# [SOLVED]compatibility condition

#### dwsmith

##### Well-known member
Given
\begin{align}
L\phi &= \lambda\phi\\
\phi_t &= M\phi
\end{align}
where $$L$$ and $$M$$ are operators and $$\lambda$$ a constant.
I want to show the compatibily condition is $$L_t + [L,M] = 0$$ where $$[,]$$ is the commutator.
$(L\phi)_t = L_t\phi + L\phi_t = \lambda\phi_t = \lambda M\phi$
That is, we have $$L_t\phi + L\phi_t - \lambda M\phi = L_t\phi + LM\phi- \lambda M\phi = 0$$.
\begin{align}
[L_t + LM - \lambda M]\phi &= 0\\
L_t + LM - \lambda M &= 0
\end{align}
Can I just let $$\lambda = L$$ which doesn't make since sense one is an operator and the other a constant? If not, how do I get the commutator $$LM - ML$$ part?

#### Ackbach

##### Indicium Physicus
Staff member
Ah, I see you are doing the Inverse Scattering Transform with Lax pairs. You have $\lambda M \phi$. Since $\lambda$ is a scalar, rearrange thus: $\lambda M \phi=M(\lambda \phi)=M L \phi$. You can go from there.