Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 2 / 35B / ii / Solution

For
$$
H=\begin{pmatrix}E_-&-i\lambda\\i\lambda&E_+\end{pmatrix},
\qquad \Delta=E_+-E_->0,
$$
the exact energies are
$$
E_{\pm}^{\rm exact}
=\frac{E_-+E_+}{2}
\pm\sqrt{\frac{\Delta^2}{4}+\lambda^2}.
$$
The perturbation series around $\lambda=0$ has radius $\Delta/2$, set by the branch points $\lambda=\pm i\Delta/2$.


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