# Choosing an Irreducible Tensor Operator Basis where the Singular Values of Each Basis Element are the Same

Let $$\mathcal{B(H)}$$ be the space of all bounded linear operators on the Hilbert space $$\mathcal{H}$$. Let $$g \rightarrow \mathcal{U}_g$$, where $$\mathcal{U}_g (\cdot):= U(g) (\cdot) U(g)^\dagger$$, be the unitary representation of a group $$G$$ on $$\mathcal{B(H)}$$.

Let $$\{ X_j^{(\lambda,\alpha)}\} \in \mathcal{B(H)}$$ be such that \begin{align} \forall g \in G: \ \mathcal{U}_g[X_j^{(\lambda,\alpha)}] = \sum_{j'} U^\lambda_{j'j} (g) X_{j'}^{(\lambda,\alpha)}, \end{align} where $$\lambda$$ labels an irrep of $$G$$, $$j$$ labels the basis vector of the irrep $$\lambda$$, and $$\alpha$$ labels any multiplicity degrees of freedom, and \begin{align} U^\lambda_{j'j} (g) := \langle \lambda,j'| U^\lambda(g) |\lambda,j\rangle , \end{align} are the matrix elements of $$U^\lambda(g)$$, a unitary irreducible representation of $$G$$ on $$\mathcal{H}$$. The $$\{X_j^{\lambda,\alpha}\}$$ are then called an \emph{irreducible tensor operator basis} of $$G$$ on $$\mathcal{B(H)}$$.

We choose this basis to be normalized such that \begin{align} \text{Tr}[ {X_j^{(\lambda,\alpha)\dagger}} X_{j'}^{(\lambda',\alpha')} ] = \delta_{\lambda,\lambda'} \delta_{j,j'} \delta_{\alpha,\alpha'}. \end{align}

Is it possible to choose an irreducible tensor operator basis such that the singular values of of $$X^{(\lambda,\alpha)}_j$$ are all $$\frac{1}{\sqrt{d_\lambda}}$$, where $$d_\lambda$$ is the dimension of the irrep $$\lambda$$?

• I really enjoy this question. Is it a known fact that one can orthonormalize the tensors while preserving their transformation properties? Also, is it known that given some $\mathcal{H}$, one can form a complete basis of tensor operators that spans the space $\mathcal{B(H)}$? Jun 25, 2021 at 4:59

## 1 Answer

Here is a tentative answer of no, crossposted from my answer on MSE 4182419.

Please comment if I am misunderstanding the question. My claim is that $$G = SU(2)$$ has a counterexample. I will first motivate the counterexample, and then give a proof that it is the only allowed tensor operator basis.

Motivation:

For $$\lambda$$ the adjoint rep, also called the spin-$$1$$ representation, one has families of tensor operators for every rep of $$SU(2)$$: $$X_{j=-1}^{(\lambda=adj)} = S^-, X_{j=0}^{(\lambda=adj)} = \sqrt{2} S^z, X_{j=1}^{(\lambda=adj)} = -S^+$$. These are normalized for the fundamental representation on $$\mathcal{H} = \mathbb{C}^2$$, and can be normalized simply for other representations.

Explicitly for $$\mathcal{H} = \mathbb{C}^2$$, one would have $$S^- = \begin{bmatrix}0 & 0\\ 1 & 0\end{bmatrix}$$, and $$S^z = \begin{bmatrix}1/\sqrt{2} & 0\\ 0 & -1/\sqrt{2}\end{bmatrix}$$, and $$-S^+ = \begin{bmatrix}0 & -1\\ 0 & 0\end{bmatrix}$$. These satisfy your orthonormalization condition for $$\lambda = \lambda' = \text{adjoint}$$, and you can tack on $$I/\sqrt{2}$$ for the $$\lambda$$ trivial representation for a complete orthonormal basis satisfying $$\text{Tr}[ {X_j^{(\lambda,\alpha)\dagger}} X_{j'}^{(\lambda',\alpha')} ] = \delta_{\lambda,\lambda'} \delta_{j,j'} \delta_{\alpha,\alpha'}$$.

To see that these are tensor operators, I prefer to use the "commutator" definition of tensor operators that states that the commutator of the generators of the group $$T_a$$ (in $$\mathcal{B(H)}$$) with a tensor operator $$X_j^{(\lambda,\alpha)}$$ is a linear combination of tensor operators weighted by the matrix elements of the generator in representation $$\lambda$$.

$$[T_a, X_j^{(\lambda,\alpha)}] = \sum_{j'} {T_a}^\lambda_{j'j} X_{j'}^{(\lambda,\alpha)}$$

In particular, we can take linear combinations of generators in the above. I will consider the complexification of the generators $$S^+ = S^x + i S^y$$ from which the rest of the algebra can be defined. It is straightforward to check that one has $$[S^+, X_{j=-1}^{(\lambda=adj)}] = 2 S^z = \sqrt{2} (X_{j=0}^{(\lambda=adj)})$$ and $$[S^+, X_{j=0}^{(\lambda=adj)}] = -\sqrt{2} S^+ = \sqrt{2} (X_{j=1}^{(\lambda=adj)})$$, which coincides correctly with the adjoint representation's $$(S^+)_{adj} = \begin{bmatrix}0 & \sqrt{2} & 0 \\ 0 & 0 & \sqrt{2}\\ 0 & 0 & 0 \end{bmatrix}$$.

Now, in this tensor operator representation, $$X_{j=-1}^{(\lambda=adj)} = S^-$$ and $$X_{j=1}^{(\lambda=adj)} = -S^+$$ don't have the right singular values. Their singular values are not the desired $$(1/\sqrt{2}, 1/\sqrt{2})$$ but are instead $$(1,0)$$.

Uniqueness proof: One might wonder if a different choice for the operators $$X_j^{(\lambda,\alpha)}$$ in $$\lambda = adjoint$$ could work for this case of $$\mathcal{H} = \mathbb{C}^2$$. The answer is no, the above $$\{X_j^{(\lambda,\alpha)}\}$$ is the only choice for tensor operators for $$\lambda = adjoint$$ on this $$\mathcal{H}$$ (up to complex phase, which leaves singular values unchanged).

To see this, note that any choice of $$\{ X_{j}^{(\lambda=adj)} \}$$ has $$[S^+, [S^+, X_{j=-1}^{(\lambda=adj)}]] = 2X_{j=1}^{(\lambda=adj)}$$, according to the commutator definition of tensor operators above.

However, for ANY $$X_{j=-1}^{(\lambda=adj)}$$, $$[S^+, [S^+, X_{j=-1}^{(\lambda=adj)}]] \propto = \begin{bmatrix}0 & -1\\ 0 & 0\end{bmatrix}$$.

Thus, $$X_{j=1}^{(\lambda=adj)} = e^{i\phi} \begin{bmatrix}0 & -1\\ 0 & 0\end{bmatrix}$$. This in turn fixes $$X_{j=0}^{(\lambda=adj)}$$ via $$X_{j=0}^{(\lambda=adj)} = \frac{1}{\sqrt{2}} [S^-, X_{j=1}^{(\lambda=adj)}]$$ and fixes $$X_{j=-1}^{(\lambda=adj)}$$ via $$X_{j=-1}^{(\lambda=adj)} = \frac{1}{2}[S^-,[S^-, X_{j=1}^{(\lambda=adj)}]]$$, fixing the form of the representation up to phase $$e^{i\phi}$$.

Thus, all tensor operator bases of $$SU(2)$$ on $$\mathcal{H} = \mathbb{C}^2$$ have two tensor operators without the desired singular values, giving the answer of no to the question.