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The determinant of a two-qubit $4 \times 4$ density matrix--that is, a Hermitian, nonnegative definite matrix with unit trace--lies between $0$ and $(\frac{1}{2})^8$. (A "pure state" has determinant zero, and the fully mixed [classical] state, with $\frac{1}{4}$'s on its diagonal, determinant $(\frac{1}{2})^8$.)

The determinant of the partial transpose (transpose in place the four $2 \times 2$ blocks) of such a matrix (nonnegative values indicating separability) lies between $-(\frac{1}{2})^4$ and $(\frac{1}{2})^8$. (The minimum is achieved by a "Bell state" and the maximum, again by the fully mixed state.)

What is the range (upper and lower limits) of the difference of these two determinants?

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    $\begingroup$ The value of (determinant of state) - (determinant of partial transpose) can vary from 0 to (1/2)^4 when the state in question is pure (rank 1). There is a rank-3 state (the rank-3 projection onto the orthogonal complement of the Bell state) that achieves a value of -1/432. Numerics suggest that the lower bound -1/432 and the upper bound (1/2)^4 are optimal, but proving that may be difficult. $\endgroup$
    – Nathaniel Johnston
    Commented Feb 19, 2014 at 0:53

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