#### Hans de Vries

Science Advisor

Gold Member

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Typically you want to link the delta function and its derivatives to the differential operators,

[tex]

\begin{aligned}

&~~\delta(t)&\ast~~& f(t) &= ~~&~~f(t) \\ \\

&~\frac{\partial \delta(t)}{\partial t}&\ast~~ &f(t) &= ~~&~\frac{\partial f(t)}{\partial t} \\ \\

&\frac{\partial^2 \delta(t)}{\partial t^2}&\ast~~& f(t) &= ~~&\frac{\partial^2 f(t)}{\partial t^2} \\ \\

\end{aligned}

[/tex]

for consistent operations, where [itex]\ast[/itex] denotes convolution. This is a symmetric definition of [itex]\delta(x)[/itex],

and corresponds with 1/2 for the half-space integral.

Note that higher order derivatives of the delta function occur in physical propagators

as soon as you work in more then three dimensions. See for instance:

http://physics-quest.org/Higher_dimensional_EM_radiation.pdf

Regards, Hans

[tex]

\begin{aligned}

&~~\delta(t)&\ast~~& f(t) &= ~~&~~f(t) \\ \\

&~\frac{\partial \delta(t)}{\partial t}&\ast~~ &f(t) &= ~~&~\frac{\partial f(t)}{\partial t} \\ \\

&\frac{\partial^2 \delta(t)}{\partial t^2}&\ast~~& f(t) &= ~~&\frac{\partial^2 f(t)}{\partial t^2} \\ \\

\end{aligned}

[/tex]

for consistent operations, where [itex]\ast[/itex] denotes convolution. This is a symmetric definition of [itex]\delta(x)[/itex],

and corresponds with 1/2 for the half-space integral.

Note that higher order derivatives of the delta function occur in physical propagators

as soon as you work in more then three dimensions. See for instance:

http://physics-quest.org/Higher_dimensional_EM_radiation.pdf

Regards, Hans

Last edited: