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?

  • $\begingroup$ 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

2 Answers 2


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

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.