Upper bound for norm of 1-body correlation tensor of qubit

Question

Any $n$-qubit state can be expressed as

$$\rho=\frac{1}{2^{n}} \sum_{\mu_{1}, \ldots, \mu_{n}=0,1,2,3} T_{\mu_{1}, \ldots, \mu_{n}} \sigma_{\mu_{1}} \otimes \ldots \otimes \sigma_{\mu_{n}}$$

where $\sigma_{\mu_k}\in \{\mathbb{1,\sigma_1,\sigma_2,\sigma_3\}}$. The coefficients $T_{\mu_{1}, \ldots, \mu_{n}}$ are real numbers in $[-1,1]$ given by the correlation function values for measurements of products of Pauli operators

$$T_{\mu_{1}, \ldots, \mu_{n}}=\left\langle\sigma_{\mu_{1}} \otimes \ldots \otimes \sigma_{\mu_{n}}\right\rangle_{\rho}=\operatorname{Tr}\left(\rho \sigma_{\mu_{1}} \otimes \ldots \otimes \sigma_{\mu_{n}}\right)$$

We will call $\mathcal T$ the correlation tensor.

The paper now says, that since $\text{Tr}(\rho^2_j)\leq 1$, all 1-body correlation tensors must obey

$$||T^{(j)}||\leq \sqrt{\frac{2(d_j-1)}{d_j}} $$ with equality iff the state is pure, where the number of superscripts denotes the order of the tensor, $||\cdot ||$ is the standard Euclidean norm for vectors and $d_j$ is the dimension of the Hilbert space $\mathcal{H}_j$.

Unfortunately, the paper does not argue, how this follows from the fact that the trace is bounded by $1$.

Does anyone know how this identity follows from the trace, or where such a proof can be found?

Thanks!

EDIT: thanks to @NorbertSchuch I now understand how the above holds for qubits with $d=2$. However, I am still unclear how the general case of $d>2$ is proven.

Answer 1

If I understand well the meaning of the various symbols, the strict inequality is already false for $n=1$. In that case, as remarked in Norbert's answer, a generic mixed state has always the form $$\rho = \frac{1}{2}\left(t_0 \sigma_0 + \sum_{k=1}^3 t_k \sigma_k \right)\:, \quad t_\mu \in \mathbb{R}$$ where $\sqrt{\sum_{k=1}^3 t^2_k}\leq 1$ and $t_0=1$. However what matters in the discussed formula seems to be instead $$||t|| = \sqrt{t_0^2 + \sum_{k=1}^3 t^2_k} = \sqrt{1+ \sum_{k=1}^3 t^2_k}\geq 1 \:. \tag{1}$$ Furthermore, $$\sqrt{\frac{2(d-1)}{d}}= 1 \quad \mbox{if $d=2$}$$ so that we should find is $$||t|| \leq 1 $$ contrarily to (1). I expect that $||T^{j}||$ is defined in a way to disregard (some? all?) $0$-components, since some of them are blocked. Indeed, $$\rho=\frac{1}{2^{n}} \sum_{\mu_{1}, \ldots, \mu_{n}=0,1,2,3} T_{\mu_{1}, \ldots, \mu_{n}} \sigma_{\mu_{1}} \otimes \ldots \otimes \sigma_{\mu_{n}}$$ implies $$Tr(\rho) = T_{00\ldots 0}$$ and this trace must be $1$ by definition.

Answer 2

This is nothing but a different way of stating the well-known fact that points on the surface of the Bloch sphere are pure:

For a point $\vec r\in\mathbb R^3$ in or on the Bloch sphere, $\rho=\tfrac12(I+\vec r\cdot\vec \sigma)$ is precisely of the form you give for $n=1$ and with $T=(1,\vec r)$.