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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} $$

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    $\begingroup$ 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. $\endgroup$
    – TK-421
    May 13, 2021 at 10:41
  • $\begingroup$ 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 $\endgroup$
    – Talmsmen
    May 13, 2021 at 17:14
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    $\begingroup$ 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. $\endgroup$
    – TK-421
    May 13, 2021 at 17:51
  • $\begingroup$ 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 $\endgroup$
    – Talmsmen
    May 13, 2021 at 18:57

1 Answer 1

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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.

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