Symmetries of a spherical harmonic basis

Question

I am interested in constructing a set of functions $Z_l^m$ which are linear combinations of spherical harmonics up to a maximum degree $L$, $$ Z_k^n(\theta,\phi) = \sum_{l=0}^L \sum_{m=-l}^l \alpha_{kl}^{nm} Y_l^m(\theta,\phi) $$ or in matrix-vector notation, $$ \mathbf{Z}(\theta,\phi) = A \mathbf{Y}(\theta,\phi) $$ where $\mathbf{Z},\mathbf{Y}$ are vectors of length $N = (L+1)^2$, and $A_{(kn),(lm)} = \alpha_{kl}^{nm}$. Now, the spherical harmonics have the following nice properties:

1. $\left< Y_k^n, Y_l^m \right> = \int d\Omega Y_k^n Y_l^{m*} = \delta_{kl} \delta_{nm}$
2. $Y_l^{-m} = (-1)^m Y_l^{m*}$

I would like the new functions $Z_l^m$ to satisfy these properties also. The first condition leads to $$ I = \int d\Omega \mathbf{Z} \mathbf{Z}^{\dagger} = A \left( \int d\Omega \mathbf{Y} \mathbf{Y}^{\dagger} \right) A^{\dagger} = A A^{\dagger} $$ which means $A$ must be a unitary matrix, $A \in U(N)$ in order for the $Z_l^m$ to be orthonormal.

The second condition is what I am having trouble with. The condition $Z_k^{-n} = (-1)^n Z_k^{n*}$ leads to the following condition on the $A$ matrix, $$ \alpha_{kl}^{-n,m} = (-1)^{n+m} \alpha_{kl}^{n,-m*} $$ My question is, does there exist some subset of $U(N)$ which satisfies this condition above? I would like to determine all possible unitary matrices $A$ which satisfy this symmetry condition above. Certainly $A = I$ satifies the condition, but I would like to find nontrivial solutions. How would one start to attack this problem?

Answer 1

This is not an answer to my question, but I have managed to show that all matrices $A \in U(N)$ which satisfy $$ \alpha_{kl}^{-n,m} = (-1)^{n+m} \alpha_{kl}^{n,-m*} $$ form a subgroup of $U(N)$. To prove closure, assume that $A,B$ are unitary matrices satisfying the above condition and let $C = AB$. Then, \begin{align} C_{(kn)(lm)} = A_{(kn)(uv)} B_{(uv)(lm)} \end{align} where I use the Einstein summation convention over $(uv)$. Then, \begin{align} C_{(k,-n)(lm)} &= A_{(k,-n)(uv)} B_{(uv)(lm)} \\ &= (-1)^{n+v} A_{(kn)(u,-v)}^{*} B_{(uv)(lm)} \\ &= (-1)^{n+v} A_{(kn)(u,-v)}^{*} (-1)^{v+m} B_{(u,-v)(l,-m)}^{*} \\ &= (-1)^{n+m} A_{(kn)(uv)}^{*} B_{(uv)(l,-m)}^{*} \\ &= (-1)^{n+m} C_{(kn)(l,-m)}^{*} \end{align} which proves that $C$ is in the subgroup. To show that the inverse is also in the subgroup, note that $A^{-1} = A^{\dagger}$. Then, \begin{align} (A^{\dagger})_{(kn)(lm)} &= A_{(lm)(kn)}^{*} \end{align} We then have, \begin{align} (A^{\dagger})_{(k,-n)(lm)} &= A_{(lm)(k,-n)}^{*} \\ &= (-1)^{m+n} A_{(l,-m)(kn)} \\ &= (-1)^{m+n} (A^{\dagger})_{(kn)(l,-m)}^{*} \end{align} which proves that $A^{-1} = A^{\dagger}$ belongs to the subgroup. So the search for matrices satisfying my condition can be restricted to subgroups of $U(N)$. I still don't know how to actually find this subgroup...

