# Solving the free Schroedinger equation in Madelung variables

Can explicit solutions to the (unforced) Linear Schrodinger equation (LSE) be found via the Madelung transformation?

(A note on motivation. I am trying to constrain the behavior of the phase of $A$ for a more complicated higher order nonlinear Schrodinger equation, to see if shocks form in $|A|$. To do this, I want to see if I understand how the phase is related to the amplitude in the much simpler LSE).

To motivate this, consider $$iA_t+A_{xx} = 0$$

for $A$ a complex valued function $t\in(0,\infty)$ and $x\in (-\infty,\infty)$. One may solve this using Fourier Transforms. That is, letting

$$A=\frac{1}{2\pi}\int_{-\infty}^{\infty} a (k,t) e^{-ikx}\ dk,$$

we have $a(k,t)=a(k,0)e^{ik^2 x}$, where the initial condition $a(k,0)$ may be found via

$$a(k,0) = \int_{-\infty}^{\infty} A(x,0)e^{ikx} \ dx.$$

For particular initial conditions, i.e. a Gaussian envelope, explicit solutions may be found.

Next, let $A= \sqrt{\rho} e^{i\theta}$ for $\rho, \theta$ real valued functions of $(x,t)$. The LSE becomes two coupled equations:

$$\rho_t +( u\rho )_x=0,$$ $$u_t+uu_x -2\frac{\partial}{\partial x} \left(\frac{1}{\sqrt{\rho}}\frac{\partial^2 \sqrt{\rho}}{\partial x^2}\right)=0,$$

where $u =2\theta_x$. The analogy with hydrodynamics (the first equation is the statement of mass conservation, while the second is momentum conservation) is now obvious. This helps aiding the physical interpretation of the LSE.

However, is it possible to then solve explicitly for $\rho$ and $u$?

It seems like the method of characteristics will yield some progress, especially for simple examples but I cannot seem to generalize my results in any kind of insightful way.

• You are just looking at free propagation, but you have managed to garble it beyond recognition as the standard Q Hamilton-Jacobi. In particular (pleeeease, use v for the velocity instead of V... there is no potential here, beyond the quantum potential) you, untypically increased the order of the equation by using v instead of θ !? ... and I suspect you got the second one wrong in the process... you surely know how to write the real Madelung equations for free plane waves, no? Dec 12, 2017 at 16:27
• You are probably referring to the transform of the Schrödinger equation introduced by the German physicist with the name Erwin Madelung. Mandelung is not correct! Dec 12, 2017 at 16:29
• @CosmasZachos Thank you very much for your comments. This is not a problem motivated by quantum mechanics, so I chose nomenclature that's less egregious in my field. I have indeed increased the order of $\theta$ to make an analogy between the LSE and hydrodynamics, motivated by Madelung (and similar to the Q Hamilton-Jacobi picture you are referring to). I do not see how the second equation is wrong. Dec 12, 2017 at 17:12
• It's probably right, but hardly functional... The Schr eqn is always linear, so L sounds redundant and puzzling... you might as well call it free. As for solutions, hint: look at the spreading of the free Gaussian wavepacket, so a(k,0) is a Gaussian in k. Dec 12, 2017 at 17:27
• @CosmasZachos Thank you again for your comment. The problem is motivated by the nonlinear Schrodinger equation and its higher order analogs, hence the specification of the linear Schrodinger equation. As I referred to in the question, I know it's straightforward to solve for a Gaussian initial condition. Dec 12, 2017 at 18:06

I'll take the "stationary" packet with its maximum always hovering over the x origin, and nondimensionalize $$\hbar=1, ~ m=1$$ out for simplicity, but I will keep the 1/2 in front of the kinetic term, egregious as it might well be in your field. Quantum physicists might well lynch me if I took $$m=1/2$$ instead. So Schroedinger's free equation is $$i\partial_t \psi = -\frac{1}{2}\partial_x^2 \psi,$$ solved by normalized Gaussian wavepackets $$\bbox[yellow]{\Large \psi= \sqrt{\frac {2}{\pi (1+2it)^2} } ~e^{-\frac{x^2}{1+2it}}=\sqrt{\rho} ~e^{iS}, \\ \rho=\psi^* \psi = \sqrt{\frac {2}{\pi (1+4t^2)} } ~~ \Large e ^{-\frac{2x^2}{1+4t^2}}, \\ S=\frac{2tx^2}{1+4t^2} -\frac{i}{4}\ln \frac{1-2it}{1+2it} }~~.$$
Note the velocity $$v\equiv j/\rho= \frac{1}{2i\rho}(\psi^* \partial_x \psi -\psi \partial_x \psi^*)=\partial_x S= \left ( \frac{4tx}{1+4t^2} \right )$$ to be used in the continuity equation $$0=\partial_t \rho + \partial_x j= \partial_t \rho + \partial_x (v \rho)=\partial_t \rho + \partial_x \left (\rho ~\frac{4tx}{1+4t^2}\right ),$$ which you may readily check. The farther away you are from the origin the faster you spread. Further check the large t limit.
The real part of the polar form of the Schroedinger equation that Madelung explored is the QHJ equation for S, $$\bbox[yellow]{ 0=\partial_t S +\frac{1}{2} (\partial_x S)^2 +Q \\ Q\equiv - \frac{1}{2} \frac{ \partial_x^2 \sqrt{\rho}}{\sqrt{\rho}}=\frac {1}{1+4t^2}\left (1-\frac{2x^2}{1+4t^2}\right) } ,$$ the Q being the quantum potential—the curvature of the wf amplitude. Note this is of lower order than the one you have: it has effectively integrated it once!
It is possible this explicit expression, and the WP translating it with group velocity $$k_0$$, instead, \begin{align} \psi &= \frac{ \sqrt{2/\pi}}{\sqrt{1 + 2it}} e^{-\frac{1}{4}k_0^2} ~ e^{-\frac{1}{1 + 2it}\left(x - \frac{ik_0}{2}\right)^2}\\ &= \frac{ \sqrt{2/\pi}}{\sqrt{1 + 2it}} e^{-\frac{1}{1 + 4t^2}(x - k_0t)^2}~ e^{i \frac{1}{1 + 4t^2}\left((k_0 + 2tx)x - \frac{1}{2}tk_0^2\right)} ~\Longrightarrow \end{align} \\ \Large \rho= \frac{ \sqrt{2/\pi}}{\sqrt{1+4t^2}}~e^{-\frac{2(x-k_0t)^2}{1+4t^2}} , might help you with bottom-rung checking of your methods...