Magnetic Moment MIT Bag Model

Question

I'm reading the book "Advances in Nuclear Physics vol 13" by J. W. Negele and Erich Vogt

In chapter 3, one wants to calculate the magnetic moment for a current loop. In page 29 how does one go from equation 3.8 to 3.9:

equation 38: $$\mu=\frac{N^2}{2}\sum_iQ_i\int_{bag}d\textbf{r} \ \textbf{r}^2[j_0(\omega r/R),-i\sigma_i\hat{r}j_1(\omega r/R)] \begin{pmatrix} 0 & \textbf{r}\times\sigma_i \\ \textbf{r}\times\sigma_i & 0 \end{pmatrix} \begin{pmatrix} j_0(\omega r/R) \\ i\sigma_i\hat{r}j_1(\omega r/R) \end{pmatrix} $$

equation 39:

$$\mu=\mu_0\sum_i \sigma_iQ_i$$

Manipulating expression 38 I arrive at:

$$\sum_iQ_i\frac{N^2}{2}\int_0^R d\textbf{r} \ \textbf{r}^2\left(ij_0\left(\frac{\omega r}{R}\right)j_1\left(\frac{\omega r}{R}\right)[\textbf{r}\times\sigma_i,\sigma_i\hat{r}]\right)$$

Where $[\textbf{r}\times\sigma_i,\sigma_i\hat{r}])$ is the commutator.

How should I proceed? I know I should manage to isolate $\sigma_i$, but I don't know how to go further.

Answer 1

I ended up expanding the commutator, using the levi-civita definition of the cross product. Arriving at the result:

$ [\textbf{r}\times \sigma_i, \sigma_i\hat{r}]=2i(\textbf{r}\sigma_i\hat{r}-\textbf{r}\sigma_i)$

From there I computed the integral and arrived at 3.9