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.

  • 2
    $\begingroup$ 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. $\endgroup$
    – Javier
    Commented Dec 18, 2018 at 12:40

1 Answer 1


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).

  • $\begingroup$ 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 $\endgroup$ Commented Oct 11, 2019 at 13:46
  • $\begingroup$ 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. $\endgroup$
    – secavara
    Commented Oct 11, 2019 at 14:35
  • $\begingroup$ I see. Could you please share what reference you were using for this? Thanks again! $\endgroup$ Commented Oct 13, 2019 at 22:22
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    $\begingroup$ 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. $\endgroup$
    – secavara
    Commented Oct 14, 2019 at 7:21

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