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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?

Pedro Lobo
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1 Answers1

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@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} $$\left [\frac{r_{i}}{r},\hat{P}_{j}\right]=i\frac{\partial }{\partial X_{j}}\left (\frac{r_{i}}{r} \right )$$

${\color{blue} {\begin{align}\left[\frac{r_{i}}{r},\hat{P}_{j}\right]&=i\frac{\partial}{\partial X_{j}}\left(\frac{r_{i}}{r} \right)\\& =i\left(\frac{1}{r}\frac{\partial r_{i}}{\partial x_{j}}+r_{i}\frac{\partial}{\partial x_{j}}\left(\frac{1}{r}\right)\right)\\& =i\left(\frac{1}{r}\frac{\partial x_{i}}{\partial x_{j}}+x_{i}\frac{\partial}{\partial x_{j}}\left(\frac{1}{r}\right)\right)\\& =i\left(\frac{\delta _{ij}}{r}+\frac{r_{i}r_{j}}{r^{3}}\right)\end{align}}}$

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Pedro Lobo
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