- #1
FatPhysicsBoy
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Homework Statement
So we know that for two operators [itex]\hat{A}[/itex] and [itex]\hat{B} + \hat{C}[/itex] we have the following rule for the commutator of the two: [itex][\hat{A},\hat{B} + \hat{C}] = [\hat{A},\hat{B}] + [\hat{A},\hat{C}][/itex]
However, if I'm commuting [itex][\hat{p_{x}}, \hat{H}][/itex] where [itex]\hat{H}[/itex] is the Hamiltonian and [itex]\hat{p_{x}}[/itex] is the momentum operator, can I use this operator rule to "expand" the commutator [itex][\hat{p_{x}}, \hat{H}][/itex] in the same way as [itex]\hat{A}[/itex] and [itex]\hat{B} + \hat{C}[/itex] even though [itex]V(x)[/itex] in the Hamiltonian is not an operator?
If this can be done then we can get [itex][\hat{p_{x}}, \hat{H}][/itex] = [itex][\hat{p_{x}}, V(x)][/itex] and expand that commutator instead of messing about with the [itex]\frac{\hat{p^{2}}_{x}}{2m}[/itex] part in the Hamiltonian since [itex][\hat{p_{x}}, \hat{p^{2}}_{x}][/itex] = 0.
Thank You