Wigner $ D $ matrix equivalent for cyclic symmetry

Question

$\newcommand{\ket}[1]{\left|#1\right\rangle}$The action of $ g \in SU(2) $ on a spin $ j $ system (with a Hilbert space of size $ 2j+1 $) is by the Wigner $ D $ matrix $ D^j(g) $. There are formulas for Wigner $ D $ matrices available in many textbooks and implemented in many computational packages. A spin $ j $ representation results from the completely symmetrized coupling of $ n=2j $ spin $ 1/2 $ representations; it is the highest-dimensional representation in that coupling. Normally a system of $ n $ spin $ 1/2 $ particles has a Hilbert space of size $ 2^n $ but for the completely symmetrized coupling there is an $ S_n $ symmetry that reduces the Hilbert space to size $ n+1 $.

What if instead of having full $ S_n $ symmetry the particles have cyclic $ C_n $ symmetry. Then the Hilbert space dimension, call it $ N $, is smaller than $ 2^n $ but still much larger than $ n+1 $. Like I say the $ n+1 \times n+1 $ Wigner D matrices for the action of $ g \in SU(2) $ on the $ S_n $ symmetrized coupling of $ n $ many spin $ 1/2 $ representations is easily found in many places. Is there somewhere to find a formula for the $ N \times N $ matrices for the action of $ g \in SU(2) $ on a system of $ n $ spin $ 1/2 $ particles with $ C_n $ cyclic symmetry?

Example:

Let me give an example of what I mean for $ n=5 $.

We start with a Hilbert space of size $ 2^5 $. But after symmetrizing with respect to $ S_5 $ we are left with only a $ 6 $ dimensional Hilbert space, the basis for which is given by the Dicke states $ \ket{D^5_0}, \ket{D^5_1},\ket{D^5_2},\ket{D^5_3},\ket{D^5_4}, \ket{D^5_5}, $ see Collective angular momentum , Dicke states and indistinguishable particles you can follows links in that post but just to be clear that by Dicke state I mean normalized symmetric superposition of all states of $ n $ spin $ 1/2 $ systems with $ k $ excitations e.g. $$ |D^5_1 \rangle= \frac{1}{\sqrt{5}} ( \ket{\uparrow\downarrow\downarrow\downarrow\downarrow} +\ket{\downarrow\uparrow\downarrow\downarrow\downarrow}+\ket{\downarrow\downarrow\uparrow\downarrow\downarrow}+\ket{\downarrow\downarrow\downarrow\uparrow\downarrow}+\ket{\downarrow\downarrow\downarrow\downarrow\uparrow}) $$ or equivalently $$ |D^5_1 \rangle= \frac{1}{\sqrt{5}} ( \ket{10000} +\ket{01000}+\ket{00100}+\ket{00010}+\ket{00001}) $$

Sticking to the same example of $ n=5 $ but now symmetrizing over $ C_5 $ instead of $ S_5 $ we get something slightly different. Again we start with a Hilbert space of size $ 2^5 $. But after symmetrizing with respect to $ C_5 $ we are left with only an $ 8 $ dimensional Hilbert space, the basis for which is given by the (generalized?) Dicke-type states $ \ket{D^5_0}, \ket{D^5_1},\ket{D^5_{2a}},\ket{D^5_{2b}},\ket{D^5_{3a}},\ket{D^5_{3b}},\ket{D^5_4}, \ket{D^5_5} $ where $ \ket{D^5_0}, \ket{D^5_1},\ket{D^5_4}, \ket{D^5_5} $ are the standard Dicke states described above while the generalized Dicke states satisfy the equations $$ \ket{D^5_2}=\frac{1}{\sqrt{2}} (\ket{D^5_{2a}}+\ket{D^5_{2b}}) $$ $$ \ket{D^5_3}=\frac{1}{\sqrt{2}} (\ket{D^5_{3a}}+\ket{D^5_{3b}}) $$ To be even more explicit $$ \ket{D^5_{2a}}=\frac{1}{\sqrt{5}} ( \ket{11000} +\ket{01100}+\ket{00110}+\ket{00011}+\ket{10001}) $$ $$ \ket{D^5_{2b}}=\frac{1}{\sqrt{5}} ( \ket{10100} +\ket{01010}+\ket{00101}+\ket{10010}+\ket{01001}) $$ $$ \ket{D^5_{3a}}=\frac{1}{\sqrt{5}} ( \ket{00111} +\ket{10011}+\ket{11001}+\ket{11100}+\ket{01110}) $$ $$ \ket{D^5_{3b}}=\frac{1}{\sqrt{5}} ( \ket{01011} +\ket{10101}+\ket{11010}+\ket{01101}+\ket{10110}) $$

