Recent content by vballpro

2nd time's a charm? -i*(h_bar)^3 [ cos(theta)/sin(theta) * d^2/d(theta)^2 + d^3/d(theta)^3 + 1/sin^2(theta) * d^3/d^3(theta) why is it that it's d/dx [ f(y)g'(y) d/dy + f(y)g(y)d^2/dy^2] instead of d/dx [ f(y)g'(y) d/dy + f'(y)g(y)d^2/dy^2]? (i added in an f ' to the second part. it might be...

do you use the product rule or are they two totally different derivations?

how do i work with 1/sin(theta)d/d(theta) * (sin(theta)d/d(theta)) part?

this is what i have... Lz(L^2) = ih_bar^3 * {[1/cos(theta) (sin(theta)) + (cos(theta))*1/sin(theta)] + 1/sin^2(theta)} simplified...ih_bar^3 [sin(theta)/cos(theta) + cos(theta)/sin(theta) + 1/sin^2(theta)] I did the exact same thing for L^2(Lz), but I realize that making a mistake for each of...

gabba, I would say that LaTeX Code: L^2(L_z\\psi) and LaTeX Code: L_z(L^2\\psi) for any wavefunction LaTeX Code: \\psi would be equivalent. My problem is more with doing the problem. Do I do the product rule when it comes to -h_bar^2 * 1/sin(theta) * d/d(theta), or is -h_bar^2 pulled out to...

Homework Statement Using the definitions of Lz and L^2, show that these two operators commute. Homework Equations Lz = -ih_bar * d/d(phi) L^2 = -(h_bar)^2 {1/sin(theta) * d/d(theta) * [sin(theta) * d/d(theta)] + 1/sin^2(theta) d^2/d(phi)^2} The Attempt at a Solution I'm actually...