A Bell diagonal state is a 2-qubit state that is diagonal in the Bell basis. In other words, it is a mixture of the four Bell states. It can be written as

p_I∣Φ⁺⟩⟨Φ⁺∣ + p_x∣Ψ⁺⟩⟨Ψ⁺∣ + p_y∣Ψ⁻⟩⟨Ψ⁻∣ + p_z∣Φ⁻⟩⟨Φ⁻∣
.

In matrix form it looks like

$$\frac{1}{2} \begin{bmatrix} p_I + p_z & 0 & 0 & p_I - p_z \\ 0 & p_x + p_y & p_x - p_y & 0 \\ 0 & p_x - p_y & p_x + p_y & 0 \\ p_I - p_z & 0 & 0 & p_I + p_z \\ \end{bmatrix}$$
where the matrix is in the computational basis.

Because of the simple structure, many questions that are difficult to answer for general 2-qubit states simplify when they are restricted to Bell-diagonal states.

Properties

• The weights (p₁, p₂, p₃, p₄) can be permuted to any other order by local unitaries. Unilateral π rotation around the x-, y- and z-axes and bilateral π/2 rotations around the same axes are sufficient for this.
• A Bell-diagonal state is separable if all the probabilities are less or equal to 1/2.
• Many entanglement measures have a simple formulas for entangled Bell-diagonal states
  • Relative entropy of entanglement: E_r = 1 − h(p_max), where h is the binary entropy function h(x) =  − xlog₂(x) − (1 − x)log₂(1 − x)quant-ph/9702027
  • Entanglement of formation: $E_f = h\left(1/2 + \sqrt{p_{max}(1-p_{max})}\right)$quant-ph/9604024
  • Negativity: N = p_max − 1/2
  • Log-negativity: E_N = log(2p_max)
• Any 2-qubit state where both qubits are maximally mixed, ρ_A = ρ_B = I/2, is bell-diagonal in some local basis. I.e. there exist local unitaries U₁, U₂ such that U₁ ⊗ U₂ρ_ABU₁^† ⊗ U₂^† is bell-diagonal.quant-ph/9607007

Visualization

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The set of Bell-diagonal states can be visualized as a tetrahedron where the four Bell states are the corners. The following change of coordinate system makes the plotting of states easy:

$$\beta_0 = \frac{1}{2} ( p_I + p_x + p_y + p_z )$$

$$\beta_1 = \frac{1}{2} ( p_I - p_x - p_y + p_z )$$

$$\beta_2 = \frac{1}{\sqrt{2}} ( p_I - p_z )$$

$$\beta_3 = \frac{1}{\sqrt{2}} ( p_x - p_y )$$
The coordinate β₀ will always be equal to 1/2, and β₁…β₃ can be plotted in 3D. In these coordinates the Bell states are located at

$$|\Phi+ \rangle: \left(\frac{1}{2},\frac{1}{\sqrt{2}}, 0 \right)$$
, $|\Psi+ \rangle: \left(-\frac{1}{2},0,\frac{1}{\sqrt{2}}\right)$, $|\Psi- \rangle: \left(-\frac{1}{2},0,-\frac{1}{\sqrt{2}}\right)$, $|\Phi-\rangle: \left(\frac{1}{2},-\frac{1}{\sqrt{2}},0\right)$

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Another useful coordinate system is the one where the corners of the tetrahedron lie in four of the corners of a cube, with the edges going along the diagonals of the cube's faces.

$$\gamma_0 = \frac{1}{2}(p_I + p_x + p_y + p_z)$$

$$\gamma_1 = \frac{1}{2}(p_I - p_x - p_y + p_z)$$

$$\gamma_2 = \frac{1}{2}(p_I - p_x + p_y - p_z)$$

$$\gamma_3 = \frac{1}{2}(p_I + p_x - p_y - p_z)$$
In these coordinates, the Bell states are situated at

$$|\Phi+ \rangle: (\frac{1}{2},\frac{1}{2},\frac{1}{2})$$
, $|\Psi+ \rangle: (-\frac{1}{2},-\frac{1}{2},\frac{1}{2})$, $|\Psi- \rangle: (-\frac{1}{2},\frac{1}{2},-\frac{1}{2})$, $|\Phi-\rangle: (\frac{1}{2},-\frac{1}{2},-\frac{1}{2})$

The β-coordinate system has the advantage that two of the edges are parallel to axes of the coordinate system. The γ-coordinate system on the other hand inherits more of the symmetry from the cube. Both coordinate transformations are orthogonal, and the transformation from p_i to γ_i is its own inverse.

Category:Quantum States