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On page 157 of Schwartz's QFT book, He write that “$\sigma_i$ do not change under rotations”.

If so, changes in $\psi$ and $B$ cancels, so we can get that $(\vec{\sigma} \cdot \vec B)\psi$ is rotationally invariant.

But why Pauli matrices are the same in any frame? Any hint or reference would be helpful!

Qmechanic
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RicknJerry
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3 Answers3

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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.

ACuriousMind
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  • I don't think this answers the question. You can view $\vec{\sigma}$ as being an object with 12 components and 3 indices (1 vector index and 2 spinor indices). When you transform this object under a passive rotation - contracting the vector index with an ordinary rotation matrix and the spinor indices with corresponding spinor rotation matrices - the result is that all 12 components just stay the same. How can this be explained? – Brian Bi Feb 25 '20 at 00:00
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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.

Urb
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Nikita
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    "Pauli matrices is sets of numbers, they don't transform under rotations" -- I'd say this depends on the context. – Norbert Schuch Feb 24 '20 at 12:42
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    Could you present example, when Pauli matrices transforms?? It is only coefficients! – Nikita Feb 24 '20 at 12:45
  • A density matrix of one qubit can be written as $\rho=\tfrac12(I+\vec r\cdot\vec\sigma)$, with the Bloch vector $|\vec r|\le1$. If I describe my state in a different basis (or rotate it) via a rotation $R$, it transforms to $R\rho R^\dagger$. This means that the Paulis transform as $R\sigma_iR^\dagger$ (since everything else is just numbers). This implies a transformation of the vector $\vec\sigma$ of Pauli matrices, which can then be moved to the Bloch vector $\vec r$. – Norbert Schuch Feb 24 '20 at 12:54
  • (Note that in that case, the Paulis are not "just coefficients", but rather a basis for the space of hermitian 2x2 matrices - which transform.) – Norbert Schuch Feb 24 '20 at 12:55
  • I disagree with you. Transforms $\vec{r}$, and sigma matrices retranslate this transformation to spinor indeces. – Nikita Feb 24 '20 at 13:09
  • Well, this is not how people in e.g. quantum information think about that. Rotating a qubit means applying $R=\exp[-i\vec\phi\cdot\vec\sigma]$ -- just plain matrix-matrix multiplication. Of course, this translates to a rotation of the vector, so you might argue that in fact the vector transforms, but you can equivalently think that the Paulis transform (which is evident in the matrix-matix multiplication). --- Put differently, the only thing which transforms in a well-defined way is $\vec r\cdot\vec\sigma$ -- how you assign this to $\vec r$ or $\vec\sigma$ is not uniquely defined. – Norbert Schuch Feb 24 '20 at 13:15
  • $\sigma_i = S^{-1}_{ij} R\sigma_j R^\dagger $, where R is spinor rotations, S vector rotation with the same angle parameters. I think that you mean this. But in my point of view it is meaningless to give some transformation law to "set of numbers". But my be in some situation it is practically useful.. – Nikita Feb 24 '20 at 13:26
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    @Nikita The Pauli matrices transform as vectors (i.e. tensor operators with angular momentum $1$) under rotations. I’m not sure what you’re trying to say but it’s not clear as you have written it. Under rotation $\sigma_i\to R \sigma_i R^{-1}\ne \sigma_i$ in general. – ZeroTheHero Feb 24 '20 at 15:44
  • Ok, let's imagine that you are true. Sigma matrices have one vector index and two spinor index. Why you ignore transformations of vector index? – Nikita Feb 24 '20 at 18:27
  • @Nikita If you think of the Paulis jointly as a three-index tensor, then the whole object transforms trivially. This does not mean it doesn't transform -- just as a singlet state of two spin-$\tfrac12$ transforms trivially: This doesn't mean it is "just a set of numbers", but it is an object in the trivial irreducible representation of the symmetry action. – Norbert Schuch Feb 24 '20 at 18:30
  • But if object doesn't transform, it is obviously set of numbers:)? – Nikita Feb 24 '20 at 18:32
  • @Nikita No. It transforms like an irreducible representation. It happens to be the trivial one. "A set of numbers" is not equipped with a group action. An object which transforms trivially *is* equipped with such an action. (Putting it differently, if I have a representation $U(g) = \bigoplus_\alpha D_\alpha(g)$ with irreps $D_\alpha$, the whole space transforms in some way -- in which way it does depend on the irrep. – Norbert Schuch Feb 24 '20 at 20:57
  • @Nikita I'm not sure why people are criticizing your answer. You are absolutely right: the Pauli matrices are just numbers. They do not "transform" in any reasonable sense of the word. So don't worry about the downvote or the comments. – AccidentalFourierTransform Feb 25 '20 at 00:06
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Pauli matrices are, as @Nikita pointed out, set of numbers. They are a $2\times2$ representation of the $SU(2)$ group. They obey: $$[\sigma_a,\sigma_b]=2i\epsilon_{abc}\sigma_c$$ $$\{\sigma_a,\sigma_b\}=2\delta_{a,b}I$$ Since even in the rotated frame they have to obey the same conditions, their form will be the same. It is analogous to choosing a reference frame where we define unit vectors $\hat x, \hat y, \hat z$ and do an active rotation where the reference frame is fixed but the state rotates. So the $\sigma$ matrixes are like these basis vectors. They are defined by maintaining their relation between each other.

Superfast Jellyfish
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    Hold on a minute here. Yes under conjugation $\sigma_i\to T\sigma_i T^{-1}$ preserves commutation but that doesn’t mean $T\sigma_i T^{-1}=\sigma_i$. You’re right that the set ${T\sigma_i T^{-1}}$ is equivalent to the Pauli matrices but that doesn’t mean they are the same. – ZeroTheHero Feb 24 '20 at 15:46