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?


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.

  • $\begingroup$ We are talking of one qubit here? Do you know about the Bloch sphere? And what is your question: Why pure states have ||T||=1? $\endgroup$ Commented Aug 8, 2019 at 10:11
  • $\begingroup$ Related: physics.stackexchange.com/questions/63718, physics.stackexchange.com/questions/189253. $\endgroup$ Commented Aug 8, 2019 at 10:40
  • $\begingroup$ @NorbertSchuch, my question is why $||T^{(j)}||\leq \sqrt{\frac{2(d_j-1)}{d_j}} $ $\endgroup$
    – Pentaquark
    Commented Aug 8, 2019 at 11:06
  • $\begingroup$ I do know about the Bloch sphere, though I do not immediately see how the result follows- $\endgroup$
    – Pentaquark
    Commented Aug 8, 2019 at 11:12
  • 1
    $\begingroup$ For a point $\vec r\in\mathbb R^3$ on the Bloch sphere, $\rho=(I + \vec r\cdot \vec\sigma)/2$ is exactly of the form above with $T=(1,\vec r)$. $\endgroup$ Commented Aug 8, 2019 at 11:17

2 Answers 2


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.

  • 1
    $\begingroup$ Good point. Indeed, pg 2 of the reference (unedited): "We will de- compose the tensor T i 1 ···i n into the m-body correlation tensors T i j , T i j i k , etc, which are tensors of order m in- dicated by the number of labels in the superscript. All the indices not labeled in the superscript are fixed to be zero while the other indices take every possible value but zero (i. e. the identity is not taken into account). For in- stance, the 1-body correlation tensor for particle 1, given (1) by T i 1 = T i 1 0···0 with i 1 6 = 0, completely characterizes the reduced state ρ 1". $\endgroup$ Commented Aug 8, 2019 at 16:07
  • $\begingroup$ Thank you both for your help, and I'm really sorry if I'm being dense here, but why is $Tr(\rho)=T_{00...0}$? Why do you know that all other $T_{ii.....i}$ are zero? $\endgroup$
    – Pentaquark
    Commented Aug 9, 2019 at 9:46
  • 1
    $\begingroup$ Because $Tr(\sigma_k)=0$ for every $k=1,2,3$ and $Tr(A_1\otimes \cdots \otimes A_n) = Tr(A_1) \cdots Tr(A_n)$ if the $A_k$ act on corresponding pairwise different Hilbert spaces. Therefore, the only surviving trace is the one which acts on a term which does not contain factors $\sigma_k$ with $k=1,2,3$. Only $\sigma_0 \otimes \cdots \otimes \sigma_0$ is of this sort. $\endgroup$ Commented Aug 9, 2019 at 9:49
  • $\begingroup$ That makes sense, thank you! $\endgroup$
    – Pentaquark
    Commented Aug 9, 2019 at 9:53

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 $T=(1,\vec r)$.

  • $\begingroup$ Thanks for your answer. I now understand the case of $d=2$, but I am still unclear how the general case of $d>2$ is proven. $\endgroup$
    – Pentaquark
    Commented Aug 8, 2019 at 11:42
  • $\begingroup$ In the general case, I don't think there is such a statement: It is non-trivial to determine the shape of the Bloch "sphere", i.e. the set of all $T$ such that $\rho$ is positive semi-definite. See e.g. the links I provided above in the comment. $\endgroup$ Commented Aug 8, 2019 at 12:06
  • $\begingroup$ The paper I am reading claims the general statement without proof: arxiv.org/pdf/1106.5756.pdf (see page three eq. (3)) $\endgroup$
    – Pentaquark
    Commented Aug 8, 2019 at 12:24
  • $\begingroup$ @Pentaquark Presumably their $T^{(j)}$ is different from just $T$. It is on you to figure out what it is. But if it is the $T$ for the reduced 1-qubit density matrix, the qubit statement holds. $\endgroup$ Commented Aug 8, 2019 at 12:28

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