TITLE: Eigenvalue perturbation theory via Feynman diagrams
QUESTION [12 upvotes]: Suppose I have a matrix given by a sum
$$A=D+\epsilon B$$
where $D$ is diagonal and $\epsilon$ is small, and I want the eigenvalues of $A$ as a power series in $\epsilon$. The first two orders in perturbation theory are well known. Third and higher orders are briefly discussed here. However, the equations become horrible. 
I hear that Feynman diagrams are an efficient way to formulate perturbation theory, but I can't find an accessible exposition of this approach. Note that I have in mind the simple matrix setting. I don't want vacuum states, quantum field theory, Dirac equation, etc. Can someone help?

REPLY [13 votes]: I think the simplest way is to use the very simple and very useful resolvant formula
$$ (A+B)^{-1}=A^{-1}-A^{-1}B(A+B)^{-1}.$$
For perturbative theory, we just iterate this formula 
$$ (A+B)^{-1}=A^{-1}-A^{-1}BA^{-1}+A^{-1}BA^{-1}B(A+B)^{-1}$$
and obtain a power series in $B$.
$$ (A+B)^{-1}=A^{-1}-A^{-1}BA^{-1}+A^{-1}BA^{-1}BA^{-1}-A^{-1}BA^{-1}BA^{-1}BA^{-1} +\cdots$$
With your notation it gives 
$$ A^{-1}=D^{-1}-\epsilon D^{-1}BD^{-1}+\epsilon^2 D^{-1}BD^{-1}BD^{-1}-\epsilon^3 D^{-1}BD^{-1}BD^{-1}BD^{-1} +\cdots$$
For the eigenvalues there is then the Cauchy formula
$$ z_0 = \frac{1}{2i\pi}\oint \frac{z}{z-z_0} dz$$
which for matrices gives
$$\lambda_0 = \frac{1}{2i\pi}\oint \lambda Tr(A-\lambda)^{-1}d\lambda$$
where the integral is a closed loop around an eigenvalue of $D$.
And we obtain the perturbative expansion:
$$ \lambda_0 =\frac{1}{2i\pi} \oint \lambda Tr(D-\lambda)^{-1} d\lambda \\-\epsilon \frac{1}{2i\pi} \oint \lambda Tr (D-\lambda)^{-1}B(D-\lambda)^{-1} \\+\epsilon^2 \frac{1}{2i\pi} \oint \lambda Tr (D-\lambda)^{-1}B(D-\lambda)^{-1}B(D-\lambda)^{-1}\\-\epsilon^3 \cdots$$
Now take a loop sufficiently close to the initial eigenvalue of $D$ and it will give what you wanted.