Solve L=2 Atom Problem: Find Min (L_x)^2 + (L_y)^2

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SUMMARY

The discussion focuses on calculating the minimum value of (L_x)^2 + (L_y)^2 for an electron in an l=2 state. The established formula for total angular momentum is L = √(l(l+1))ħ, leading to a maximum value of L^2 = 6ħ^2 when l=2. The maximum value of (L_z)^2 is determined to be 4ħ^2, which implies that the minimum value of (L_x)^2 + (L_y)^2 is 2ħ^2. The relationship between L_x, L_y, and L_z is governed by the conservation of total angular momentum.

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Homework Statement


For an electron in an atom that is an l=2 state, find the smallest value of (L_x)^2+(L_y)^2.


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The Attempt at a Solution


okay so L = (l(l+2))^(1/2)

and that squared with l =2 would be 6h(bar)^2

kind of running into a wall here..
Or is it the maximum value of (L_x)^2 + (L_y)^2 occurs when (L_z)^2 is a maximum value? and L_z = m_l*h(bar).. which would have a maximum value of 4h_bar^2 so therefore L_x^2 + L_y^2 would be 2h(bar)^2?

Ps I need to learn LaTex X_x
 
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l=2 tells you the Lz state.
How do the Lx and Ly states relate to that?
Hint: conservation of total angular momentum.
 

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