$\left< \frac{\partial (xp)}{\partial t} \right> = 0$ When is this true?
Answers
Answered by
0
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
⟨∂(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
Similar questions