- #1

- 71

- 0

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Trave11er
- Start date

- #1

- 71

- 0

- #2

- 269

- 24

- #3

arildno

Science Advisor

Homework Helper

Gold Member

Dearly Missed

- 10,025

- 135

"How will this look like when I integrate the expression?"

I haven't seen the result you post before, but do remember the two following results:

[tex]\int_{-\infty}^{\infty}f(x)\delta{(x-y)}dx=f(y)[/tex]

[tex]\int_{-\infty}^{\infty}f(y)\delta{(x-y)}dy=f(x)[/tex]

Most likely, your result follows from some clever manipulation of these two basic results.

- #4

- 71

- 0

Thanks for the replies.

Have found the answer in Parr and Yang's book on Density Functional Theory:

For a functional F=F[f] have

[tex]\delta F = \int \frac {\delta F} {\delta f(y)} \delta f(y) \,dy[/tex]

In a special case that F=F(f), i.e. F is just some function of f it is required that:

[tex]\frac {\delta F(f(x))} {\delta f(y)} = \frac {dF}{df} \delta(x-y) [/tex] in order to have:

[tex] \delta F = \int \frac {\delta F} {\delta f(y)} \delta f(y) \,dx = \frac {dF} {df} \delta f(x) [/tex]

So that taking F = f, get:

[tex] \frac {\delta f(x)} {\delta f(y)} = \delta (x-y) [/tex]

Have found the answer in Parr and Yang's book on Density Functional Theory:

For a functional F=F[f] have

[tex]\delta F = \int \frac {\delta F} {\delta f(y)} \delta f(y) \,dy[/tex]

In a special case that F=F(f), i.e. F is just some function of f it is required that:

[tex]\frac {\delta F(f(x))} {\delta f(y)} = \frac {dF}{df} \delta(x-y) [/tex] in order to have:

[tex] \delta F = \int \frac {\delta F} {\delta f(y)} \delta f(y) \,dx = \frac {dF} {df} \delta f(x) [/tex]

So that taking F = f, get:

[tex] \frac {\delta f(x)} {\delta f(y)} = \delta (x-y) [/tex]

Last edited:

Share: