Kramers-Kronig Relations: Principal Value

[CharlieCW]

Homework Statement

Using Kramers-Kronig relations, find the imaginary part of $\epsilon(\omega)$ when the real part of the the dielectric function is that of a plasma, that is, $\epsilon(\omega)=1-\omega^2_p/\omega^2$. What type of absorption does this describe?

Relevant Equations

Kramers-Kronig relations

$$\chi_1(\omega)=\frac{1}{\pi}P\int_{-\infty}^{\infty}d\Omega\frac{\chi_2(\Omega)}{\Omega-\omega}$$

$$\chi_2(\omega)=-\frac{1}{\pi}P\int_{-\infty}^{\infty}d\Omega\frac{\chi_1(\Omega)}{\Omega-\omega}$$

Principal value definition

$$P\int_{-\infty}^{\infty}dx f(x)=\lim_{c\rightarrow 0^+}\left[ \int_{-\infty}^{\eta-c/2}dx f(x)+\int_{\eta+c/2}^{\infty}dx f(x)\right ]$$

I'm kind of confused on how to evaluate the principal value as it's a topic I've never seen in complex analysis and all the literature I've read so far only deals with the formal definition, not providing an example on how to calculate it properly. Therefore, I think just understanding at least a couple of examples will be enough to understand in how to deal with these.

From the dielectric function, we can see that,

$$\chi_1(\omega)=\omega_p^2/\omega^2$$

Introducing it into the Kramers-Kronig relations,

$$\chi_2(\omega)=-\frac{\omega_p^2}{\pi}P\int_{-\infty}^{\infty}\frac{d\Omega}{\Omega^2(\Omega-\omega)}$$

If I try to use the definition of the principal value, we note that we have three poles, one in $\Omega=0$ and two in $\Omega=\pm \omega$. Let's consider the first one. Substituting and using Mathematica,

$$\lim_{c\rightarrow 0^+}\left[ \int_{-\infty}^{-c/2}\frac{d\Omega}{\Omega^2(\Omega-\omega)} +\int_{+c/2}^{\infty}\frac{d\Omega}{\Omega^2(\Omega-\omega)}\right ]=-sgn(\omega)(\infty)$$

Which obviously doesn't make sense. A similar result is yield when evaluating the other poles, so clearly I must be doing something wrong as I have no idea how to evaluate these integrals with their proper limits.

I also tried using the residue theorem just out of curiosity, but I only get that $\chi_2(\omega)=-2i\omega_p^2/\omega^2$ which makes less sense as now I'm saying that a real number is imaginary, so clearly this is now the way to go.

I would like to understand how to calculate these relations, so I'll really appreciate any help.

 

[jasonRF]

This is a tricky question. As you have seen, you get nasty singularities along the integration path. When this happens, one approach is to "work backwards" and see if you can be clever enough to figure it out. Start with the reverse,
$$
\chi_1 = \frac{1}{\pi} P\int_{-\infty}^\infty \frac{\chi_2(\Omega)}{\Omega-\omega}
$$

If you had $\chi_1=-\omega_0/\omega$, what would $\chi_2$ need to be? EDIT: just to be explicit, I am asking you to basically use the "method of guessing", as opposed to using fancy math to solve the integral equation.

If you can figure this out, going the next step to solving your problem should be pretty straightforward.

Jason

 

[CharlieCW]

This is a tricky question. As you have seen, you get nasty singularities along the integration path. When this happens, one approach is to "work backwards" and see if you can be clever enough to figure it out. Start with the reverse,
$$
\chi_1 = \frac{1}{\pi} P\int_{-\infty}^\infty \frac{\chi_2(\Omega)}{\Omega-\omega}
$$

If you had $\chi_1=-\omega_0/\omega$, what would $\chi_2$ need to be? EDIT: just to be explicit, I am asking you to basically use the "method of guessing", as opposed to using fancy math to solve the integral equation.

