Hamiltonian-momentum commutator

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SUMMARY

The discussion centers on the computation of the Hamiltonian-momentum commutator \(\left[H, \mathbf{p}\right]\) for a potential of \(-\frac{1}{r}\). The result obtained is \(i \hbar \left(\frac{1}{r^{2}}, 0, 0\right)\). The method involves using the momentum operator \(\mathbf{p} = -i \hbar \nabla\) and the Hamiltonian \(H = \frac{-\hbar^{2}}{2m} \nabla^{2} + U\), where \(U\) is the potential energy. The commutation is simplified by noting that the kinetic term commutes with \(\mathbf{p}\), allowing focus on the potential term.

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jarvinen
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I have a potential of -1/r and I need to compute [itex]\left[H , \ \mathbf{p} \right][/itex].

I got the result of [itex]i \hbar \left( \frac{1}{r^{2}}, \ 0 , \ 0 \right)[/itex].

Am I wrong about this?
 
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Can you post your method of computing the commutator ?
 
Just used that [itex]\mathbf{p} = -i \hbar \nabla[/itex] and [itex]H = \frac{- \hbar ^{2}}{2m} \nabla ^{2} + U[/itex]

Hence [itex]H \mathbf{p} \psi - \mathbf{p} H \psi[/itex] can be written but note that the [itex]\mathbf{p} \dot \mathbf{p}[/itex] part of H will commute with [itex]\mathbf{p}[/itex], hence only consider [itex]U \mathbf{p} \psi - \mathbf{p} U \psi = \left( -i \hbar \right) \left( - \psi \nabla U \right)[/itex] then substitute for the given U.
 

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