# Change of basis in non-linear Schrodinger equation

At the mean-field level, the dynamics of a polariton condensate can be described by a type of nonlinear Schrodinger equation (Gross-Pitaevskii-type), for a classical (complex-number) wavefunction $\psi_{LP}$. Its form in momentum space reads:

\begin{multline} i \frac{d}{dt}\psi_{LP}(k) =\left[\epsilon(k) -i\frac{\gamma(k)}{2}\right] \psi_{LP}(k) +F_{p}(k)\,\, e^{-i\omega_{p}t} \\ + \sum_{q_1,q_2} g_{k,q_1,q_2}\, \psi^{\star}_{LP}(q_1+q_2-k) \, \psi_{LP}(q_1)\, \psi_{LP}(q_2). \end{multline}

The function $\epsilon(k)$ is the dispersion of the particles (polaritons). The polaritons are a non-equilibrium system, due to their finite lifetime (damping rate $\gamma$). Therefore, they need continuous pumping with amplitude $F_p$ at energy $\omega_p$. Finally, there exists a momentum-dependent nonlinear interaction $g_{k,q_1,q_2}$ that depends of the so-called Hopfield coefficients $X$ (simple functions of momentum) as:

$$g_{k,q_1,q_2}=g\, X^{\star}(k)\, X^{\star}(q_1+q_2-k)\, X(q_1)\, X(q_2)$$

How can one transform the equation for $\psi$ to real-space?

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My intuition and first attempt would be to write $\psi_{LP}(k) = \int dx e^{ikx} \psi_{LP}(x)$ and substitute that in the equation. – Lagerbaer May 17 '13 at 20:12
Thanks @Lagerbaer, I followed your suggestion and got $\int dxe^{-ikx}i\frac{d}{dt}\psi_{LP}(x)=\int dx\left[\epsilon(k)-i\frac{\gamma(k)}{2}\right]e^{-ikx}\psi_{LP}(x)+F_{p}(k)\,\, e^{-i\omega_{p}t}+\int dx_{3}dx_{1}dx_{2}\sum_{q_{1},q_{2}}g_{k,q_{1},q_{2}}\, e^{i(q_{1}+q_{2}-k)x_{3}}\psi_{LP}^{\star}(x_{3})\, e^{-iq_{1}x_{1}}\psi_{LP}(x_{1})\, e^{-iq_{2}x_{2}}\psi_{LP}(x_{2})$. Now what :) ? – Andrei May 18 '13 at 9:41
The general idea is then to "compare coefficients": Because the plain waves are linearly independent functions, the left-hand side and ride-hand side must match coefficient-wise. But looking at what you've got I'm not so sure if my naive approach works :-( – Lagerbaer May 18 '13 at 14:15

The linear terms it seems you can handle. As piece of general advice, the meaning of these terms are always clearly if integrate over the momentum coordinates of each of the fields, using delta functions to preserve the value. So the non-linear term would be $$\sum_{q_1,q_2,q_3} g(q_1,q_2,q_3) \psi(q_1)^*\psi(q_2)\psi(q_3)\delta(-q_1+q_2+q_3 -k)$$

Maybe you can also see this way that the structure is determined by momentum conservation/translation invariance. Now when I integrate this by $\int\!dk\,e^{ikr}$, the $k$ integral is resolved trivially and I'm left with fourier transforms over the $q$s. Since Fourier transforms take multiplication to convolution, you can calculate that we get $$\int dr_{123}\,\tilde{g}(r-r_1,r-r_2,r-r_3)\tilde{\psi}^*\!(r_1)\tilde{\psi}(r_2)\tilde{\psi}(r_3)$$

Which is more or less the most general third order nonlinear term you can write. In your case you can also reduce this further by using the real space transforms of $X$, either by plugging in directly to the first equation I wrote, or by calculating $\tilde{g}$ and plugging into the second.

