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+---
+title: "Nonlinear Schrödinger equation"
+sort_title: "Nonlinear Schrodinger equation" # sic
+date: 2024-09-15
+categories:
+- Physics
+- Mathematics
+- Fiber optics
+- Nonlinear optics
+layout: "concept"
+---
+
+The **nonlinear Schrödinger (NLS) equation**
+is a nonlinear 1+1D partial differential equation
+that appears in many areas of physics.
+It is used to describe pulses in fiber optics (as derived below),
+waves over deep water, local opening of DNA chains, and more.
+It is often given as:
+
+$$\begin{aligned}
+    \boxed{
+        i \pdv{u}{z} + \pdvn{2}{u}{t} + |u|^2 u
+        = 0
+    }
+\end{aligned}$$
+
+Which is its dimensionless form,
+governing the envelope $$u(z, t)$$
+of an underlying carrier wave,
+with $$t$$ being the transverse coordinate.
+Notably, the NLS equation has **soliton** solutions,
+where $$u$$ maintains its shape over great distances.
+
+
+
+## Derivation
+
+We only consider fiber optics here;
+the NLS equation can be derived in many other ways.
+We start from the most general form of the
+[electromagnetic wave equation](/know/concept/electromagnetic-wave-equation/),
+after assuming the medium cannot be magnetized ($$\mu_r = 1$$):
+
+$$\begin{aligned}
+    \nabla \cross \big( \nabla \cross \vb{E} \big)
+    = - \mu_0 \varepsilon_0 \pdvn{2}{\vb{E}}{t} - \mu_0 \pdvn{2}{\vb{P}}{t}
+\end{aligned}$$
+
+Using the vector identity
+$$\nabla \cross (\nabla \cross \vb{E}) = \nabla (\nabla \cdot \vb{E}) - \nabla^2 \vb{E}$$
+and [Gauss's law](/know/concept/maxwells-equations/) $$\nabla \cdot \vb{E} = 0$$,
+and splitting the polarization $$\vb{P}$$
+into linear and nonlinear contributions
+$$\vb{P}_\mathrm{L}$$ and $$\vb{P}_\mathrm{NL}$$:
+
+$$\begin{aligned}
+    \nabla^2 \vb{E} - \mu_0 \varepsilon_0 \pdvn{2}{\vb{E}}{t}
+    &= \mu_0 \pdvn{2}{\vb{P}_\mathrm{L}}{t} + \mu_0 \pdvn{2}{\vb{P}_\mathrm{NL}}{t}
+\end{aligned}$$
+
+In general, $$\vb{P}_\mathrm{L}$$ is given by the convolution
+of $$\vb{E}$$ with a second-rank response tensor $$\chi^{(1)}$$:
+
+$$\begin{aligned}
+    \vb{P}_\mathrm{L}(\vb{r}, t)
+    = \varepsilon_0 \int_{-\infty}^\infty \chi^{(1)}(t - t') \cdot \vb{E}(\vb{r}, t') \dd{t'}
+\end{aligned}$$
+
+In $$\vb{P}_\mathrm{NL}$$ we only include third-order nonlinearities,
+since higher orders are usually negligible,
+and second-order nonlinear effects only exist in very specific crystals.
+So we "only" need to deal with a fourth-rank response tensor $$\chi^{(3)}$$:
+
+$$\begin{aligned}
+    \vb{P}_\mathrm{NL}(\vb{r}, t)
+    = \varepsilon_0 \iiint_{-\infty}^\infty \chi^{(3)}(t \!-\! t_1, t \!-\! t_2, t \!-\! t_3)
+    \:\vdots\: \vb{E}(\vb{r}, t_1) \vb{E}(\vb{r}, t_2) \vb{E}(\vb{r}, t_3) \dd{t_1} \dd{t_2} \dd{t_3}
+\end{aligned}$$
+
+In practice, two phenomena contribute to $$\chi^{(3)}$$:
+the *Kerr effect* due to electrons' response to $$\vb{E}$$,
+and *Raman scattering* due to nuclei's response,
+which is slower because of their mass.
