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.

  • $\begingroup$ How are the wavefunctions represented in the large-N limit? $\endgroup$
    – davidhigh
    Commented May 8, 2023 at 6:08
  • $\begingroup$ The wavefunctions are well-defined functions in the position basis. $\endgroup$ Commented May 12, 2023 at 10:30
  • $\begingroup$ 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? $\endgroup$
    – davidhigh
    Commented May 12, 2023 at 17:41
  • $\begingroup$ Yes, they are. The wavefunctions are well-defined for all N. $\endgroup$ Commented May 14, 2023 at 9:07

1 Answer 1


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).

None of this of course will get you around the curse of dimensionality.


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