$\left< \frac{\partial (xp)}{\partial t} \right> = 0$ When is this true?
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Is this always true in quantum mechanics?
⟨∂(xp)∂t⟩=0⟨∂(xp)∂t⟩=0
I encountered this when working problem 3.31 in Griffiths Introduction to Quantum Mechanics II.
He does this in his solutions manual:
ddt⟨Q⟩=iℏ⟨[H^,Q^]⟩+⟨∂Q^∂t⟩ddt⟨Q⟩=iℏ⟨[H^,Q^]⟩+⟨∂Q^∂t⟩
ddt⟨xp⟩=iℏ⟨[H,xp]⟩ddt⟨xp⟩=iℏ⟨[H,xp]⟩
I'm not sure why ⟨∂(xp)∂t⟩=0⟨∂(xp)∂t⟩=0. It sort of seems plausible that the expectation of a change in xpxpwith respect to time would be 00 due to symmetry but is there more to it
⟨∂(xp)∂t⟩=0⟨∂(xp)∂t⟩=0
I encountered this when working problem 3.31 in Griffiths Introduction to Quantum Mechanics II.
He does this in his solutions manual:
ddt⟨Q⟩=iℏ⟨[H^,Q^]⟩+⟨∂Q^∂t⟩ddt⟨Q⟩=iℏ⟨[H^,Q^]⟩+⟨∂Q^∂t⟩
ddt⟨xp⟩=iℏ⟨[H,xp]⟩ddt⟨xp⟩=iℏ⟨[H,xp]⟩
I'm not sure why ⟨∂(xp)∂t⟩=0⟨∂(xp)∂t⟩=0. It sort of seems plausible that the expectation of a change in xpxpwith respect to time would be 00 due to symmetry but is there more to it
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∂Q^∂t∂Q^∂t means that one should take a derivative of the operator Q^Q^ keeping xx and pp fixed, i.e. it keeps track of explicit time dependences that operators map have. The operator xpxp has no explicit time dependence. Hence ∂(xp)∂t=0∂(xp)∂t=0.
∂Q^∂t∂Q^∂t means that one should take a derivative of the operator Q^Q^ keeping xx and pp fixed, i.e. it keeps track of explicit time dependences that operators map have. The operator xpxp has no explicit time dependence. Hence ∂(xp)∂t=0∂(xp)∂t=0.
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