in free scalar field, the momentum operator is $$P=-\int d^3 x \pi \nabla \phi$$. If we write it with creation and annihilation operator, then we can get the apparently time independent form,$$P=\int d ^3 p \vec p a_p^\dagger a_p$$. My question is that can we just use the commutation relation of $\pi$ ,$\phi$ and Hamiltonian to show that momentum operator is time is time independent?
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
1
Yes. Just take the time derivative of $P$ in its original form and use equations of motion. We have $$ \begin{align} \partial_t P &= - \int d^3 x \left( \partial_t \pi \nabla \phi + \pi \nabla \partial_t \phi \right) \\ &= - \int d^3 x \left( \nabla^2 \phi \nabla \phi + \pi \nabla \pi \right) \\ &= - \frac{1}{2} \int d^3 x \nabla \left( \nabla \phi \nabla \phi + \pi^2 \right) \\ &= 0 \end{align} $$
-
$\begingroup$ Just add an comment, $\nabla^2 \phi \nabla \phi=\frac{1}{2}\nabla(\nabla \phi\nabla \phi)$ since $\nabla \phi$ commutes with $\nabla \nabla \phi$ $\endgroup$– xjtanCommented Oct 20, 2015 at 17:13