# Monte Carlo simulation of many-body wave function overlaps

Consider two wavefunctions $$\psi_{1}$$ and $$\psi_{2}$$ over $$\otimes_{i=1}^{N}S$$. I want to evaluate the overlap between these two functions numerically: $$\int d\tau \psi_{2}^{\star}\psi_{1}$$ in the large $$N$$ limit. Given that $$\psi_{2}^{\star}\psi_{1}$$ is not a valid PDF, we cannot directly apply the MH algorithm for sampling. However, we can always introduce a distribution which we expect to have better convergence properties than uniform sampling, such as $$\psi_{1}^{\star}\psi_{1}$$ to evaluate this integral and then calculate the expectation of $$\frac{\psi_{2}^{\star}}{\psi_{1}^{\star}}$$. My question is the following: The typical spaces for $$\vert \psi_{2}^{\star}\psi_{1} \vert$$ and $$\psi_{1}^{\star}\psi_{1}$$ are in general going to be different. How can we be sure that sampling according to $$\psi_{1}^{\star}\psi_{1}$$ or $$\psi_{2}^{\star}\psi_{2}$$ will explore the sample space well enough? One way would be to just sample according to $$\vert \psi_{2}^{\star}\psi_{1}\vert$$. But these would become prohibitive if say, I want the overlap between $$m$$ wavefunctions as I would have to run at least $$\binom{m}{2}$$ chains.

• How are the wavefunctions represented in the large-N limit? Commented May 8, 2023 at 6:08
• The wavefunctions are well-defined functions in the position basis. Commented May 12, 2023 at 10:30
• Yes, but numerically there's an upper limit at about N=3,4,5 (depending on the system and the accuracy). So, as you're talking about the large N limit, are the wavefunctions analytically given? Commented May 12, 2023 at 17:41
• Yes, they are. The wavefunctions are well-defined for all N. Commented May 14, 2023 at 9:07

You are right that $$\psi_2^\ast\psi_1$$ is not a probability distribution (not even a non-normalized one) because it is complex-valued and possibly negative. But $$p(r)=|\psi_2(r)^\ast\psi_1(r)|$$ can serve that purpose, and so if you compute samples $$r_i$$ based on $$p(r)$$, then you can approximate $$\int dr \; \psi_2(r)^\ast\psi_1(r) = \int dr \; \frac{\psi_2(r)^\ast\psi_1(r)}{|\psi_2(r)^\ast\psi_1(r)|} p(r) \approx \frac{1}{N} \sum_i \frac{\psi_2(r_i)^\ast\psi_1(r_i)}{|\psi_2(r_i)^\ast\psi_1(r_i)|}.$$ Importantly, the denominator will never be zero in this formula because samples that would make it zero have zero probability and so will never be drawn (assuming you start with a sample with non-zero probability).