# Why is $(\vec \sigma \cdot \vec B)\psi$ in Schrodinger-Pauli equation rotationally invariant?

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.

• The term is not invariant, but it is covariant; i.e., it transforms in the same way as the other terms. This makes the whole equation invariant. Commented Dec 18, 2018 at 12:40

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 $$$$B_a (\vec{x},t) \, (\sigma^a)^{\beta}_{\ \ \alpha} \, \psi^{\alpha}(\vec{x},t) \rightarrow \left[ (R_{\mathrm{Vector}})^{b}_{\ \ a} \, B_b (\vec{x}',t) \right] \, (\sigma^a)^{\beta}_{\ \ \alpha} \, \left[ (R_{\mathrm{Spinor}})^{\alpha}_{\ \ \gamma} \, \psi^{\gamma}(\vec{x}',t)\right] \, .$$$$

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 $$$$(R_{\mathrm{Vector}})^{b}_{\ \ a} \, (\sigma^a)^{\beta}_{\ \ \alpha} \, (R_{\mathrm{Spinor}})^{\alpha}_{\ \ \gamma} \, = \, (R_{\mathrm{Spinor}})^{\beta}_{\ \ \delta} \, (\sigma^b)^{\delta}_{\ \ \gamma} \, .$$$$ This implies that the transformation of our term becomes

$$$$B_a (\vec{x},t) \, (\sigma^a)^{\beta}_{\ \ \alpha} \, \psi^{\alpha}(\vec{x},t) \rightarrow (R_{\mathrm{Spinor}})^{\beta}_{\ \ \delta} \, B_b (\vec{x}',t) \, (\sigma^b)^{\delta}_{\ \ \gamma} \, \psi^{\gamma}(\vec{x}',t) \, .$$$$

On the left hand side of the equation we have instead $$i \, \partial_t\psi^\beta (\vec{x},t)$$ which indeed transforms as $$$$i \, \partial_t \psi^\beta (\vec{x},t) \rightarrow (R_{\mathrm{Spinor}})^{\beta}_{\ \ \delta} \, i \, \partial_t \psi^\delta (\vec{x}',t) \, ,$$$$ hence preserving the form of the equations of motion (you can check this for the remaining term as well).

• I thought the pauli vector $\sigma_a$ should also transform as $R^{-1}\sigma_a R$. Why do we leave it alone in your first equation? Thank you Commented Oct 11, 2019 at 13:46
• I think in some conventions it is indeed convenient to see the transformation as acting over all objects from the beginning, including the Pauli matrices and other coordinate- and field-independent quantities with indices. My personal preference is to stick to this version that I have seen in the references I am familiar with. Commented Oct 11, 2019 at 14:35
• I see. Could you please share what reference you were using for this? Thanks again! Commented Oct 13, 2019 at 22:22
• What I did here is analogue to the discussion of Lorentz invariance of the Dirac equation in the Peskin and Schroeder QFT book, section 3.2, in the paragraphs following equation 3.28. Commented Oct 14, 2019 at 7:21