- #1

JhonDoe

- 7

- 0

- Homework Statement
- This is not homework

- Relevant Equations
- ##E_k=\int \frac{\nabla \rho(\mathbf{r}) \cdot \nabla \rho(\mathbf{r})}{\rho(\mathbf{r})}d\mathbf{r}##

I have some doubts with respect on how the functional derivative for the kinetic energy in density functional theory is obtained.

I have been looking at this article in wikipedia: https://en.wikipedia.org/wiki/Functional_derivative

In particular, I'm interested in how to get the result##\frac{\delta E_k}{\delta \rho(\mathbf{r})}=-\frac{1}{4} \frac{\nabla^2 \rho(\mathbf{r})}{\rho(\mathbf{r})}+\frac{1}{8}\frac{\nabla \rho(\mathbf{r}) \cdot \nabla \rho(\mathbf{r})}{\rho(\mathbf{r})^2}##

I understand that the procedure starts by Taylor expansion of the argument in the integral.

Taylor expanding at ##\mathbf{r}## gives, for a point ##\mathbf{r}'## close to ##\mathbf{r}##

##\frac{\nabla \rho(\mathbf{r}') \cdot \nabla \rho(\mathbf{r}')}{\rho(\mathbf{r}')} \simeq \frac{\nabla \rho(\mathbf{r}) \cdot \nabla \rho(\mathbf{r})}{\rho(\mathbf{r})} - \frac{1}{4} \frac{\nabla^2 \rho(\mathbf{r})}{\rho(\mathbf{r})} \delta \rho(\mathbf{r}') + \frac{1}{8}\frac{\nabla \rho(\mathbf{r}) \cdot \nabla \rho(\mathbf{r})}{\rho(\mathbf{r})^2}\delta \rho(\mathbf{r}')##

That is the result reported everywhere, including wikipedia, so I know it is correct. However, I have no problem with the first linear term, this is to say the term ##\frac{1}{4} \frac{\nabla^2 \rho(\mathbf{r})}{\rho(\mathbf{r})} \delta \rho## which is obtained by simply taking the derivative with respect to ##\rho## for the denominator. The term I have trouble with is the one that involves the derivative ##\frac{\partial }{\partial \rho} \nabla \rho(\mathbf{r}) \cdot \nabla \rho(\mathbf{r})## I don't really understand how to get the derivative of the gradient with respect to the function ##\rho##.

I have been looking at this article in wikipedia: https://en.wikipedia.org/wiki/Functional_derivative

In particular, I'm interested in how to get the result##\frac{\delta E_k}{\delta \rho(\mathbf{r})}=-\frac{1}{4} \frac{\nabla^2 \rho(\mathbf{r})}{\rho(\mathbf{r})}+\frac{1}{8}\frac{\nabla \rho(\mathbf{r}) \cdot \nabla \rho(\mathbf{r})}{\rho(\mathbf{r})^2}##

I understand that the procedure starts by Taylor expansion of the argument in the integral.

Taylor expanding at ##\mathbf{r}## gives, for a point ##\mathbf{r}'## close to ##\mathbf{r}##

##\frac{\nabla \rho(\mathbf{r}') \cdot \nabla \rho(\mathbf{r}')}{\rho(\mathbf{r}')} \simeq \frac{\nabla \rho(\mathbf{r}) \cdot \nabla \rho(\mathbf{r})}{\rho(\mathbf{r})} - \frac{1}{4} \frac{\nabla^2 \rho(\mathbf{r})}{\rho(\mathbf{r})} \delta \rho(\mathbf{r}') + \frac{1}{8}\frac{\nabla \rho(\mathbf{r}) \cdot \nabla \rho(\mathbf{r})}{\rho(\mathbf{r})^2}\delta \rho(\mathbf{r}')##

That is the result reported everywhere, including wikipedia, so I know it is correct. However, I have no problem with the first linear term, this is to say the term ##\frac{1}{4} \frac{\nabla^2 \rho(\mathbf{r})}{\rho(\mathbf{r})} \delta \rho## which is obtained by simply taking the derivative with respect to ##\rho## for the denominator. The term I have trouble with is the one that involves the derivative ##\frac{\partial }{\partial \rho} \nabla \rho(\mathbf{r}) \cdot \nabla \rho(\mathbf{r})## I don't really understand how to get the derivative of the gradient with respect to the function ##\rho##.