The Peres-Horodecki criteria for a 2*2 state states that if the smallest eigenvalue of the partial transpose of the state is negative, it is entangled, else it is separable.

According to this paper (page 4, left side), the following is an equivalent formulation to express the above criterion.

Assume the matrix in question looks like this:

$$\begin{bmatrix} \rho_{00,00} & \rho_{00,01} & \rho_{00,10} & \rho_{00,11} \\ \rho_{01,00} & \rho_{01,01} & \rho_{01,10} & \rho_{01,11} \\ \rho_{10,00} & \rho_{10,01} & \rho_{10,10} & \rho_{10,11} \\ \rho_{11,00} & \rho_{11,01} & \rho_{11,10} & \rho_{11,11} \\ \end{bmatrix}$$

Consider the following three determinants:

$$ W_2 = \begin{vmatrix} \rho_{00,00} & \rho_{01,00} \\ \rho_{00,01} & \rho_{01,01} \\ \end{vmatrix} $$

$$W_3 = \begin{vmatrix} \rho_{00,00} & \rho_{01,00} & \rho_{00,10} \\ \rho_{00,01} & \rho_{01,01} & \rho_{00,11} \\ \rho_{10,00} & \rho_{11,00} & \rho_{10,10} \\ \end{vmatrix}$$

$$W_4 = \begin{vmatrix} \rho_{00,00} & \rho_{01,00} & \rho_{00,10} & \rho_{01,10} \\ \rho_{00,01} & \rho_{01,01} & \rho_{00,11} & \rho_{01,11}\\ \rho_{10,00} & \rho_{11,00} & \rho_{10,10} & \rho_{11,10}\\ \rho_{10,01} & \rho_{11,01} & \rho_{10,11} & \rho_{11,11}\\ \end{vmatrix}$$

Notice that $W_4$ is the determinant of the partial transpose of the matrix and $W_3$ and $W_2$ are the first 3*3 and 2*2 elements of the partial transpose.

The condition is if $W_2 \geq 0$ and ($W_3 < 0$ or $W_4 < 0$), then the state is entangled. If not, it is separable.

How are these two equivalent?

Also, can this method be extended to ensure the smallest eigenvalue is greater than any $x$, where $x$ is not necessarily 0?

  • $\begingroup$ You're stating this as an equivalent to the standard formulation. What's your source? $\endgroup$
    – DaftWullie
    Jun 15, 2019 at 7:11
  • $\begingroup$ Incidentally, the fact that you've diagonalised $W_4$ does not mean that you've diagonalised $W_2$ or $W_3$. For example, you cannot just take a diagonal density matrix - obviously it's not entangled and therefore certainly should not fit the criteria. $\endgroup$
    – DaftWullie
    Jun 15, 2019 at 7:12
  • $\begingroup$ @DaftWullie, added source, and removed the diagonalisation. My bad. A diagonal density matrix in a separable basis should be separable, and is therefore correctly identified by this criteria, right? $\endgroup$ Jun 15, 2019 at 10:39

1 Answer 1


This is called Sylvester's Criterion. There's plenty of information available once you have the name. The linked wikipedia article contains a proof.

Strictly, Sylvester's Criterion requires that $W_2,W_3,W_4> 0$ for the state to be positive under the partial transpose. However, for a density matrix, $W_2$ is always positive semi-definite. This is because you could certainly define a density matrix of the form $$ \left(\begin{array}{cccc} \rho_{00,00} & \rho_{01,00} & 0 & 0 \\ \rho_{00,10} & \rho_{01,01} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{array}\right) $$ up to some rescaling. This has the same determinant $W_2$ but is separable, and hence unaffected by the partial transpose. Hence, non-positivity is determined by looking at a violation caused by $W_3$ or $W_4$.

I agree with a query in the comments that it is not immediately obvious that it is relevant for detecting negativity (I will continue to think about it...). If it is a problem, we should be able to come up with a counter-example.

The method is easily extended to any smallest eigenvalue. Simply reduce the matrix by $x$ times the appropriately sized identity matrix, shifting the min eigenvalue from $x$ to 0.

  • $\begingroup$ According to the Wikipedia page, a Hermitian matrix is positive semi-definite if all principal minors are non-negative, not just the leading ones. If we look at a density matrix, there are 10 principal minors, 4 1*1 ones, but they're all $\geq$ 0, 3 2*2 ones, the extremes of which are $\geq$ 0 , 2 3*3 ones, and 1 4*4 one. By this condition, if we want to find whether the matrix is entangled, we need to check if any of the 4 determinants (1 2*2, 2 3*3 and 1 4*4 are negative), right? This isn't equal to the condition given in the paper? $\endgroup$ Jun 17, 2019 at 11:10

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