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Pulse energy of a given frequency bandwidth, $$\Delta\mathcal{E} = \frac{\intop_{\lambda_1}^{\lambda_2}I(\lambda)\mathrm{d}\lambda}{\intop_{\lambda_\mathrm{min}}^{\lambda_\mathrm{max}}I(\lambda)\mathrm{d}\lambda}\mathcal{E}.$$

Pulse energy is obtained by dividing the optical power \( P \) by the pulse repetition rate \( f \), $$\mathcal{E}=\frac{P}{f}.$$

Optical power, given by a powermeter. Since pulse spectral density \( I(\lambda) \) is given in arbitrary units, value of \( P \) is used to obtain the spectral density scaling factor \( s \), for which \(sI(\lambda) \to I(\lambda)\) and $$\intop_{\lambda_\mathrm{min}}^{\lambda_\mathrm{max}}I(\lambda)\mathrm{d}\lambda = P.$$

Product of pulse duration and spectral width frequency (both in FWHM). Has its minimum for ideal transform-limited pulses:

• Gaussian, \(I(t)\propto \exp\left[-(4\ln 2)t^2/\Delta t^2\right]\):$$\Delta t\cdot \Delta\nu = \frac{2\ln 2}{\pi}\approx0.441.$$
• \(\mathrm{sech}^2\), \(I(t)\propto\left[\exp(2t/\Delta t)+\exp(-2t/\Delta t)\right]^{-1}\):$$\Delta t\cdot \Delta\nu = \frac{4\ln^2(\sqrt{2}+1)}{\pi^2}\approx0.315.$$
• Lorentzian, \(I(t)\propto \left[1+4\left(\sqrt{2}-1\right)\left(t/\Delta t\right)^{2}\right]^{-2}\):$$\Delta t\cdot \Delta\nu = \frac{\ln 2\sqrt{\sqrt{2}-1}}{\pi}\approx0.142.$$

Divergence angle \( \vartheta \) describes how Gaussian beam diameter spreads in the far field (\(z\gg z_\mathrm{R} \)). For beam quality factor \( M^2 \), $$\vartheta = 2M^2\frac{\lambda}{\pi w_0}.$$ Beam divergece half-angle \( \theta = \vartheta/2 \) is often used.

Rayleigh length is distance from beam waist to the point, where beam diameter is \( 2\sqrt{2}w_0 \). for beam with quality factor \( M^2 \) is $$ z_\mathrm{R} = \frac{\pi w_0^2}{M^2 \lambda}. $$ Rayleigh length is equal to confocal parameter \( b \) divided by 2.

Beam parameter product (BPP) is product of divergence half-angle \( \vartheta/2 \) and radius at waist \( w_0 \), $$ \mathrm{BPP} = M^2 \frac{\lambda}{\pi},$$ usually measured in mm\(\cdot\)mrad.

Phase matching condition: $$ \frac{n_\mathrm{e}(\vartheta,\lambda_3)}{\lambda_3} = \left( \frac{n_\mathrm{o}(\lambda_1)}{\lambda_1} + \frac{n_\mathrm{o}(\lambda_2)}{\lambda_2} \right)\cos\vartheta_0. $$ Here \( \vartheta_0 \) is the angle of incidence. Phase matching angle: $$ \vartheta =\arcsin\sqrt{\frac{\frac{(\lambda_{1}+\lambda_{2})^{2}}{\left(n_\mathrm{o}(\lambda_{1})\lambda_{2}+n_\mathrm{o}(\lambda_{2})\lambda_{1}\right)^{2}\cos^{2}\vartheta_{0}}-\frac{1}{n^2_\mathrm{o}(\lambda_{3})}}{\frac{1}{n_\mathrm{e}^{2}(\lambda_{3})}-\frac{1}{n_\mathrm{o}^{2}(\lambda_{3})}}} $$

Phase matching condition: $$ \frac{n_\mathrm{e}(\vartheta,\lambda_3)}{\lambda_3} = \left( \frac{n_\mathrm{e}(\vartheta,\lambda_1)}{\lambda_1} + \frac{n_\mathrm{o}(\lambda_2)}{\lambda_2} \right)\cos\vartheta_0. $$ Here \( \vartheta_0 \) is the angle of incidence.

