0
$\begingroup$

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.

$\endgroup$
0

2 Answers 2

3
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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
1
$\begingroup$

the Cartesian sphere components are:

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

thus:

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

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

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

and

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

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 $a_\vartheta$

$\endgroup$
1
  • $\begingroup$ 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

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.