# Recovering the nonlinear Schrödinger equation from its Lax pair

My question is less concerned with the physical aspects of the nonlinear Schrödinger equation and more with the mathematical mechanics of using a Lax pair.

I am considering how to recover the defocusing nonlinear Schrödinger equation $$iq_t + q_{xx} - 2|q|^2 q = 0$$ from its Lax pair. As I understand the equation arises from the requirement that the mixed second partials of the wave function must be equivalent which generates the requirement discussed in Lemma 1. $$\Psi(x,t,k)$$ is a 2x2 matrix.

$$\Psi_x + ik\sigma_3 \Psi = Q \Psi \text{ for } k \in \mathbb{C}$$ $$\Psi_t + 2ik^2 \sigma_3 \Psi = (2k Q - iQ_x \sigma_3 - i|q|^2 \sigma_3)\Psi$$

$$\sigma_3= \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$$

$$Q = \begin{pmatrix} 0 & q(x,t) \\ \overline{q}(x,t) & 0 \end{pmatrix}$$

The operators $$U$$ and $$V$$ are defined by the following equations. $${\Psi}_{{{x}}}={U}{\Psi}={\left({Q}-{i}{k}{\sigma}_{{{3}}}\right)}{\Psi}$$ $${\Psi}_{{{t}}}={V}{\Psi}={\left({2}{k}{Q}-{i}{Q}_{{{x}}}{\sigma}_{{{3}}}-{i}{\left|{q}\right|}^{{2}}{\sigma}_{{{3}}}-{2}{i}{k}^{{2}}{\sigma}_{{{3}}}\right)}\Psi$$

Lemma 1(Compatibility of the Lax Pair): Imposing the requirement on mixed partials yields the following relationship, $${\Psi}_{{{x}{t}}}={\Psi}_{{{t}{x}}}\Leftrightarrow{U}_{{{t}}}-{V}_{{{x}}}+{\left[{U},{V}\right]}={0}$$.

$${\Psi}_{{{x}{t}}}={U}_{{{t}}}{\Psi}+{U}{\Psi}_{{{t}}}$$

$${\Psi}_{{{t}{x}}}={V}_{{{x}}}{\Psi}+{V}{\Psi}_{{{x}}}$$

\begin{align} &{\Psi}_{{{x}{t}}}-{\Psi}_{{{t}{x}}}=\\ &{\left({U}_{{{t}}}-{V}_{{{x}}}\right)}{\Psi}+{U}{\Psi}_{{{t}}}-{V}{\Psi}_{{{x}}}= \\ &{\left({U}_{{{t}}}-{V}_{{{x}}}\right)}{\Psi}+{\left({U}{V}-{V}{U}\right)}{\Psi}= \\ &{\left({U}_{{{t}}}-{V}_{{{x}}}+{\left[{U},{V}\right]}\right)}{\Psi}= {0} \end{align}

The proof is concluded as biconditionality follows from algebraic equivalency.

Point of Difficulty: Substituting the original Lax pair into the condition $${U}_{{{t}}}-{V}_{{{x}}}+{\left[{U},{V}\right]}={0}$$ should yield the original nonlinear Schrödinger equation.