+But if the light pulses are sufficiently long (>1ps in silica),
+both effects can be treated as fast, so:
+
+$$\begin{aligned}
+    \chi^{(3)}(t \!-\! t_1, t \!-\! t_2, t \!-\! t_3)
+    &= \chi^{(3)} \delta(t - t_1) \delta(t - t_2) \delta(t - t_3)
+\end{aligned}$$
+
+Where $$\delta$$ is the [Dirac delta function](/know/concept/dirac-delta-function/).
+To keep things simple,
+we consider linearly $$x$$-polarized light $$\vb{E} = \vu{x} |\vb{E}|$$,
+such that the tensor can be replaced with its scalar element $$\chi^{(3)}_{xxxx}$$.
+Then:
+
+$$\begin{aligned}
+    \vb{P}_\mathrm{NL}
+    = \varepsilon_0 \chi^{(3)}_{xxxx} \big( \vb{E} \cdot \vb{E} \big) \vb{E}
+\end{aligned}$$
+
+For the same reasons, the linear polarization is reduced to:
+
+$$\begin{aligned}
+    \vb{P}_\mathrm{L}
+    &= \varepsilon_0 \chi^{(1)}_{xx} \vb{E}
+\end{aligned}$$
+
+Next, we decompose $$\vb{E}$$ as follows,
+consisting of a carrier wave $$e^{-i \omega_0 t}$$
+at a constant frequency $$\omega_0$$,
+modulated by an envelope $$E$$
+that is assumed to be slowly-varying compared to the carrier,
+plus the complex conjugate $$E^* e^{i \omega_0 t}$$:
+
+$$\begin{aligned}
+    \vb{E}(\vb{r}, t)
+    &= \vu{x} \frac{1}{2} \Big( E(\vb{r}, t) e^{- i \omega_0 t} + E^*(\vb{r}, t) e^{i \omega_0 t} \Big)
+\end{aligned}$$
+
+Note that no generality has been lost in this step.
+Inserting it into the polarizations:
+
+$$\begin{aligned}
+    \mathrm{P}_\mathrm{L}
+    &= \vu{x} \frac{1}{2} \varepsilon_0 \chi^{(1)}_{xx} \Big( E(\vb{r}, t) e^{- i \omega_0 t} + E^*(\vb{r}, t) e^{i \omega_0 t} \Big)
+    \\
+    \vb{P}_\mathrm{NL}
+    &= \vu{x} \frac{1}{8} \varepsilon_0 \chi^{(3)}_{xxxx} \Big( E e^{- i \omega_0 t} + E^* e^{i \omega_0 t} \Big)^{3}
+    \\
+    &= \vu{x} \frac{1}{8} \varepsilon_0 \chi^{(3)}_{xxxx}
+    \Big( E^3 e^{- i 3 \omega_0 t} + 3 E^2 E^* e^{- i \omega_0 t} + 3 E (E^*)^2 e^{i \omega_0 t} + (E^*)^3 e^{i 3 \omega_0 t} \Big)
+\end{aligned}$$
+
+The terms with $$3 \omega_0$$ represent *third-harmonic generation*,
+and only matter if the carrier is phase-matched
+to the tripled wave, which is generally not the case,
+so they can be ignored.
+Now, if we decompose the polarizations in the same was as $$\vb{E}$$:
+
+$$\begin{aligned}
+    \vb{P}_\mathrm{L}(\vb{r}, t)
+    &= \vu{x} \frac{1}{2} \Big( P_\mathrm{L}(\vb{r}, t) e^{- i \omega_0 t} + P_\mathrm{L}^*(\vb{r}, t) e^{i \omega_0 t} \Big)
+    \\
+    \vb{P}_\mathrm{NL}(\vb{r}, t)
+    &= \vu{x} \frac{1}{2} \Big( P_\mathrm{NL}(\vb{r}, t) e^{- i \omega_0 t} + P_\mathrm{NL}^*(\vb{r}, t) e^{i \omega_0 t} \Big)
+\end{aligned}$$
+
+Then it is straightforward to see that their envelope functions are given by:
+
+$$\begin{aligned}
+    P_\mathrm{L}
+    &= \varepsilon_0 \chi^{(1)}_{xx} E
+    \\
+    P_\mathrm{NL}
+    &= \frac{3}{4} \varepsilon_0 \chi^{(3)}_{xxxx} |E|^2 E
+\end{aligned}$$
+
+The forward carrier $$e^{- i \omega_0 t}$$
+and the backward carrier $$e^{i \omega_0 t}$$
+can be regarded as separate channels,
+which only interact via $$P_\mathrm{NL}$$.
