Deriving the L+-L- Formula in Quantum Mechanics

AI Thread Summary
The discussion revolves around deriving the formula L+-L- =-(h/2π)^2[d²/dθ²+cotθ d/dθ +cot²θ d²/dθ²+id/dφ] from the initial expression L+-L-=±(h/2π)e^±(iθ)[d/dθ + i cotθ d/dφ]. The original poster expresses confusion about the derivation process and the apparent contradiction in the formula's equality. They later clarify their mistake and confirm that they found the correct answer. The conversation highlights the challenges of working with quantum mechanics equations and the importance of precision in mathematical expressions. Ultimately, the poster successfully resolves their issue.
venusfly
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Hello, I have a quantum mechanics problem. Here is the problem statement. I must derive this formula,

L+-L- =-(h/2π)^2[d²/dθ²+cotθ d/dθ +cot²θ d²/dθ²+id/dφ]

from this formula,

L+-L-=±(h/2π)e^±(iθ)[d/dθ + i cotθ d/dφ]

I have no idea how to approach this problem. Can anyone give me a suggestion on how to begin the proof? Thanks.
 
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Something is wrong here: L+-L- can't be equal to both! The first is clearly a second derivative. Is it supposed to be something like (L+-L-)2?
 
Done

Sorry, wrote it wrong. Got the answer though. Thanks.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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