Soft sphere trial wave function

At short distances the scattering solution $f(r)$ (1.55) is expected to proved a good approximation for the two-body Bijl-Jastrow term $f_2(r)$ (2.37) in a gas with the soft sphere interaction potential (1.52). At the larger distances $f_2(r)$ should saturate to a constant in a smooth way. We choose an exponential type of decay:

$$f_2(r) = \left\{ {\begin{array}{ll} \displaystyle \frac{A\mathop{... ...C \exp\left(-\frac{r}{\alpha}\right),&R_{m}\le \frac L2\\ \end{array}} \right.$$ (2.73)

The description of a continuous matching at the point $R$ is described in Sec. 1.3.2.3. Now we shall discuss the matching procedure at the point $R_{m}$. As usual we have three matching conditions:

1. Continuity of the function $f_2(r)$:
   

   $$f(R_{m}) = \frac{B\sin(kR_{m}+\delta)}{R_{m}} = 1-C \exp\left(-\frac{R_{m}}{\alpha}\right)$$ (2.74)

   
   

2. Continuity of the logarithmic derivative $f_2'(r)/f_2(r)$ which fixes the value of the parameter $C$:

   
   

   $$C = \exp(R_{m}/\alpha) \frac{k\mathop{\rm ctg}\nolimits (kR_{m}+\delta)-1/R_{m}}{k\mathop{\rm ctg}\nolimits (kR_{m}+\delta)-1/R_{m}+1/\alpha}$$ (2.75)

   
   

   Substitution of (2.75) into (2.74) fixes value of $B$
   

   $$B = \frac{R_{m}/\alpha}{\sin(kR_{m}+\delta)} \frac{1}{k\mathop{\rm ctg}\nolimits (kR_{m}+\delta)-1/R_{m}+1/\alpha}$$ (2.76)

   
   

3. The kinetic energy must be continuous at $r = R_{m}$. This condition yields
   

   $$-\left(\frac{f''(R_{m})}{f(R_{m})}+\frac{2}{R_{m}}\frac{f'(R_{m})}{f(R_{m})}\right) = k^2$$ (2.77)

   
   

After some mathematics the following procedure is obtained:

1. By choosing the value of the scattering length $a_{3D}$ and the range of the potential $R$ define the value of $\varkappa$ (i.e. height of the potential $V_0$ as related by (1.54)) by solving the transcendental equation (1.59)
2. Introduce $x = k R_{m}$, $y = R_{m}/\alpha$ and $\bar\delta = \delta/x$. Both equations (1.57) and
   

   $$1+\bar\delta = \frac{1}{x}\mathop{\rm arctg}\nolimits \left(\frac{x(y-2)}{x^2+y-2}\right)$$ (2.78)

   
   
   has to be satisfied in order to loop to determine $x = k R_{m}$ and $\delta$. This can be done using iterative procedure:
   1. Choose the value for the $\bar\delta_a^{(i)}$ (the first time it is initialized with $\bar\delta_a^{(0)} = -a/R_{m}$) and obtain the value of $x^{(i)}$ as the solution of Eq. 2.78
   2. Fix the scattering momentum $k^{(i)} = x^{(i)}/R_{m}$ and obtain the phase $\delta_b^{(i)}$ as a solution of (1.57)
   3. Iterate $(\bar\delta_a^{(i+1)} = (\delta_a^{(i)}+\delta_b^{(i)})/2$ until both numbers converge to the same value by repeating steps (a-c)
3. Then the constants (A,B,C) are given by formulae
   1. $C = \exp(y)x^2/(x^2+y^2-2y)$
   2. $B = R_{m}/\sin(x+\delta)$
   3. $A = B\sin(kR_{m}+\delta)/\mathop{\rm sh}\nolimits (kR_{m})$

Once all parameters are fixed, the two-body term $f_2(r)$ is given by (2.73), the drift force contribution (2.39) is given by

$${{\cal F}_2}(r) = \left\{ \begin{array}{lll} \displaystyle \sqrt{... ...ac{r}{\alpha}-C\right)\right)^{-1},& \vert x\vert > R_{m}\\ \end{array}\right.$$ (2.79)

Local energy (2.40) equals to

$${{\cal E}_2}(r) = \left\{ \begin{array}{ll} \displaystyle {\cal E... ...ac{1}{\exp\left(r/\alpha\right)-C},& \vert r\vert > R_{m}\\ \end{array}\right.$$ (2.80)