+From now on, we only consider the forward-propagating wave,
+so all terms containing $$e^{i \omega_0 t}$$ are dropped;
+by taking the complex conjugate of the resulting equations,
+the backward-propagating counterparts can always be recovered,
+so no information is really lost.
+Therefore, the main wave equation becomes:
+
+$$\begin{aligned}
+    0
+    &= \bigg(
+        \nabla^2 E - \mu_0 \varepsilon_0 \pdvn{2}{E}{t} - \mu_0 \pdvn{2}{P_\mathrm{L}}{t} - \mu_0 \pdvn{2}{P_\mathrm{NL}}{t}
+    \bigg) e^{-i \omega_0 t}
+    \\
+    &= \bigg(
+        \nabla^2 E - \Big( 1 + \chi^{(1)}_{xx} + \frac{3}{4} \chi^{(3)}_{xxxx} |E|^2 \Big) \mu_0 \varepsilon_0 \pdvn{2}{E}{t}
+    \bigg) e^{-i \omega_0 t}
+\end{aligned}$$
+
+Where we have used our assumption that $$E$$ is slowly-varying
+to treat $$|E|^2$$ as a constant,
+in order to move it outside the $$t$$-derivative.
+We thus arrive at:
+
+$$\begin{aligned}
+    0
+    &= \bigg( \nabla^2 E - \frac{\varepsilon_r}{c^2} \pdvn{2}{E}{t} \bigg) e^{-i \omega_0 t}
+\end{aligned}$$
+
+Where $$c = 1 / \sqrt{\mu_0 \varepsilon_0}$$ is the phase velocity of light in a vacuum,
+and the relative permittivity $$\varepsilon_r$$ is defined as shown below.
+Note that this is a mild abuse of notation,
+since the symbol $$\varepsilon_r$$ is usually reserved for linear materials:
+
+$$\begin{aligned}
+    \varepsilon_r
+    \equiv 1 + \chi^{(1)}_{xx} + \frac{3}{4} \chi^{(3)}_{xxxx} |E|^2
+\end{aligned}$$
+
+Next, we take the [Fourier transform](/know/concept/fourier-transform/)
+$$t \to (\omega\!-\!\omega_0)$$ of the wave equation,
+once again treating $$|E|^2$$ (inside $$\varepsilon_r$$) as a constant.
+The constant $$s = \pm 1$$ is included here
+to deal with the fact that different authors use different sign conventions:
+
+$$\begin{aligned}
+    0
+    &= \hat{\mathcal{F}}\bigg\{ \nabla^2 E - \frac{\varepsilon_r}{c^2} \pdvn{2}{E}{t} \bigg\}
+    \\
+    &= \int_{-\infty}^\infty
+    \bigg( \nabla^2 E - \frac{\varepsilon_r}{c^2} \pdvn{2}{E}{t} \bigg)
+    e^{i s (\omega - \omega_0) t} \dd{t}
+    \\
+    &= \nabla^2 E + s^2 \frac{\varepsilon_r}{c^2} (\omega - \omega_0)^2 E
+\end{aligned}$$
+
+We use $$s^2 = 1$$ and define $$\Omega \equiv \omega - \omega_0$$
+as the frequency shift relative to the carrier wave:
+
+$$\begin{aligned}
+    0
+    &= \nabla^2 E + \frac{\Omega^2 \varepsilon_r}{c^2} E
+\end{aligned}$$
+
+This is a so-called *Helmholtz equation* in 3D,
+which we will solve using separation of variables,
+by assuming that its solution can be written as:
+
+$$\begin{aligned}
+    E(\vb{r}, \Omega)
+    &= F(x, y) \: A(z, \Omega) \: e^{i \beta_0 z}
+\end{aligned}$$
+
+Where $$\beta_0$$ is the wavenumber of the carrier,
+which will be determined later.
