The two-qubit binegativity lower bound at fixed negativity

Abstract

For every two-qubit density matrix, the binegativity is bounded below by the Girard–Gour curve at its negativity. The negative eigenvector of the partial transpose has a Schmidt product constrained from below by positivity of the density matrix.

Definition 1.1 (The lower bound of Girard and Gour).

$$(claim)\iff (\forall \sigma :\mathrm{Matrix}\left((\mathrm{Fin}\left(2\right)\times \mathrm{Fin}\left(2\right)),(\mathrm{Fin}\left(2\right)\times \mathrm{Fin}\left(2\right)),\mathbb{C}\right),(\mathrm{IsDensity}\left(\sigma \right))\Rightarrow (\mathrm{let}\nu :\mathbb{R}=2\cdot \Re \mathrm{Tr}\left(((\sigma {)}^{\Gamma }{)}_{-}\right);\mathrm{let}c:\mathbb{R}=\sqrt{2\cdot \nu \cdot (\nu +1)}-\nu ;\frac{\nu }{2}\cdot (1+\frac{c}{\sqrt{(1-c{)}^2+(c{)}^2}})\le \left(N_2\right)\left(\sigma \right)))$$

Formalization. D5/S3/Quantum/Entanglement/TwoQubitBinegativityLowerBound.claim (✓ std3).

Citation. Mark W. Girard; Gilad Gour (2017). The binegativity of two qubits. DOI: 10.48550/arXiv.1701.02724. URL: https://arxiv.org/abs/1701.02724v3.

Commentary.

Girard and Gour, Eq. (8), conjecture the displayed bound for every two-qubit density matrix sigma. A density matrix is positive semidefinite with trace one. Here nu is its negativity, c = sqrt(2 nu (nu + 1)) - nu, and N_2 is its binegativity. The symbol c denotes this function of nu, rather than the concurrence of an arbitrary state.

For a self-adjoint matrix X, X_- is its positive semidefinite negative component, with X = X_+ - X_- and X_+ X_- = 0. All traces below are taken through their real parts, as in the negativity appearing in the statement. The binegativity is $\forall \sigma :\mathrm{Matrix}\left((\mathrm{Fin}\left(2\right)\times \mathrm{Fin}\left(2\right)),(\mathrm{Fin}\left(2\right)\times \mathrm{Fin}\left(2\right)),\mathbb{C}\right),\left(N_2\right)\left(\sigma \right)=\Re \mathrm{Tr}\left(((\sigma {)}^{\Gamma }{)}_{-}\right)+2\cdot \Re \mathrm{Tr}\left(((((\sigma {)}^{\Gamma }{)}_{-}{)}^{\Gamma }{)}_{-}\right)$.

The partial transpose is on the second qubit: $\forall \sigma :\mathrm{Matrix}\left((\mathrm{Fin}\left(2\right)\times \mathrm{Fin}\left(2\right)),(\mathrm{Fin}\left(2\right)\times \mathrm{Fin}\left(2\right)),\mathbb{C}\right),\forall i:\mathrm{Fin}\left(2\right),\forall j:\mathrm{Fin}\left(2\right),\forall k:\mathrm{Fin}\left(2\right),\forall l:\mathrm{Fin}\left(2\right),((\sigma {)}^{\Gamma }{)}_{(i,j),(k,l)}=(\sigma {)}_{(i,l),(k,j)}$.

Theorem 1.2 (The lower bound holds for every two-qubit state).

$$\forall \sigma :\mathrm{Matrix}\left((\mathrm{Fin}\left(2\right)\times \mathrm{Fin}\left(2\right)),(\mathrm{Fin}\left(2\right)\times \mathrm{Fin}\left(2\right)),\mathbb{C}\right),(\mathrm{IsDensity}\left(\sigma \right))\Rightarrow (\mathrm{let}\nu :\mathbb{R}=2\cdot \Re \mathrm{Tr}\left(((\sigma {)}^{\Gamma }{)}_{-}\right);\mathrm{let}c:\mathbb{R}=\sqrt{2\cdot \nu \cdot (\nu +1)}-\nu ;\frac{\nu }{2}\cdot (1+\frac{c}{\sqrt{(1-c{)}^2+(c{)}^2}})\le \left(N_2\right)\left(\sigma \right))$$

Proof. Machine-checked in Lean as D5/S3/Quantum/Entanglement/TwoQubitBinegativityLowerBound.result (✓ std3). ∎

Resolves. Problems/girard-2017-binegativity-lower-bound (proved) by D5/S3/Quantum/Entanglement/TwoQubitBinegativityLowerBound.result.

Source. Repository-derived.

Acknowledgement. Mark W. Girard; Gilad Gour (2017). The binegativity of two qubits. DOI: 10.48550/arXiv.1701.02724. URL: https://arxiv.org/abs/1701.02724v3.

Commentary.

The partial transpose sigma^Gamma has at most one negative eigenvalue: every two-dimensional subspace of the two-qubit space contains a nonzero product vector, and the quadratic form of sigma^Gamma is nonnegative on product vectors. Thus a two-dimensional negative eigenspace is impossible. If there is a negative eigenvalue -lambda, local unitaries put its unit eigenvector in Schmidt form alpha |00> + beta |11>, with alpha and beta positive and alpha^2 + beta^2 = 1. These unitaries preserve both the negativity N and the binegativity N_2. In these coordinates write a = sigma_(00,00), d = sigma_(11,11), x = a + d, and u = -Re sigma_(01,10). The eigenvalue equations give u^2 = (a + lambda)(d + lambda), nu = 2 lambda, and (2 alpha beta)(x + nu) = 2u. Positivity on |01> + |10> gives 2u <= 1 - x; equivalently, one may use |01> - |10> after a local phase change. Consequently nu (2x + nu) <= (1 - x)^2, so x <= 1 - c. The same eigenvalue equations imply (2 alpha beta)^2 (x + nu)^2 >= nu (2x + nu), which forces 2 alpha beta >= c/sqrt((1 - c)^2 + c^2). Since (sigma^Gamma)_- = lambda |psi><psi|, its partial transpose has the single negative eigenvalue -lambda alpha beta. Therefore N_2(sigma) = (nu/2)(1 + 2 alpha beta), and the lower bound follows. When sigma^Gamma has no negative eigenvalue, its negative component is zero, so nu = c = N_2(sigma) = 0 and both sides vanish.

References

• Truth anchor: D5/S3/Quantum/Entanglement/TwoQubitBinegativityLowerBound.claim
• Truth anchor: D5/S3/Quantum/Entanglement/TwoQubitBinegativityLowerBound.result
• Dependency: D5/S3/Quantum/Dynamics/KickedIsingNegativityRefutation
• Dependency: D5/S3/Quantum/Entanglement/SeparableNumericalRangeRatio
• Dependency: D5/S3/Quantum/Entanglement/TwoQubitBinegativityUpperBoundRefutation
• Dependency: D5/S3/Quantum/PureState/PureStateHandshake