Answer 1

$\newcommand{\ket}[1]{\left|#1\right\rangle}$Ok now that I think about it more, maybe the answer is obvious. The Hilbert space for the $ C_5 $ symmetrized problem is indeed $ 8 $ dimensional but a more convenient basis is actually $ \ket{D^5_0}, \ket{D^5_1},\ket{D^5_{2}},\ket{D^5_{3}},\ket{D^5_4}, \ket{D^5_5}, \ket{D^5_{2,-}},\ket{D^5_{3,-}} $ where $ \ket{D^5_0}, \ket{D^5_1},\ket{D^5_{2}},\ket{D^5_{3}},\ket{D^5_4}, \ket{D^5_5} $ are standard Dicke states while I define $$ \ket{D^5_{2,-}}=\frac{1}{\sqrt{2}} (\ket{D^5_{2a}}-\ket{D^5_{2b}}) $$ $$ \ket{D^5_{3,-}}=\frac{1}{\sqrt{2}} (\ket{D^5_{3a}}-\ket{D^5_{3b}}) $$ So as irreps of $ SU(2) $ the $ 8 $ dimensional $ C_5 $ symmetrized space decomposes as $ \textbf{6}+\textbf{2} $. Thus, in this basis, the action of $ g \in SU(2) $ is given by the matrix $$ \begin{bmatrix} D^{5/2}(g) & 0 \\ 0 & D^{1/2}(g) \end{bmatrix} $$ where $ D^{5/2}(g) $ is a $ 6 \times 6 $ block and $ D^{1/2}(g) $ is a $ 2 \times 2 $ block.

Indeed such an approach should always work since the action of $ C_ n $ on $ 2^n $ by cycling the spins commutes with the action of $ g \in SU(2) $ on $ 2^n $ by $ g^{\otimes n} $ and thus the space of $ C_n $ invariant states will always be closed under the $ SU(2) $ action, in other words the space of $ C_n $ symmetrized states will always be an $ SU(2) $ subrepresentation of $ 2^n $. It will always include the $ n+1 $ dimensional largest irrep and also some other parts. The trick is then to find a basis for this $ N $ dimensional cyclic symmetrized space that block diagonalizes the $ SU(2) $ action into irreps, as above. I did it by hand in the small case of $ n=5 $ but for larger $ n $ this can be done in a systematic way.

To do this in a systematic way for larger $ n $ you would proceed as follows. The top irrep is always in there and always the same, so always include one copy of spin $ n/2 $ irrep with basis just given by the standard Dicke states. Call the the projector onto the cyclic symmetrized space $ \Pi_{C_n} $. Then start taking products of $ \Pi_{C_n} $ with projectors onto the isotypic subspaces for spin $ (n-1)/2 $ irrep, then for spin $ (n-2)/2 $ irrep, $\dots$ and finally for spin $ 1/2 $ irrep. Then find ONB for all these orthogonal $ SU(2) $ irreps.

In the case of $ n=5 $ what you find is that the isotypic projector for the $ 5/2 $ spin irrep has multiplicity $ 1 $ i.e. rank $ 6 $ and multiplying by $ \Pi_{C_5} $ of course does not change this, and so an orthogonal basis for the 6d image of this projector is given by $ \ket{D^5_0}, \ket{D^5_1},\ket{D^5_{2}},\ket{D^5_{3}},\ket{D^5_4}, \ket{D^5_5}$. On the other hand the isotypic projector for the $ 3/2 $ spin irrep has multiplicity $ 4 $ i.e. rank $ 16 $ but when you multiply it by $ \Pi_{C_5} $ you get the zero matrix. Finally the isotypic projector for the $ 1/2 $ spin irrep has multiplicity $ 5 $ i.e. rank $ 10 $ but when you multiply it by $ \Pi_{C_5} $ you get a projector with rank $ 2 $, and an orthogonal basis for the 2d image of this projector is exactly given by the states $ \ket{D^5_{2,-}} $ and $ \ket{D^5_{3,-}} $ I described above.

