# Why Pauli matrices are the same in any frame?

Pauli matrices is sets of numbers, they don't transform under rotations in contrast of vector $$\vec{B}$$ or field $$\psi$$!

See for details An introduction to spinors around (31).

Another useful reference Spin, topology, SU(2)$$\to$$ SO(3). See around (7).

Main idea: Using therms like $$(\sigma^i B^i)$$ one can convert rotation of vector to rotations of spinor indeces:

$$(\sigma^i (B^i)^\prime) = (\sigma^i e^{i\alpha J}B^i) = e^{i\alpha\sigma/2 }(\sigma^i B^i) e^{-i\alpha\sigma/2 }$$

And due to transformation of $$\psi$$:

$$\psi^{\prime } = e^{i\alpha\sigma/2 } \psi$$

One have:

$$(\sigma^i (B^i)^\prime) \psi^\prime = e^{i\alpha\sigma/2 }(\sigma^i B^i) \psi$$

So this therm transforms like $$\psi$$. So Schrödinger–Pauli equation is rotationally invariant. Author is not quite clear make statement about it.

The Pauli matrices are a two-dimensional representation of the generators of the rotation algebra $$\mathfrak{so}(3)$$. The generators of the rotation algebra can, in principle, be acted upon by rotations via conjugation since the Lie algebra forms the adjoint representation. It is hence confusing to claim that "the Pauli matrices do not transform under rotations" - they certainly can transform under rotations and are not invariant in general. It is merely the case that there are "rotations" that simply don't act on them:

In your particular context, the expression $$(\vec \sigma \cdot \vec B$$) is supposed to be the infinitesimal version of the rotation around the axis $$\vec B$$ (the finite rotation by an angle $$\phi$$ would be $$\mathrm{e}^{\mathrm{i}\phi\vec \sigma\cdot \vec B}$$), and the rotation the $$\sigma_i$$ are supposed to be invariant under here is an active rotation, i.e. one that rotates the objects (like $$\vec B$$ and $$\psi$$) and not the axes. The $$\sigma_i$$ are the infinitesimal rotations around the $$i$$-th axis, and they do not change under such an active rotation. They do change under passive rotations - the $$\sigma_z$$ in one frame is not the same as the $$\sigma_z$$ in another frame (unless, of course, the frames' z-axes align).

Another way to think about this - without using the sometimes confusing notions of "active" and "passive" transformations - is to consider that we usually define a physical transformation (that might be a symmetry) e.g. in the context of Noether's theorem as something that acts on the dynamical content of our theory, i.e. the fields/generalized coordinates. The $$\sigma_i$$ are not dynamical fields, so they do not transform under such a transformation. So in a context such as the one you quote where we care for rotational invariance e.g. in the sense of Noether's theorem, they do not transform. In other contexts, we might want to care about their transformation in the adjoint representation.