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I'm trying to follow the exposition of spin coherent state path integral presented in Condensed Matter Field Theory by Altland and Simons (section 3.3, Page 134-142), and I have a problem with the derivation.

Question: In the paragraph under Eq. (3.55), authors write \begin{equation} \mathbf{B}_{m} = \nabla \times \mathbf{A} = \hat{e}_{r}\,, \quad{} \text{with $\mathbf{A}=\frac{1-\cos{\theta}}{\sin{\theta}}\hat{e}_{\phi}$}\,. \end{equation} I have tried the calculation following the standard formula (wikipedia) and got a different result, \begin{equation} \mathbf{B}_{m} = \hat{e}_{r} + \frac{\cos{\theta}-1}{\sin{\theta}} \hat{e}_{\theta}\,, \end{equation} where $r=1$ has been set. Then, why the $\hat{e}_{\theta}$ component is abandoned in the textbook?

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  • $\begingroup$ Hi @Qamber. Welcome to Phys.SE. Tip: Answers should be in answer boxes; not in the question box. $\endgroup$
    – Qmechanic
    Jun 10, 2021 at 7:31

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You must have made a mistake. The $B$ field is purely radial. Did you remember that ${\rm curl} {\bf A}$ is not the same as $\nabla\times {\bf A}$ in any but cartesian coordinates?

To be specific for spherical polar coordinates $$ {\rm grad\,} \varphi={\bf e}_r \frac{\partial \varphi}{\partial r} +{\bf e}_\theta\frac 1 r \frac{\partial \varphi}{\partial\theta}+{\bf e}_\phi \frac{1} {r\sin\theta} \frac{\partial \varphi}{\partial \phi}\\ {\rm curl\,} {\bf A}= {\bf e}_r \frac{1}{r\sin\theta} \left (\frac{\partial \sin\theta A_\phi}{\partial \theta}-\frac{\partial A_\theta}{\partial \phi }\right) +{\bf e}_\theta\left(\frac{1}{r\sin\theta} \frac{\partial A_r}{\partial\phi }-\frac 1 r \frac{\partial\, rA_\phi }{\partial r}\right) +{\bf e}_\phi \frac 1 r \left(\frac{\partial \,r A_\theta}{\partial r}- \frac{\partial A_r}{\partial \theta}\right)\\ {\rm div\,} {\bf A} = \frac 1 {r^2} \frac{\partial \,r^2A_r}{\partial r}+ \frac 1 {r\sin\theta} \frac{\partial \sin\theta A_\theta}{\partial \theta}+\frac{1}{r\sin\theta} \frac{\partial A_\phi}{\partial z}\nonumber $$

For cylindrical coordinates: $$ {\rm grad\,} \varphi={\bf e}_r \frac{\partial \varphi}{\partial r} +{\bf e}_\theta\frac 1 r \frac{\partial \varphi}{\partial\theta}+{\bf e}_z \frac{\partial \varphi}{\partial z}\\ {\rm curl\,} {\bf A}= {\bf e}_r \left (\frac 1 r \frac{\partial A_z}{\partial \theta}-\frac{\partial A_\theta}{\partial z}\right) +{\bf e}_\theta\left(\frac{\partial A_r}{\partial z}-\frac{\partial A_z}{\partial r}\right) +{\bf e}_z \frac 1 r \left(\frac{\partial r A_\theta}{\partial r}- \frac{\partial A_r}{\partial \theta}\right)\\ {\rm div\,} {\bf A} = \frac 1 r \frac{\partial rA_r}{\partial r}+ \frac 1 r \frac{\partial A_\theta}{\partial \theta}+\frac{\partial A_z}{\partial z} $$

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  • $\begingroup$ Thanks for your kind answer. However, I still do not understand the sentence, "${\rm curl}\mathbf{A}$ is not the same as $\nabla\times\mathbf{A}$ in any but cartesian coordinates". I want to know in what sense that they(curl and del cross product) are different? Can you give more explanation or provide any reference for that? $\endgroup$
    – Qamber
    Jun 10, 2021 at 5:57
  • $\begingroup$ Just compare the equations for $\nabla\times {\bf A}$ in spherical coordinaes with the actual expressions given above and see that they are different! For details of their derivation you can look at page 298 in our book: goldbart.gatech.edu/PostScript/MS_PG_book/bookmaster.pdf $\endgroup$
    – mike stone
    Jun 10, 2021 at 15:48
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I find that there should be a factor $r$ in the denominator of $\mathbf{A}$, which is omitted for $r=1$ in this case. For Eq. (3.54) in the textbook by Altland and Simons, \begin{align} S_{\rm top}[\phi,\theta] & = iS \int_{0}^{\beta}\! {\rm d}\tau\, \dot{\phi} (1-\cos{\theta}) \\ & = iS \oint_{\gamma}\! {\rm d}\tau\, \dot{\mathbf{n}} \cdot \hat{e}_{\phi} \frac{1-\cos{\theta}}{\sin{\theta}} \\ & = iS \oint_{\gamma}\! \frac{1-\cos{\theta}}{\sin{\theta}} \hat{e}_{\phi} \cdot {\rm d}\mathbf{n} \\ & = iS \oint_{\gamma}\! \frac{1-\cos{\theta}}{r\sin{\theta}} \hat{e}_{\phi} \cdot {\rm d}\mathbf{l} \\ & = iS \int_{\rm surface} (\nabla \times \mathbf{A}) \cdot {\rm d}\mathbf{\sigma}\,, \end{align} where $\mathbf{A} \equiv (1-\cos{\theta})/(r\sin{\theta})$, which can reproduce the correct result $\mathbf{B}_{m} = \hat{e}_{r}/r^{2} = \hat{e}_{r}$, and the question is solved.

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