Phase matching condition: $$ \frac{n_\mathrm{e}(\vartheta,\lambda_3)}{\lambda_3} = \left( \frac{n_\mathrm{o}(\lambda_1)}{\lambda_1} + \frac{n_\mathrm{e}(\vartheta,\lambda_2)}{\lambda_2} \right)\cos\vartheta_0. $$ Here \( \vartheta_0 \) is the angle of incidence.

Phase matching condition: $$ \frac{n_\mathrm{o}(\lambda_3)}{\lambda_3} = \left( \frac{n_\mathrm{e}(\vartheta,\lambda_1)}{\lambda_1} + \frac{n_\mathrm{e}(\vartheta,\lambda_2)}{\lambda_2} \right)\cos\vartheta_0. $$ Here \( \vartheta_0 \) is the angle of incidence.

Phase matching condition: $$ \frac{n_\mathrm{o}(\lambda_3)}{\lambda_3} = \left( \frac{n_\mathrm{e}(\vartheta,\lambda_1)}{\lambda_1} + \frac{n_\mathrm{o}(\lambda_2)}{\lambda_2} \right)\cos\vartheta_0. $$ Here \( \vartheta_0 \) is the angle of incidence. Phase matching angle: $$ \vartheta =\arcsin\sqrt{\frac{\frac{\lambda_{2}^{2}\cos^2\vartheta_0}{\left(n_\mathrm{o}(\lambda_3)\lambda_3-n_\mathrm{o}(\lambda_{2})\lambda_1\cos\vartheta_0\right)^{2}\cos^{2}\vartheta_{0}}-\frac{1}{n^2_\mathrm{o}(\lambda_{1})}}{\frac{1}{n_\mathrm{e}^{2}(\lambda_1})}-\frac{1}{n_\mathrm{o}^{2}(\lambda_{1})}}} $$

Phase matching condition: $$ \frac{n_\mathrm{o}(\lambda_3)}{\lambda_3} = \left( \frac{n_\mathrm{e}(\vartheta,\lambda_1)}{\lambda_1} + \frac{n_\mathrm{o}(\lambda_2)}{\lambda_2} \right)\cos\vartheta_0. $$ Here \( \vartheta_0 \) is the angle of incidence. Phase matching angle: $$ \vartheta =\arcsin\sqrt{\frac{\frac{\lambda_{1}^{2}\cos^2\vartheta_0}{\left(n_\mathrm{o}(\lambda_3)\lambda_3-n_\mathrm{o}(\lambda_{1})\lambda_2\cos\vartheta_0\right)^{2}\cos^{2}\vartheta_{0}}-\frac{1}{n^2_\mathrm{o}(\lambda_{2})}}{\frac{1}{n_\mathrm{e}^{2}(\lambda_2})}-\frac{1}{n_\mathrm{o}^{2}(\lambda_{2})}}} $$

Angle \( \rho_i \) (\(i=1,2,3\)) between the wave vector \( \vec{k}_i \) and direction of maximum beam intensity (Poyinting vector) of extraordinary ray: $$ \rho_i = -\frac{1}{n_\mathrm{e}(\lambda_i,\vartheta_i)}\cdot\frac{\partial n_\mathrm{e}(\lambda_i,\vartheta_i)}{\partial\vartheta_i}. $$ Ordinary rays do not have spatial walk-off.

Difference between \( m=-1 \) diffraction angle (\( \vartheta_{-1} \)) and AOI (\( \vartheta_0 \)) $$ \vartheta_\mathrm{d} = \arcsin\left(\frac{\lambda}{d}-\sin{\vartheta_0}\right) - \vartheta_0 . $$

If deviation angle \(\vartheta_\mathrm{d}\) is given, AOI is obtained from equation $$ \sin^2\vartheta_0\left(1+\cos\vartheta_\mathrm{d}\right)-\frac{\lambda}{d}\sin\vartheta_0\left(1+\cos\vartheta_\mathrm{d}\right)+\frac{\lambda^2}{2d^2}-\frac{\sin^2\vartheta_\mathrm{d}}{2} = 0 $$