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Here is my attempt at an answer, following the suggestion of @Lagerbaer. We first subtitute the Fourier Transform for $\psi_{LP}(k)$,

$$\psi_{LP}(k)=\int dxe^{-ikx}\psi_{LP}(x),$$

and get

\begin{multline} \int dxe^{-ikx}i\frac{d}{dt}\psi_{LP}(x)=\int dx\left[\epsilon(k)-i\frac{\gamma(k)}{2}\right]e^{-ikx}\psi_{LP}(x)\\+F_{p}(k)\,\, e^{-i\omega_{p}t} +\int dx_{3}dx_{1}dx_{2}\\ \times \sum_{q_{1},q_{2}}g_{k,q_{1},q_{2}}\, e^{i(q_{1}+q_{2}-k)x_{3}}\psi_{LP}^{\star}(x_{3})\, e^{-iq_{1}x_{1}}\psi_{LP}(x_{1})\, e^{-iq_{2}x_{2}}\psi_{LP}(x_{2}). \end{multline}

In order to simplify this expression, we need to apply the inverse transform to both sides (multiply by $\int dke^{ikx}$), obtaining

\begin{multline} i\frac{d}{dt}\psi_{LP}(x)=\int dx^{\prime}\psi_{LP}(x^{\prime})\int dk\left[\epsilon(k)-i\frac{\gamma(k)}{2}\right]e^{ik(x-x^{\prime})}+F_{p}e^{i(k_{p}x-\omega_{p}t)}\\ +\int dke^{ikx}\int dx_{3}dx_{1}dx_{2}\\ \times\sum_{q_{1},q_{2}}g_{k,q_{1},q_{2}}\, e^{i(q_{1}+q_{2}-k)x_{3}}\psi_{LP}^{\star}(x_{3})\, e^{-iq_{1}x_{1}}\psi_{LP}(x_{1})\, e^{-iq_{2}x_{2}}\psi_{LP}(x_{2}). \end{multline}

Now we look at an integral of the form $\int dx^{\prime}\psi_{LP}(x^{\prime})\int dkf(k)e^{ik(x-x^{\prime})}$, for a generic function $f(k)$. We expand our generic function in a Taylor Series around $k=0$, and get:

$$\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}\int dx^{\prime}\psi_{LP}(x^{\prime})\int dkk^{n}e^{ik(x-x^{\prime})}=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}\left(-i\right)^{n}\int dx^{\prime}\psi_{LP}(x^{\prime})\partial_{x}^{n}\delta(x-x^{\prime})$$

where we have used the well-known relation $\int dkk{}^{n}e^{ik(x-x^{\prime})}=\left(-i\right)^{n}\partial_{x}^{n}\delta(x-x^{\prime})$.

Moving the derivative operator to act on $\psi$ (integration by parts), we finally obtain

$$\int dx^{\prime}\psi_{LP}(x^{\prime})\int dkf(k)e^{ik(x-x^{\prime})}=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}(i\partial_{x})^{n}\psi_{LP}(x)=f(i\partial_{x})\psi_{LP}(x)$$

Therefore, the first term on the RHS of our equation simplifies to $\left[\epsilon(i\partial_{x})-i\frac{\gamma(i\partial_{x})}{2}\right]\psi_{LP}(x)$ and we are left dealing with the interaction term.

$$\int dx_{3}\psi_{LP}^{\star}(x_{3})\int dx_{1}\psi_{LP}(x_{1})\int dq_{1}e^{iq_{1}(x_{3}-x_{1})}\int dx_{2}\psi_{LP}(x_{2})\int dq_{2}e^{iq_{2}(x_{3}-x_{2})}\int dkg(k,q_{1},q_{2})e^{ik(x-x_{3})}$$

Substituting the form of $g(k,q_{1},q_{2})$, we get

$$g\int dx_{3}\psi_{LP}^{\star}(x_{3})\int dx_{1}\psi_{LP}(x_{1})\int dq_{1}X(q_{1})e^{iq_{1}(x_{3}-x_{1})}\int dx_{2}\psi_{LP}(x_{2})\int dq_{2}X(q_{2})e^{iq_{2}(x_{3}-x_{2})}\int dk\, X^{\star}(k)\, X^{\star}(q_{1}+q_{2}-k)\, e^{ik(x-x_{3})}$$

And here is where I got stuck..

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