How I can prove the Commutation between hamiltonian and Runge-Lenz vector? I am a undergraduate student in physics. I found this page that shows a way to prove the commutator between Runge-Lenz vector and Hamiltonian
.$\left [\hat{A}_{i},\hat{H}\right]=0$
I believe he did a good job on it. But for some reason, I don't know why $$\left[ \frac{r_i}{r},\hat{p}_l \right]=i\left( \frac{\delta_{il}}{r}-\frac{r_i r_l}{r^3} \right)\,.$$
Then, I changed the notation and I did some calculations to try to understand how he made it. Take a look.
For $\hbar=1$, we have $\hat{P}_{l}=-i\frac{\partial}{\partial X_{i}}$, then:
\begin{align}\left[ \frac{r_i}{r},\hat{P}_l \right]&=\frac{r_{i}}{r}\left(-i\frac{\partial}{\partial X_{l}}\right)+i\frac{\partial}{\partial X_{l}}\left( \frac{r_{i}}{r} \right )\\&
=i\frac{\partial}{\partial X_{l}}\left( \frac{r_{i}}{r} \right )-i\frac{r_{i}}{r}\left(\frac{\partial}{\partial X_{l}}\right)\\
&=i\left(\frac{\partial}{\partial x}+\frac{\partial}{\partial y}+\frac{\partial}{\partial z}  \right) \left(\frac{x}{\sqrt{x^{2}+y^{2}+z^{2}}}+\ldots +\ldots\right)-i\frac{r_{i}}{r}\left(\frac{\partial}{\partial X_{l}}\right)\\&
=i\left(\frac{y^{2}+z^{2}}{\left({x^{2}+y^{2}+z^{2}}\right)^{3/2}}+\frac{x^{2}+z^{2}}{\left({x^{2}+y^{2}+z^{2}}\right)^{3/2}}+\frac{x^{2}+y^{2}}{\left({x^{2}+y^{2}+z^{2}}\right)^{3/2}}\right)-i\frac{r_{i}}{r}\left(\frac{\partial}{\partial X_{l}}\right)\\&
=2i\left(\frac{x^{2}+y^{2}+z^{2}}{r^{3}}\right)-i\frac{r_{i}}{r}\left(\frac{\partial}{\partial X_{l}}\right)\\&
=2i\left(\frac{r^2}{r^{3}}\right)-i\frac{r_{i}}{r}\left(\frac{\partial}{\partial X_{l}}\right)\end{align}
What have I done wrong? If not, what is the next step? 
 A: @ACuriousMind thank you; I'll read the rules as soon as possible.
I was wrong. Look at this: 
If $\hbar=1$, then
\begin{align}\hat{P}_{j}&=-i\frac{\partial}{\partial X_{j}}\\ 
\left[ \frac{r_i}{r},\hat{P}_j \right]f&=\frac{r_{i}}{r}\left(-i\frac{\partial}{\partial X_{j}}\right)f+i\frac{\partial}{\partial X_{j}}\left( \frac{r_{i}}{r} \right )f\\&=-i\frac{r_{i}}{r}\frac{\partial f}{\partial X_{j}}+if\frac{\partial }{\partial X_{j}}\left (\frac{r_{i}}{r} \right )+i\frac{r_{i}}{r}\frac{\partial f}{\partial X_{j}}\\&=if\frac{\partial }{\partial X_{j}}\left (\frac{r_{i}}{r} \right )\end{align}
$$\left [\frac{r_{i}}{r},\hat{P}_{j}\right]=i\frac{\partial }{\partial X_{j}}\left (\frac{r_{i}}{r} \right )$$
EDIT: I messed up again. The text in red is my old calculation. I have also been helped by my advicer and by author.
${\color{red} {\begin{align}\frac{\partial}{\partial X_{j}}\left(\frac{r_{i}}{r}\right)&=\nabla \left(\frac{x+y+z}{\left(x^2+y^2+z^2\right)^{1/2}}   \right)\\&
=\left(\frac{y^{2}+z^{2}}{\left({x^{2}+y^{2}+z^{2}}\right)^{3/2}}+\frac{x^{2}+z^{2}}{\left({x^{2}+y^{2}+z^{2}}\right)^{3/2}}+\frac{x^{2}+y^{2}}{\left({x^{2}+y^{2}+z^{2}}\right)^{3/2}}\right)\\&
=2\frac{r^2}{r^3}\end{align}}}$
$$
{\color{red}{\left [\frac{r_{i}}{r},\hat{P}_{j}\right]=2i\frac{r^2}{r^3}}}$$
EDIT: This part is correct ,but useless to our purpose.
Is $$2i\frac{r^2}{r^3}=2i\left(\frac{\delta _{ij}}{r} -\frac{r_{i}r_{j}\delta _{ij}}{r^{3}}\right)\,?$$  We shall check. 
$$r^{2}=\vec{r}\cdot\vec{r}=r_{i}r_{j}\hat{e}_{i}\cdot \hat{e}_{j}=r_{i}r_{j}\delta _{ij}$$
Therefore if $i=j\,,$ then
$2i\frac{r^{2}}{r^{3}}=-2i\left(\frac{1}{r} -\frac{3r_{i}r_{i}}{r^{3}}\right)=-2i\left(\frac{r^2}{r^3} -\frac{3r^{2}}{r^{3}}\right)=2i\frac{r^2}{r^3}$
if $i\ne j\,,$ then
$$-2i\left(0- \frac{r_{i}r_{j}\delta _{ij}}{r^3} \right)=2i\frac{r^2}{r^3}$$
Therefore,
$$
{\color{red}{\left [\frac{r_{i}}{r},\hat{P}_{j}\right]=2i\frac{r^2}{r^3}=\left(\frac{\delta _{ij}}{r} -\frac{r_{i}r_{j}\delta _{ij}}{r^{3}}\right)}}$$
Looks like, but it is not the expected result. :/ 
EDIT:
Let me write the correct answer: 
\begin{align}\hat{P}_{j}&=-i\frac{\partial}{\partial X_{j}}\\ 
\left[ \frac{r_i}{r},\hat{P}_j \right]f&=\frac{r_{i}}{r}\left(-i\frac{\partial}{\partial X_{j}}\right)f+i\frac{\partial}{\partial X_{j}}\left( \frac{r_{i}}{r} \right )f\\&=-i\frac{r_{i}}{r}\frac{\partial f}{\partial X_{j}}+if\frac{\partial }{\partial X_{j}}\left (\frac{r_{i}}{r} \right )+i\frac{r_{i}}{r}\frac{\partial f}{\partial X_{j}}\\&=if\frac{\partial }{\partial X_{j}}\left (\frac{r_{i}}{r} \right )\end{align}
$$\left [\frac{r_{i}}{r},\hat{P}_{j}\right]=i\frac{\partial }{\partial X_{j}}\left (\frac{r_{i}}{r} \right )$$
${\color{blue} {\begin{align}\left[\frac{r_{i}}{r},\hat{P}_{j}\right]&=i\frac{\partial}{\partial X_{j}}\left(\frac{r_{i}}{r} \right)\\&
=i\left(\frac{1}{r}\frac{\partial r_{i}}{\partial x_{j}}+r_{i}\frac{\partial}{\partial x_{j}}\left(\frac{1}{r}\right)\right)\\&
=i\left(\frac{1}{r}\frac{\partial x_{i}}{\partial x_{j}}+x_{i}\frac{\partial}{\partial x_{j}}\left(\frac{1}{r}\right)\right)\\&
=i\left(\frac{\delta _{ij}}{r}+\frac{r_{i}r_{j}}{r^{3}}\right)\end{align}}}$
