TITLE: Why is $(\vec \sigma \cdot \vec B)\psi$ in Schrodinger-Pauli equation rotationally invariant?
QUESTION [5 upvotes]: In the Schrodinger-Pauli equation:
$$ i\partial _{t} \psi = \left[(\frac{1}{2m}(i\vec \nabla - e \vec A)^{2} - e A_{0}) 1_{2\times2} + \mu_{B} \vec B \cdot \vec \sigma\right]\psi $$
Why is $(\vec \sigma \cdot \vec B)\psi$ rotationally invariant?
This is from page 157 of Schwartz's QFT book. He write that it is because $[\sigma_{i}, \sigma_{j}] = 2i\epsilon_{ijk}\sigma_{k}$. But I still don't understand.

REPLY [2 votes]: I cannot go into too much detail right now, but this is the gist of it... sorry. Try to take a look of how rotations work for vectors and spinors so that things become clearer.
Under rotations, $\vec{B}$ transforms as a vector and $\psi$ as a spinor. This means that under rotations the term you mention (up to the factor of $\mu_B$) on the right hand side does something like
\begin{equation}
B_a (\vec{x},t) \, (\sigma^a)^{\beta}_{\ \ \alpha} \,  \psi^{\alpha}(\vec{x},t) \rightarrow  \left[ (R_{\mathrm{Vector}})^{b}_{\ \ a} \, B_b (\vec{x}',t) \right] \, (\sigma^a)^{\beta}_{\ \ \alpha} \, \left[ (R_{\mathrm{Spinor}})^{\alpha}_{\ \ \gamma} \, \psi^{\gamma}(\vec{x}',t)\right] \, .
\end{equation}
But, as implied in the book, the Pauli matrices play an important role in the rotation of 2D spinors. They have one vector index and two spinorial indices and satisfy 
\begin{equation}
(R_{\mathrm{Vector}})^{b}_{\ \ a} \, (\sigma^a)^{\beta}_{\ \ \alpha} \,  (R_{\mathrm{Spinor}})^{\alpha}_{\ \ \gamma} \, = \, (R_{\mathrm{Spinor}})^{\beta}_{\ \ \delta} \, (\sigma^b)^{\delta}_{\ \ \gamma} \, .
\end{equation}
 This implies that the transformation of our term becomes
\begin{equation}
B_a (\vec{x},t) \, (\sigma^a)^{\beta}_{\ \ \alpha} \,  \psi^{\alpha}(\vec{x},t) \rightarrow (R_{\mathrm{Spinor}})^{\beta}_{\ \ \delta}  \,  B_b (\vec{x}',t) \, (\sigma^b)^{\delta}_{\ \ \gamma} \, \psi^{\gamma}(\vec{x}',t) \, .
\end{equation}
On the left hand side of the equation we have instead $ i \, \partial_t\psi^\beta (\vec{x},t)$ which indeed transforms as
\begin{equation}
i \, \partial_t \psi^\beta (\vec{x},t) \rightarrow (R_{\mathrm{Spinor}})^{\beta}_{\ \ \delta} \, i \, \partial_t \psi^\delta (\vec{x}',t) \, ,
\end{equation}
hence preserving the form of the equations of motion (you can check this for the remaining term as well).
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