I was reading the paper device independent outlook on quantum mechanics. The author defines a generic two qubit density matrix as $$ \rho=\frac{1}{4}\left( I \otimes I + \vec{r_{\rho}} \cdot \vec{\sigma}\otimes I + I \otimes \vec{s_{\rho}} \cdot \vec{\sigma} + \sum_{i,j=x,y,z}T^{ij}_{\rho} \sigma_i \otimes \sigma_j \right) \, . \tag{1} $$

How is it obtained and what are the constraints over $T^{ij}_{\rho}$ ? Also, seeing that it has some symmetry can a general 3 qubit density matrix be written as

\begin{align} \rho = \frac{1}{8} &\left( I \otimes I \otimes I + \vec{r_{\rho}} \cdot \vec{\sigma} \otimes I \otimes I + I \otimes \vec{s_{\rho}}\cdot \vec{\sigma} \otimes I + I \otimes I \otimes \vec{t_{\rho}}.\vec{\sigma} \right. \\ &+ \sum_{i,j=x,y,z}T^{ij}_{\rho} \sigma_i \otimes \sigma_j \otimes I + \sum_{i,j=x,y,z}U^{ij}_{\rho} \sigma_i \otimes I \otimes \sigma_j \\ &\left. + \sum_{i,j=x,y,z}W^{ij}_{\rho} I \otimes \sigma_i \otimes \sigma_j +\sum_{i,j,k=x,y,z}X^{ij}_{\rho} \sigma_i \otimes \sigma_j \otimes \sigma_k \right) ? \end{align} Here $\vec{r_{\rho}}, \vec{s_{\rho}}, and \vec{t_{\rho}}$ are 3 dimensional vectors with real components and each having magnitude $\le 1$.


I get the fact that tensor of pauli matrices acts as a basis but can't get the condition on $T_{ij}$. I was able to work backwards to see that $T_{\rho}^tT_{\rho}$ has to be such that its maximum eigen value is $\le 1$ so that CHSH inequality is only violated at maximum upto $2\sqrt{2}$. So if this condition is not followed then $(1)$ should not be a valid density matrix. The form given in $(1)$ is already hermitian and has trace 1. So for $T_{\rho}^tT_{\rho}$ having maximum eigen value $\ge1$ $(1)$ might not be a positive operator but I am unable to prove that.

  • $\begingroup$ What is your question? The conditions on $T$ for 2 qubits or the form of the density matrix for general qubits (and if the latter, what exactly is your question)? $\endgroup$ Jun 16, 2015 at 6:51
  • $\begingroup$ @NorbertSchuch Yes I am looking for conditions on matrix $T$. The single condition I know about it that maximum eigen value of $T^tT$ should be 1 and I cannot prove even that condition. So basically my question if what conditions are imposed on $T$ if (1) is a valid density matrix. $\endgroup$ Jun 16, 2015 at 8:13
  • $\begingroup$ Talking about the eigenvalues of $T$ probably does not give a criterion: $T$ does not even have to be diagonalizable; and conversely, one can easily write an upper triangular $T$ with zero diagonal (i.e. eigenvalues $0$) which does not describe a positive $\rho$. $\endgroup$ Jun 18, 2015 at 18:55

1 Answer 1


http://arxiv.org/abs/quant-ph/9607007 discusses necessary conditions on $T$ (more precisely, on its singular values) for $\rho$ to be positive. They don't seem to derive sufficient conditions, however.

The basic idea is that one can perform a rotation $U_A$ and $U_B$ on the two qubits, respectively, which correspondingly transforms $r\mapsto O_Ar$, $s\mapsto O_Bs$, and $T\mapsto O_A T O_B^T$. By choosing $O_A$ and $O_B$ which give the singular value decomposition of $T$, one finds that any $\rho$ in your form (1) can be replaced by one with a diagonal $T$, with the singular values of the original $T$ on the diagonal.

Now, one can use different "trial states" $\vert\psi\rangle$ and check if $\langle\psi\vert \rho\vert\psi\rangle\ge0$ (which is necessary for positivity of $\rho$). By using the Bell states (for which the $r$ and $s$ part vanish), one obtains non-trivial constraints on $T$.


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