If angle of incidence \( \vartheta_0 \) is equal to the Littrow angle \( \vartheta_\mathrm{L} \), \( m=-1 \) reflection angle \( \vartheta_{-1} \) is equal to \( \vartheta_0 \): $$ \vartheta_\mathrm{L}=\arcsin\left(\frac{\lambda}{2d}\right) . $$

Wavenumber $$ k = \frac{1}{\lambda} \Longrightarrow k\mathrm{[cm^{-1}]} = \frac{10^{7}}{\lambda\mathrm{[nm]}} $$ Optical period $$ T = \frac{\lambda}{c} \Longrightarrow T[\mathrm{fs}] \approx \frac{\lambda[\mathrm{nm}]}{299.792} $$ Angular frequency $$\omega = \frac{2\pi c}{\lambda} \Longrightarrow \omega[\mathrm{fs^{-1}}] \approx \frac{1883.652}{\lambda[\mathrm{nm}]} $$ Energy $$ E = \frac{2\pi c\hbar}{\lambda} \Longrightarrow E[\mathrm{eV}] \approx \frac{1239.841}{\lambda[\mathrm{nm}]} $$ Frequency $$ f = \frac{c}{\lambda} \Longrightarrow f[\mathrm{THz}] \approx \frac{299792.458}{\lambda[\mathrm{nm}]} $$

Wavelength $$ \lambda = \frac{1}{k} \Longrightarrow \lambda[\mathrm{nm}] = \frac{10^7}{k[\mathrm{cm^{-1}}]} $$ Optical period $$ T = \frac{1}{ck} \Longrightarrow T[\mathrm{fs}] \approx \frac{3.336\cdot 10^4}{k[\mathrm{cm^{-1}}]} $$ Angular frequency $$\omega = 2\pi c k \Longrightarrow \omega[\mathrm{fs^{-1}}] \approx \frac{k[\mathrm{cm^{-1}}]}{5308.837} $$ Energy $$ E = 2\pi c\hbar k \Longrightarrow E[\mathrm{eV}] \approx \frac{k[\mathrm{cm^{-1}}]}{8065.550} $$ Frequency $$ f = ck \Longrightarrow f[\mathrm{THz}] \approx \frac{k[\mathrm{cm^{-1}}]}{33.356} $$

Wavelength $$ \lambda = Tc \Longrightarrow \lambda[\mathrm{nm}] \approx T[\mathrm{fs}] \cdot 299.792$$ Wavenumber $$ k = \frac{1}{Tc} \Longrightarrow k[\mathrm{cm^{-1}}] \approx \frac{3.335\cdot 10^4}{T[\mathrm{fs}]} $$ Angular frequency $$\omega = \frac{2\pi}{T} \Longrightarrow \omega[\mathrm{fs^{-1}}] \approx \frac{6.283}{T[\mathrm{fs}]} $$ Energy $$ E = \frac{2\pi\hbar}{T} \Longrightarrow E[\mathrm{eV}] \approx \frac{4.136}{T[\mathrm{fs}]} $$ Frequency $$ f = \frac{1}{T} \Longrightarrow f[\mathrm{THz}] = \frac{10^3}{T[\mathrm{fs}]} $$

Wavelength $$ \lambda = \frac{2\pi c}{\omega} \Longrightarrow \lambda[\mathrm{nm}] \approx \frac{1883.652}{\omega[\mathrm{fs^{-1}}]} $$ Wavenumber $$ k = \frac{\omega}{2\pi c} \Longrightarrow k[\mathrm{cm^{-1}}] \approx 5308.837 \cdot \omega[\mathrm{fs^{-1}}] $$ Optical period $$ T = \frac{2\pi}{\omega} \Longrightarrow T[\mathrm{fs}] \approx \frac{6.283}{\omega[\mathrm{fs^{-1}}]} $$ Energy $$ E = \hbar\omega \Longrightarrow E[\mathrm{eV}] \approx \frac{\omega[\mathrm{fs^{-1}}]}{1.519} $$ Frequency $$ f = \frac{\omega}{2\pi} \Longrightarrow f[\mathrm{THz}] \approx 159.160 \cdot \omega[\mathrm{fs^{-1}}] $$

