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I posted this question in MathOverflow but was not lucky with the answers, so wil try here.

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 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, path integrals, many body, etc. Can someone help?

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    $\begingroup$ Very relevant: physics.stackexchange.com/q/241896 $\endgroup$ – JamalS Oct 23 '17 at 20:09
  • $\begingroup$ @JamalS I did not find that "very relevant", since it is not about eigenvalues. The question was general but all the answers are about field theories. $\endgroup$ – thedude Oct 24 '17 at 0:19
  • $\begingroup$ It's relevant for simply bringing to attention reasons why a diagrammatic expansion for a perturbation theory may not be feasible in all cases. $\endgroup$ – JamalS Oct 24 '17 at 14:44
  • $\begingroup$ relevant too: Higher orders in perturbation theory $\endgroup$ – AccidentalFourierTransform Oct 24 '17 at 17:02
  • $\begingroup$ @AccidentalFourierTransform Indeed, that was a question of my own. User Qmechanic gave an answer which suggests he knows the stuff I am asking about, but unfortunately he was very succint and did not provide any references... What I want to know is exactly what he was talking about, but I would like a more careful explanation. $\endgroup$ – thedude Oct 24 '17 at 17:12
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If you want to find the eigenvalues of a finite dimensional matrix $A$ as a Taylor series on $\epsilon$ there are well known procedures to do that. If your object $A$ is infinite dimensional things become more complicated but can be carried out in principle. (Almost) Everything you want to know on the subject can be found in Kato’s book perturbation theory for linear operators.

The Feynman diagrams are essentially simply a convenient way to write a particular term of the perturbation expansion when your objects are (quantum) field theories. For example, it turns out that, because of the form of the interaction term (your $B$) you can already say that some terms in the perturbation expansion are going to be zero. In general you have a diagrammatic way to write such terms which is useful because it provides an easy way to write down these terms. At the same time it also provides a physical intuition for what these terms do and this is probably even more important.

In a simple matrix setting Feynman diagrams are of no use as they cannot even be defined.

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  • $\begingroup$ Your statement that "in a simple matrix setting Feynman diagrams cannot even be defined" is in strong disagreement with the answer by Qmechanic on this post: physics.stackexchange.com/questions/232574/… $\endgroup$ – thedude Oct 24 '17 at 19:21
  • $\begingroup$ @thedude : Qmechanics can call those mnemonics the way he wants. Feynman diagrams traditionally are something else, and if you look at the Wikipedia page you can convince yourself. Anyway this is terminology. If you are interested in the set of rules to figure out the term at a given particular order one can probably come up with something. But there are actually simpler algebraic expressions. If you look at Kato’s book you will find more that what you want $\endgroup$ – lcv Oct 25 '17 at 2:08
  • $\begingroup$ If your question is “I would like to know what is the particular form of the n-th order term of the perturbation series” you can post a new question and I can give you a separate answer $\endgroup$ – lcv Oct 25 '17 at 2:15
  • $\begingroup$ I saw your other question now and I can understand your point. The answer to the other question is “yes one can write down the general term” and you were right: the answer involves some kind of sums over partitions. $\endgroup$ – lcv Oct 25 '17 at 2:24
  • $\begingroup$ Ok, I will accept this answer and post a new question. Thanks. $\endgroup$ – thedude Oct 25 '17 at 10:43

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