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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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No, the only thing you can conclude is that $\langle\psi|[H,A]|\psi \rangle =0$. Example, for some real constants $a,b$ and for a particle described in $L^2(\mathbb R^3)$ $A= aL_x$, $H=bL_z$, $$|\psi\rangle = |\phi(r)\rangle \otimes|l=0,m_z=0\rangle\:.$$ In this case $[H,A] \neq 0$ but $\langle \psi(t)|A|\psi(t) \rangle =0$ for every $t \in \mathbb R$ since ...

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I think $[A,C]=[B,C]=0$ with $C=[A,B]$ is an assumption, because there exist counterexamples: for $A=\sigma_x$ are $B=\sigma_y$ Pauli matrix along $x, y$ directions respectively. then $C=2i\sigma_z$ is Pauli matrix along z direction. Obviously $[A,C]\neq 0$ and $[B,C]\neq 0$.

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Take the commutator acting on a function $f$. Then \begin{split} [ P_i , P_j ] f &= [ - i \partial_i - q A_i , - i \partial_j - q A_j ]f \\ &= ( i \partial_i + q A_i )( i \partial_j + q A_j ) f -( i \partial_j + q A_j ) ( i \partial_i + q A_i ) f \\ &= - \partial_i \partial_j + i q A_i \partial_j \, f + i q \partial_i ( ...

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I think your confusion is arising from the fact that you are imagining operators as matrices. This is mostly fine, but in this case, the operator itself being a vector is what is causing the confusion - so let me elaborate. ${\bf A}$ is a vector of operators. For example $${\bf A} = \pmatrix{ A_1 \\ A_2 \\ A_3}$$ We can denote this collectively as $A_i$. ...

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Answer to this question should start from why we want the physical observables to be represented by linear operators. Theoretical physics is about constructing a mathematical model which we hope describes the phenomena it's being modeled for and hence helps predicting stuff. In classical physics this mathematical model is based simply on the real numbers ...

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Even in one dimension the operator $p_r=-i\partial_r$ on the half line $r>0$ has deficiency indices $(0,1)$. There is thus no way to define it it as a self-adjoint operator. In practical terms this abstract mathematical statement means that there is no set of boundary conditions thta we can impose on the wavefunction $\psi(r)$ that lead to a ...

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Actually, if the energy of particles is high enough to take relativity into consideration, the concept of particles in quantum mechanics is no longer as valid. For example, the uncertainty relationship $\Delta E \cdot \Delta t \approx \hbar$ and energy-mass relation $E=mc^2$ suggest that there will be new particles created and annihilated in those cases. So ...

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The quantization prescription $$[\hat{x},\hat{y}] := \mathrm{i}\hbar\widehat{\{x,y\}}\tag{1}$$ for $x,y$ two classical phase space coordinates does have its subtleties. In particular, as the answer in the linked question says, it leads to inconsistent results when applied to e.g. polar coordinates. The reason for this is two-fold: For the radial ...

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You have mixed indices in the end of your line. Correctly: $$-\epsilon_{iab}\epsilon_{jcd}(x_ap_d\delta _{bc} - x_cp_b\delta _{ad}).$$ So further, \$-\epsilon_{iab}\epsilon_{jcd}(x_ap_d\delta _{bc} - x_cp_b\delta _{ad})=\epsilon_{iab}\epsilon_{jcd}x_cp_b\delta _{ad}-\epsilon_{iab}\epsilon_{jcd}x_ap_d\delta _{bc}=\\=\epsilon_{idb}\epsilon_{jcd}x_cp_b - ...

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