+Inserting this ansatz into the Helmholtz equation yields:
+
+$$\begin{aligned}
+    0
+    &= \Big( \pdvn{2}{F}{x} + \pdvn{2}{F}{y} \Big) A e^{i \beta_0 z}
+    + \pdvn{2}{}{z} \Big( A e^{i \beta_0 z} \Big) F
+    + \frac{\Omega^2 \varepsilon_r}{c^2} F A e^{i \beta_0 z}
+    \\
+    &= \bigg( \Big( \pdvn{2}{F}{x} + \pdvn{2}{F}{y} \Big) A
+    + \Big( \pdvn{2}{A}{z} + 2 i \beta_0 \pdv{A}{z} - \beta_0^2 A \Big) F
+    + \frac{\Omega^2 \varepsilon_r}{c^2} F A \bigg) e^{i \beta_0 z}
+\end{aligned}$$
+
+We divide by $$F A \: e^{i \beta_0 z}$$
+and rearrange the terms in a specific way:
+
+$$\begin{aligned}
+    \Big( \pdvn{2}{F}{x} + \pdvn{2}{F}{y} \Big) \frac{1}{F} + \frac{\Omega^2 \varepsilon_r}{c^2}
+    &= - 2 i \beta_0 \pdv{A}{z} \frac{1}{A} + \beta_0^2
+\end{aligned}$$
+
+Now all the $$x$$- and $$y$$-dependence is on the left,
+and the $$z$$-dependence is on the right.
+We have placed the $$\varepsilon_r$$-term on the left too
+because it depends relatively strongly on $$(x, y)$$
+to describe the fiber's internal structure,
+and weakly on $$z$$ due to nonlinear effects.
+Meanwhile, $$\beta_0$$ is on the right because that will lead to
+a nicer equation for $$A$$ later.
+
+Note that both sides are functions of $$\Omega$$.
+Based on the aforementioned dependences,
+in order for this equation to have a solution for all $$(x, y, z)$$,
+there must exist a quantity $$\beta(\Omega)$$ that is constant in space,
+such that we obtain two separated equations for $$F$$ and $$A$$:
+
+$$\begin{aligned}
+    \beta(\omega)
+    &= \bigg( \pdvn{2}{F}{x} + \pdvn{2}{F}{y} \bigg) \frac{1}{F} + \frac{\omega^2 \varepsilon_r}{c^2}
+    \\
+    \beta(\Omega)
+    &= - 2 i \beta_0 \pdv{A}{z} \frac{1}{A} + \beta_0^2
+\end{aligned}$$
+
+Note that we replaced $$\Omega$$ with $$\omega$$ in $$F$$'s equation
+(and redefined $$\beta$$ and $$\varepsilon_r$$ accordingly).
+This is not an innocent detail:
+the idea is that $$\omega \sqrt{\varepsilon_r} / c$$
+would be the light's wavenumber if it had not been trapped in a waveguide,
+and that $$\beta$$ is the *confined* wavenumber,
+also known as the **propagation constant**.
+If we had kept $$\Omega$$,
+the meaning of $$\beta$$ would not be so straightforward.
+
+The difference between $$\beta(\omega)$$ and $$\beta_0$$
+is simply that $$\beta_0 \equiv \beta(\omega_0)$$.
+Our ansatz for separating the variables contained $$\beta_0$$,
+such that the full carrier wave $$e^{i \beta_0 z - i \omega_0 t}$$ was represented
+(with $$e^{- i \omega_0 t}$$ now hidden inside the Fourier transform).
+But later, to properly describe how light behaves inside the fiber,
+the full dispersion relation $$\beta(\omega)$$ will be needed.
+
+Multiplying by $$F$$ and $$A$$,
+we get the following set of equations,
+implicitly coupled via $$\beta$$:
+
+$$\begin{aligned}
+    \boxed{
+        \begin{aligned}
+            0
+            &= \pdvn{2}{F}{x} + \pdvn{2}{F}{y} + \bigg( \frac{\omega^2 \varepsilon_r}{c^2} - \beta^2 \bigg) F
+            \\
+            0
+            &= 2 i \beta_0 \pdv{A}{z} + \big( \beta^2 - \beta_0^2 \big) A
+        \end{aligned}
+    }
+\end{aligned}$$
+
+The equation for $$F$$ must be solved first.
+To do so, we treat the nonlinearity as a perturbation
+to be neglected initially.
