The Commutator of M⁺ with H_op

Remembering that

$$H_{op} = \frac{p^2}{2m} = - \frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}$$

after some effort, one has

$$[H,M^+]= - \frac{\hbar^2}{2m} \left ( - \left (\frac{\pi}{L}... ...{L} \right )\frac{\hbar^2}{2m}\frac{\partial^2 }{\partial x^2}$$ (5)