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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?

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  • $\begingroup$ Related: physics.stackexchange.com/q/328439/2451 $\endgroup$
    – Qmechanic
    Commented Dec 26, 2018 at 14:11
  • $\begingroup$ 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. $\endgroup$
    – Sunyam
    Commented Dec 26, 2018 at 15:05

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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.

Note in particular problem 2.16. This writes the matrix element that you want as a continued fraction, and also comments on the connection to Haydock recusion.

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