+In other words, we first solve the following eigenvalue problem for $$\beta^2$$,
+where $$n(x, y)$$ is the linear refractive index,
+with $$n^2 = 1 + \Real\{\chi^{(1)}_{xx}\} \approx \varepsilon_r$$:
+
+$$\begin{aligned}
+    \pdvn{2}{F}{x} + \pdvn{2}{F}{y} + \bigg( \frac{\omega^2 n^2}{c^2} - \beta^2 \bigg) F
+    = 0
+\end{aligned}$$
+
+This gives us the allowed values of $$\beta$$;
+see [step-index fiber](/know/concept/step-index-fiber/) for an example solution.
+Now we add the small index change $$\Delta{n}(x, y)$$ due to nonlinear effects:
+
+$$\begin{aligned}
+    \varepsilon_r
+    = (n + \Delta{n})^2
+    \approx n^2 + 2 n \: \Delta{n}
+\end{aligned}$$
+
+Then it can be shown using first-order
+[perturbation theory](/know/concept/time-independent-perturbation-theory/)
+that the eigenfunction $$F$$ is not really affected,
+and the eigenvalue $$\beta^2$$ is shifted by $$\Delta(\beta^2)$$, given by:
+
+$$\begin{aligned}
+    \Delta(\beta^2)
+    = \frac{2 \omega^2}{c^2} \frac{\displaystyle \iint_{-\infty}^\infty n \: \Delta{n} \: |F|^2 \dd{x} \dd{y}}
+    {\displaystyle \iint_{-\infty}^\infty |F|^2 \dd{x} \dd{y}}
+\end{aligned}$$
+
+But we are more interested in the *wavenumber* shift $$\Delta{\beta}$$
+than the *eigenvalue* shift $$\Delta(\beta^2)$$.
+They are related to one another as follows:
+
+$$\begin{aligned}
+    \beta^2 + \Delta(\beta^2)
+    = (\beta + \Delta{\beta})^2
+    \approx \beta^2 + 2 \beta \Delta{\beta}
+\end{aligned}$$
+
+Furthermore, we assume that the fiber only consists of materials
+with similar refractive indices, or in other words,
+that it confines the light using only a small index difference,
+in which case we can treat $$n$$ as a constant and move it outside the integral.
+Then $$\Delta{\beta}$$ becomes:
+
+$$\begin{aligned}
+    \Delta{\beta}
+    = \frac{\omega^2 n}{\beta c^2} \frac{\displaystyle \iint_{-\infty}^\infty \Delta{n} \: |F|^2 \dd{x} \dd{y}}
+    {\displaystyle \iint_{-\infty}^\infty |F|^2 \dd{x} \dd{y}}
+\end{aligned}$$
+
+Recall that $$\beta$$ is the wavenumber of the confined mode:
+by solving the unperturbed $$F$$-equation,
+it can be shown that $$\beta$$'s value is somewhere
+between the bulk wavenumbers of the fiber materials.
+Since we just approximated $$n$$ as a constant,
+this means that $$\omega n / c \approx \beta$$, leading us to
+the general "final" form of $$\Delta{\beta}$$,
+with all the arguments shown for clarity:
+
+$$\begin{aligned}
+    \boxed{
+        \Delta{\beta}(\omega)
+        = \frac{\omega}{c \mathcal{A}_\mathrm{mode}(\omega)}
+        \iint_{-\infty}^\infty \Delta{n}(x, y, \omega) \: |F(x, y, \omega)|^2 \dd{x} \dd{y}
+    }
+\end{aligned}$$
+
+Where we have defined the *mode area* $$\mathcal{A}_\mathrm{mode}$$ as shown below.
+In order for $$\mathcal{A}_\mathrm{mode}$$ to be in units of area,
+$$F$$ must be dimensionless,
+and consequently $$A$$ has (SI) units of an electric field.
+
+$$\begin{aligned}
+    \mathcal{A}_\mathrm{mode}(\omega)
+    \equiv \iint_{-\infty}^\infty |F(x, y, \omega)|^2 \dd{x} \dd{y}
+\end{aligned}$$
+
+Now we finally turn our attention to the equation for $$A$$.
+Before perturbation, it was:
+
+$$\begin{aligned}
+    0
+    &= 2 i \beta_0 \pdv{A}{z} + \big( \beta^2 - \beta_0^2 \big) A
+\end{aligned}$$
+
+Where $$\beta \approx \beta_0$$, so we can replace
+$$\beta^2 - \beta_0^2$$ with $$2 \beta_0 (\beta - \beta_0)$$.
