Challenging Quantum Problem

  • Thread starter Tuneman
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In summary, the question asks for a formula to calculate the time derivative of the expectation value of the angular momentum. The Simplified Problem states that d/dt(L) = i/(hbar) ( p^2/2m + V , -i(hbar) x (grad) ) + (r x dp/dt) which can be simplified to d/dt(L) = i/(hbar) ( p^2/2m + V , -i(hbar) x (grad) ) + (r x dp/dt) where dr/dt x p is 0.
  • #36
Yep. The quick way of seeing this intuitively is by recalling that [itex]\vec \nabla V[/itex] is the vector which, at eacht point, points in the direction of maximum increase of V. If V is spherically symmetric, what other direction can this be but radially outward? In the tangential direction it must be zero.
So [itex]\vec \nabla V(r)[/itex] points in the direction of [itex]\vec r[/itex] everywhere, so its cross product with [itex]\vec r[/itex] is zero everywhere.
 
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  • #37
Actually, you could have used the equation is suggested. The commutator [H,L] is zero, and <dl/dt> is obviously zero. Its a nifty little formula, and its very useful.

- harsh
 

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