Recognizing Entanglement from the Density Matrix in 2-qubit case

Question

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.

Answer 1

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.