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[SOLVED] compatibility condition


Well-known member
Feb 1, 2012
L\phi &= \lambda\phi\\
\phi_t &= M\phi
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 \(L_t\phi + L\phi_t - \lambda M\phi = L_t\phi + LM\phi- \lambda M\phi = 0\).
[L_t + LM - \lambda M]\phi &= 0\\
L_t + LM - \lambda M &= 0
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?


Indicium Physicus
Staff member
Jan 26, 2012
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.