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The harmonic spectra are calculated as $|FT(\frac{d}{dt}\mathcal{J}(t))|^2$, where $FT$ si the Fourier Transform and $\mathcal{J}(t)$ is the current. We need to identify which multiple of incident frequency is contained in the harmonic spectra. Now as much as I know the helicity is defined as $\frac{I_{\circlearrowright} - I_{\circlearrowleft}}{I_{\circlearrowright} + I_{\circlearrowleft}}$ where $I_{\circlearrowright} (I_{\circlearrowleft})$ is the component of the harmonic intensity rotating clockwise (anticlockwise).

Here, I am confused how to perform the calculation of helicity. If I have two component data of $\mathcal{J}(t)=\mathcal{J}_x(t)\hat{i}+\mathcal{J}_y(t)\hat{j}$, i.e. for each time I have $\mathcal{J}_x(t)$ and $\mathcal{J}_y(t)$. I can calculate harmonic spectra from this data by doing the FFT of the first-time derivative of two current components. However, I do not have an idea how to calculate helicities from this data. How to determine $I_{\circlearrowright} (I_{\circlearrowleft})$? If someone wants to know what is the helicity of third harmonics how to find that?

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Typically, when you are calculating the emission from a given source, the output will be a time-dependent current. You wrote this as $\mathcal{J}(t)$, but I will rewrite this as $\mathbf J(t)$ in order to emphasize that the current (or other such quantity, such as the derivative of a dipole moment) is a vector quantity.

As you rightly point out, if we want to calculate the harmonic spectrum, the first thing to do is to take a Fourier transform, i.e., we write $$ \tilde{\mathbf{J}}(\omega) = \frac{1}{\sqrt{2\pi}} \int \mathbf J(t) e^{i\omega t} \mathrm dt $$ in order to get the (vector) spectral amplitude $\tilde{\mathbf{J}}(\omega)$ of the current, and then we know that we can decompose the time-dependent current as a superposition of monochromatic oscillations in the form $$ \mathbf J(t) = \frac{1}{\sqrt{2\pi}} \int \tilde{\mathbf{J}}(\omega) e^{-i\omega t} \mathrm d\omega . $$

So, what about the helicity? Well, that only makes sense on a harmonic-per-harmonic basis, because the full time-dependent $\mathbf J(t)$ might have a rather more complicated shape than just an ellipse. But if you filter down to the contribution of just one of the monochromatic oscillations,$$ \mathbf J_\omega(t) = \operatorname{Re}\mathopen{}\left[ \tilde{\mathbf{J}}(\omega) e^{-i\omega t} \right]\mathclose{} = \frac12\left[ \tilde{\mathbf{J}}(\omega) e^{-i\omega t} + \tilde{\mathbf{J}}(-\omega) e^{+i\omega t} \right] $$ (where necessarily $\tilde{\mathbf{J}}(-\omega) = \tilde{\mathbf{J}}(\omega)^*$ to keep $\mathbf J(t)$ real-valued), then this is a vector-valued monochromatic wave which can indeed be elliptical.

So, this is where we define the ellipticity (and, with it, the helicity), by taking the components of this vector oscillation along the right- and left-handed circular basis. Thus, we define \begin{align} I_\circlearrowright(\omega) & = |\hat{\mathbf{e}}_\circlearrowright \cdot \tilde{\mathbf{J}}(\omega)|^2 \\ I_\circlearrowleft(\omega) & = |\hat{\mathbf{e}}_\circlearrowleft\cdot \tilde{\mathbf{J}}(\omega)|^2, \end{align} where $\hat{\mathbf{e}}_\circlearrowright = \tfrac{1}{\sqrt{2}}(\hat{\mathbf{e}}_x+ i\hat{\mathbf{e}}_y)$ and $\hat{\mathbf{e}}_\circlearrowleft = \tfrac{1}{\sqrt{2}}( \hat{\mathbf{e}}_x- i\hat{\mathbf{e}}_y)$. These are the quantities used to define the spectral ellipticity (or spectral helicity) as $$ \varepsilon(\omega) = \frac{I_{\circlearrowright}(\omega) - I_{\circlearrowleft}(\omega)}{I_{\circlearrowright}(\omega) + I_{\circlearrowleft}(\omega)} . $$

(And, finally, about the name: there isn't a particularly strongly accepted convention about what to call ellipticity / signed ellipticity / helicity. Just use one that feels right to you, but use it consistently. My approach would be to call the above quantity the ellipticity, allowing it to be both positive and negative, and then denote the helicity as $h=\operatorname{sgn}(\varepsilon)$. But there's plenty of other possible approaches!)

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  • $\begingroup$ In this case the $I$ is related to the current, or the first time derivative of the current? I found "doi.org/10.1103/PhysRevB.102.134115" In this paper, they have defined this quantity in eq. 46 for HHG, which is first time derivative of the current, not only current! @Emilio Pisanty $\endgroup$ Commented Nov 15, 2023 at 10:10
  • $\begingroup$ It's quite interesting to use "they" for an author list that includes me. The material in this answer works equally well for the current, the dipole, and any of their time derivatives. $\endgroup$ Commented Nov 15, 2023 at 18:03
  • $\begingroup$ Apologies! I just overlooked the coincidence. Just one more doubt is that the definition of circular dichroism in your paper is $\frac{I^+_{k,RCP}-I^-_{k,LCP}}{I^+_{k,RCP}+I^-_{k,LCP}}$, I hope that is what you defined as $\epsilon(\omega)$ in the answer. In your answer, you have not mentioned how the current is being generated, i.e. with LCP light puls or RCP light puls, but this driving puls ellipticity is essential in that definition of CD as in the paper. $\endgroup$ Commented Nov 16, 2023 at 6:41
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You need to specify the field of application (i.e., the area in physics) as helicity may be defined differently.

General approach:

  1. Figure out what helicity is and how it is derived from $I_x(t), I_y(t)$ for your specific application.
  2. Then you can see how it behaves under FT.

Typically, helicity-related things are treated using complex amplitudes: $$\vec{A} = \vec{A}_\mathrm{complex} = \vec{A}_x \pm i\vec{A}_y.$$ Assuming harmonic evolution, this would result in $\vec{A}_\mathrm{observed}(t) = \mathrm{Re}[\vec{A} e^{i\omega t}]$. You can see how this would result in rotation: $\vec{A}_\mathrm{observed}(0) = \vec{A}_x, \vec{A}_\mathrm{observed}(\pi/2\omega)=\mp\vec{A}_y$ and so on.

Also $I_↻$ and $I_↺$ together contain the same amount of information as $I_x$ together with $I_y$. But $|I|$ and other expressions with modulus contain less information, so it is impossible to restore both $I_↻$ and $I_↺$ from it.

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  • $\begingroup$ This is not quite right. Typically, the notation $I_x$ (etc) would indicate that it is an intensity, i.e. that the modulus-square has already been taken. (And, indeed, if that were not the case, then there is no guarantee that $I_{\circlearrowright} + I_{\circlearrowleft} \neq 0$, which you need in order to be able to divide by it.) $\endgroup$ Commented Sep 4, 2023 at 15:15

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