Wavelength $$ \lambda = \frac{2\pi c\hbar}{E} \Longrightarrow \lambda[\mathrm{nm}] \approx \frac{1239.841}{E[\mathrm{eV}]} $$ Wavenumber $$ k = \frac{E}{2\pi c\hbar} \Longrightarrow k[\mathrm{cm^{-1}}] \approx 8065.550 \cdot E[\mathrm{eV}] $$ Optical period $$ T = \frac{2\pi\hbar}{E} \Longrightarrow T[\mathrm{fs}] \approx \frac{4.136}{E[\mathrm{eV}]} $$ Angular frequency $$ \omega = \frac{E}{\hbar} \Longrightarrow \omega \approx 1.519\cdot E[\mathrm{eV}] $$ Frequency $$ f = \frac{E}{2\pi\hbar} \Longrightarrow f[\mathrm{THz}] \approx 241.764 \cdot E[\mathrm{eV}] $$

Wavelength $$ \lambda = \frac{c}{f} \Longrightarrow \lambda[\mathrm{nm}] \approx \frac {299792.458}{f[\mathrm{THz}]} $$ Wavenumber $$ k = \frac{f}{c} \Longrightarrow \approx 33.356 \cdot f[\mathrm{THz}] $$ Optical period $$ T = \frac{1}{f} \Longrightarrow T[\mathrm{fs}] = \frac{10^3}{f[\mathrm{THz}]} $$ Angular frequency $$ \omega = 2\pi f \Longrightarrow \omega[\mathrm{cm^{-1}}] \approx \frac{f[\mathrm{THz}]}{159.160} $$ Energy $$ E = 2\pi\hbar f \Longrightarrow E[\mathrm{eV}] \approx \frac{f[\mathrm{THz}]}{241.764} $$

Carrier-envelope phase \( \varphi_\mathsf{CE} \) is the phase difference between the maxima of (i) oscillating field intensity and (ii) carrier envelope. After propagating distance \( L \) in medium, the CE phase changes due to diffence of phase and group velocities, $$ \Delta\varphi_\mathsf{CE} = \omega_0 \sum_{i=1}^N\left(\frac{1}{v_{\mathsf{g},i}} - \frac{1}{v_{\mathsf{p},i}} \right) h_i . $$ CE phase shift is proportional to the first derivative of refractive index over the wavelength, $$ \Delta\varphi_\mathsf{CE} = -2\pi \sum_{i=1}^N h_i \frac{\partial n_i(\lambda)}{\partial \lambda} . $$

Lateral shift of optical axis after passing through a slab of thickness \( h \), refractive index \( n=n(\lambda) \) at angle of indicence \( \vartheta_0 \), $$ d = h \sin\vartheta_0\left( 1 - \sqrt{\frac{1-\sin^2\vartheta_0}{n^2-\sin^2\vartheta_0}}\right).$$

Optical path in system of two slabs, characterized by distance \( L \), angle of incidence \( \vartheta_0 \) and group velocity at material \( v_\mathrm{g} \), $$t = \frac{2l}{v_\mathrm{g}} + \frac{L-2\sqrt{l^2-d^2}}{c}. $$ Here \( d \) is displacement of optical path and optical path length within a slab is $$l = \frac{nh}{\sqrt{n^2-\sin^2\vartheta_0}}.$$

Time of flight of Gaussian beam through optical path length \( L \), $$ t = \sum_{i=1}^N\frac{h_i}{v_{\mathsf{g},i}} . $$

Optical path length \( L \), $$ L = \sum_{i=1}^N h_i n_i. $$

Exact and approximate relations between the bandwidth in wavelength and wavenumber units is given by: $$ \Delta\lambda = \frac{4\pi c}{\Delta \omega} \left( \sqrt{1+\frac{\lambda_0^2\Delta \omega^2}{4\pi^2 c^2}} - 1 \right) \approx \frac{\Delta \omega\lambda_0^2}{2\pi c} = \Delta k \lambda_0^2. $$ If bandwidth \( \Delta k \) is given in inverse centimeters, bandwidth in nanometers is approximately $$ \Delta\lambda\mathrm{[nm]} \approx 10^{-7} \cdot \Delta k\mathrm{[cm^{-1}]}\cdot(\lambda_0\mathrm{[nm]})^2. $$

