Matrix Diagonalisation#

The matrix \(A\) in equation (12.9) can be written in a different basis such that it has only nonzero diagonal elements. This is called matrix diagonalisation. It is a type of similarity transform. See appendix sec:similarity for more on similarity transforms. Matrix diagonalisation is a similarity transform that takes the unit vectors \(\hat{\vec{x}}\) and \(\hat{\vec{y}}\) (i.e. the basis vectors in which \(A\) was written) and turns them into the two eigenvectors. The matrix that does this similarity transform can be constructed by taking the column vector eigenvectors to give \(S = (\vec{u}_+ ~ \vec{u}_-)\), i.e.

\[\begin{split}\begin{aligned} S = \begin{pmatrix} 3 & 3 \\ 3+\sqrt{15} & 3-\sqrt{15} \end{pmatrix}\, . \end{aligned}\end{split}\]

This is takes the basis states to the eigenvectors. The eigenvalue equation can now be written in the form

\[AS=SA^{\prime}\]

where \(A^{\prime}\) is the diagonal matrix given in equation eq:diagonal. Therefore to calculate the diagonal matrix we have

\[A^{\prime}=S^{-1}AS\]

so we need to calculate the inverse of \(S\) (see section Inverse of a matrix). In our example this is

\[\begin{split}\begin{aligned} S^{-1} = \begin{pmatrix} \frac{-3+\sqrt{15}}{6\sqrt{15}} & \frac{1}{2\sqrt{15}} \\ \frac{3+\sqrt{15}}{6\sqrt{15}}& -\frac{1}{2\sqrt{15}} \end{pmatrix}\, . \end{aligned}\end{split}\]

We can then verify that

\[\begin{split}\begin{aligned} S^{-1}AS = \begin{pmatrix} \frac{-3+\sqrt{15}}{6\sqrt{15}} & \frac{1}{2\sqrt{15}} \\ \frac{3+\sqrt{15}}{6\sqrt{15}}& -\frac{1}{2\sqrt{15}} \end{pmatrix} \begin{pmatrix} 1 & 3 \\ 2 & 7 \end{pmatrix} \begin{pmatrix} 3 & 3 \\ 3+\sqrt{15} & 3-\sqrt{15} \end{pmatrix} = \begin{pmatrix} 4+\sqrt{15} & 0 \\ 0 & 4-\sqrt{15} \end{pmatrix} \end{aligned}\end{split}\]

as expected.

In summary the general rule for matrix diagonalisation is to create the similarity transform matrix \(S\) by collecting the (column) eigenvectors of \(A\), and calculating \(S^{-1} A S\). eigenvectors are orthonormal, the similarity transform is unitary, then \(S^{-1} = S^\dagger\) and \(A' = S^\dagger AS\).