Dirac delta from poles of a function
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Suppose we are given the simple expression
$$
F(k) = frac1E^2-E(k)^2
$$
which has a pole when $E^2 = E(k)^2$ and where $E, E(k)$ are real numbers. When working with this expression (e.g. inside an integral) we can add some infinitesimal constant $epsilon >0$ to take care of the poles and let this $epsilon$ go to zero in the end.
What I want to show now is that the imaginary part of the square of $F$ is proportional to a delta function $deltaleft( E^2 - E(k)^2 right)$. What I tried was simply to calculate
$$
F^2 = lim_epsilonto 0frac1(E^2+E(k)^2+i epsilon)^2 =
lim_etato 0 frac1(E^2-E(k)^2)^2 + ieta(E^2-E(k)^2)
$$
where $eta = 2epsilon$ and where I have set $epsilon^2 = 0$. My idea was to use the representation of the delta function
$$
pidelta(x) = lim_eta to 0 fracetaeta^2+x^2
$$
but I failed to bring $F^2$ to this form. Is there any way to do this anyway? For example by introducing an infinitesimal parameter somewhere else in the fraction? Or maybe by doing something completely different?
limits fractions dirac-delta regularization analytic-continuation
asked Aug 14 at 8:55
MeMeansMe
1354
add a comment |Â
up vote
3
down vote
favorite
Suppose we are given the simple expression
$$
F(k) = frac1E^2-E(k)^2
$$
which has a pole when $E^2 = E(k)^2$ and where $E, E(k)$ are real numbers. When working with this expression (e.g. inside an integral) we can add some infinitesimal constant $epsilon >0$ to take care of the poles and let this $epsilon$ go to zero in the end.
What I want to show now is that the imaginary part of the square of $F$ is proportional to a delta function $deltaleft( E^2 - E(k)^2 right)$. What I tried was simply to calculate
$$
F^2 = lim_epsilonto 0frac1(E^2+E(k)^2+i epsilon)^2 =
lim_etato 0 frac1(E^2-E(k)^2)^2 + ieta(E^2-E(k)^2)
$$
where $eta = 2epsilon$ and where I have set $epsilon^2 = 0$. My idea was to use the representation of the delta function
$$
pidelta(x) = lim_eta to 0 fracetaeta^2+x^2
$$
but I failed to bring $F^2$ to this form. Is there any way to do this anyway? For example by introducing an infinitesimal parameter somewhere else in the fraction? Or maybe by doing something completely different?
limits fractions dirac-delta regularization analytic-continuation
asked Aug 14 at 8:55
MeMeansMe
1354
I would like to know if there is any particular for which you need to consider the square of $F$. As a matter of fact, $F$ itself can be analytically continued as the (Fantappié) indicatrix of the distribution $deltabig(E-E(k)big)$: this seems to rule out the possibility to do the same thing for its square $F^2$.
– Daniele Tampieri
Aug 14 at 17:55
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Suppose we are given the simple expression
$$
F(k) = frac1E^2-E(k)^2
$$
which has a pole when $E^2 = E(k)^2$ and where $E, E(k)$ are real numbers. When working with this expression (e.g. inside an integral) we can add some infinitesimal constant $epsilon >0$ to take care of the poles and let this $epsilon$ go to zero in the end.
What I want to show now is that the imaginary part of the square of $F$ is proportional to a delta function $deltaleft( E^2 - E(k)^2 right)$. What I tried was simply to calculate
$$
F^2 = lim_epsilonto 0frac1(E^2+E(k)^2+i epsilon)^2 =
lim_etato 0 frac1(E^2-E(k)^2)^2 + ieta(E^2-E(k)^2)
$$
where $eta = 2epsilon$ and where I have set $epsilon^2 = 0$. My idea was to use the representation of the delta function
$$
pidelta(x) = lim_eta to 0 fracetaeta^2+x^2
$$
but I failed to bring $F^2$ to this form. Is there any way to do this anyway? For example by introducing an infinitesimal parameter somewhere else in the fraction? Or maybe by doing something completely different?
limits fractions dirac-delta regularization analytic-continuation
asked Aug 14 at 8:55
MeMeansMe
1354
Suppose we are given the simple expression
$$
F(k) = frac1E^2-E(k)^2
$$
which has a pole when $E^2 = E(k)^2$ and where $E, E(k)$ are real numbers. When working with this expression (e.g. inside an integral) we can add some infinitesimal constant $epsilon >0$ to take care of the poles and let this $epsilon$ go to zero in the end.
What I want to show now is that the imaginary part of the square of $F$ is proportional to a delta function $deltaleft( E^2 - E(k)^2 right)$. What I tried was simply to calculate
$$
F^2 = lim_epsilonto 0frac1(E^2+E(k)^2+i epsilon)^2 =
lim_etato 0 frac1(E^2-E(k)^2)^2 + ieta(E^2-E(k)^2)
$$
where $eta = 2epsilon$ and where I have set $epsilon^2 = 0$. My idea was to use the representation of the delta function
$$
pidelta(x) = lim_eta to 0 fracetaeta^2+x^2
$$
but I failed to bring $F^2$ to this form. Is there any way to do this anyway? For example by introducing an infinitesimal parameter somewhere else in the fraction? Or maybe by doing something completely different?
limits fractions dirac-delta regularization analytic-continuation
asked Aug 14 at 8:55
MeMeansMe
1354
asked Aug 14 at 8:55
MeMeansMe
1354
asked Aug 14 at 8:55
MeMeansMe
1354
asked Aug 14 at 8:55
MeMeansMe
1354
1354
I would like to know if there is any particular for which you need to consider the square of $F$. As a matter of fact, $F$ itself can be analytically continued as the (Fantappié) indicatrix of the distribution $deltabig(E-E(k)big)$: this seems to rule out the possibility to do the same thing for its square $F^2$.
– Daniele Tampieri
Aug 14 at 17:55
add a comment |Â
I would like to know if there is any particular for which you need to consider the square of $F$. As a matter of fact, $F$ itself can be analytically continued as the (Fantappié) indicatrix of the distribution $deltabig(E-E(k)big)$: this seems to rule out the possibility to do the same thing for its square $F^2$.
– Daniele Tampieri
Aug 14 at 17:55
I would like to know if there is any particular for which you need to consider the square of $F$. As a matter of fact, $F$ itself can be analytically continued as the (Fantappié) indicatrix of the distribution $deltabig(E-E(k)big)$: this seems to rule out the possibility to do the same thing for its square $F^2$.
– Daniele Tampieri
Aug 14 at 17:55
I would like to know if there is any particular for which you need to consider the square of $F$. As a matter of fact, $F$ itself can be analytically continued as the (Fantappié) indicatrix of the distribution $deltabig(E-E(k)big)$: this seems to rule out the possibility to do the same thing for its square $F^2$.
– Daniele Tampieri
Aug 14 at 17:55
add a comment |Â
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I would like to know if there is any particular for which you need to consider the square of $F$. As a matter of fact, $F$ itself can be analytically continued as the (Fantappié) indicatrix of the distribution $deltabig(E-E(k)big)$: this seems to rule out the possibility to do the same thing for its square $F^2$.
– Daniele Tampieri
Aug 14 at 17:55