# Inverse of a matrix in a Path Integral

Good morning! I can't make sense of an inverse of a matrix appearing in a calculation for a Wiener Path Integral. In discretized form: $$\int \prod_{i=1}^N \frac{dx_i}{\sqrt{\pi \epsilon}} e^{-\frac{1}{\epsilon} \sum_{i=1}^N \left( x_i-x_{i-1} \right)^2-\sum_{i=1}^N p_i x^2_i} \delta(x_N-x)$$ where $$p_j=p(j\epsilon)$$ and $$p(\tau)$$ is a real function. Now i wrote the delta function as an integral and in the end I have a gaussian integral where i need the determinant of the matrix $$a = \begin{bmatrix} a_1 & -\frac{1}{\epsilon} & 0 & \dots & \dots & 0 \\ -\frac{1}{\epsilon} & a_2 & \frac{-1}{\epsilon} & \dots & \dots & 0 \\ 0 & -\frac{1}{\epsilon} & a_3 & -\frac{1}{\epsilon} & \cdots & 0 \\ \vdots & \dots & -\frac{1}{\epsilon} & \dots & \dots & \vdots \\ \vdots & & & & & -\frac{1}{\epsilon} \\ 0 & \dots & \dots & 0 & -\frac{1}{\epsilon} & a_N \end{bmatrix}$$ where $$a_i=p_i \epsilon + \frac{2}{\epsilon} \quad j \neq N$$ and $$a_N=p_N \epsilon + \frac{1}{\epsilon}$$. I made sense of the determinant thanks to the Gelfand Yaglom method but I have no idea how to compute the inverse matrix element $$a^{-1}_{N,N}$$. Any clue?

• Dec 26, 2018 at 14:11
• Quick hint : Matrix elements of inverse matrix, i.e, $\left[a_{}^{-1}\right]_{ij}^{}$ is related to the determinant of the $\left(N-1\right)\times\left(N-1\right)$, obtained from matrix $a$ by removing element of $i_{}^{th}$ column and $j_{}^{th}$ row. Dec 26, 2018 at 15:05

It's a tridagonal matrix, so the inverse is found as the Green function of the associated three-term recurrence relation. There is no closed-form solution for general $$p_i$$, but a detailed description of the related math is in the excercises starting on page 86 of my lecture notes at https://courses.physics.illinois.edu/phys508/fa2018/amaster.pdf.