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My exam is tomorrow... Help will be much appreciated...

Say l=0 is given. My hemiltonian is [itex] \bar {H}=\frac {1} {\hbar} \Omega (\vec {n} \cdot \vec {L})^2 [/itex]

And we use "base states" that are [itex] L_z [/itex] eigenstates, |1 m>

n is a general direction unit vector.

Now, I am looking for all the possible energies I can measure.

What I thought was the answer is - sincs [itex] L^2=l(l+1)\hbar^2 |lm> [/itex]

There is only one option- since l=1. But the solution shows a different thing:

They refer to the possible m's: since l=1, m=-1,0,1

So they actually use [itex] L_z^2 [/itex] instead of [itex] L^2 [/itex]

Why is that true?

In the next section of the question, the vector [itex] \vec {n}= \frac {1} {\sqrt 2} (1,1,0) [/itex] is given.

Now, we put the system(for t=0) in such a way so that [itex] |\psi>= |1,-1> [/itex] and we are looking for the wave function for any time t.

So now is what I don't understand: in the first section, they used [itex] L_z^2 [/itex] in the hemiltonian, instead of [itex] L^2 [/itex] .

But now, using the given n, in the hemiltonian, since n does not have a z component, the Hemiltinian should be zero!

What an I not getting right?

Thank you so much!!

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# Angular Mumentum: Question for my exam tomorrow

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