A similar analysis reveals that by choosing an appropriate basis for the $ 20 $ dimensional Hilbert space of $ C_7 $ symmetrized states for $ n=7 $ we can write the action of $ g \in SU(2) $ as a block diagonal matrix given by $$ \begin{bmatrix} D^{7/2}(g) & 0 & 0 & 0 & 0 \\ 0 & D^{3/2}(g) & 0 & 0 & 0\\ 0 & 0 & D^{3/2}(g) & 0 & 0\\ 0 & 0 & 0 & D^{1/2}(g) & 0 \\ 0 & 0 & 0 & 0 & D^{1/2}(g) \end{bmatrix} $$

For the $ n=5 $ case there is a unique decomposition as $\textbf{6}+\textbf{2}$ with an essentially canonical choice of basis given by $\ket{D^5_0}, \ket{D^5_1},\ket{D^5_{2}},\ket{D^5_{3}},\ket{D^5_4}, \ket{D^5_5}$ and $ \ket{D^5_{2,-}},\ket{D^5_{3,-}} $. However for the general case, and already even for the $ n=7 $ case, there is not a totally canonical choice of basis.

For at least $ 8 $ of the $ 20 $ basis vectors there is a canonical choice, given by $\ket{D^7_0}, \ket{D^7_1},\ket{D^7_{2}},\ket{D^7_{3}},\ket{D^7_4}, \ket{D^7_5}, \ket{D^7_6}, \ket{D^7_7}$. This follows from the canonical choice of $ \ket{D^7_7} $ and then applying the lowering operator $ 7 $ times (and normalizing). I'll denote the lowering operator here by $ J_- $ although $ S_- $ is also reasonable notation.

$ \ket{D^7_7} $ spans the $ 7/2 $ eigenspace of $ J_Z $, $ \ket{D^7_6} $ spans the $ 5/2 $ eigenspace of $ J_Z $.

But the $ 3/2 $ eigenspace of $ J_Z $ is $ 3 $ dimensional, so the orthogonal complement of $ \ket{D^7_5} $ in the $ 3/2 $ eigenspace is 2 dimensional. Let $ \ket{D^7_{5a}} $ denote the uniform superposition over all $ 7 $ cyclic permutations of $ 0011111 $, $ \ket{D^7_{5b}} $ denote the uniform superposition over all $ 7 $ cyclic permutations of $ 0101111 $ and let $ \ket{D^7_{5c}} $ denote the uniform superposition over all $ 7 $ cyclic permutations of $ 0110111 $. Then one reasonable choice for an orthonormal basis of the $ 3/2 $ eigenspace of $ J_z $ is $$ \frac{1}{\sqrt{3}} (\ket{D^7_{5a}}+\ket{D^7_{5b}}+\ket{D^7_{5c}})= \ket{D^7_5} $$ $$ \frac{1}{\sqrt{3}} (\ket{D^7_{5a}}+ \omega \ket{D^7_{5b}}+ \omega^2 \ket{D^7_{5c}}) $$ $$ \frac{1}{\sqrt{3}} (\ket{D^7_{5a}}+ \omega^2 \ket{D^7_{5b}}+ \omega \ket{D^7_{5c}}) $$ where $ \omega=e^{2 \pi i/3} $. Note the first basis vector is exactly $ \ket{D^7_5} $.

If you prefer real coefficients, another reasonable choice for an orthonormal basis of the $ 3/2 $ eigenspace of $ J_z $ is $$ \frac{1}{\sqrt{3}} (\ket{D^7_{5a}}+\ket{D^7_{5b}}+\ket{D^7_{5c}}) $$ $$ \frac{1}{\sqrt{2}} (\ket{D^7_{5a}}- \ket{D^7_{5b}}) $$ $$ \frac{1}{\sqrt{6}} (\ket{D^7_{5a}}+\ket{D^7_{5b}}-2 \ket{D^7_{5c}}) $$ Again, the first basis vector is exactly $ \ket{D^7_5} $.