\begin{align} &{Q}_{{{t}}}-{2}{k}{Q}_{{{x}}}+{i}{Q}_{{{x}{x}}}{\sigma}_{{{3}}}+{i}{\sigma}_{{{3}}}{\left({\left|{q}\right|}^{{2}}\right)}_{{{x}}}+ {\left({Q}-{i}{k}{\sigma}_{{{3}}}\right)}{\left({2}{k}{Q}-{i}{Q}_{{{x}}}{\sigma}_{{{3}}}-{i}{\left|{q}\right|}^{{2}}{\sigma}_{{{3}}}-{2}{i}{k}^{{2}}{\sigma}_{{{3}}}\right)}- {\left({2}{k}{Q}-{i}{Q}_{{{x}}}{\sigma}_{{{3}}}-{i}{\left|{q}\right|}^{{2}}{}{\sigma}_{{{3}}}-{2}{i}{k}^{{2}}{\sigma}_{{{3}}}\right)}(Q -ik\sigma_3)= \\ &{Q}_{{{t}}}-{2}{k}{Q}_{{{x}}}+{i}{Q}_{{{x}{x}}}{\sigma}_{{{3}}}+{i}{\sigma}_{{{3}}}{\left({\left|{q}\right|}^{{2}}\right)}_{{{x}}}+{2}{k}{Q}^{{2}}- {i}QQ_{{{x}}}{\sigma}_{{{3}}}-{i}{\left|{q}\right|}^{{2}}{Q}{\sigma}_{{{3}}}-{2}{i}{k}^{{2}}{Q}{\sigma}_{{{3}}}-{2}{i}{k}^{{2}}{\sigma}_{{{3}}}{Q}-{k}{\sigma}_{{{3}}}{Q}{\sigma}_{{{3}}}-{k}{\left|{q}\right|}^{{2}}{{\sigma}_{{{3}}}^{{2}}}-{2}{k}^{{3}}{{\sigma}_{{{3}}}^{{2}}}-{2}{k}{Q}^{{2}}+{i}{Q}_{{{x}}}{\sigma}_{{{3}}}{Q}+{i}{\left|{q}\right|}^{{2}}{\sigma}_{{{3}}}{Q}+{2}{i}{k}^{{2}}{\sigma}_{{{3}}}{Q}+{2}{i}{k}^{{2}}{Q}{\sigma}_{{{3}}}+{k}{Q}{{\sigma}_{{{3}}}^{{2}}}+{k}{\left|{q}\right|}^{{2}}{{\sigma}_{{{3}}}^{{2}}}+{2}{k}^{{3}}{{\sigma}_{{{3}}}^{{2}}}= \\ & {Q}_{{{t}}}-{2}{k}{Q}_{{{x}}}+{i}{Q}_{{{x}{x}}}{\sigma}_{{{3}}}+{i}{\sigma}_{{{3}}}{\left({\left|{q}\right|}^{{2}}\right)}_{{{x}}}+{i}{\left[{Q}_{{{x}}}{\sigma}_{{{3}}},{Q}\right]}+{i}{\left|{q}\right|}^{{2}}{\left[{\sigma}_{{{3}}},{Q}\right]}+{2}{i}{k}^{{2}}{\left[{\sigma}_{{{3}}},{Q}\right]}+{2}{i}{k}^{{2}}{\left[{Q},{\sigma}_{{{3}}}\right]}+{k}{\left[{Q}{\sigma}_{{{3}}},{\sigma}_{{{3}}}\right]}= \\ & {Q}_{{{t}}}-{2}{k}{Q}_{{{x}}}+{i}{\left\lbrace{Q}_{{{x}{x}}}{\sigma}_{{{3}}}+{\sigma}_{{{3}}}{\left({\left|{q}\right|}^{{2}}\right)}_{{{x}}}+{\left[{Q}_{{{x}{x}}}{\sigma}_{{{3}}},{Q}\right]}+{\left({\left|{q}\right|}^{{2}}\right)}{\left[{\sigma}_{{{3}}},{Q}\right]}+{k}{\left[{Q}{\sigma}_{{{3}}},{\sigma}_{{{3}}}\right]}\right\rbrace}=\\ &\dots \text{ I do not understand how to proceed.} \end{align}

• Hello! I think that 1) you would find it helpful to organize your computations into pieces rather than doing it all at once, 2) there appears to be a typo for your V - the term involving |q|^2 is missing a factor of 1/2, and 3) you forgot to capitalize U_t and V_x in the lemma and its proof. May 13, 2021 at 10:41
• Thank you, I fixed the capitalization issue. I'm sorry, I don't follow what you mean with "$|q|^2$ in $V$ is missing a factor of $\frac{1}{2}$". Do you want me to define $V(x)=2kQ-iQ_{x}\sigma_3 - i \frac{1}{2}|q|^2 \sigma_3 - 2ik^2 \sigma_3$? Thanks again May 13, 2021 at 17:14
• Yeah! At least, when I worked out some details that was my conclusion. It's perhaps worth trying again with that new V. If it still doesn't work out, I'll gladly double-check my scratch notes and, if correct, post an answer with more detail. May 13, 2021 at 17:51
• Yes, please! I would appreciate any help regarding the problem. Yesterday, I found the following article which provides a parallel development of the problem on page 4. The author states that it is possible but does not provide the particulars of proving it. Thank you again. iopscience.iop.org/article/10.1088/0951-7715/18/4/019 May 13, 2021 at 18:57

Okay, here's the computation in detail. Really, this is a straightforward bracket computation, though it appears my comment earlier about a factor of 1/2 was in error (oops!) so no need to make a correction about that in your post.

