- #1
Dustinsfl
- 2,281
- 5
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?
\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?