# Recognizing Entanglement from the Density Matrix in 2-qubit case

We know that the density matrix for a 2-qubit system can be written in the Pauli representation as :
$$\rho = \frac{1}{4}\sum_{ij}t_{ij}\sigma_i\otimes\sigma_j$$
where $\sigma_i$ are the Pauli operators with $\sigma_0 = I$ and $t_{ij} = \langle\sigma_i\otimes\sigma_j\rangle = Tr(\rho\sigma_i\otimes\sigma_j)$.
Recently, I read in a book that if the qubits are entangled then $|t_{11}| + |t_{22}| + |t_{33}| > 1$, so the condition for the state to be written as a product of two 1-qubit states is that the sum must be less than or equal to 1. Is there any elementary proof for it ?
The easiest and the most straightforward test for a 2-qubit state given by the column vector \begin{pmatrix} a \\ b \\ c \\ d \\ \end{pmatrix} w.r.t. the basis $\{|00\rangle, |01\rangle, |10\rangle, |11\rangle\}$, to be separable is that $ad = bc$. How is this related to the one given above ?
Thanks.

• Which book states this? Commented Jun 1, 2017 at 16:32
• @lalala Quantum Computing Explained by David McMahon, Wiley Interscience Commented Jun 1, 2017 at 16:37
• Your first paragraph is about pure state entanglement, the second about mixed state entanglement. You should first learn about those concepts before asking this question. Commented Jun 18, 2017 at 15:47

Just to be clear, you want to prove that $|t_{11}|+|t_{22}|+|t_{33}|>1$ $\Rightarrow$ ($\rho$ is entangled), or equivalently ($\rho$ is separable) $\Rightarrow$ $|t_{11}|+|t_{22}|+|t_{33}|\leq 1$. The other direction is not true.
Let's prove this for a product state $\rho=\rho_A\otimes \rho_B$ first. In this case $$t_{ij}=(\mathrm{tr} \rho_A \sigma_i) (\mathrm{tr} \rho_B \sigma_j)=u_i v_j.$$ So $$|t_{11}|+|t_{22}|+|t_{33}|=|u_1| |v_1|+|u_2| |v_2|+|u_3||v_3| \leq |\vec{u}| |\vec{v}| \leq 1,$$ where we used the Cauchy-Schwarz inequality and that the Bloch vectors of $\rho_A$ and $\rho_B$ have norm less or equal one (otherwise the reduced states wouldn't be positive).
Now for a general separable state, a mixture of product states $\rho=\sum_k p_k \rho_A^{(k)}\otimes \rho_B^{(k)}$ with probabilities $p_k$, we find that $$\sum_i |t_{ii}|=\sum_i |\sum_k p_k t_{ii}^{(k)}|\leq\sum_k p_k \sum_i |t_{ii}^{(k)}|\leq 1.$$ In the last step we used that we proved the relation for product states already and that the probabilities sum up to one.