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The issue here is whether or not you have a sum over $l$. If you want $r_l^2$ to mean any of $r_1^2, r_2^2, r_3^2$, then when you write $r_l^2 = r_l r_l$ you should not be summing over $l$. So $[r_l^2,L_i] = 2i\hbar \epsilon_{ijl}r_jr_l$ is correct as long as you sum over $j$ but not over $l$. On the other hand, if in $r_lr_l$ you sum over $l$ you get the ...

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but keep getting confused if I should sum over each term separately. It's not particularly clear what you mean here. Your expression uses the Einstein summation convention, which means that every repeated index is summed over. In principle this should be done for each term separately, so your expression reads $${R_{00}}= ... 0 Yes it is wrong because multiplication of matrices, you know it, gives matrices and I don't think it makes sense to put a matrice inside a ket or even a bra vector. Actually even with a constant (complex number) if you have k |v\rangle it does not make any sense to put it inside the ket vector like |kv \rangle. However if you have a constant k in the ... 6 Dirac notation is ill-suited for non-self-adjoint operators. Here's why: Let (-,-) be the inner product on our Hilbert space. The expectation value of AB is then$$ \langle AB \rangle_\psi = (\psi,AB\psi)$$by definition, and Dirac notation writes \langle \psi \vert AB \vert \psi \rangle. for this. But, in this notation, it is no longer clear to which ... 0  \langle\psi|AB|\psi\rangle  is a complex number (as opposed to a matrix), so taking its transpose gives you back the same thing, i.e.$$ \langle\psi|AB|\psi\rangle^{\dagger} = \langle\psi|AB|\psi\rangle^*, $$and therefore$$ = \langle\psi|B^\dagger A^\dagger|\psi\rangle . $$EDIT I just realised that you then equated this to  \langle\psi|B^\dagger ... 5 Comments to the question (v5): In this quantum case the overline/bar notation \bar{A}=\langle A\rangle is borrowed from statistics and it denotes a quantum expectation value of a quantity A. See also Ehrenfest theorem. The problem from Ref. 1 considers a harmonic oscillator with Hamiltonian operator$$\tag{A} H~=~\frac{p^2}{2m} ...

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$\frac{dM}{dt} = \frac{\partial{M}}{\partial{t}}+\frac{\partial{M}}{\partial{x}}\frac{d{x}}{d{t}} = \frac{\partial{M}}{\partial{t}}+v\cdot\nabla{M}$ (with no assumption on what is M) . So if $v\cdot\nabla{M} \neq0$ you can have one of $\frac{dM}{dt}$ and $\frac{\partial{M}}{\partial{t}}$ that is zero when the other is not. ...

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Your last expression is not valid, for two reasons: first, any given index can only occur twice per term in the Einstein convention, once as an upper index and once as a lower index. Remember that when an index is repeated, it means you sum over it with the metric: $$T^a T_a = \sum_{a,b} g_{ab} T^a T^b$$ You have terms like ...

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