Trial wave function in the BCS limit

In the construction of the trial function the antisymmitrization is included through the Slater determinant ${\cal D}{\bf (R)}$

$$\psi_T({\bf R}) = {\cal D}{\bf (R)}\prod\limits_{i=1}^Nf_1({\vec r}_i) \prod\limits_{j<k}^Nf_2(\vert{{\vec r}_j-{\vec r}_k}\vert)$$ (2.92)

Thus at the variational trial move one has to calculate the ratio of two determinants in addition to usual one- and two- body correlation terms present in the bosonic VMC algorithm (compare with (2.37) and see Sec. 2.2). An element of the Slater matrix is given by ${\cal D}_{i\alpha} = \varphi_\alpha({\vec r}_i)$, where $\varphi_\alpha(\r)$ is a single particle orbital. In further latin indices will always refer to particle number and the greek indices to orbital number. During a trial move in which position of only one particle get changed, just one row of the Slater matrix changes. This means that instead of a direct calculation of the Slater determinant a more efficient method can be used. Before doing the trial move one should calculate the inverse matrix ${\cal\overline D}$ such that ${\cal D\overline D} = I$ or in terms of the matrix elements

$$\sum\limits_{\alpha=1}^N{\cal D}_{i\alpha}{\cal \overline D}_{j\alpha} = \delta_{ij}$$ (2.93)

If we denote the matrix with coordinate of the $i^{th}$ particle changed as ${\cal D'}$ then the ratio of interest becomes

$$\frac{\vert{\cal D'}\vert^2}{\vert{\cal D}\vert^2} = \frac{\vert{... ...\overline DD'}\vert^2}{\vert{\cal D}\vert^2} = \vert{\cal D'\overline D}\vert^2$$ (2.94)

The matrix ${\cal D'\overline D}$ is almost diagonal. Indeed, only $i^{th}$ row is different from the one of a unitary matrix. It means that the determinant of such a matrix equals to the $i^{th}$ element of this row, i.e.

$$q = \frac{\vert{\cal D'}\vert^2}{\vert{\cal D}\vert} = \sum\limits_{\alpha=1}^N\varphi_\alpha({\vec r}_i'){\cal \overline D}_{i\alpha}$$ (2.95)

After the move is accepted the inverse matrix must be updated. There is a fast way of doing it. Instead of direct inversion of the determinant matrix one can use $q$ from eq.(2.95):

$${\cal \overline D}_{j\alpha} = \left\{ {\begin{array}{ll} {\cal \... ...a({\vec r}_i'){\cal \overline D}_{j\beta}}{q} ,&j \ne i\\ \end{array}} \right.$$ (2.96)

Differentiating the trial wave function (2.92) one finds the expression for the kinetic energy. It is equal to

$$T^{loc}({\bf R}) = \frac{\hbar^2}{2m}\left\{ \sum\limits_{i=1}^N{... ...j-{\vec r}_k}\vert) -\sum\limits_{i=1}^N \vert\vec F_i({\bf R})\vert^2\right\},$$ (2.97)

where the two-body contribution to the local energy is the same as in the bosonic case

$${\cal E}^{loc}_2(r) =-\frac{f_2''(r)}{f_2(r)}+\left(\frac{f_2'(r)}{f_2(r)}\right)^2$$ (2.98)

and there is an additional term coming from the determinant part of the trial wave function.

$${\cal E}^{i}_{\cal D}(\vec r) = -\frac{\Delta_i {\cal D}({\bf R})... ...({\bf R})} +\left(\frac{\nabla_i {\cal D}({\bf R})}{{\cal D}({\bf R})}\right)^2$$ (2.99)

The drift force (2.15) appearing in (2.97) is given by

$$\vec F_i({\bf R}) = \frac{\nabla_i {\cal D}({\bf R})}{{\cal D}({\... ...c r}_k}\vert)}\frac{{{\vec r}_i-{\vec r}_k}}{\vert{{\vec r}_i-{\vec r}_k}\vert}$$ (2.100)

The derivatives of the determinant are related to the derivatives of the orbitals in an easy way (see formula (2.95))

$$\frac{\nabla_i {\cal D}({\bf R})}{{\cal D}({\bf R})} =\sum\limits_{\alpha=1}^N {\cal \overline D}_{i\alpha} \nabla_i \varphi_\alpha({\vec r}_i)$$ (2.101)