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First, I'm no expert in symmetry analysis of evolution equations and so I apologize if this post is a bit of a cobble. The question I have is about the evolution of $\psi: \mathbb{R}^{1+1}\to \mathbb{C}$ given by

$i \psi_{t} + \psi_{xx} + \frac{1}{2}|\psi|^{2} \psi + \lambda \left(\left[|\psi|^{2} \psi\right]_{xx} + |\psi|^{4} \psi \right) =0, \, \, \lambda \in \mathbb{R}$,

which could be thought of as a fully nonlinear perturbation of the cubic focusing nonlinear Schrodinger equation, which has a well-known and rich structure endowed by complete integrability.

I've given this equation and its complex conjugate to the SYM software for Mathematica. To ease the computational burden, I've run the procedure for various $\lambda$. Doing so reveals that $\partial_{t}$ and $\partial_{x}$ are Lie point symmetries for the evolution. Given the form of the PDE, this seems sensible. I've found in several locations that time translation invariance conserves the system "energy/Hamiltonian" and that spatial translation invariance conserves "mass/momentum." Decoding these statements is where I'm having problems.

Let's concentrate on the time translation symmetry (TTS). I'm aware of Geometric structure of NLS evolution and can generalize the process to include the power term, but I don't see a way to incorperate the fully nonlinear term.

  1. What is the notion of "energy/Hamiltonian" that is conserved by TTS?
  2. Supposing that I could derive a Hamiltonian functional, is this the quantity which is conserved by TTS?
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Perhaps part of your question could be phrased as follows. Does there exists a (formally defined) functional $S_1[\psi,\bar{\psi}] = \int dt\,dx\, L_1[\psi,\bar{\psi}]$ such that its Euler-Lagrange derivative gives $E[\psi,\bar\psi] = \frac{\delta S_1}{\delta\bar{\psi}} = \left[ |\psi|^2 \psi \right]_{xx}$, and by conjugation symmetry $\bar{E}[\psi,\bar\psi] = \frac{\delta S_1}{\delta\psi} = \left[ |\psi|^2 \bar{\psi} \right]_{xx}$? The Euler-Lagrange derivative has the general form $\frac{\delta S_1[\psi,\bar{\psi}]}{\delta \bar{\psi}} = \frac{\partial L_1}{\partial \bar{\psi}} - \partial_i \frac{\partial L_1}{\partial(\partial_i\bar{\psi})} + \partial_i\partial_j \frac{\partial L_1}{\partial(\partial_i\partial_j\bar{\psi})} - \cdots$, and analogously for $\psi$.

If that were the case, then the linearizations of $E[\psi,\bar{\psi}]$ and $\bar{E}[\psi,\bar\psi]$ would constitute a formally self-adjoint linear differential operator (exercise!). That is, we would have \begin{multline*} \int dx\,dt\, \left(\bar{\phi} \dot{E}[\psi,\bar{\psi}; \xi,\bar{\xi}] + \phi \dot{\bar{E}}[\psi,\bar{\psi}; \xi,\bar{\xi}]\right) \\ = \int dx\,dt\, \left(\dot{E}[\psi,\bar{\psi}; \phi,\bar{\phi}] \bar{\xi} + \dot{\bar{E}}[\psi,\bar{\psi}; \phi,\bar{\phi}] \xi\right) , \end{multline*} whenever $\xi,\bar{\xi}, \phi, \bar{\phi}$ are all smooth and compactly supported, where \begin{align*} E[\psi+\epsilon\xi,\bar{\psi}+\epsilon\bar{\xi}] &= E[\psi,\bar{\psi}] + \epsilon \dot{E}[\psi,\bar{\psi}; \xi,\bar{\xi}] + O(\epsilon^2) , \\ \bar{E}[\psi+\epsilon\xi,\bar{\psi}+\epsilon\bar{\xi}] &= \bar{E}[\psi,\bar{\psi}] + \epsilon \dot{\bar{E}}[\psi,\bar{\psi}; \xi,\bar{\xi}] + O(\epsilon^2) . \end{align*}

In your case, \begin{align*} \left[ |\psi+\epsilon\xi|^2 (\psi+\epsilon\xi) \right]_{xx} &= \left[ |\psi|^2 \psi \right]_{xx} + \epsilon \left[ 2|\psi|^2 \xi + \bar{\xi} \psi^2 \right]_{xx} + O(\epsilon^2), \\ \left[ |\psi+\epsilon\xi|^2 (\bar{\psi}+\epsilon\bar{\xi}) \right]_{xx} &= \left[ |\psi|^2 \bar{\psi} \right]_{xx} + \epsilon \left[ 2|\psi|^2 \bar{\xi} + \xi \bar{\psi}^2 \right]_{xx} + O(\epsilon^2) . \end{align*} But then, \begin{multline*} \int dx\,dt\, \left( \bar{\phi} \left[ 2|\psi|^2 \xi + \bar{\xi} \psi^2 \right]_{xx} + \phi \left[ 2|\psi|^2 \bar{\xi} + \xi \bar{\psi}^2 \right]_{xx} \right) \\ = \int dx\,dt\, \left( [2|\psi|^2 \phi_{xx} + \psi^2 \bar{\phi}_{xx}] \bar{\xi} + [2|\psi|^2 \bar{\phi}_{xx} + \bar{\psi}^2 \phi_{xx}] \xi \right) \end{multline*}

Obviously, (the linearizations of) your $E[\psi,\bar{\psi}]$ and $\bar{E}[\psi,\bar{\psi}]$ do not satisfy the necessary formal self-adjointness condition. Hence, no such $S_1[\psi,\phi]$ functional exists.

If your modified Schroedinger equation cannot be obtained from a variational principle, then Noether's theorem no longer applies and you cannot deduce the existence of conserved currents from the existence of symmetries.

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  • $\begingroup$ I've worked out similar mathematical statements and have grown to the opinion that it is impossible to encode this term into a functional. However, it was extremely helpful to read your exposition. So, thanks a bunch for taking the time to write all of this out! $\endgroup$
    – user153764
    Commented May 19, 2017 at 15:08

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