Doubt regarding functional derivative for the Thomas Fermi kinetic energy

[JhonDoe]

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 $\rho$.

 

[etotheipi]

Starting from$$E_k[\rho] = \int d\mathbf{r} \frac{\nabla \rho \cdot \nabla \rho}{8\rho} : = \int d\mathbf{r} f \quad \mathrm{where} \quad f := \frac{\nabla \rho \cdot \nabla \rho }{8\rho}$$Using the result in the Wikipedia article for the functional derivative of a functional of the form $F[\rho] = \int d\mathbf{r} f(\mathbf{r}, \rho{(\mathbf{r})}, \nabla \rho(\mathbf{r}))$, you have$$\frac{\delta E_k}{\delta \rho} = \frac{\partial f}{\partial \rho} - \nabla \cdot \frac{\partial f}{\partial \nabla \rho} \ \ \ (1)$$You then have for the partial derivative with respect to $\rho$\begin{align*}

\frac{\partial f}{\partial \rho} = \frac{- \nabla \rho \cdot \nabla \rho}{8 \rho^2}\end{align*}and with respect to $\nabla \rho$,\begin{align*}

\frac{\partial f}{\partial \nabla \rho} = \frac{1}{8 \rho} \left( 2\nabla \rho \right) = \frac{\nabla \rho}{4 \rho}\end{align*}from which it follows that, denoting $\Delta : = \nabla^2$,\begin{align*}

\nabla \cdot \frac{\partial f}{\partial \nabla \rho} = \nabla \cdot \frac{\nabla \rho}{4 \rho} &= \frac{\rho \Delta \rho - \nabla \rho \cdot \nabla \rho}{4\rho^2}\end{align*}Inserting those results in $(1)$ gives the quoted functional derivative

 

[JhonDoe]

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 with the identity given in wikipedia, your post clarified it.

How should I proceed to obtain the derivative $\frac{\partial }{\partial \rho} \nabla \rho(\mathbf{r}) \cdot \nabla \rho(\mathbf{r})$ if I wanted to obtain the result in that way?

 

[etotheipi]

No problem. I'm not completely sure what you're trying to do in your last sentence, because $\frac{\partial}{ \partial \rho} \nabla \rho \cdot \nabla \rho$ ought to just be zero. Perhaps someone else understands better what your method is, and will chip in

 

[JhonDoe]

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 $E_k[\rho]=\int f d\mathbf{r}$. Given that wikipedia uses that formula, which I don't know how to demonstrate, I wanted to follow their approach by Taylor expanding the kernel of the integral to the first order. The Taylor expansion should give:

$\frac{\nabla \rho(\mathbf{r}') \cdot \nabla \rho(\mathbf{r}')}{\rho(\mathbf{r}')} \simeq \frac{\nabla \rho(\mathbf{r}) \cdot \nabla \rho(\mathbf{r})}{\rho(\mathbf{r})} - \frac{1}{4} \frac{\nabla^2 \rho(\mathbf{r})}{\rho(\mathbf{r})} \delta \rho(\mathbf{r}') + \frac{1}{8}\frac{\nabla \rho(\mathbf{r}) \cdot \nabla \rho(\mathbf{r})}{\rho(\mathbf{r})^2}\delta \rho(\mathbf{r}')+\ldots$
In that way, keeping the taylor expansion to first order gives the desired result for the functional derivative. The first term in the right hand side goes to the lhs, and the desired result is obtained.

So, the term involving the Laplacian comes from the derivative $\frac{\partial}{ \partial \rho} \nabla \rho \cdot \nabla \rho$. But now, as you pointed out, that can not be, because that partial derivative should be zero, so, maybe in the Taylor expansion I should consider $\nabla \rho$ as an independent variable. I think that may give the correct result. However, the book doesn't make explicit the dependence on $\nabla \rho$ for the functional. It gives the functional $E_k[\rho]$, which only depends explicitly on $\rho$. So, I am trying to guess, but maybe instead of the partial derivative I should have the total derivative $\frac{d}{ d \rho} \nabla \rho \cdot \nabla \rho= \frac{\partial}{ \partial \nabla \rho} \nabla \rho \cdot \nabla \rho \frac{\partial \nabla \rho}{ \partial \rho}$, the term $\frac{ \partial \nabla \rho} { \partial \rho}$ must have a dependence on $\rho$, doesn't it? how should I evaluate that term? I think it just has to give the divergence appearing at the end in the formula. Maybe using index notation for the i-th component of the gradient $\frac{ \partial \nabla \rho} { \partial \rho}_i$ should give in some way the divergence, but I'm a bit confused.

By the way, how do you evaluate the derivative $\frac{\partial}{ \partial \nabla \rho} \nabla \rho \cdot \nabla \rho$? In index notation should it be $\frac{\partial}{ \partial \nabla \rho} \nabla \rho \cdot \nabla \rho|_i=\frac{\partial}{ \partial \nabla \rho_i} \frac{\partial \rho }{ \partial x_j} \frac{\partial \rho }{ \partial x_j}$ ? how do you treat the derivative in the denominator with respect to the gradient?

 

[JhonDoe]

What is the result for this derivative: $\frac{d }{d\rho} (\nabla \rho)^2$?

