Delta function and derivative of function wrt itself

  • Thread starter Trave11er
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  • #1
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Can someone explain the following profound truth: [tex]\frac {\partial f(x)}{\partial f(y)} =\delta(x-y)[/tex]
 

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  • #2
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The expression [itex]\dfrac{\partial f(x)}{\partial g(y)}[/itex] is really just a short-hand, but there are a handful of rules about which manipulations "usually work" with the short-hand. It turns out that the convention you've named---that [itex]\dfrac{\partial f(x)}{\partial f(y)}[/itex] is [itex]1[/itex] if [itex]x=y[/itex] and [itex]0[/itex] if [itex]x\neq y[/itex]---gives us a slightly broader notion of "usually".
 
  • #3
arildno
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When you see the Dirac delta function in action, you ought to remind yourself:
"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
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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]
 
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