The Lie bracket of $$U$$ with $$V$$ is a combination of the non-trivial Lie brackets: $$[Q,Q_x\sigma_3],\,[Q,\sigma_3],$$ and $$[\sigma_3, Q_x\sigma_3]$$. In order to compute these brackets we'll write out the relevant matrix products:

\begin{equation*} \begin{aligned} Q_x\sigma_3&=\begin{pmatrix} 0 & -q_x\\ \bar{q}_x & 0\end{pmatrix},\\ QQ_x\sigma_3&=\begin{pmatrix}q\bar{q}_x & 0\\ 0 & -q_x\bar{q} \end{pmatrix},\\ Q_x\sigma_3Q&=\begin{pmatrix}-q_x\bar{q} & 0\\ 0 & \bar{q}_xq \end{pmatrix},\\ [Q,Q_x\sigma_3]&=\begin{pmatrix} q\bar{q}_x+q_x\bar{q} & 0\\ 0 & -(q_x\bar{q}+\bar{q}_xq) \end{pmatrix},\\ \,&=|q|^2_x\sigma_3. \end{aligned} \end{equation*}

Also, notice that left multiplication by $$\sigma_3$$ of a $$2\times 2$$ matrix $$A$$ returns $$A$$ with negative the bottom row and right multiplication by $$\sigma_3$$ of $$A$$ returns $$A$$ but with negative 2nd column. Thus,

\begin{equation*} \begin{aligned} \left[Q,\sigma_3\right]&=\begin{pmatrix} 0 & -2q\\ 2\bar{q} & 0\end{pmatrix},\\ [\sigma_3,Q_x\sigma_3]&=\sigma_3Q_x\sigma_3-Q_x,\\ &=\begin{pmatrix} 0 & -2q_x\\ -2\bar{q}_x & 0 \end{pmatrix},\\ &=-2Q_x. \end{aligned} \end{equation*}

Now, looking at the Lie bracket of $$U$$ and $$V$$ we find \begin{equation*} \begin{aligned} \left[U,V\right]&=\left[Q-ik\sigma_3,\,2kQ-iQ_x\sigma_3-i|q|^2\sigma_3-2ik^2\sigma_3\right],\\ \,&=-i\left[Q, Q_x\sigma_3\right]-i(|q|^2+2k^2)\left[Q,\sigma_3\right]+2ik^2\left[Q,\sigma_3\right]-k\left[\sigma_3,Q_x\sigma_3\right],\\ &=-i|q|^2_x\sigma_3-i|q|^2\begin{pmatrix} 0 & -2q\\2\bar{q} & 0\end{pmatrix}+2kQ_x. \end{aligned} \end{equation*} So, in combination with the derivatives \begin{equation*} \begin{aligned} U_t&=Q_t,\\ V_x&=2kQ_x-iQ_{xx}\sigma_3-i|q|^2_x\sigma_3, \end{aligned} \end{equation*} we find that \begin{equation*} \begin{aligned} U_t-V_x+\left[U,V\right]&=Q_t+iQ_{xx}\sigma_3-i|q|^2\begin{pmatrix} 0 & -2q\\2\bar{q} & 0\end{pmatrix}\\ &=\begin{pmatrix} 0 & q_t-iq_{xx}+2i|q|^2q\\ \bar{q}_t-i\bar{q}_{xx}-2i|q|^2\bar{q} & 0 \end{pmatrix}. \end{aligned} \end{equation*}

Remark: The version of the Lax pair formulation you see in this problem with $$U_t-V_x+\left[U,V \right]=0$$ may also be referred to as a "zero curvature condition" and is slightly different from, but equivalent to, the Lax eqaution $$L_t=[B,L]$$ for appropriately chosen $$L$$ and $$B$$ operators.