+Also including $$\Delta{\beta}$$, we get:
+
+$$\begin{aligned}
+    0
+    &= i \pdv{A}{z} + \big( \beta + \Delta{\beta} - \beta_0 \big) A
+\end{aligned}$$
+
+Usually, we do not know a full expression for $$\beta(\omega)$$,
+so it makes sense to expand it around the carrier frequency $$\omega_0$$ as follows,
+where $$\beta_n = \idvn{n}{\beta}{\omega} |_{\omega = \omega_0}$$:
+
+$$\begin{aligned}
+    \beta(\omega)
+    &= \beta_0
+    + (\omega - \omega_0) \beta_1
+    + (\omega - \omega_0)^2 \frac{\beta_2}{2}
+    + (\omega - \omega_0)^3 \frac{\beta_3}{6}
+    + \: ...
+\end{aligned}$$
+
+Spectrally, the broader the light pulse, the more terms must be included.
+Recall that earlier, in order to treat $$\chi^{(3)}$$ as instantaneous,
+we already assumed a temporally broad
+(spectrally narrow) pulse.
+Hence, for simplicity, we can cut off this Taylor series at $$\beta_2$$,
+which is good enough for many cases.
+Inserting the expansion into $$A$$'s equation:
+
+$$\begin{aligned}
+    0
+    &= i \pdv{A}{z} + i \frac{\beta_1}{s} (-i s \Omega) A - \frac{\beta_2}{2 s^2} (- i s \Omega)^2 A + \Delta{\beta}_0 A
+\end{aligned}$$
+
+Which we have rewritten as preparation for taking the inverse Fourier transform,
+by introducing $$s$$ and by replacing $$\Delta{\beta}(\omega)$$
+with $$\Delta{\beta_0} \equiv \Delta{\beta}(\omega_0)$$
+in order to remove all explicit dependence on $$\omega$$.
+After transforming and using $$s^2 = 1$$,
+we get the following equation for $$A(z, t)$$:
+
+$$\begin{aligned}
+    0
+    &= i \pdv{A}{z} + i s \beta_1 \pdv{A}{t} - \frac{\beta_2}{2} \pdvn{2}{A}{t} + \Delta{\beta}_0 A
+\end{aligned}$$
+
+The next step is to insert our expression for $$\Delta{\beta}_0$$,
+for which we must first choose a specific form for $$\Delta{n}$$
+according to which effects we want to include.
+Earlier, we approximated $$\varepsilon_r \approx n^2$$,
+so if we instead say that $$\varepsilon_r = (n \!+\! \Delta{n})^2$$,
+then $$\Delta{n}$$ should include absorption and nonlinearity.
+A simple and commonly used form for $$\Delta{n}$$ is therefore:
+
+$$\begin{aligned}
+    \Delta{n}
+    = n_2 I + i \frac{\alpha c}{2 \omega}
+\end{aligned}$$
+
+Where $$I$$ is the intensity (i.e. power per unit area) of the light,
+$$n_2$$ is the material's *Kerr coefficient* in units of inverse intensity,
+and $$\alpha$$ is the attenuation coefficient
+consisting of linear and nonlinear contributions
+(see [multi-photon absorption](/know/concept/multi-photon-absorption/)).
+Specifically, they are given by:
+
+$$\begin{aligned}
+    n_2
+    = \frac{3 \Real\{\chi^{(3)}_{xxxx}\}}{4 \varepsilon_0 c n^2}
+    \qquad
+    \alpha
+    = \frac{\omega \Imag\{\chi^{(1)}_{xx}\}}{c n}
+    + \frac{3 \omega \Imag\{\chi^{(3)}_{xxxx}\}}{2 \varepsilon_0 c^2 n^2} I
+    \qquad
+    I
+    = \frac{\varepsilon_0 c n}{2} |E|^2
+\end{aligned}$$
+
+For simplicity, we set $$\Imag\{\chi^{(3)}_{xxxx}\} = 0$$,
+which is a good approximation for fibers made of silica.