From such a basis one can produce a basis for both of the $ \textbf{4} $ irreps by just applying lowering operator $ J_- $ three successive times.

Having done this, one obtains a basis for both the $ \textbf{4} $ irreps in the decomposition $ \textbf{8}+ \textbf{4}+\textbf{4}+\textbf{2}+\textbf{2}$ of the $ 20 $ dimensional $ C_7 $ symmetrized space. Thus we have $ 16 $ basis vectors and all that remains is finding a basis for the two $ \textbf{2} $ irreps. This can be done by applying $ J_- $ to the chosen basis for the $ 3/2 $ eigenspace of $ J_z $ and then taking the three resulting vectors in the $ 1/2 $ eigenspace of $ J_z $ and extending that to a basis if the entire $ 5 $ dimensional $ 1/2 $ eigenspace by finding two more vectors. These final two vectors then generate the two copies of $ \textbf{2} $ by applying $ J_- $ to them once each.

Answer 2

You don't need to think of representations of the rotation group SU(2) as necessarily in the D-matrix spherical basis, but you may. When you tensor n such, you get a reducible $2^n \times 2^n$ matrix. It normally reduces to a direct sum of irreducible representations of varying dimensionalities. (My favorite accounting for this reduction is summarized here.)

The correct statement supplanting that of yours, above, is: "A spin representation results from the completely symmetrized coupling of =2 spin 1/2 representations; it is the highest-dimensional representation in the reduction of that coupling".

let me illustrate it for n=3 . $$ 1/2\otimes 1/2 \otimes 1/2= 3/2\oplus 1/2\oplus 1/2, $$ as summarized in elementary QM texts, corresponding to dimensionalities $$ 8=2^3= 4+2 +2, $$ as you worked out in your question. Too the original 8-dim Hilbert space reduces ("chops up") to three subspaces: as linked in the answer, the 4-dim one is fully symmetric (spin 3/2), and the two spin 1/2 ones may be classified by the cyclic group.

It should be evident from the respective Young tableaux, when you get to them, that there are all types of symmetries for general n, not just the groups you mentioned in your question.

Note this discussion deals with the state vectors. The corresponding rotation generators are listed in WP, in the standard notation substantially superior to that of D-matrices...

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Edit on n=5 comment

Indeed, $2^5=32=6+4+4+4+4+2+2+2+2+2$, so your reduction has a fully symmetric spin 5/2; four spin 3/2; and five spin 1/2 s, all mutually orthogonal for common m.

Given your "Dicke state" $\newcommand{\ket}[1]{\left|#1\right\rangle}$ $$ |D^5_1 \rangle= \frac{1}{\sqrt{5}} ( \ket{\uparrow\downarrow\downarrow\downarrow\downarrow} +\ket{\downarrow\uparrow\downarrow\downarrow\downarrow}+\ket{\downarrow\downarrow\uparrow\downarrow\downarrow}+\ket{\downarrow\downarrow\downarrow\uparrow\downarrow}+\ket{\downarrow\downarrow\downarrow\downarrow\uparrow}), $$ consider for example the state $$ | \psi \rangle= \frac{1}{\sqrt{5}} ( \ket{\uparrow\downarrow\downarrow\downarrow\downarrow} +\omega\ket{\downarrow\uparrow\downarrow\downarrow\downarrow}+\omega^2\ket{\downarrow\downarrow\uparrow\downarrow\downarrow}+\omega^3\ket{\downarrow\downarrow\downarrow\uparrow\downarrow}+\omega^4\ket{\downarrow\downarrow\downarrow\downarrow\uparrow}), $$ where $$ \omega=\exp(2\pi i /5) ~~~~~\leadsto ~~~~ 1+\omega+\omega^2+\omega^3+\omega^4=0, $$ whence $$ S_-|\psi\rangle=0, ~~~~\hbox{and} ~~~ \langle D^5_1|\psi\rangle =0. $$ So ψ is spin 3/2, the m=-3/2 part of the quartet, with eigenvalue ω under $C_5$; you can likewise construct the other spin 3/2 states. Further, explore the symmetry of the spin 1/2 s.