Peak width relations: $$ \mathrm{FWHM} = 2\sqrt{2\ln2}\sigma,$$ $$D_{1/\mathrm{e}^2} = 4\sigma = \sqrt{\frac{2}{\ln2}}\mathrm{FWHM},$$ $$D_{1/\mathrm{e}} = 2\sqrt{2}\sigma = \frac{\mathrm{FWHM}}{\sqrt{\ln 2}}.$$

Exact and approximate relations between the bandwidth in wavenumber and wavelength units is given by: $$ \Delta k = \frac{\Delta\lambda}{\lambda_0^2 - \frac{\Delta\lambda^2}{4}} \approx \frac{\Delta\lambda}{\lambda_0^2} .$$ If bandwidth \( \Delta \lambda \) is given in nanometers, bandwidth in inverse centimeters is approximately $$ \Delta k\mathrm{[cm^{-1}]} \approx 10^7 \cdot \frac{\Delta\lambda\mathrm{[nm]}}{(\lambda_0\mathrm{[nm]})^2}.$$

Peak width relations: $$ \mathrm{FWHM} = 2\sqrt{2\ln2}\sigma,$$ $$D_{1/\mathrm{e}^2} = 4\sigma = \sqrt{\frac{2}{\ln2}}\mathrm{FWHM},$$ $$D_{1/\mathrm{e}} = 2\sqrt{2}\sigma = \frac{\mathrm{FWHM}}{\sqrt{\ln 2}}.$$

Carrier-envelope phase \( \varphi_\mathsf{CE} \) is the phase difference between the maxima of (i) oscillating field intensity and (ii) carrier envelope. After propagating distance \( L \) in medium, the CE phase changes due to diffence of phase and group velocities, $$\Delta\varphi_\mathsf{CE} = \omega_0 \left(\frac{1}{v_\mathsf{g}} - \frac{1}{v_\mathsf{p}} \right) L. $$ CE phase shift is proportional to the first derivative of refractive index over the wavelength, $$\Delta\varphi_\mathsf{CE} = -2\pi L \frac{\partial n(\lambda)}{\partial \lambda} . $$

Time of flight of Gaussian beam through optical path length \( L \), $$ t = \frac{L}{v_\mathsf{g}}=\frac{L}{c}\left( n(\lambda) - \lambda \frac{\partial n(\lambda)}{\partial \lambda} \right). $$

Group velocity dispersion (GVD) in material with refraction index \(n(\lambda)\): $$ \mathrm{GVD}(\lambda) = \frac{\lambda^3}{2\pi c^2}\frac{\partial^2 n(\lambda)}{\partial \lambda^2}. $$

Third-order dispersion (TOD) in material with refraction index \(n(\lambda)\): $$ \mathrm{TOD}(\lambda) = -\frac{\lambda^{4}}{4\pi^{2}c^{3}}\left[3\frac{\mathrm{d}^{2}n}{\mathrm{d}\lambda^{2}}+\lambda\frac{\mathrm{d}^{3}n}{\mathrm{d}\lambda^{^{3}}}\right]. $$

$$ n_\mathrm{g} = \frac{c}{v_\mathrm{g}} = n(\lambda) - \lambda \frac{\partial n(\lambda)}{\partial \lambda} $$

$$R = \left| \frac{f}{n-1} \right|. $$

Maximal pulse intensity (at beam center). For temporally Gaussian pulse, peak intensity is related to peak fluence as $$I_0 =\frac{2F_{0}}{\Delta t}\sqrt{\frac{\ln2}{\pi}}\approx\frac{0.94F_0}{\Delta t}. $$ Here \(\Delta t\) is pulse length (FWHM).

