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 (
*p*_{1},*p*_{2},*p*_{3},*p*_{4}) 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*) = −*x*log_{2}(*x*) − (1 −*x*)log_{2}(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(2*p*_{max})

- Relative entropy of entanglement:
- 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*U*_{1},*U*_{2}such that*U*_{1}⊗*U*_{2}*ρ*_{AB}*U*_{1}^{ † }⊗*U*_{2}^{ † }is bell-diagonal.quant-ph/9607007

### Visualization

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 *β*_{0} will always be equal to 1/2, and *β*_{1}…*β*_{3} can be plotted in 3D. In these coordinates the Bell states are located at

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

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

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.