If you can figure this out, going the next step to solving your problem should be pretty straightforward.

Jason

Thanks for your reply, Jason. Just from the form of the equations, if I had $\chi_1=-\omega_0/\omega$, then $\chi_2=\pi\omega_0/2$, as when taking the limits only the evaluation on zero will lead to two non-zero identical terms. Here I have no problem evaluating the limits of the integral.

Now I tried going in reverse and finding $\chi_1$ given $\chi_2=\pi\omega_0/2$ but when I try to evaluate the integral directly I get infinity due to the additional singularity in $\omega=0$, and I don't know how to recover the $\pi$ factor on the numerator.

Is there a way to connect the principal value to a contour integral so I can apply the residue theorem?

 

[jasonRF]

Thanks for your reply, Jason. Just from the form of the equations, if I had $\chi_1=-\omega_0/\omega$, then $\chi_2=\pi\omega_0/2$

That is not correct. Try again

Hint: think generalized functions. The answer to your question is not an analytic function.

 

[jasonRF]

I realized I didn't answer all of your questions above...

Is there a way to connect the principal value to a contour integral so I can apply the residue theorem?

Not really. The real part of your susceptibility has a double pole at the origin, and the trick of indenting the contour at a simple pole and taking the limit does not extend to a double pole. You need to do something different.

Jason

 

[CharlieCW]

That is not correct. Try again

Hint: think generalized functions. The answer to your question is not an analytic function.

You're right, I had taken the wrong limits of integration. Sorry for the late reply by the way, I had an exam this morning.

Now continuing with the equations, let's start from the beginning. Assuming $\chi_1=\omega^2_p/\omega$, then we can substitute in the Kramers-Kronig relation as,

$$\frac{\omega_p^2}{\omega}=\frac{1}{\pi}P\int_{-\infty}^{\infty}\frac{\chi_2(\Omega)d\Omega}{\Omega-\omega}$$

We can see that at least $\chi_2(\Omega)$ should be proportional to $\pi\omega_p^2$ to cancel the proper terms. This leaves us with,

$$1=P\int_{-\infty}^{\infty}\frac{\chi'_2(\Omega)d\Omega}{\Omega-\omega}$$

Since the only function that satisfies that being integrated from $-\infty$ to $+\infty$ is a Dirac delta, then I should have $\chi'_2(\Omega)=\delta(\Omega)(\Omega-\omega)$ so that,

$$1=P\int_{-\infty}^{\infty}\delta(\Omega)d\Omega$$

Therefore we finally get that $\chi_2(\omega)=\pi\omega_p^2\delta(\omega)(\omega-\omega')$. However, the result seems a bit weird as $\chi_2$ should only depend on $\omega$, and instead it depends on two variables. Would this be correct?

I also checked about the notion of generalized functions, as I've never seen heard about them before, and while the mathematical jargon is a bit complex, I understand it's just a generalization of the notion of functions and applied in cases of making discontinuous functions more like smooth functions. Under this logic, the Dirac delta $\delta(x)$ seems to be a type of generalized function (correct me if I'm wrong). I also found in a rather obscure reference the following identity,

$$\lim_{\epsilon\rightarrow 0}\frac{\epsilon}{\epsilon^2+\omega^2}=\pi\delta(\omega)$$

which could be useful to solve directly the integral.