+Inserting this form of $$\Delta{n}$$ into $$\Delta{\beta_0}$$ then yields:
+
+$$\begin{aligned}
+    \Delta{\beta}_0
+    &= i \frac{\alpha}{2} \frac{\mathcal{A}_\mathrm{mode}}{\mathcal{A}_\mathrm{mode}}
+    + \frac{\omega_0 \varepsilon_0 c n n_2}{2 c \mathcal{A}_\mathrm{mode}} |A|^2 \iint_{-\infty}^\infty |F|^4 \dd{x} \dd{y}
+    \\
+    &= i \frac{\alpha}{2}
+    + \gamma_0 \frac{\varepsilon_0 c n}{2} \mathcal{A}_\mathrm{mode} |A|^2
+\end{aligned}$$
+
+Where we have defined the nonlinear parameter $$\gamma_0$$ like so,
+involving the **effective mode area** $$\mathcal{A}_\mathrm{eff}$$,
+which contains all information about $$F$$ needed for solving $$A$$'s equation:
+
+$$\begin{aligned}
+    \boxed{
+        \gamma_0
+        = \gamma(\omega_0)
+        \equiv \frac{\omega_0 n_2}{c \mathcal{A}_\mathrm{eff}}
+    }
+    \qquad \qquad
+    \boxed{
+        \mathcal{A}_\mathrm{eff}(\omega_0)
+        \equiv \frac{\displaystyle \bigg( \iint_{-\infty}^\infty |F|^2 \dd{x} \dd{y} \bigg)^2}
+        {\displaystyle \iint_{-\infty}^\infty |F|^4 \dd{x} \dd{y}}
+    }
+\end{aligned}$$
+
+Substituting $$\Delta{\beta_0}$$ into the main problem
+yields a prototype of the NLS equation:
+
+$$\begin{aligned}
+    0
+    &= i \pdv{A}{z} + i s \beta_1 \pdv{A}{t} - \frac{\beta_2}{2} \pdvn{2}{A}{t} + i \frac{\alpha}{2} A
+    + \gamma_0 \frac{\varepsilon_0 c n}{2} \mathcal{A}_\mathrm{mode} |A|^2 A
+\end{aligned}$$
+
+The factor $$\varepsilon_0 c n / 2$$ looks familiar from the intensity $$I$$.
+This, combined with $$\mathcal{A}_\mathrm{mode}$$
+and the fact that $$A$$ is an electric field,
+suggests that we can redefine $$A \to A'$$
+such that $$|A'|^2$$ is the optical power in watts.
+Hence we make the following transformation:
+
+$$\begin{aligned}
+    \frac{\varepsilon_0 c n}{2} \mathcal{A}_\mathrm{mode} |A|^2
+    \:\:\to\:\:
+    |A|^2
+\end{aligned}$$
+
+We can divide away the transformation factors
+from all other terms in the equation, since they are linear,
+leading to the full *nonlinear Schrödinger equation*:
+
+$$\begin{aligned}
+    \boxed{
+        0
+        = i \pdv{A}{z} + i s \beta_1 \pdv{A}{t} - \frac{\beta_2}{2} \pdvn{2}{A}{t} + i \frac{\alpha}{2} A + \gamma_0 |A|^2 A
+    }
+\end{aligned}$$
+
+This can be reduced by switching to a coordinate system
+where the time axis slides along the propagation axis at a speed $$s v$$,
+so we define $$Z \equiv z$$ and $$T \equiv t - s z / v$$ such that:
+
+$$\begin{aligned}
+    \pdv{A}{z}
+    &= \pdv{A}{Z} \pdv{Z}{z} + \pdv{A}{T} \pdv{T}{z}
+    = \pdv{A}{Z} - \frac{s}{v} \pdv{A}{T}
+    \\
+    \pdv{A}{t}
+    &= \pdv{A}{Z} \pdv{Z}{t} + \pdv{A}{T} \pdv{T}{t}
+    = \pdv{A}{T}
+\end{aligned}$$
+
+We insert this and set $$v = v_g$$,
+where $$v_g = 1 / \beta_1$$ is the light's group velocity:
+
+$$\begin{aligned}
+    \boxed{
+        0
+        = i \pdv{A}{Z} - \frac{\beta_2}{2} \pdvn{2}{A}{T} + i \frac{\alpha}{2} A + \gamma_0 |A|^2 A
+    }
+\end{aligned}$$
+
+The NLS equation's name is due to its similarity
+to the Schrödinger equation of quantum physics,
+if you set $$\alpha = 0$$ and treat $$\gamma_0 |A|^2$$ as a potential.
