}a\\b\end{pmatrix} $$ Now you can diagonalize the 2-by-2 matrix and express the new operators in terms of its eigenvectors. E.g., we could introduce transformation matrix $S$, satisfying $S^\dagger=S^{-1}$ a, $\det S=1$, so that the new operators are $$ \begin{pmatrix}a\\ b\end{pmatrix}=S\begin{pmatrix}c_1\\ c_2\end{pmatrix}\Leftrightarrow \begin{pmatrix}c_1\\ c_2\end{pmatrix}=S^\dagger\begin{pmatrix}a\\ b\end{pmatrix} $$ and $$ S^\dagger \begin{pmatrix}\hbar\Omega&\hbar u^*\\\hbar u&\hbar\omega\end{pmatrix} S= \begin{pmatrix}\hbar\omega_1&0\\ 0&\hbar\omega_2\end{pmatrix} $$ Note that simple form $$ S=\begin{pmatrix}\cos\alpha&\sin\alpha\\ -\sin\alpha&\cos\alpha\end{pmatrix} $$ works only for real matrix. In case of complex values of $u,u^*$ one needs to use a more general expression.

Remark: Term Bogoliubov transformation is usually applied to transformation that mixes the annihilation and creation operators, as we do in case of superconductors. In this case I would simply call it just a canonical transformation or transformation of basis/rotation in Hilbert space.

$\endgroup$

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.