 

[jasonRF]

Now continuing with the equations, let's start from the beginning. Assuming $\chi_1=\omega^2_p/\omega$, then we can substitute in the Kramers-Kronig relation as,

$$\frac{\omega_p^2}{\omega}=\frac{1}{\pi}P\int_{-\infty}^{\infty}\frac{\chi_2(\Omega)d\Omega}{\Omega-\omega}$$

We can see that at least $\chi_2(\Omega)$ should be proportional to $\pi\omega_p^2$ to cancel the proper terms. This leaves us with,

$$1=P\int_{-\infty}^{\infty}\frac{\chi'_2(\Omega)d\Omega}{\Omega-\omega}$$

Since the only function that satisfies that being integrated from $-\infty$ to $+\infty$ is a Dirac delta, then I should have $\chi'_2(\Omega)=\delta(\Omega)(\Omega-\omega)$ so that,

$$1=P\int_{-\infty}^{\infty}\delta(\Omega)d\Omega$$

Therefore we finally get that $\chi_2(\omega)=\pi\omega_p^2\delta(\omega)(\omega-\omega')$. However, the result seems a bit weird as $\chi_2$ should only depend on $\omega$, and instead it depends on two variables. Would this be correct?

Which problem are you doing now - your original $\chi_1=-\omega^2_p/\omega^2$ or the "toy" problem I gave with $\chi_1=-\omega_0/\omega$?

Assuming the later, With this, using your definitions of the K-K relations I get,
$$-\frac{\omega_0}{\omega}= \frac{1}{\pi}P\int_{-\infty}^{\infty}\frac{\chi_2(\Omega)d\Omega}{\Omega-\omega}$$
or, using your notation with $\chi_2 = \pi \omega_0 \chi_2^\prime$,
$$-\frac{1}{\omega}= P\int_{-\infty}^{\infty}\frac{\chi_2^\prime(\Omega)d\Omega}{\Omega-\omega}.$$
This is not quite the same thing that you found, but you clearly know enough to solve this by inspection.

The same general approach works for the original problem you posted.

I also checked about the notion of generalized functions, as I've never seen heard about them before, and while the mathematical jargon is a bit complex, I understand it's just a generalization of the notion of functions and applied in cases of making discontinuous functions more like smooth functions. Under this logic, the Dirac delta $\delta(x)$ seems to be a type of generalized function (correct me if I'm wrong). I also found in a rather obscure reference the following identity,

$$\lim_{\epsilon\rightarrow 0}\frac{\epsilon}{\epsilon^2+\omega^2}=\pi\delta(\omega)$$

which could be useful to solve directly the integral.

If you know about Dirac delta functions and their derivatives, then you know enough about generalized functions to do this problem. Don't worry if you haven't learned about generalized functions in any other way. From your original question it wasn't obvious what math you knew or were allowed to use on this problem.

The identity you found actually isn't so obscure. It turns out that you can also differentiate both sides of that identity and swap the order of the limit and the differentiation (while this does not always hold for analytic functions, it is always allowed in the theory of generalized functions). In any case, yes, those kinds of identities can be useful for solving some of these kinds of problems. An identity that is sometimes more useful is the Plemelj,
$$ \lim_{\epsilon\rightarrow 0^+} \frac{1}{\omega \pm i \epsilon} = P\frac{1}{\omega}\mp i \pi \delta(\omega)$$
which can be used to derive the Kramers-Kronig relations.

jason

 

[jasonRF]

By the way, the Kramers-Kronig relations also hold for $\omega \, \chi(\omega)$, which is proportional to the conductivity. A classic book, Waves in Plasmas by Stix, does it this way. With this approach, you can allow a simple pole in the conductivity (double pole in $\chi$, like in your case) and still do standard principle value kinds of math.

jason

 

[Charles Link]

The mathematics is quite difficult, but you may find this of interest: https://www.spaceacademy.net.au/spacelink/blackout.htm
Index of refraction $n=\sqrt{\epsilon (\omega)}$ becomes imaginary for $\omega < \omega_p$.
That means the wave $E(x,t)=Re[ Ae^{i(nk_ox-\omega t)}]=Re[A e^{i ((n_r+in_i)k_o x-\omega t)}] =Re[Ae^{-n_i k_o x} e^{i(n_r k_o-\omega t)}]$ does not propagate.
$n_r=0$, and $n_i$ is quite large.
The wave number $k_o=\frac{2 \pi}{\lambda_{vacuum}}$