I am looking for the Pauli vector in spherical coordinate basis, like so:

$$\vec{\sigma} = \sigma_r \vec e_r + \sigma_{\theta} \vec e_{\theta} + \sigma_{\phi} \vec e_{\phi}$$

instead of

$$\vec{\sigma} = \sigma_x \vec{x} + \sigma_{y} \vec{y} + \sigma_{z} \vec{z}$$

in cartesian coordinates

At the end of a day I want calculate $ (\vec{\sigma} \vec{\nabla})\frac{K_1(r)}{r}$ where $K_n(r)$ is the modified Bessel function of second kind.

Thus, I want to use the $\nabla$-operator in spherical coordinates. Therefore I also need the Pauli vector in these coordinates.


2 Answers 2


You just need to use the following set of simple relations which connect unit vectors in spherical and cartesian coordinate systems.

$$\hat e_r = \frac{x\hat x + y\hat y + z\hat z}{r} = \hat x \sin \theta \cos \phi + \hat y \sin \theta \sin \phi + \hat z \cos \theta$$ $$\hat e_\phi = \frac{\hat z \times \hat e_r}{\sin \theta} = -\hat x \sin \phi + \hat y \cos \phi$$ $$\hat e_\theta = \hat e_\phi \times \hat e_r = \hat x \cos \theta \cos \phi + \hat y \cos \theta \sin \phi - \hat z \sin \theta$$

Then you need to use the fact that $$\boxed{\sigma_i = \hat i \cdot \vec \sigma}$$ So, $\sigma_r = \hat e_r \cdot \vec \sigma$ and so on.

Hope this helps in calculating the Pauli vector in spherical coordinates.

  • $\begingroup$ Thank you, I also thought about doing like this - but then I was wondering about $\theta$ and $\phi$ - the angles between $\vec{\sigma}$ and $\vec{\nabla}$? Which value should I take for them? $\endgroup$
    – nuemlouno
    Commented May 17, 2020 at 11:40
  • $\begingroup$ @nuemlouno $\theta$ and $\phi$ are the position angles: the same quantities that $K_n$ depends on. $\endgroup$
    – Javier
    Commented May 17, 2020 at 18:57
  • $\begingroup$ @Javier and if $K_n$ is independet on any angles? $\endgroup$
    – nuemlouno
    Commented May 17, 2020 at 19:12
  • $\begingroup$ @nuemlouno Then the corresponding derivatives will be zero. $\endgroup$
    – Javier
    Commented May 17, 2020 at 19:35

the Cartesian sphere components are:

$$\vec{R}=\begin{bmatrix} x \\ y \\ z \\ \end{bmatrix}= r\,\left[ \begin {array}{c} \sin \left( \vartheta \right) \cos \left( \varphi \right) \\ \sin \left( \vartheta \right) \sin \left( \varphi \right) \\ \cos \left( \vartheta \right) \end {array} \right] $$


$$\vec{e}_r=\frac{1}{||\frac{\partial \vec{R}}{\partial r}||}\,\frac{\partial \vec{R}}{\partial r}= \left[ \begin {array}{c} \sin \left( \vartheta \right) \cos \left( \varphi \right) \\ \sin \left( \vartheta \right) \sin \left( \varphi \right) \\ \cos \left( \vartheta \right) \end {array} \right] $$

$$\vec{e}_\varphi=\frac{1}{||\frac{\partial \vec{R}}{\partial \varphi}||}\,\frac{\partial \vec{R}}{\partial \varphi}= \left[ \begin {array}{c} -\sin \left( \varphi \right) \\ \cos \left( \varphi \right) \\ 0\end {array} \right] $$

$$\vec{e}_\vartheta= \frac{1}{||\frac{\partial \vec{R}}{\partial \vartheta}||}\,\frac{\partial \vec{R}}{\partial \vartheta}=\left[ \begin {array}{c} \cos \left( \vartheta \right) \cos \left( \varphi \right) \\ \cos \left( \vartheta \right) \sin \left( \varphi \right) \\ -\sin \left( \vartheta \right) \end {array} \right] $$


$$\vec{\sigma}=a_\varphi\,\vec{e}_\varphi+ a_\theta\,\vec{e}_\theta+a_r\,\vec{e}_r$$

where $$a_r=\vec{e}_r\cdot \vec{\sigma}=\sigma_{{x}}\sin \left( \vartheta \right) \cos \left( \varphi \right) +\sigma_{{y}}\sin \left( \vartheta \right) \sin \left( \varphi \right) +\sigma_{{z}}\cos \left( \vartheta \right) $$

analog $a_\varphi$ and $a_\vartheta$

  • $\begingroup$ Thus applying $\vec{\sigma} \vec{\nabla}$ on a function which depends only on $r$ (and not $\theta,\phi$) will lead to following expression: $\vec{\sigma} \vec{\nabla} f(r) = \sigma_x \frac{1}{r^2} \frac{\partial( r^2f(r) )}{\partial r} $ $\endgroup$
    – nuemlouno
    Commented May 17, 2020 at 16:01

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