# The anticommutativity of $K=\gamma_4 (\vec\Sigma \cdot \vec L +\hbar)$ and $\frac{\vec\sigma\cdot \vec x}{r}$

I have been trying to follow the solution to the Dirac equation for the hydrogen atom.

There is a claim from an online source that the following two operators anticommute: $$K=\gamma_4 (\vec\Sigma \cdot \vec L +\hbar)$$ and $$\frac{\vec \sigma\cdot \vec x}{r}$$ where $\vec\Sigma = (\Sigma_x, \Sigma_y, \Sigma_z)$ is the vector of spin operators, $\vec L$ is the vector of angular momentum operators and $\gamma_4$ is the Dirac matrix of the 4th component with $x_4 = ict$.

The source points to Paul Strange's Relativistic Quantum Mechanics, which unfortunately I have no access to the original text.

While I am trying to verify this claim I am apparently stuck: Using the fact that $\frac{\vec\sigma\cdot\vec x}{r}$ commutes with $\vec J$, $J^2$ and $S^2=\frac{3\hbar^2}{4}$, I am able to show that

$$\left[ \frac{\vec\sigma \cdot \vec x}{r} , \, K \right] = -\frac{1}{\hbar r} \gamma_4 \vec\sigma \cdot \left[ \vec x , \, L^2 \right] + \frac{1}{r} ([\vec\sigma,\gamma_4]\cdot x) (\vec\Sigma \cdot \vec L + \hbar)$$

Then I can show that $[\vec\sigma,\gamma_4]=2\vec\sigma \gamma_4$, so the second term gives $2(\frac{\vec\sigma\cdot \vec x}{r})K$. But I am not able to show that the first term gives $0$:

$$[x_n, L^2] = L_k[x_n, \, L_k] + [x_n, \, L_k]L_k = i\hbar \epsilon_{kni} (L_kx_i+x_iL_k)$$

It is from here I do not know how to proceed.

So as nobody viewing this question even had a try, I guess I shall answer my own question. The trick lies in observing the result for a specific $n$, e.g. $n=3$, and then rotating the indices:
$$[x_3, L^2] = i\hbar\epsilon_{k3i} (L_k x_i + x_i L_k) = i\hbar (L_2x_1 - L_1 x_2 + x_1L_2 - x_2 L_1 ) = -[x_1,L_2] + [x_2,L_1]$$
Now use the equation $[x_n, L_k] = \epsilon_{kni}x_i$ to give $$[x_3, L^2] = i\hbar (x_3-x_3)=0$$
This is a scalar, so there is no indices to rotate, hence $[x_n,L^2]=0$ for each $n$.
Supposedly this can be done in pure tensor notation and not for a specific $n$, but I do not know how.