Slater determinants

In this section, we explicity compute the gradient and Laplacian of the determinant of the trial wave function. In particular, we seek

$$\nabla [\ln \det(A)],$$ (19)

where A is either the up or down matrix. To simplify the analysis, we will initially work again in terms of components. Let $\partial_i$ represent the derivative with respect to a single component of the 3N dimensional R. Trivially, then,

$$\partial_i \ln[\det(A)] = \frac{1}{\det A} \partial_i \det (A)$$ (20)

At first glance, taking the derivative of a determinant appears a daunting task. Considerable simplification is possible with the following simple relation, which we state without proof:

$$\partial_i \det(A) = \det(A) \,\text{Tr}\,[A^{-1} \partial_i A]$$ (21)

Then

$$\partial_i \ln [\det(A)] = \,\text{Tr}\,[A^{-1} \partial_i A]$$ (22)

Given this form, we are left with the calculation of the the elements of $\partial_i A$. The elements of A are just

$$A_{kl} = \phi_k({\mathbf r}_l)$$ (23)

It is quite easy to see that the elements of the derivative matrix will be zero unless $i \in {x_l, y_l, z_l}$. Then the derivative matrix will have a single non-zero column. By taking advantage of symmetry, we can calculate the x, y, and z components simultaneously by directly calculating $\nabla_l \phi_k({\mathbf r}_l)$. If we make the substitution,

$$\tilde{{\mathbf r}}\equiv {\mathbf r}_l - {\mathbf c}_k$$ (24)

then $\phi_k(\tilde{{\mathbf r}})$ will be radially symmetric in $\tilde{{\mathbf r}}$. Since these vector operators are translationally invariant,

$$\nabla_{\tilde{{\mathbf r}}} = \nabla_{r_l}.$$ (25)

If we work in spherical coordinates, $\phi_k(\tilde{r}, \tilde{\theta}, \tilde{\phi})$ will be independent of $\tilde{\theta}$ and $\tilde{\phi}$. We can then express the gradient as

$$\nabla_{\tilde{{\mathbf r}}} \phi_k(\tilde{{\mathbf r}}) = \partial_{\tilde{r}} \phi_k(\tilde{r}) \hat{\tilde{{\mathbf r}}},$$ (26)

where $\hat{\tilde{{\mathbf r}}}$ represent the unit vector in the direction of $\tilde{{\mathbf r}}$. Written in terms of $\tilde{{\mathbf r}}$, the orbitals take the form,

$$\phi_k(\tilde{{\mathbf r}}) = \exp \left( \frac{ -\tilde{r}^2}{\omega_k^2 + \nu_k \tilde{r}} \right)$$ (27)

The derivative takes the form

$$\partial_{\tilde{r}} \phi_k(\tilde{r}) \phi_k(\tilde{r}) \partial_{\tilde{r}} \left[ \frac{- \tilde{r}^2}{\omega_k^2 + \nu_k \tilde{r}}\right]$$ =

$$\phi_k(\tilde{r}) \left[\frac{-2 \tilde{r}}{\omega_k^2 + \nu_k \tilde{r}} + \frac{\nu_k \tilde{r}^2}{(\omega_k^2 + \nu_k \tilde{r})^2} \right]$$ =

$$\phi_k(\tilde{r}) \left[\frac{-2\tilde{r}(\omega^2 + \nu_k \tilde{r}) + \nu_k \tilde{r}^2} {(\omega_k^2 + \nu_k)^2} \right]$$ =

$$\phi_k(\tilde{r}) \left[ \frac{ -\tilde{r}(2\omega_k^2 + \nu_k \tilde{r})}{(\omega_k^2 + \nu_k \tilde{r})^2} \right]$$ = (28)

and furthermore,

$$\nabla_{\tilde{{\mathbf r}}} \phi_k(\tilde{{\mathbf r}}) \phi_k(\tilde{r}) \left[ \frac{ -\tilde{r}(2\omega_k^2 + \nu_k \tilde{r})}{(\omega_k^2 + \nu_k \tilde{r})^2} \right] \hat{\tilde{{\mathbf r}}}$$ =

$$\phi_k(\tilde{r}) \left[ \frac{ -\tilde{r}(2\omega_k^2 + \nu_k \t... ...k \tilde{r})^2} \right] \left(\frac{{\mathbf r}- {\mathbf c}}{\tilde{r}}\right)$$ =

$$\phi_k(\tilde{r}) \left[ \frac{ -(2\omega_k^2 + \nu_k \tilde{r})}{(\omega_k^2 + \nu_k \tilde{r})^2} \right] ({\mathbf r}- {\mathbf c}).$$ = (29)

