Consider the "first passage problem"
A random walk proceeds on a graph of connected points. On this graph, there is one "end" point $j$ meaning that if the random walker lands on this point the process ends. Suppose we wish to know the mean number of steps taken by the walk before it ends, given that it started on point $i$.
Let us denote the probability that the random walk lands on point $j$ for the first time after $n$ steps, given that it started on point $i$, by the symbol $f_{ji}(n)$. In other words, $f_{ji}(n)$ is the probability that the walk ends after $n$ steps. The solution to the question is then given by $\sum_{n=1}^\infty n f_{ji}(n)$. This is, however, not generally easy to compute because $f_{ji}(n)$ are difficult to compute.
Amazingly, we can relate the $f_{ji}$ to the unconstrained random walk probabilities Define the probability that, in the absence of any termination, the random walk lands on point $j$ after $n$ steps, having started at $i$, by the symbol $p_{ji}(n)$. In other words, $p_{ji}(n)$ gives the probabilities in the case where we've just removed the fact that point $j$ ends the walk. Now define two transforms $$ F_{ji}(z) \equiv \sum_{n=0}^\infty f_{ji}(n) z^n \qquad P_{ji}(z) \equiv \sum_{n=0}^\infty p_{ji}(n) z^n \, . $$ It turns out that $$ F_{ji}(z) = P_{ji}(z) / P_{jj}(z) \, .$$ This looks suspiciously like something having to do with self-energy in QFT. Is there a mathematical, or even better a physical connection between self-energy and the "recurrence" term $P_{jj}(z)$? To put it another way, is there a way to think of self-energy as the transform of a recurrence probability (amplitude)?