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}
$$



Recovering the nonlinear Schrödinger equation from its Lax pair
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.