I have trouble when deriving inside the gradient: $\frac{d }{d\rho} (\nabla \rho)^2=2\nabla \rho \times \frac{d }{d\rho} (\nabla \rho)=2\nabla \rho \times(\nabla \frac{d }{d\rho} \rho)$ ?

 

[etotheipi]

I would have thought that's again zero, and it's probably best to use the "$\cdot$" product instead of a $\times$ symbol, considering that $(\nabla \rho)^2$ is shorthand for $\nabla \rho \cdot \nabla \rho$ and is a scalar.

I'll admit that I'm not completely following what you're trying to do here, so I'm going to ask and see if anyone else can help. Hold tight!

 

[JhonDoe]

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

 

[etotheipi]

Ok. Thanks, but I think the gradient has a dependence on $\rho$, so I think it shouldn't be zero.

In this context however, it looks like we're considering a function of both $\rho$ and $\nabla \rho$, which are themselves both functions of $\mathbf{r}$, in which case we ought to treat them as independent variables when taking partial derivatives.

 

[JhonDoe]

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
$\frac{d }{d\rho} \frac{\partial}{\partial x_i} \rho \frac{\partial}{\partial x_i} \rho=\frac{\partial}{\partial x_i} \frac{d \rho}{d\rho} \times \frac{\partial}{\partial x_i} \rho +\frac{\partial}{\partial x_i} \frac{d \rho}{d\rho} \times \frac{\partial}{\partial x_i} \rho$

Then
$\frac{\partial}{\partial x_i} \frac{d }{d\rho} \rho \frac{\partial}{\partial x_i} \rho=2 \frac{\partial}{\partial x_i} \frac{\partial}{\partial x_i}\rho$?

Which means: $\frac{d }{d\rho} (\nabla \rho)^2=2\nabla^2 \rho$

I would like to know if that is correct. And if it isn't, where I am committing the mistake.

 

[etotheipi]

I really can't follow what you did there, I'm not sure why you kept interchanging the orders of the $d/d \rho$ and $\partial / \partial x^i$ operators. I don't think you can simplify past $\frac{d}{d\rho} (\nabla \rho)^2 = \frac{d}{d\rho} \left( \frac{\partial \rho}{\partial x^i} \frac{\partial \rho}{\partial x^i} \right) = 2 \frac{\partial \rho}{\partial x^i} \frac{d}{d\rho} \left(\frac{\partial \rho}{\partial x^i} \right)$, with summation understood over $i$.

 

[hilbert2]

I'm not sure if this is of any help, but in Section 3 of this article there's a justification for that kinetic energy functional in the case of 1-electron system, i.e. when $\rho (\mathbb{r}) = |\psi (\mathbb{r})|^2$.

http://przyrbwn.icm.edu.pl/APP/PDF/115/a115z310.pdf

 

[JhonDoe]

I really can't follow what you did there, I'm not sure why you kept interchanging the orders of the $d/d \rho$ and $\partial / \partial x^i$ operators. I don't think you can simplify past $\frac{d}{d\rho} (\nabla \rho)^2 = \frac{d}{d\rho} \left( \frac{\partial \rho}{\partial x^i} \frac{\partial \rho}{\partial x^i} \right) = 2 \frac{\partial \rho}{\partial x^i} \frac{d}{d\rho} \left(\frac{\partial \rho}{\partial x^i} \right)$, with summation understood over $i$.

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

 

[TeethWhitener]

I’m not a mathematician, so this argument will be VERY heuristic, but nonentheless, here’s how I justify to myself that a function and its derivative in a functional are to be treated as independent variables:

We start with $\frac{d}{d\rho} = \frac{d}{dx}\frac{dx}{d\rho}$ and when we plug this back into our original expression, we find that we have pieces that look like $\frac{d}{dx}\frac{dx}{d\rho}\frac{d\rho}{dx}$. Continuing to be VERY HEURISTIC AND NOT AT ALL RIGOROUS, we note that with the right assumptions and well-behaved enough functions, $\frac{dx}{d\rho}\frac{d\rho}{dx} =1$. Then taking the derivative of 1 gives you the zero that you need.

Edit: here’s a much more rigorous explanation for your question:
https://math.stackexchange.com/ques...-and-y-be-independent-in-variational-calculus

 

[kimbyd]

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 haven't read this thread in too much detail, but it really seems like there's some confusing language in this thread about how functional derivatives are distinguished from normal derivatives. In particular, I don't think Taylor expansion is the way to go. What you probably want to do is use the formula described at this section of the linked Wikipedia page:
https://en.wikipedia.org/wiki/Functional_derivative#Determining_functional_derivatives

The key point is to consider the integral to be a function of two independent parameters: $\rho$ and $\nabla \rho$. Then the functional derivative becomes:
$${\delta f(\rho, \nabla \rho) \over \delta \rho} = {\partial f \over \partial \rho} - \nabla \cdot {\partial f \over \partial \nabla \rho}$$

So to answer your question at the bottom, $\frac{\partial }{\partial \rho} \nabla \rho(\mathbf{r}) \cdot \nabla \rho(\mathbf{r}) = 0$. The other term comes from taking the partial with respect to $\nabla \rho$.