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My question may be stupid, so please correct me if you find anything which is obviously erroneous.

In the following I will place a question mark (?) besides points/steps I consider doubtful.

My observations:

  1. In density functional theory (DFT) (at zero temperature), the energy of the system $E$ is broken up as $E = \int_{V} d v \left[ e_{K}(n(\vec{r})) + e_{H}(n(\vec{r})) + e_{XC}(n(\vec{r})) \right]$, where $V$ is the system size, $e$ the energy density per unit volume, $n(\vec{r})$ the electronic density, subscript "$K$" for "kinetic", "$H$" for "Hartree" and "$XC$" for "exchange-correlation".

  2. In the generalized gradient approximation (gga), $e$ also depends on $\nabla n(\vec{r})$. Thus write $e(n(\vec{r}), \nabla n(\vec{r}))$.

Assumptions: 1. spherical symmetry (closed-shell)

By spherical symmetry $e_{XC}(n(\vec{r}), \nabla n(\vec{r})) = e(n(r), \nabla n(r))$. Of course in writing this we are restricting the variation in density $\delta n$ in the radial direction only. I'm doubtful about the validity of this step.

Anyway, proceed as usual. The exchange-correlation potential is defined as the functional derivative of $e_{XC}$ w.r.t. $\delta n$. Now please point out if there are any mistakes in the following steps:

$$\begin{eqnarray} \delta \int dv e_{XC}(n(r), n^{\prime}(r)) &=& \int dv \left[ \frac{\partial e_{XC}}{\partial n(r)} \delta n(r) + \frac{\partial e_{XC}}{\partial n^{\prime}(r)} \delta n^{\prime}(r) \right] \end{eqnarray}$$,

where $n^{\prime}(r) = \partial_r n(r)$. The first term is done; consider the second term. $\delta n^{\prime} (r) = \partial_r \delta n(r)$. Thus we integrate the second term on the right hand side by parts, and obtain

$4 \pi \int_0^{r_0} dr \partial_r \left[ r^2 \frac{\partial e_{XC}}{\partial n^{\prime}} \delta n(r) \right] - 4 \pi \int_0^{r_0} dr \left[ \partial_r \left( r^2 \frac{e_{XC}}{n^{\prime}}\right) \delta n(r) \right]$. The first term is the boundary term; ignore it for the time being. The second term seems to give the correction to the exchange correlation potential due to the introduction of gradient-dependence. Note that the second term is equivalent to $\int_0^{r_0} dr 4 \pi r^2 \left[ - \frac{1}{r^2} \partial_r \left( r^2 \frac{\partial e_{XC}}{\partial n^{\prime}} \right)\right]$, the square-bracketed term seems to give the exchange-correlation potential (?).

Now $- \frac{1}{r^2} \partial_r \left( r^2 \frac{\partial e_{XC}}{\partial n^{\prime}} \right) = - \frac{2}{r} \frac{\partial e_{XC}}{\partial n^{\prime}} - \partial_r \frac{\partial e_{XC}}{\partial n^{\prime}}$. What I worry about is the first term, which seems to diverge at $r=0$ (but it shouldn't; or is it cancelled by a piece in the second term?)

A expression for $e_{XC}$ may be the becke88 functional (exchange only) (PRA, 38, 3098). $$ \begin{eqnarray}e^{Becke}_{X} &=& e^{LDA}_{X}(n(r)) - \beta \sum_{\sigma} \int n_{\sigma}^{4/3} \frac{x_{\sigma}^2}{1+6 \beta x_{\sigma} \sinh^{-1}x_{\sigma}} d\vec{r} \end{eqnarray}$$, where $\sigma$ is the spin-index (For closed-shell system $n_{+} = n_{-} = n/2$ (?)), $\beta$ is a parameter (=0.0042 a.u.), and $x_{\sigma} = | \nabla n_{\sigma}|/n_{\sigma}^{4/3}$. In spherically symmetric system, can we change $| \nabla n_{\sigma}|$ to $|n^{\prime}(r)|$?. Now I am not sure about the following claims

  1. $\frac{\partial e_{X}^{Becke}}{\partial n^{\prime}(r)}$ is non-zero at the origin. (?) The reason is that, while it seems that $n^{\prime}(0)$ should be zero, it may not be. Consider helium. When solving the schroedinger eq. for the 1s state under arbitrary potential which tends to $2/r$ at $r=0$, the slope of the radial wave function (excluding the $r$-factor from the volume element) is non-zero (Consider for example the radial wave function of the hydrogen ground state, which behaves as $e^{-r}$). This means $n^{\prime}(0) \neq 0$. But then $\frac{\partial e_{X}^{Becke}}{\partial n^{\prime}(r)}$ will be non-zero at the origin and thus $- \frac{2}{r} \frac{\partial e_{XC}}{\partial n^{\prime}} $ will tend to infinity at $r=0$.

This should not be the case because the hartree (i.e. "classical" contribution by all electrons) contribution to the potential is small as $r \rightarrow 0$, and the exchange contribution should cancel the "self-action" in the hartree part.

Now, sorry for the long post. Note that I might have done something very wrong, such that the whole argument does not stand. Thanks for your time.

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    $\begingroup$ There is too much going on here. Please enumerate your questions (all the question marks), so that they can be individually addressed. You might also want to collect your assumptions, so that the circumstances of their validity can be addressed. $\endgroup$ – Deathbreath Jun 29 '16 at 17:01

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