The total Laplacian will be a sum of the Laplacian's with respect to each electronic coordinate, r_i. Let i represent a particular component of the 3N dimensional vector, R, eg. the x component of r₄. Then we have that

$$\partial_i \ln [\det(A)] \,\text{Tr}\,[A^{-1} \partial_i A]$$ =

$$\partial^2_i \ln [\det(A)] \partial_i \,\text{Tr}\,[A^{-1} \partial_i A]$$ =

$$\,\text{Tr}\,[A^{-1} \partial^2_i A] + \,\text{Tr}\,[(\partial_i A^{-1})(\partial_i A)]$$ = (30)

where $\partial^2_i A$ actually denotes the Laplacian of the components of the matrix A. For the latter term, we utilize the fundamental property of inverses,

A^-1 A = 1

$$(\partial_i A^{-1})A + A^{-1} \partial_i A$$ = 0

$$\partial_i A^{-1} - A^{-1} (\partial_i A) A^{-1}$$ = (31)

Using this new relation

$$\partial_i^2 \ln[\det(A)] \,\text{Tr}\,[A^{-1} \partial^2_i A] - \,\text{Tr}\,[A^{-1}(\partial_i A) A^{-1} (\partial_i A)]$$ =

$$\,\text{Tr}\,[A^{-1} \partial^2_i A] - \,\text{Tr}\,[(A^{-1}(\partial_i A))^2]$$ = (32)

Now, to calculate the Laplacian with respect to r_m, we sum over the three components, j, of the r_m.

$$\nabla^2_{{\mathbf r}_m} \ln[\det(A)] = \,\text{Tr}\,[A^{-1} ... ...,z\}} \,\text{Tr}\,[(A^{-1}(\partial_{{\mathbf r}_m^j} A))^2],$$ (33)

where r_m^j refers to the $j^{\text{th}}$ component of r_m.

We already have the first derivative of A, which we found in the gradient part of the calculation. Next we calculate the Laplacian of the compoents of A. Specifically, we seek $\nabla_{{\mathbf r}_m}^2 A_{kl}$, where $A_{kl} = \phi_k({\mathbf r}{l})$. We have that

$$\nabla_{{\mathbf r}_m}^2 A_{kl} = \delta_{lm} \nabla^2_{{\mathbf r}_l} \phi_k({\mathbf r}_{l})$$ (34)

What remains is the Laplacian of the molecular orbitals, $\phi_k$, with respect to r_l. If we make the definition, ${\mathbf r}_{kl} \equiv {\mathbf r}_l - {\mathbf c}_k$, we recognize that $\phi_k({\mathbf r}_{kl})$ will spherically symmetric. Differential operators are invariant under translations, so we simplify the calculation our calculation by exploiting this translational invariance

$$\nabla_{{\mathbf r}_m}^2 A_{kl} \frac{1}{r_{kl}^2} \partial_{r_{kl}} r_{kl}^2 \partial_{r_{kl}} \phi_k(r_{kl})$$ =

$$\frac{2}{r_{kl}}\partial_{r_{kl}} \phi_k(r_{kl}) + \partial^2_{r_{kl}} \phi(r_{kl})$$ = (35)

The first derivative of the molecular orbitals were computed earlier in this paper, so we finally turn to computing the second derivative of the $\phi_k$'s.

$$\partial^2_{\tilde{r}} \phi_k(\tilde{r}) \partial_{\tilde{r}} \phi_k(\tilde{r}) \left[ \frac{ -\tilde{r}(2\omega_k^2 + \nu_k \tilde{r})}{(\omega_k^2 + \nu_k \tilde{r})^2} \right]$$ =

$$\left[ \frac{ -\tilde{r}(2\omega_k^2 + \nu_k \tilde{r})}{(\omega_... ...de{r}(2\omega_k^2 + \nu_k \tilde{r})}{(\omega_k^2 + \nu_k \tilde{r})^2} \right]$$ =

$$\phi_k(\tilde{r}) \frac{\left[\tilde{r}(2\omega_k^2 + \nu_k \tilde{r})\right]^2}{(\omega_k^2 + \nu_k \tilde{r})^4}$$ =

$$- 2 \phi_k(\tilde{r}) \left[\frac{(\omega_k^2 + \nu_k \tilde{r})^... ...de{r})(2\omega_k^2 + \nu_k \tilde{r})}{(\omega_k^2 + \nu_k \tilde{r})^4}\right]$$ (36)