+In fiber optics, the equation is usually rearranged
+to highlight that $$Z$$ (or $$z$$) is the propagation direction:
+
+$$\begin{aligned}
+    \pdv{A}{Z}
+    = - i \frac{\beta_2}{2} \pdvn{2}{A}{T} - \frac{\alpha}{2} A + i \gamma_0 |A|^2 A
+\end{aligned}$$
+
+Next, we want to reduce the equation to its dimensionless form.
+To do so, we make the following coordinate transformation,
+where $$\tilde{A}$$, $$\tilde{Z}$$ and $$\tilde{T}$$ are unitless,
+and $$A_c$$, $$Z_c$$ and $$T_c$$ are dimensioned scale parameters
+to be determined later:
+
+$$\begin{aligned}
+    \tilde{A}(\tilde{Z}, \tilde{T})
+    = \frac{A(Z, T)}{A_c}
+    \qquad\qquad
+    \tilde{Z}
+    = \frac{Z}{Z_c}
+    \qquad\qquad
+    \tilde{T}
+    = \frac{T}{T_c}
+\end{aligned}$$
+
+We insert this into the NLS equation,
+after setting $$\alpha = 0$$ according to convention:
+
+$$\begin{aligned}
+    0
+    = i \frac{A_c}{Z_c} \pdv{\tilde{A}}{\tilde{Z}}
+    - \frac{\beta_2}{2} \frac{A_c}{T_c^2} \pdvn{2}{\tilde{A}}{\tilde{T}}
+    + \gamma_0 A_c^3 \big|\tilde{A}\big|^2 \tilde{A}
+\end{aligned}$$
+
+Multiplying by $$Z_c / A_c$$ to make all terms dimensionless leads us to:
+
+$$\begin{aligned}
+    0
+    = i \pdv{\tilde{A}}{\tilde{Z}}
+    - \frac{\beta_2 Z_c}{2 T_c^2} \pdvn{2}{\tilde{A}}{\tilde{T}}
+    + \gamma_0 A_c^2 Z_c \big|\tilde{A}\big|^2 \tilde{A}
+\end{aligned}$$
+
+The goal is to remove those constant factors.
+In other words, we demand:
+
+$$\begin{aligned}
+    \frac{\beta_2 Z_c}{2 T_c^2}
+    = \mp 1
+    \qquad\qquad
+    \gamma_0 A_c^2 Z_c
+    = 1
+\end{aligned}$$
+
+Where the choice of $$\mp$$ will be explained shortly.
+Note that we only have two equations for three unknowns
+($$A_c$$, $$Z_c$$ and $$T_c$$),
+so one of the parameters needs to fixed manually.
+For example, we could choose $$Z_c = 1\:\mathrm{m}$$, and then:
+
+$$\begin{aligned}
+    A_c
+    = \frac{1}{\sqrt{\gamma Z_c}}
+    \qquad\qquad
+    T_c
+    = \sqrt{\frac{\mp \beta_2 Z_c}{2}}
+\end{aligned}$$
+
+Note that this requires that $$\gamma_0 > 0$$,
+which is true for the vast majority of materials,
+and that we choose the sign $$\mp$$ such that $$\mp \beta_2 > 0$$.
+We thus arrive at:
+
+$$\begin{aligned}
+    \boxed{
+        0
+        = i \pdv{\tilde{A}}{\tilde{Z}}
+        \pm \pdvn{2}{\tilde{A}}{\tilde{T}}
+        + \big|\tilde{A}\big|^2 \tilde{A}
+    }
+\end{aligned}$$
+
+Because soliton solutions only exist
+in the *anomalous dispersion* regime $$\beta_2 < 0$$,
+most authors just write $$+$$.
+There are still plenty of interesting effects
+in the *normal dispersion* regime $$\beta_2 > 0$$,
+hence we write $$\pm$$ for the sake of completeness.
+
+
+
+## References
+
+1.  G.P. Agrawal,
+    *Nonlinear fiber optics*, 6th edition,
+    Elsevier.
+2.  O. Bang,
+    *Nonlinear mathematical physics: lecture notes*,
+    2020, unpublished.