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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)?

DanielSank
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1 Answers1

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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.

DanielSank
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