Confused about multipole expansion of vector potential

  • #1
kelly0303
561
33
Hello! I found an expression in this paper (eq. 1) for the multipole expansion of the vector potential. I am not sure I understand what form do the vector spherical harmonics (VSH) have. Also, for example, the usual hyperfine interaction operator is given by ##\frac{\mathbf{\mu}\cdot(\mathbf{r}\times \mathbf{\alpha})}{r^3}##. I am not sure how to get back to this expression using equation 1 (or 2), for k=1. On Wikipedia it seems like VSH are defined as ##Y_{lm}\hat{r}##, while in the reference they mention in the paper it would be ##\frac{1}{\sqrt{J(J+1)}}\mathbf{L}Y_{JM}##, where ##\mathbf{L}## is the orbital angular momentum operator. I tried using both and still didn't get back the original formula. Can someone help me with this? Thank you!
 
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  • #2
For a multipole decomposition of a general vector field you need three kinds of vector-spherical harmonics. Written in standard spherical coordinates they are
$$\vec{\Psi}_{lm}(\vartheta,\varphi)=r \vec{\nabla} \mathrm{Y}_{lm}(\vartheta,\varphi),$$
$$\vec{\Phi}_{lm}(\vartheta,\varphi)=\vec{r} \times \vec{\nabla} \mathrm{Y}_{lm}(\vartheta,\varphi),$$
and
$$\vec{\mathrm{Y}}_{lm}(\vartheta,\varphi)=\vec{e}_r \text{Y}_{lm}(\vartheta,\varphi).$$
These are all mutually orthogonal to each other under the scalar product on the unit sphere
$$\langle \vec{V}_1|\vec{V}_2 \rangle=\int_{0}^{\pi} \mathrm{d} \vartheta \int_0^{2 \pi} \mathrm{d} \varphi \sin \vartheta \vec{V}_1^*(\vartheta,\varphi) \cdot \vec{V}_2(\vartheta,\varphi)$$
and normlized according to
$$\langle \vec{\Psi}_{lm}|\vec{\Psi}_{l'm'} \rangle=l(l+1) \delta_{ll'} \delta_{mm'},$$
$$\langle \vec{\Phi}_{lm}|\vec{\Phi}_{l'm'} \rangle=l(l+1) \delta_{ll'} \delta_{mm'},$$
$$\langle \vec{\mathrm{Y}}_{lm}|\vec{\mathrm{Y}}_{l'm'} \rangle=\delta_{ll'} \delta_{mm'}.$$
 
  • #3
vanhees71 said:
For a multipole decomposition of a general vector field you need three kinds of vector-spherical harmonics. Written in standard spherical coordinates they are
$$\vec{\Psi}_{lm}(\vartheta,\varphi)=r \vec{\nabla} \mathrm{Y}_{lm}(\vartheta,\varphi),$$
$$\vec{\Phi}_{lm}(\vartheta,\varphi)=\vec{r} \times \vec{\nabla} \mathrm{Y}_{lm}(\vartheta,\varphi),$$
and
$$\vec{\mathrm{Y}}_{lm}(\vartheta,\varphi)=\vec{e}_r \text{Y}_{lm}(\vartheta,\varphi).$$
These are all mutually orthogonal to each other under the scalar product on the unit sphere
$$\langle \vec{V}_1|\vec{V}_2 \rangle=\int_{0}^{\pi} \mathrm{d} \vartheta \int_0^{2 \pi} \mathrm{d} \varphi \sin \vartheta \vec{V}_1^*(\vartheta,\varphi) \cdot \vec{V}_2(\vartheta,\varphi)$$
and normlized according to
$$\langle \vec{\Psi}_{lm}|\vec{\Psi}_{l'm'} \rangle=l(l+1) \delta_{ll'} \delta_{mm'},$$
$$\langle \vec{\Phi}_{lm}|\vec{\Phi}_{l'm'} \rangle=l(l+1) \delta_{ll'} \delta_{mm'},$$
$$\langle \vec{\mathrm{Y}}_{lm}|\vec{\mathrm{Y}}_{l'm'} \rangle=\delta_{ll'} \delta_{mm'}.$$
Thank you for the reply. But I am not sure I understand how to use this for the given expression. Is ##C_{k,\mu}^{(0)}(\hat{r})## a linear combination of the 3 terms you mentioned above? Also, what is the ##(0)## standing for?
 
  • #4
I hate non-selfcontained papers :-(. Obviously they expect that you have the cited book at hand and look it up. Just laziness! I don't have the book at hand unfortunately.
 
  • #5
vanhees71 said:
I hate non-selfcontained papers :-(. Obviously they expect that you have the cited book at hand and look it up. Just laziness! I don't have the book at hand unfortunately.
Ah I see it's not even a universal definition... I found the book here (please let me know if you can't access it). The section is 1.5.2, I would appreciate any insight from you as I am still confused after reading it.
 

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