Kramers-Kronig, parity and delta function

[happyparticle]

Homework Statement

Let $Im[\epsilon_r(\omega)] = \delta(\omega - \omega_0)$
Add a peak to the parity relation at $\omega = -\omega_0$

Find $Re[\epsilon_r(\omega)]$ using kramers-kronig

Relevant Equations

$Im[\epsilon_r(\omega)] = \delta(\omega - \omega_0)$

$Re[\epsilon_r(\omega)] = 1 + \frac{P}{\pi} \int_{-\infty}^{\infty} \frac{Re[\epsilon_r(\omega ') ]}{\omega ' - \omega} d \omega '$

Hi,
First of all, I'm not sure to understand what he Kramers-kronig do exactly. It is used to get the Real part of a function using the imaginary part?

Then, when asked to add a peak to the parity at $\omega = -\omega_0$, is $Im[\epsilon_r(\omega)] = \delta(\omega^2 - \omega_0 ^2)$ correct?
Then should I plug the relation above in the Kramers-kronig relation?
I hope my questions are clear.

Thanks

 

[vanhees71]

First, the "relevant equation" should read
$$\mathrm{Re}[\epsilon_r(\omega)]=1+\frac{1}{\pi} \mathcal{P} \int_{\mathbb{R}} \mathrm{d} \omega' \frac{\text{Im} \epsilon_r(\omega')}{\omega'-\omega}.$$
This formula, of course cannot be applied directly to your case, where $\mathrm{Im} \epsilon_r$ consists of $\delta$-distributions.

Here the trick is to take the "causal realization" of the $\delta$-distributions, i.e., using the fact that the response functions must be analytic in the upper half-plain of the complex $\omega'$ plane (when using the physicists' convention for the Fourier integral, i.e.,
$$\tilde{f}(\omega)=\int_{\mathbb{R}} \mathrm{d} t f(t) \exp(\mathrm{i} \omega t), \quad f(t)=\int_{\mathbb{R}} \mathrm{d} \omega \exp(-\mathrm{i} \omega t) \tilde{f}(\omega),$$
because then $f$ is "retarded" in the sense that $f(t) \propto \Theta(t)$, because you have to close the contour when calculating the Fourier integral wrt. $\omega$ in the upper (lower) plane for $t<0$ ($t>0$), and due to the theorem of residues then this integral vanishes for $t<0$.

From this analyticity you derive the Kramers-Kronig relations in the first place. Here you realize the $\delta$ distribution in the "causal way", i.e., by
$$\delta_{\epsilon}(\omega-\omega_0)=\frac{1}{\pi} \mathrm{Im} \frac{1}{\omega-\omega_0+\mathrm{i} \epsilon}, \quad \epsilon>0,$$
which only has a simple pole in the lower complex plane at $\omega_0-\mathrm{i} \epsilon$, and then
$$\delta(\omega-\omega_0) = \lim_{\epsilon \rightarrow 0^+} \mathrm{im} \frac{1}{\omega-\omega_0+\mathrm{i} \epsilon},$$
where the limit is to be understood in the sense of distributions.

Of course this already gives you the solution, i.e.,
$$\epsilon_r(\omega)=1+\frac{1}{\omega-\omega_0 + \mathrm{i} 0^+}$$
and then of course the analogous calculation for the other pole at $-\omega_0$.

A pretty nice summary about the Kramers-Kronig relation is here:

https://webhome.phy.duke.edu/~rgb/Class/phy319/phy319/node56.html

 

[happyparticle]

From this analyticity you derive the Kramers-Kronig relations in the first place. Here you realize the δ distribution in the "causal way", i.e., by
δϵ(ω−ω0)=1πIm1ω−ω0+iϵ,ϵ>0,

I'm not really sure where this came from.
I don't really see the link between the questions asked in the homework and your explanation.
I thought that since I have to find an expression for $Im[\epsilon_r(\omega)]$ for both $\delta(\omega - \omega_0)$ and $\omega = - \omega_0$ I could use KK relation directly.
I guess, I don't understand what it asked me to do.

 

[vanhees71]

The problem is that there's no definition for the principal value of an integral applied to a $\delta$ distribution. That's why I proposed to use an adequate limiting procedure. "Adequate" in this case means that you have to use an approximation of the $\delta$ distribution that respects the analyticity properties, which guarantee causality, i.e., the use of the retarded Green's function of the d'Alembert operator. At least that's the usual handling applied in the real-time formalism of QFT. The complete spectral function is something like
$$\mathrm{im}[\epsilon_r(\omega)] \propto \delta(\omega-\omega_0)-\delta(\omega+\omega_0),$$
which obeys the reality condition of the em. field in the time domain, i.e.,
$$\epsilon(\omega)=\epsilon^*(-\omega), \Rightarrow \mathrm{Re} \epsilon(\omega)=\mathrm{Re} \epsilon(-\omega), \quad \mathrm{Im} \epsilon(\omega)=-\mathrm{Im} \epsilon(-\omega).$$
Now use the "regularized" $\delta$ distributions $\delta_{\epsilon}(\omega \pm \omega_0)$, calculate $\epsilon_r(\omega)$ from the Kramers-Kronig relation for this regularized function and then take the limit $\epsilon \rightarrow +0$.

 

[TSny]

Then, when asked to add a peak to the parity at $\omega = -\omega_0$, is $Im[\epsilon_r(\omega)] = \delta(\omega^2 - \omega_0 ^2)$ correct?

In general, is $Im[\epsilon_r(\omega)]$ even or odd in $\omega$?

Is $\delta(\omega^2 - \omega_0 ^2)$ even or odd in $\omega$?

Then should I plug the relation above in the Kramers-kronig relation?

Once you get the correct expression for $Im[\epsilon_r(\omega)]$ that includes the peak at $\omega = -\omega_0$, then I think you can just plug into the KK relation. [EDIT: I just saw that @vanhees71 has essentially given you the correct expression for $Im[\epsilon_r(\omega)]$ in post #4.]

Now, vanhees71 is bringing up some issues with applying the KK relation to a dirac delta function. This is above my pay grade. However, in Jackson's E&M book, there is an example where the KK relation is used to find $Re[\epsilon_r(\omega)]$ from $Im[\epsilon_r(\omega)]$ for a case where $Im[\epsilon_r(\omega)]$ contains dirac delta functions. They just plug the expression for $Im[\epsilon_r(\omega)]$ into the integrand without worrying and get a result straight away.

 

[vanhees71]

Perhaps I'm too careful with the $\delta$ distribution. I'm not sure whether one can take a principal-value integral in such a case, but usually indeed the Dirac distribution is pretty benevolent ;-)).