Answer 2

I wanted to be sure there exist nontrivial elements of the subgroup I seek. If we define $$ G_L := \left\{ A \in U((L+1)^2) : A_{(k,-n)(lm)} = (-1)^{n+m} A_{(kn)(l,-m)}^{*}\right\} $$ then it is easy to see that $G_0 = \left\{ \pm I \right\}$ so the $L=0$ case is trivial. However, $L=1$ is more interesting. Using the symmetry condition, we find that for $n=0, m=0$, $$ A_{(k0)(l0)} = A_{(k0)(l0)}^{*} $$ so these matrix entries must be real. The structure of a general element of $G_1$ will look like: $$ G_1 = \begin{pmatrix} a & \alpha & c & -\alpha^{*} \\ \gamma & \epsilon & \delta & \eta \\ d & \beta & b & -\beta^{*} \\ -\gamma^{*} & \eta^{*} & -\delta^{*} & \epsilon^{*} \end{pmatrix} $$ where $a,b,c,d \in \mathbb{R}$ and $\alpha,\beta,\gamma,\epsilon,\delta,\eta \in \mathbb{C}$. Here, I have ordered the matrix entries as $(0,0), (1,-1), (1,0), (1,1)$. There is also the additional constrant that the above matrix must be unitary which is a bit more complicated. However, if we consider diagonal matrices, we immediately see that the following matrix is in $G_1$, $$ \begin{pmatrix} \pm 1 & 0 & 0 & 0 \\ 0 & e^{i \theta} & 0 & 0 \\ 0 & 0 & \pm 1 & 0 \\ 0 & 0 & 0 & e^{-i \theta} \end{pmatrix} $$ for any $\theta \in \mathbb{R}$. This result shows that there are indeed nontrivial elements of $G_L$ for $L > 0$.

Answer 3

Any three-dimensional rotation of space should satisfy the constraints, because a rotation is unitary, it doesn't mix $l$ subspaces, and the result within each $l$ subspace is to make a new set of spherical harmonics corresponding to a different axis, and hence, by transforming $A=I$ under the transformation, you get a non-trivial $A$. To effect the transformation, we rely on the definition of the Wigner-$D$ matrices, which are unitary matrices satisfying $$ D^{(l)}_{m_1m_2}=(-1)^{m1-m2}(D^{(l)}_{-m_1,-m_2})^*, $$ where $l$, $m_1$, and $m_2$ are angular momentum quantum numbers. In particular, for $l=0$ and $l=1$, these matrices are $$ D^{(0)}=[1], $$ and $$ D^{(1)} = \begin{bmatrix} e^{-i \phi -i \psi } \cos ^2\left(\frac{\theta }{2}\right) & -\sqrt{2} e^{-i \psi } \cos \left(\frac{\theta }{2}\right) \sin \left(\frac{\theta }{2}\right) & e^{i \phi -i \psi } \sin ^2\left(\frac{\theta }{2}\right) \\ \sqrt{2} e^{-i \phi } \cos \left(\frac{\theta }{2}\right) \sin \left(\frac{\theta }{2}\right) & \cos (\theta ) & -\sqrt{2} e^{i \phi } \cos \left(\frac{\theta }{2}\right) \sin \left(\frac{\theta }{2}\right) \\ e^{i \psi -i \phi } \sin ^2\left(\frac{\theta }{2}\right) & \sqrt{2} e^{i \psi } \cos \left(\frac{\theta }{2}\right) \sin \left(\frac{\theta }{2}\right) & e^{i \phi +i \psi } \cos ^2\left(\frac{\theta }{2}\right) \\ \end{bmatrix}. $$ The quantities $\psi$, $\theta$, and $\phi$ are Euler angles, yielding a parameterization of all three-dimensional rotations. These $D$-matrices are unitary representations of both $SU(2)$ and $SO(3)$, and this implies (I think), that these groups are at least subgroups of the group that you are looking for, and I suspect that this is actually all such transformations (up to including an inversion in space, which might work.

In general, one can make the matrix you are looking for by just constructing the block-diagonal matrix with $D^{(0)}$ in the first block, $D^{(1)}$ in the second block, and so on.

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Note that for the case of $L=1$, the matrix constructed here has eigenvalues 1, 1, and $e^{\pm i\theta}$ if $\psi=0$ and $\phi=0$, so one of the matrices constructed in one of the other answers is a special case of this. The other is likely the corresponding one from $O(3)$ that also has an inversion in space.

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Now, I suspect that this is somewhat uninteresting, in the sense that all we're really doing is constructing new spherical harmonics about a different axis and then just adding them directly. In other words, the functions thus created are essentially the same as the case where $A=I$, except that we've rotated the $z$-axis. What would be interesting if there existed such matrices for which spherical harmonics of different $l$'s mixed with each other, i.e. $A_{kl,mn}\neq0$ when $k\neq l$, but I have been unable to construct those.

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Incidentally, if you have access to Mathematica, the Wigner $D$ matrices are implemented as

WignerD[{l, m1, m2}, ψ, θ, ϕ]

where of course, l is a non-negative integer, and m1 and m2 are integers between -l and l.