Ladder operators for angular momentum

[Tvdmeer]

This might be a basic question, but I'm having some difficulty understanding expectation values and ladder operators for angular momentum.

<L+> = ?

I know that L+ = Lx+iLy, but I don't know what the expectation value would be?

Someone told me something that looked like this:

<lml|L+|lml>=<lml|lml+1

But I don't really get how that works, nor do I know if that is the same as <L+>.

Any help would be appreciated. Thanks in advance!

[Meir Achuz]

Since L+ changes one state to another, its expectation value is zero unless the state is a mixture.

[Tvdmeer]

Thanks for getting back to me. I understand conceptually why the expectation value of L+ is equal to zero. I guess I'm more asking what the proof is for that. Obviously since it is just an operator it will have a zero likelihood of being anywhere, but I just don't know why in this equation my teacher new to sandwich l+ with lms on either side:

<lml|L+|lml>

I just don't understand the setup

[Meir Achuz]

L+|l,m> gives |l,m+1>, which iis orthogonal to |l,m>.

[ccesare]

If a system is in the state \left|\psi\right>, the expectation value of an Hermitian operator A is given by

\left<\psi\right|A\left|\psi\right>

This has nothing to do with an operator "being anywhere," but rather with the average value you would see if you performed an experiment many times and measured the physical quantity represented by the operator A.

Of course the operator L_+ is NOT Hermitian, and the interpretation of its expectation value as an expected average over many experimental trials is nonsense. In this particular case, the expectation value vanishes for the reason already mentioned, namely

L_+ \left|l,m_l\right> \propto \left|l,m_l +1 \right>

and the states \left|l,m_l\right> are orthonormal, i.e. \left<l,m_l\right|k,n_k\left> = \delta_{l,k} \delta_{m_l,n_k}