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Soft sphere trial wave function
At short distances the scattering solution
(1.55) is expected
to proved a good approximation for the two-body Bijl-Jastrow term
(2.37) in a gas with the soft sphere interaction potential (1.52). At
the larger distances
should saturate to a constant in a smooth way. We
choose an exponential type of decay:
![$\displaystyle f_2(r) =
\left\{
{\begin{array}{ll}
\displaystyle
\frac{A\mathop{...
...C \exp\left(-\frac{r}{\alpha}\right),&R_{m}\le \frac L2\\
\end{array}}
\right.$](img903.gif) |
|
|
(2.73) |
The description of a continuous matching at the point
is described in
Sec. 1.3.2.3. Now we shall discuss the matching procedure at the point
.
As usual we have three matching conditions:
- Continuity of the function
:
![$\displaystyle f(R_{m}) = \frac{B\sin(kR_{m}+\delta)}{R_{m}} = 1-C \exp\left(-\frac{R_{m}}{\alpha}\right)$](img904.gif) |
|
|
(2.74) |
- Continuity of the logarithmic derivative
which fixes the value
of the parameter
:
![$\displaystyle C = \exp(R_{m}/\alpha)
\frac{k\mathop{\rm ctg}\nolimits (kR_{m}+\delta)-1/R_{m}}{k\mathop{\rm ctg}\nolimits (kR_{m}+\delta)-1/R_{m}+1/\alpha}$](img906.gif) |
|
|
(2.75) |
Substitution of (2.75) into (2.74) fixes value of
![$\displaystyle B = \frac{R_{m}/\alpha}{\sin(kR_{m}+\delta)}
\frac{1}{k\mathop{\rm ctg}\nolimits (kR_{m}+\delta)-1/R_{m}+1/\alpha}$](img907.gif) |
|
|
(2.76) |
- The kinetic energy must be continuous at
. This condition yields
![$\displaystyle -\left(\frac{f''(R_{m})}{f(R_{m})}+\frac{2}{R_{m}}\frac{f'(R_{m})}{f(R_{m})}\right)
= k^2$](img909.gif) |
|
|
(2.77) |
After some mathematics the following procedure is obtained:
- By choosing the value of the scattering length
and the range of the
potential
define the value of
(i.e. height of the potential
as
related by (1.54)) by solving the transcendental equation (1.59)
- Introduce
,
and
.
Both equations (1.57) and
![$\displaystyle 1+\bar\delta = \frac{1}{x}\mathop{\rm arctg}\nolimits \left(\frac{x(y-2)}{x^2+y-2}\right)$](img914.gif) |
|
|
(2.78) |
has to be satisfied in order to loop to determine
and
. This
can be done using iterative procedure:
- Choose the value for the
(the first time it is
initialized with
) and obtain the value of
as
the solution of Eq. 2.78
- Fix the scattering momentum
and obtain the phase
as a solution of (1.57)
- Iterate
until
both numbers converge to the same value by repeating steps (a-c)
- Then the constants (A,B,C) are given by formulae
-
-
-
Once all parameters are fixed, the two-body term
is given by (2.73),
the drift force contribution (2.39) is given by
![$\displaystyle {{\cal F}_2}(r) =
\left\{
\begin{array}{lll}
\displaystyle \sqrt{...
...ac{r}{\alpha}-C\right)\right)^{-1},& \vert x\vert > R_{m}\\
\end{array}\right.$](img924.gif) |
|
|
(2.79) |
Local energy (2.40) equals to
![$\displaystyle {{\cal E}_2}(r) =
\left\{
\begin{array}{ll}
\displaystyle {\cal E...
...ac{1}{\exp\left(r/\alpha\right)-C},& \vert r\vert > R_{m}\\
\end{array}\right.$](img925.gif) |
|
|
(2.80) |
Next: Trial wave function of
Up: Three-dimensional wave functions
Previous: Hard sphere trial wave
  Contents
G.E. Astrakharchik
15-th of December 2004