# Parity of the vector spherical harmonics?

The vector spherical harmonics can be defined as $$\textbf{Y}_{j, \ell, 1}^m(\theta, \phi) = \sum_{m_{\ell}=-\ell}^{+\ell} \sum_{m_s=-1}^{+1}C_{\ell, m_{\ell}, 1, m_s}^{j, m_j} Y_{\ell, m_{\ell}}(\theta, \phi)\hat{\textbf{e}}_{m_s},$$ where $$Y_{\ell, m}(\theta, \phi)$$ is a usual spherical harmonic, $$\hat{\textbf{e}}_{m_s}$$ is a spherical coordinate vector, and $$C_{\ell, m_{\ell}, 1, m_s}^{j, m}$$ is a Clebsch-Gordan coefficient.

The question is what is the parity of $$Y_{j, \ell, 1}^{m}(\theta, \phi)$$? Under inversion the regular spherical harmonics collect a sign of $$(-1)^{\ell}$$. My understanding is that, since the $$\hat{\textbf{e}}_{m_s}$$ are vectors, they will collect a sign of $$-1$$. I would then expect the total parity of the vector spherical harmonics to be given by $$(-1)^{\ell + 1}$$. But the references I've looked at have quoted the spherical harmonics as having a parity of $$(-1)^{\ell}$$ just like the regular spherical harmonics. Am I missing something?

One reference is Blatt and Weisskopf, Theoretical Nuclear Physics, Appendix B.

• I think what's going on is that the unit vectors do no transform under parity the way normal vectors do. Under parity, a true vector like ${\bf r}$ goes as ${\bf r}\rightarrow-{\bf r}$; however, a coordinate unit vector will instead transform as $\hat{\bf e}_{r}\rightarrow\hat{\bf e}_{r}$, because it has to point along the radial coordinate direction in the new coordinates.
– Buzz
Commented Apr 14, 2022 at 23:11
• @Buzz I guess $\hat{e}_{\theta}\rightarrow \hat{e}_{\theta}$ also holds in this case? I don't see what makes coordinate unit vectors not transform... maybe it's a active vs. passive thing? Like in this case we're thinking of parity as an active transformation so vectors themselves transform but coordinate vectors do not transform? Commented May 15, 2022 at 21:10
• Yes, exactly. We are talking about active transformations, because the real transformations that are related to things like parity conservation or violation have to be active transformations.
– Buzz
Commented May 15, 2022 at 21:28
• A nice comparison of various associated harmonics is given by Thorne in his 1980 Rev. Modern Phys.. His $\mathbf{Y}^{l',lm}$ [Eq. (2.16)] looks essentially the same as your expression, and he does indeed list its parity as $(-1)^{l'+1}$. Of course, that's just one of several forms of vector spherical harmonics found in the literature that he reviewed, and some have different parities.
– Mike
Commented May 20, 2022 at 3:03

There is not complete uniformity in the definition of the vector spherical harmonics, so it is possible that different definitions may actually refer to expressions with different parities. However, if they are defined they way they are in Jackson's Classical Electrodynamics, $${\bf X}_{l,m}(\theta,\phi)=\frac{1}{\sqrt{l(l+1)}}{\bf L}Y_{l,m}(\theta,\phi),$$ where $${\bf L}$$ is the operator $$\frac{1}{i}({\bf x}\times{\nabla})$$ (which would be the angular momentum divided by $$\hbar$$ in quantum mechanics), then the parity of $${\bf X}_{l,m}$$ is indeed equal to $$(-1)^{l}$$. This can be discerned fairly straightforwardly from the known parities of $${\bf L}$$ and $$Y_{l,m}$$. Angular momentum, as an axial vector (or an element of a second-rank tensor) is even under parity, which leaves $${\bf X}_{l,m}$$ with the same parity as $$Y_{l,m}$$.
Whenever one talks about the transformation properties of a quantity, the real physical content is associated with how the properties under active transformations. So if you want to find the behaviors of the electrostatic under parity/space inversion, you would consider a situation in which all the charges have their positions relocated—so there is a new charge distribution $$\rho'(\mathbf{x})=\rho(-\mathbf{x})$$. (Passive transformations, in contrast, are just relabelings of coordinates.) The electric field of $$\rho'$$ is clearly $$\mathbf{E}'(\mathbf{x})=-\mathbf{E}(-\mathbf{x})$$, exhibiting the odd parity. On the other hand, a magnetostatic field transforms as $$\mathbf{B}'(\mathbf{x})=\mathbf{B}(-\mathbf{x})$$, because the source $$\mathbf{J}'$$ has both its location and direction inverted, $$\mathbf{J}'(\mathbf{x})=-\mathbf{J}(-\mathbf{x})$$; that is, the current density is polar vector, like the electric field, whereas the magnetic field is an axial vector.
So when looking at the transformation of a vector like $$\mathbf{E}$$ or $$\mathbf{X}_{l,m}$$, when you decompose it into components, the components of $$\mathbf{E}'$$ are opposite those of $$\mathbf{E}$$: $$E'_{x}(\mathbf{x})=-E_{x}(-\mathbf{x})$$, $$E_{y}'(\mathbf{x})=-E_{y}(-\mathbf{x})$$, and $$E_{z}'(\mathbf{x})=-E_{z}(-\mathbf{x})$$, because $$\mathbf{E}'$$ the field of the inverted charge distribution $$\rho'$$ in uninverted coordinates. Since the coordinates are not inverted, the basis vectors are unaffected, and this gives rise to the correct overall transformation $$\mathbf{E}'(\mathbf{x})=E_{x}'(\mathbf{x})\hat{\mathbf{e}}_{x}+E_{y}'(\mathbf{x})\hat{\mathbf{e}}_{y}+E_{z}'(\mathbf{x})\hat{\mathbf{e}}_{z} \\ =-E_{x}(-\mathbf{x})\hat{\mathbf{e}}_{x}-E_{y}(-\mathbf{x})\hat{\mathbf{e}}_{y}-E_{z}(-\mathbf{x})\hat{\mathbf{e}}_{z}=-\mathbf{E}(-\mathbf{x}).$$ The same thing applies to the unit vectors in the angular directions, $$\hat{\mathbf{e}}_{r}$$, $$\hat{\mathbf{e}}_{\theta}$$, and $$\hat{\mathbf{e}}_{\phi}$$. They are unchanged under a parity transformations, but the three corresponding components of the vector spherical harmonic $$\mathbf{X}_{l,m}(\theta,\phi)$$ do change their signs.
• The Jackson expression for $\textbf{X}_{\ell,m}(\theta, \phi)$ is equivalent to my $\textbf{Y}_{\ell, \ell, 1}^m(\theta, \phi)$. In fact Jackson cites Blatt and Weisskopf which is how I came to the latter reference. This is exactly why I deeply dislike how parity is defined. The two expressions are equal to each other, but the calculation of parity comes out different depending on which definition you use. Commented Apr 11, 2022 at 6:05