Trial Wavefunction

The trial wavefunction $\psi_T$ should be chosen as close to the true wave function $\psi$ of the system as possible. If we were able to approximate the system wavefunction with satisfying accuracy, then the sampling over the corresponding distribution (for example with the help of Metropolis algorithm) would give us all properties of the system. However, the problem is that very often it is impossible to find the wavefunction of the system using analytical methods. Here enters the Diffusion Monte Carlo method, which compensates our lack of knowledge and corrects the trial wavefunction provided that the projection of the trial wavefunction on the true system wavefunction of the system differs from zero. As the use of the trial wavefunction lies at the heart of the method, it has to be expressed in a way that is fast to calculate or it has to be tabulated. The common way to construct many-body wavefunctions is to use the Jastrow function consisting of the product of an uncorrelated state and a correlation factor, which is a product of two-body wavefunctions.

$$\psi_T = \prod\limits_i \phi(\vec r_i) \prod\limits_{i<j} g(\vec r_i,\vec r_j)$$ (3.30)

The one-body term describes the effect of an external field and is absent in the case of a homogeneous system. For a homogeneous system the trial wavefunction can be written in general as

$$\psi_T(r_1,...,r_N) = \prod\limits_{i=1\atop j=1}^N g(\vert\vec r_i-\vec r_j\vert)$$ (3.31)

Since we are mainly interested to dilute system a possible way to obtain the pair function $g(r)$ is through the solution of the two-body Schrödinger equation.

$$\left(-\frac{\hbar^2}{2\mu}\triangle+V(\vec r\,) \right)g = {\cal E} g,$$ (3.32)

here $\mu = m_1 m_2 / (m_1+m_2) = 2m$ is the reduced mass and $r$ is the interparticle distance. Let's search for the $l = 0$ solution in spherical coordinates

$$-\frac{\hbar^2}{2\mu} \left(g''+\frac{2}{r}g'\right) + V(r) g = {\cal E} g$$ (3.33)

The particles are modeled by hard spheres of diameter $a$ and the interaction potential is

$$V({\bf r}) = \left\{ {\begin{array}{ll} +\infty,&\vert r\vert \le a\\ 0,&\vert r\vert > a\\ \end{array}} \right.$$ (3.34)

The dimensionless Schrödinger equation is obtained by expressing all distances in units of $a$ and energy in units of $\hbar^2/(2ma^2)$

$$\left\{ {\begin{array}{ll} \displaystyle g(x) = 0,&\vert x\vert \... ...ystyle g''+\frac{2}{x}g' - 2E g = 0,& \vert x\vert > 1 \\ \end{array}} \right.$$ (3.35)

So, it is necessary to solve the differential equation

$$\left\{ {\begin{array}{l} \displaystyle g''+\frac{2}{x}g'- 2E g = 0\\ \displaystyle g(1) = 0\\ \end{array}} \right.$$ (3.36)

The solution of equation (3.36) is $g(x) = A\sin\left(\sqrt{2E}(x-1)\right)\Bigl/x\Bigr.$, where $A$ is an arbitrary constant. In dilute systems for small interparticle distance $r$ the correlation factor is well approximated by the function $g(r)$, i.e. by the wavefunction of a pair of particles in vacuum. At large distances the pair wavefunction should be constant, which corresponds to uncorrelated particles. Taking these facts into account let us introduce the trial function in the following way [24]

$$g(x) = \left\{ {\begin{array}{ll} \displaystyle \frac{A}{x} \sin(... ...\exp\left(-\frac{x}{\alpha}\,\right),& \vert x\vert > R\\ \end{array}} \right.$$ (3.37)

This function has to be smooth at the matching point $R$, i.e. 1) the $f(x)$ must be continuous

$$\frac{A}{R} \sin(\sqrt{2E}(R-1)) = 1 - B \exp\left(-\frac{R}{\alpha}\,\right)$$ (3.38)

2) its derivative $f'(x)$ must be continuous

$$-\frac{A}{R^2} \sin(\sqrt{E}(R-1)) +\frac{A\sqrt{2E}}{R} \cos(\sqrt{2E}(R-1)) = \frac{B}{\alpha} \exp\left(-\frac{R}{\alpha}\,\right)$$ (3.39)

3) the local energy $E_{L}(x) = \psi_T^{-1}\hat H\psi_T$ must be continuous

$$2E = \frac{\left(\displaystyle\frac{1}{\alpha^2}-\frac{2}{R\alpha... ...ac{R}{\alpha}\,\right) }{1-B \exp\left(-\displaystyle\frac{R}{\alpha}\,\right)}$$ (3.40)

The solution of this system is

$$\left\{ {\begin{array}{l} A =\displaystyle\frac{R}{\sin(u(1-1/R))... ...\\ B =\displaystyle\frac{u^2 \exp(\xi)}{\xi^2-2\xi+ u^2}, \end{array}} \right.$$ (3.41)

where we used the notation $u = \sqrt{2E}R$ and $\xi = R/\alpha$. The value of $\xi$ is obtained from the equation

$$1-\frac{1}{R} = \frac{1}{u} \mathop{\rm atan}\nolimits \frac{u(\xi-2)}{u^2+\xi-2}$$ (3.42)

There are three conditions for the determination of five unknown parameters, consequently two parameters are left free. The usual way to define them is minimize the variational energy in Variational Monte Carlo which yields an optimized trial wavefunction.