Yes, the claim is true.
Reason:
From the principle of Quantum Mechanics any two physical quantity can be measured simultaneously if their Quantum Poisson's bracket corresponding to the operator of those quantities is zero i.e it's commutes.
Now,
"\\hat T=-\\frac{\\hbar}{2m}\\nabla^2"And
"\\hat p=-i\\hbar\\nabla"Thus,
"[\\hat T,\\hat p]=\\hat T\\hat p-\\hat p\\hat T\\\\\n\\implies [\\hat T,\\hat p]=-\\frac{\\hbar}{2m}\\nabla^2(-i\\hbar\\nabla)-i\\hbar\\nabla(\\frac{\\hbar}{2m}\\nabla^2)=0"Hence, we are done.
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