... wavefunction^3.1

One of the reasons for using the product of wavefunctions as the probability distribution instead of sampling $\psi$ is that the average over the latter is ill defined $\langle A\rangle = \int A\psi\,dR/\int\psi\,dR$, on the contrary the average over the product of wavefunctions has the meaning of the mixed estimator $\langle A\rangle = \int \psi_T A \psi\,dR/ \int\psi_T\psi\,dR$

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... distribution^3.1

The formula (3.26) should be understood in the statistical sense, the average of any value $A$ over the l.h.s.and r.h.s distributions are equal to each other in the limit when size of the population $N_W$ tends to infinity $\int A({\bf R}) f({\bf R},t)\,{\bf dR} = \lim\limits_{N_W\to\infty} \int A({\bf R}) \sum\limits_{i=1}^{N_W} C \delta ({\bf R - R_i}(t))\,{\bf dR}$

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... energy^3.1

In real DMC simulations the upper limit of the integrals is truncated by $L/2$. The ``tail'' energy, which is small and is typically much less than $1\%$ of the total energy can be approximated by the formula $E_{tail} = n \int_{L/2}^{\infty}E_L(r) 4\pi r^2 dr$. The idea is that at large distances, where the integral is evaluated, one can safely assume uniform distribution of particles.

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...$D_0$^3.1

It is necessary to note that here the ``diffusion'' occurs in imaginary time and it has nothing to do with diffusion in real space.

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... general^4.1

The only exception[11] known to us takes place in the vicinity of the $\lambda$-transition in liquid $^4$He, where $N_0 \propto(T_\lambda -T)^{2\beta}$ while $\rho_s \propto(T_\lambda-T)^{(2-\alpha)/3}$, and $2\beta =(2-\alpha)(1-\zeta)/3$ with $\zeta$ positive. Still, for $^4$He both indices $\alpha$ and $\zeta$ are very small, so with very good accuracy $\rho_s \sim N_0 \sim (T_\lambda -T)^{2/3}$ and the difference is so small that there is no hope to measure it experimentally.

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... squared^6.1

The $1/r^2$ dependence can be calculated directly from (1.45) by taking the limit $p \to 0$. Then $N_p \to mc/2p$ and integral (1.45) takes the form of a Fourier transform of a Coulomb potential $\int\exp(i{\bf kr})/r\,{\bf dr} = 4\pi/k^2$ and produces the first term of eq. (B.5).

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