Maximal pulse intensity (at beam center). For temporally sech² pulse, peak intensity is related to peak fluence as $$I_{0}=\frac{\mathrm{arccosh}\sqrt{2}F_{0}}{\Delta t}\approx\frac{0.88F_{0}}{\Delta t}. $$ Here \(\Delta t\) is pulse length (FWHM).

Maximal pulse power. For temporally Gaussian pulse, peak power is related to pulse energy \( \mathcal{E} \) and length \( \Delta t\) (FWHM) as $$P_0 =\frac{2\mathcal{E}}{\Delta t}\sqrt{\frac{\ln2}{\pi}}\approx\frac{0.94\mathcal{E}}{\Delta t}. $$

Maximal pulse power. For temporally sech² pulse, peak power is related to pulse energy \( \mathcal{E} \) and length \( \Delta t\) (FWHM) as $$P_0 =\frac{\mathrm{arccosh}\sqrt{2}\mathcal{E}}{\Delta t}\approx\frac{0.88\mathcal{E}}{\Delta t}. $$

Peak fluence \(F_0\) - maximal energy density per unit area (at beam center). Pulse energy \(\mathcal{E}\) is equal to the integrated fluence \(F\), $$\mathcal{E}=\intop F(r)\mathrm{d}S. $$ If fluence and beam intensity is super-Gaussian function, $$F(r)=F_0\left[-2\left(\frac{r}{w_{0}}\right)^{2n}\right],$$ peak fluence is obtained as $$F_0 = \mathcal{E}\frac{2^{\frac{1}{n}}n}{\pi w_{0}^{2}\Gamma\left(\frac{1}{n}\right)}. $$ Here \(\Gamma\) is gamma function, \(w_0\) - half width of the peak at \(1/\mathrm{e}^2\) intensity. If \(n=1\) (Gaussian beam), $$F_0 = \mathcal{E}\frac{2}{\pi w_{0}^{2}}. $$

Coefficient \(n\) of normalized super-Gaussian function $$ f_\mathrm{SG}=\left(\frac{n2^{1/n}}{\pi w_{0}^{2}\Gamma(1/n)}\right)\exp\left[-2\left(\frac{r}{w_{0}}\right)^{2n}\right]. $$ If \(n=1\), function is Gaussian. Here \(\Gamma\) is gamma function, \(w_0\) - half width of the peak at \(1/\mathrm{e}^2\) intensity.

$$ \vartheta_1 = \arcsin \left[ n \sin \left( \alpha - \arcsin \frac{\vartheta_0}{n} \right) \right] $$

$$ \delta = \vartheta_0 + \arcsin \left[ n \sin \left( \alpha - \arcsin \frac{\vartheta_0}{n} \right) \right] - \alpha$$

$$ R_\mathrm{s} = \frac{|E_\mathrm{r}^\mathrm{s}|^2}{|E_\mathrm{i}^\mathrm{s}|^2}=\frac{|\cos\vartheta_0-n\cos\vartheta_1|^2}{|\cos\vartheta_0+n\cos\vartheta_1|^2}. $$ Here \( \vartheta_0 \) is AOI and $$ \vartheta_1 = \arcsin\frac{\sin\vartheta_0}{n} $$ is angle of refraction.

$$ R_\mathrm{p} = \frac{|E_\mathrm{r}^\mathrm{p}|^2}{|E_\mathrm{i}^\mathrm{p}|^2}=\frac{|\cos\vartheta_1-n\cos\vartheta_0|^2}{|\cos\vartheta_1+n\cos\vartheta_0|^2}. $$ Here \( \vartheta_0 \) is AOI and $$ \vartheta_1 = \arcsin\frac{\sin\vartheta_0}{n} $$ is angle of refraction.

Reflectance of p-polarized beam is minimal when angle of incidence is equal to Brewster's angle $$ \vartheta_\mathrm{Br}=\arctan(n)$$.

For given angle of incidence \(\vartheta_0\), prism with apex angle $$\alpha_0=2\arcsin\frac{\sin\vartheta_0}{n}$$ would cause minimal possible deviation angle \(\delta\). In that case the refraction angle is equal to the angle of incidence, \( \vartheta_0=\vartheta_1 \).