Evaluation of the commutator $[\frac{\partial}{\partial x_j},\frac{x_ix_j-r^2\delta_{ij}}{r^3}]=-2x_i/r^3$.

Question

I am trying to evaluate a commutator of the form $[\frac{\partial}{\partial x_j},\frac{x_ix_j-r^2\delta_{ij}}{r^3}]=-2x_i/r^3$. Bt acting it on a function $f(\vec{r})$, I get that the commuator is equal to zero. Can you help?

Answer 1

Calculate first this version

\begin{eqnarray} \require{cancel} \left[\frac{\partial}{\partial x_k}, \frac{x_i x_j - r^2\delta_{ij}}{r^3}\right]f &=& \frac{\partial}{\partial x_k}\left(\frac{x_i x_j - r^2\delta_{ij}}{r^3}f \right) - \frac{x_i x_j - r^2\delta_{ij}}{r^3}\frac{\partial f}{\partial x_k} \\ &=& \left(\frac{\partial}{\partial x_k}\frac{x_i x_j - r^2\delta_{ij}}{r^3}\right)f + \cancel{\frac{x_i x_j - r^2\delta_{ij}}{r^3}\frac{\partial f}{\partial x_k}} - \cancel{\frac{x_i x_j - r^2\delta_{ij}}{r^3}\frac{\partial f}{\partial x_k}} \end{eqnarray}

So the problem reduces to

\begin{eqnarray} \left[\frac{\partial}{\partial x_k}, \frac{x_i x_j - r^2\delta_{ij}}{r^3}\right] &=&\frac{\partial}{\partial x_k}\frac{x_i x_j - r^2\delta_{ij}}{r^3} \\ &=& \frac{\partial}{\partial x_k} \frac{x_i x_j}{r^3} - \delta_{ij}\frac{\partial}{\partial x_k}\frac{1}{r} \\ &=& \left(\frac{\delta_{ik}x_j}{r^3} + \frac{x_i\delta_{jk}}{r^3}- 3\frac{x_ix_j x_k}{r^5} \right) + \delta_{ij}\frac{x_k}{r^3} \\ &=& \frac{1}{r^3}\left(\delta_{ik} x_j + x_i\delta_{jk} + \delta_{ij} x_k\right) - 3 \frac{x_i x_j x_k}{r^5} \tag{1} \end{eqnarray}

Now replace $k\to j$ and sum over $j$

\begin{eqnarray} \left[\frac{\partial}{\partial x_j}, \frac{x_i x_j - r^2\delta_{ij}}{r^3}\right] &=& \frac{1}{r^3} \left(\delta_{ij} x_j + x_i\delta_{jj} + \delta_{ij} x_j\right) - 3 \frac{x_i x_j x_j}{r^5} \\ &=& \frac{1}{r^3} \left(2x_i + 3x_i\right) - 3 \frac{x_i r^2}{r^5} \\ &=& 2\frac{x_i}{r^3} \end{eqnarray}

Answer 2

Consider any mathematical Expression $A$ and apply the commutator on it. Then

$[\frac{\partial}{\partial x_j},f]A = \frac{\partial}{\partial x_j}(fA)-f \frac{\partial}{\partial x_j}A$

Note that in this calculation, the first term on the right Hand side is a composition of two operations:

First multiplication of $f$ with A because the $f$ is directly left next to the $A$

then applying the differential Operator,since it is left next to all other expressions.

Finally use product rule and lookwhich factor is in front of $A$. So the result of the commutator $R = [\frac{\partial}{\partial x_j},f]$ is

$[\frac{\partial}{\partial x_j},f]A = RA$