Decoupling theory by diagonalising the Hamiltonian

Question

I have a Hamiltonian of the form
$H = 2k(alpha alpha^* -beta beta^*) -2lambda (alphabeta^* + beta alpha^* )$
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.

ZeroTheHero · Answer

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}
Mcdot 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".

Decoupling theory by diagonalising the Hamiltonian

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