# Traceless symmetric tensor's transformation under a Lie group

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?

• Suggestion: if you use the index $i$ to label $L$’s, don’t use it to label $a^\dagger$’s to avoid confusion. See Rowe DJ, Le Blanc R, Repka J. A rotor expansion of the su (3) Lie algebra. Journal of Physics A: Mathematical and General. 1989 Apr 21;22(8):L309. for an explicit construction of the $L=2$ tensor. Jul 3, 2021 at 12:03
• Jul 3, 2021 at 17:07
• @ZeroTheHero I think the indices of ladder operators are spacetime indices, as the above transformations caused by $L$ indicates. Indeed, the ladder operators corresponds to 3-dimensional H.O. so the indices corresponds to $x, y, z$ in a proper system. Could you clarify the reason why we have to distinguish them in more detail? Jul 4, 2021 at 4:22
• The indices can be whatever you want (nothing to do with space time) but why do you insist on using $i$ in both $L_i$ and $a_i^\dagger a_j$ with $i$ repeated? Although $su(2)\sim so(3)$ they are embedded differently and inequivalently in su(3); see the Rowe paper above, where the $L=1$ tensor (the so(3) generators) and the $L=2$ ( quadrupole ) tensor are given explicitly. The $L=0$ part is $N=\sum_k a_k^\dagger a_k$. Jul 4, 2021 at 12:39
• It doesn’t matter. The point is if the index on $i$ on $L_i$ need not be repeated in $a_i^\dagger a_k$. $L_i$ trivially commutes with $a_i^\dagger a_i$ so it looks confusing as written. Jul 4, 2021 at 23:02

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.

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.
• Then, my mistake is not sandwiching the operator between $a^\dagger a$? Jul 4, 2021 at 3:51
• I didn't understand why I have to do so. Actually, my operator is all written in terms of ladder operators. Hence, I think we don't have to do so repeatedly. However, I suppose my actual mistake is the way to take its trace over this Fock space. The trace of $(a^\dagger_i a_j + a^\dagger_j a_i)$ is not just $2 \sum_i a^\dagger_i a_i$ but we have to calculate $\sum_{{\bf n}} \langle {\bf n} | (a^\dagger_i a_j + a^\dagger_j a_i) | {\bf n} \rangle$. Jul 4, 2021 at 4:10
• Then I can say the traceless symetric tensor transfoms as ${\bf 5}$ since the number of independent components is 5 and its tracelessness and symmetry are conserved under the transformations. Furthermore, there is no fixed tensor with such properties. This fact corresponds to existence of quintuplets in some Hilbert space. Is this correct thought? Jul 4, 2021 at 4:18
• Thank you very much. Now I understand. I forgot the indice of $L$ has nothing to do with the trace...sorry. In addition, the above result implies the commutator of $L$ and symmetric tensor, not necessary traceless, becomes traceless. Indeed, in the above discussion, we know $[L, {\rm Tr}(a^\dagger a)] = 0$ hence the minus trace term does not affect the commutator of $N$. Jul 5, 2021 at 10:30