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)?

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    $\begingroup$ An old colleague of mine wrote his PhD on a method to compute elements of the many-body perturbation expansion in terms of walks on graphs. Your question reminds me quite a lot of this. The basic idea is to represent quantum dynamics as a random walk on the graph whose adjacency matrix is the Hamiltonian. Unfortunately, his thesis is apparently not available online, but you might find some insights in his papers, here, here and here (I helped a bit with the last one, but just with numerics). $\endgroup$ Commented Jul 2, 2015 at 16:30
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    $\begingroup$ @MarkMitchison every single time you post something I learn a lot. You're one of the users on the site that seems to know a lot about the topics that most interest me. May your use of this site be long-lived :-) $\endgroup$
    – DanielSank
    Commented Jul 2, 2015 at 16:32
  • $\begingroup$ Feelings 100% mutual :) I do try to encourage more people with an interest in quantum stochastic problems to join SE whenever I can, it's a shame there are not more users here who share our interests. $\endgroup$ Commented Jul 2, 2015 at 16:33
  • $\begingroup$ Sorry, the third link I posted was incorrect: here is the paper I meant. Its value is probably more in the slightly more physical discussion of what path-sums actually mean in the context of an example model. By the way, these papers are seriously, seriously mathematical. They will probably need some commitment if you want to understand them properly (I do not myself). $\endgroup$ Commented Jul 2, 2015 at 16:35
  • $\begingroup$ @MarkMitchison I see. Still, it's actually useful even just to see by what words these ideas go in the literature. This is really helpful in finding further references. $\endgroup$
    – DanielSank
    Commented Jul 2, 2015 at 16:39

1 Answer 1


There is a connection between QFT and random walks. It turns out the generating function $P_{ji}(z)$ is equivalent to the correlation function for the free scalar field on a Euclidean lattice. The parameter $z$ in the generating function effectively ends up being related to the mass of the scalar field (actually to a combination of mass and temperature).

The details are in most books on statistical field theory (for instance Giorgio Parisi's book) when they talk about the Gaussian model, which is essentially the free scalar field with a different interpretation of the parameters.

I'll give a brief argument....

Let's consider the partition function for a free scalar field $$Z = \int D\phi \exp\left[-\beta\int (\partial \phi)^2+m^2\phi^2\right],$$ and discretize it on a lattice with spacing $a$ (I'll do a 1d lattice for simplicity) $$Z = \int \prod_i d\phi_i \exp\left[-\beta\sum_i a \left(\frac{(\phi_{i+1}-\phi_i)^2}{a^2}+m^2\phi_i^2\right)\right],$$ $$= \int \prod_i d\phi_i \exp\left[\sum_{i,j}-A\delta_{ij}\phi_i\phi_j + B J_{ij}\phi_i\phi_j\right]$$

The coefficients $A$ and $B$ depend on $m,a,\beta$. And $J$ is matrix that is nonzero only for nearest neighbors. Usually the field is renormalized so that $A=1$ (so $B$ also picks up dependence on $m$).

Now if you know about Gaussian integrals, the correlation function or propagator $G_{ij}=\langle \phi_i \phi_j\rangle$ is the inverse of the matrix in the exponent (I've set $A=1$).

$$\sum_k(\delta_{i k}-B J_{ik})G_{kj} = \delta_{ij}.$$

Now let's go to random walks. There is a recursion relation for the probability $p_{ji}(n)$ to go to $j$ from $i$ in n steps, $$p_{ji}(n+1) = \sum_k J_{jk} p_{ki}(n)$$ where $J$ is now some matrix which gives the probability of going from $k$ to $j$ in one step.

Now consider the expression $\sum_k J_{jk} P_{ki}(z)$, \begin{align} \sum_k J_{jk} P_{ki}(z) &= \sum_k \sum_{n=0}J_{jk} z^n p_{ki}(n) \\ &= \sum_{n=0} z^n p_{ji}(n+1) \\ &= z^{-1} \sum_{n=0} z^{n+1} p_{ji}(n+1) \\ &= z^{-1}(P_{ji}(z) - p_{ji}(0)) \end{align} Rearranging... $$\sum_k(\delta_{k j} - z J_{j k}) P_{ki}(z) = \delta_{ij},$$ which is same equation as for the propagator, with $z$ taking the role of $B$ which depends on the other parameters.

Given this interpretation, $P_{jj}(z)$ is essentially giving you the variance of the field, and so $F_{ji}$ looks a bit like a normalized correlation function.

But I should point out (I didn't in the first edit) that partition function $Z$ for a statistical field theory in $d$ Euclidean dimensions is related to the path integral for a QFT in $d-1$ spatial dimensions. So $P_{jj}(z)$ is related to the vacuum fluctuations at the point $j$, $\langle 0|\hat{\phi}(j)\hat{\phi}(j) |0\rangle$. So if this is what you are thinking of as self energy, that is valid I think.

  • $\begingroup$ Could you give any of the mathematical details? Not being immersed in any of these topics I don't really understand much of what's written here. $\endgroup$
    – DanielSank
    Commented Jul 2, 2015 at 17:38
  • $\begingroup$ @DanielSank, No problem, I added some details. $\endgroup$
    – octonion
    Commented Jul 2, 2015 at 18:27
  • $\begingroup$ In going from the second to third equations, did the order of the terms in the $\exp$ switch? $\endgroup$
    – DanielSank
    Commented Jul 2, 2015 at 18:57
  • $\begingroup$ Yes there is a lot of rearranging, the term with $A$ is everything like $\phi_i^2$ and the term with $B$ like $\phi_i \phi_{i+1}$ $\endgroup$
    – octonion
    Commented Jul 2, 2015 at 18:59
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    $\begingroup$ @DanielSank, If you are thinking of vacuum fluctuations when you say self energy I think there is a relation (I thought you were talking about renormalization of mass). I edited a statement at the end. $\endgroup$
    – octonion
    Commented Jul 3, 2015 at 2:14

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