A 2-qubit state's density matrix can be written in Pauli representation as:

$$ \rho = \frac{1}{4} \sum_{i, j = 0}^{3} c_{i, j}\; \sigma_i \otimes \sigma_j $$

For a given $\rho$, to compute the factors $c_{i, j}$, we construct the tensor product of the Pauli matrices, multiply that with the density matrix, and compute the trace, which is then the factor $c_{i, j}$. In Python (with '*' being the tensor product):

    paulis = [ops.Identity(), ops.PauliX(), ops.PauliY(), ops.PauliZ()]
    c = np.zeros((4, 4), dtype=np.complex64)
    for i in range(4):
      for j in range(4):
        tprod = paulis[i] * paulis[j]
        c[i][j] = np.trace(rho @ tprod)

Here is the question: To test for entanglement, add up the absolute values of the lower three diagonal elements (exclude $c_{0, 0}$). If this sum is < 1.0, then the state is separable, else it is entangled. I don't understand where this comes from? (Thanks for any pointers).

An example: For the entangled state $|\psi\rangle = \frac{|00\rangle + |11\rangle}{\sqrt{2}}$ the density matrix $\rho$ is

$$\rho = \begin{bmatrix} \frac{1}{2} & 0 & 0 & \frac{1}{2} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ \frac{1}{2} & 0 & 0 & \frac{1}{2}\end{bmatrix}$$

with a trace of 1. However, we are talking about the Pauli representation of this matrix. The coefficient matrix computed with the Python snippet above is:

$$ c = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & 1\end{bmatrix}$$

This corresponds to a Pauli representation of:

$$ \rho = \sigma_0 \otimes \sigma_0 + \sigma_1 \otimes \sigma_1 - \sigma_2 \otimes \sigma_2 + \sigma_3 \otimes \sigma_3 $$

We add the absolute values of elements $c_{1, 1}$, $c_{2, 2}$, $c_{3, 3}$, which is 3, which is 3.0 > 1.0, showing that the state was entangled. I don't understand where this last derivation comes from.

  • $\begingroup$ Not sure I understood this. Can you link the source? Because if you have the state $|\psi \rangle = \dfrac{|00\rangle + |11\rangle}{\sqrt{2}}$ then the sum of the sum of the lower three diagonal elements of the matrix $\rho = |\psi \rangle \langle \psi |$ is $1/2 < 1$, but this state is obviously an entangled state. $\endgroup$
    – KAJ226
    May 31, 2022 at 18:32
  • $\begingroup$ Thanks. I've added your example to the original question to clarify. $\endgroup$
    – rhundt
    May 31, 2022 at 20:53
  • $\begingroup$ I see. My bad. That was a stupid comment. I didn't read the question carefully. $\endgroup$
    – KAJ226
    Jun 1, 2022 at 0:56

1 Answer 1


We want to prove that is the sum is 1 or larger, the state is entangled. I'm just going to do it for larger than 1. To do this, observe that under local operations and classical communication, we can convert the state $\rho$ to $$ \rho'=\frac14\left(\rho+X\otimes X\rho X\otimes X+Y\otimes Y\rho Y\otimes Y+Z\otimes Z\rho Z\otimes Z\right), $$ and the entanglement is non-increasing. So, if I can prove that this state is entangled, $\rho$ is also entangled. However, $\rho'$ has the same values of $c_{ii}$ as $\rho$ did, but all off-diagonal elements are 0. Now, we know the eigenvalues of $\rho'$ are non-negative (because it's a state). These are $$ 1+c_{33}\pm(c_{11}-c_{22}), \qquad 1-c_{33}\pm(c_{11}+c_{22}) $$ which are 4 of the 8 possible ways of combining the $c_{ii}$ with different signs. Next, we take the partial transpose of $\rho'$. The state is entangled if this has a negative eigenvalue. But the partial transpose just changes $c_{22}\mapsto-c_{22}$. So the eigenvalues it gives us are the other 4 ways of combining the $c_{ii}$. So, if the smallest number from combining the $c_{ii}$ is negative, the state is entangled. The smallest number is always $$ 1-|c_{11}|-|c_{22}|-|c_{33}|<0, $$ which should look familiar! On the other hand, if that number is positive or 0, $\rho'$ is separable. Note that this does not imply that $\rho$ is separable.

The flip side of this is to try and prove that the state is separable when $\sum|c_{ii}|<1$. In fact, this calim if false, as I can prove by counter-example. Consider the state $$ \rho=\frac12\begin{bmatrix} 1-p & 0 & 0 & i(1-p) \\ 0 & p & ip & 0 \\ 0 & -ip & p & 0 \\ -i(1-p) & 0 & 0 & 1-p \end{bmatrix}. $$ This state has $c_{33}=1-2p$ and $c_{11}=c_{22}=0$. Hence, by your criteria, the state is not entangled (with the possible exception of $p=0,1$). However, if you apply the partial transpose criterion, you will see that it is entangled for all $p\neq \frac12$. Thus, your condition is wrong.

How might you fix this? I believe that there always exists local unitaries $U$ and $V$ which can transform the two-body terms into a standard form which means that the bottom-right $3\times 3$ matrix of the $c$ coeficients is diagonal. If my memory is correct on this matter, then you can replace the three $|c_{ii}|$ terms with the three singular values of that bottom-right $3\times 3$ matrix. The proof for detection of etnanglement follows as before. Since you almost have a form $\rho'$ for your state, you might believe the detection of separability at this point as well. Particularly as the only variation is in local terms which tend to add to the separable nature of the state. But I don't immediately hae a proof.

  • $\begingroup$ Thank you so much. I'm a bit puzzled by the first part of your answer, but the counter example is convincing! $\endgroup$
    – rhundt
    Jun 2, 2022 at 14:35

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