# Decoupling theory by diagonalising the Hamiltonian

I have a Hamiltonian of the form

$$H = 2k(\alpha \alpha^* -\beta \beta^*) -2\lambda (\alpha\beta^* + \beta \alpha^* )$$

and I'd like to decouple the $$\alpha$$'s and $$\beta$$'s if possible. I know I need to diagonalise the Hamiltonian, like this:

$$H = (\alpha, \alpha^*, \beta, \beta^* ) \begin{pmatrix} 0& k & 0 &-\lambda \\ k & 0 & -\lambda & 0 \\ 0 & -\lambda & 0 & -k\\ -\lambda &0 &-k &0 \end{pmatrix} \begin{pmatrix} \alpha \\ \alpha^* \\ \beta\\ \beta^* \end{pmatrix}$$

Then use $$H = PDP^{-1}$$. I can use Mathematica to find the diagonal matrix and $$P$$ but then I'm stuck how to use this to decouple the $$\alpha$$'s and $$\beta$$'s. I end up with

$$H =(\alpha, \alpha^*, \beta, \beta^* ) P D P^{-1} \begin{pmatrix} \alpha \\ \alpha^* \\ \beta\\ \beta^* \end{pmatrix}$$,

but of course multiplying this out gets me back where I started. I think I'm misunderstanding something in the process. Any help is much appreciated.

• Let $\left(\begin{array}{c} a\\a^*\\b\\b^*\end{array} \right) = P^{-1}\left(\begin{array}{c} \alpha\\\alpha^*\\\beta\\\beta^*\end{array} \right)$ and your Hamiltonian is diagonal in the $a$ and $b$ operators – By Symmetry Jul 23 at 10:58

Write \begin{align} H=(\alpha,\alpha^*,\beta,\beta^*)M (\alpha,\alpha^*,\beta,\beta^*)^\top \end{align}
You need first to find the matrix $$P$$ that will diagonalize your $$M$$ so that \begin{align} P^{-1}MP=D\, ,\qquad \Rightarrow \qquad M=P D P^{-1} \end{align} where $$D$$ is diagonal and contains the eigenvalues of $$M$$ on the diagonal.
Next, you go to a new basis $$P^{-1}(\alpha,\alpha^*,\beta,\beta^*)^\top= (a,a^*,b,b^*)^\top$$ so that \begin{align} H&=(\alpha,\alpha^*,\beta,\beta^*)PDP^{-1} (\alpha,\alpha^*,\beta,\beta^*)^\top \nonumber \\ &= (a,a^*,b,b^*)^\top D (a,a^*,b,b^*)^\top \end{align} will be diagonal in the new variables.
The matrix $$P$$ is obtained by arranging the orthonormal eigenvectors of $$M$$ in columns, and you need the eigenvalues of $$M$$ to get the eigenvectors. Luckily for you, notice that \begin{align} M\cdot M= (k^2+\lambda^2)\mathbb{I} \end{align} so the eigenvalues of $$M$$ are $$\pm \sqrt{k^2+\lambda^2}$$, each eigenvalue occurring twice. The job of finding each eigenvector is systematic, with the twist that you will need to make sure that both eigenvectors with eigenvalue $$+\sqrt{k^2+\lambda^2}$$ are orthogonal to each other, and both eigenvectors with eigenvalue $$-\sqrt{k^2+\lambda^2}$$ are orthogonal to each other as orthogonality of eigenvectors corresponding with the same eigenvalue must be enforced "by hand".