Recent content by JhonDoe

Why not? its a linear operator. Maybe you are right, but I'm not really sure about that.

I think that shouldn't be necessary. I've tried this. In index notation: ##\frac{d }{d\rho} (\nabla \rho)^2=\frac{d }{d\rho} \frac{\partial}{\partial x_i} \rho \frac{\partial}{\partial x_i} \rho## Then ##\frac{d }{d\rho} \frac{\partial}{\partial x_i} \rho \frac{\partial}{\partial x_i}...

Ok. Thanks. I agree its a scalar, but the argument has a dependence on ##\rho##, so I think it shouldn't be zero.

What is the result for this derivative: ##\frac{d }{d\rho} (\nabla \rho)^2##? I have trouble when deriving inside the gradient: ##\frac{d }{d\rho} (\nabla \rho)^2=2\nabla \rho \times \frac{d }{d\rho} (\nabla \rho)=2\nabla \rho \times(\nabla \frac{d }{d\rho} \rho) ## ?

You are absolutely right, let me explain. I was just following the book "Density functional theory of atoms and molecules" by Parr and Weitago. In the appendix of that book the authors obtain the functional derivatives by making a Taylor expansion of the function in the kernel ##f## in...

Hi. Thank you for your useful reply. I realized that I made a wrong statement in the first post. The term that gives me confusion is the one that involves the Laplacian, the term ##\frac{1}{4} \frac{\nabla^2 \rho(\mathbf{r})}{\rho(\mathbf{r})} \delta \rho##. I see now that it can be obtained...

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...