Traceless symmetric tensor's transformation under a Lie group

Question

Usually, one decomposes a tensor product whose elements are transformed under a Lie group into its trace part, traceless symmetric part and antisymmetric part to obtain an irreducible representation of the Lie group.

For example, the $\bf{3} \otimes 3$ of $\mathfrak{su}(2)$ is represented by a sum of irreducible reps, ${\bf 1, 3, 5}$. However, I don't understand this result in terms of tensor components.

Let me focus on a specific example to clarify my question. Let $L_i~(i = 1, 2, 3)$ be Hermitian generators of $\mathrm{SO}(3)$ which obey the $\mathfrak{su}(2)$ Lie algebra and $a^{\dagger}_{i}, a_{i}$ be tensor operators which transform as ${\bf 3}$ under the algebra.

In other words, $$ [L_i, a^{\dagger}_{j}] = i \varepsilon_{ijk} a^{\dagger}_{k}, \\ [L_i, a_{j}] = i \varepsilon_{ijk} a_{k}. \\ $$ One can think these $a^{\dagger}_{i}, a_{i}$ as creation/annihilation operators of a three-dimensional isotropic harmonic oscillator.

Then we can construct a symmetric tensor product with the form $a^{\dagger}_{i}a_{j} + a^{\dagger}_{j} a_{i}$.

Since $(\bf{3} \otimes \bf{3})_{\rm sym} = {\bf 5} + {\bf 1}$, we should show that we can derive two operators transform under ${\bf 5}$ and ${\bf 1}$ respectively from the symmetric product.

As I said, we usually find the trace part as ${\bf 1}$ (singlet) and indeed $$ [L_i, \sum_{j} a^{\dagger}_{j} a_{j}] = 0. $$

Thus what we have to do is showing the traceless symmetric tensor $$ \frac{1}{2} (a^{\dagger}_{i}a_{j} + a^{\dagger}_{j} a_{i}) - \frac{1}{3} \delta_{ij} \sum_{k} a^{\dagger}_{k} a_{k} $$ transforms as ${\bf 5}$.

However, calculating the commutator gives us $$ \left[L_i, \frac{1}{2} (a^{\dagger}_{j}a_{k} + a^{\dagger}_{k} a_{j}) - \frac{1}{3} \delta_{jk} \sum_{l} a^{\dagger}_{l} a_{l}\right] = \left[L_i, \frac{1}{2} (a^{\dagger}_{j}a_{k} + a^{\dagger}_{k} a_{j})\right] $$ so that we can check only this is transforms under ${\bf 3} \otimes {\bf 3}$.

Since $e^B A e^{-B} = A + [B, A] + \cdots$, I want to show this transformation leaves its traceless and symmetric properties invariant but how can I do that?

Any comments are appreciated.

Answer 1

You are trying to investigate the transformation of symmetric traceless matrices $M_{jl}$, in the spin 2 irrep of rotations, under rotations, orthogonal matrices $e^{\theta L^i}$, where $L^i_{jk}= \epsilon^{ijk}$, real antisymmetric. You wish to translate this to the Jordan-Schwinger realization, which you are misconfiguring and misapplying, so let's defer that for the time being.

So, how do you see $RMR^T=RMR^{-1}$ is symmetric traceless just like $M$? Here: $$ M'_{jl}=R_{jk}M_{kr}R^T_{rl}= R_{jk}R_{lr} M_{kr} ~~~\leadsto \\ M'_{jj} = R_{jk}R_{jr} M_{kr} =\delta_{kr} M_{kr} =0, $$ From the orthogonality of R and the tracelessness of M.

So M' is symmetric traceless, also a 5.

If, in your Hadamard identity commutator expansion, you repeated this to first order in $L^i$, or any antisymmetric $\theta^iL^i$, for that matter, your'd find the term linear in θ is likewise symmetric traceless, and then, recursively, likewise for the terms of any order in θ !

The Jordan construction is guaranteed to work, but it's wild-wild overkill. You'd have to convince yourself that the $O(\theta)$ term $$ [a^\dagger_j L^i_{jk}a_k, a^\dagger_r M_{rs} a_s] $$ is a traceless symmetric matrix sandwiched between a triplet of $a^\dagger$s and $a$s, fairly straightforward: $\propto a^\dagger_j (\epsilon^{ijk} M_{kl} + \epsilon^{ilk} M_{kj} )a_l$.

Note that this is not quite how the $\mathfrak{su}(2)~~~$ 5 presents in Wikipedia, cited, but that's not your concern given your setup.

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Note in response to comments

Evidently the J-S realization bridge did not help. Simply consider your traceless symmetric tensor $$ N_{ij}=\frac{1}{2} (a^{\dagger}_{i}a_{j} + a^{\dagger}_{j} a_{i}) - \frac{1}{3} \delta_{ij} \sum_{k} a^{\dagger}_{k} a_{k} $$ and confirm your second term in the Hadamard expansion, $[L^i,N]$ is traceless symmetric, $$ \epsilon^{ijk} N_{kl} - N_{jk}\epsilon^{ikl}, $$ which it is, in the free indices j and l, while i is an inert label index! Summation over repeated indices is implied, and the contraction of a symmetric and an antisymmetric tensor vanishes.