# Bell-diagonal state

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

pI∣Φ + ⟩⟨Φ + ∣ + px∣Ψ + ⟩⟨Ψ + ∣ + py∣Ψ − ⟩⟨Ψ − ∣ + pz∣Φ − ⟩⟨Φ −
.

In matrix form it looks like

$$\frac{1}{2} \begin{bmatrix} p_I + p_z &amp; 0 &amp; 0 &amp; p_I - p_z \\ 0 &amp; p_x + p_y &amp; p_x - p_y &amp; 0 \\ 0 &amp; p_x - p_y &amp; p_x + p_y &amp; 0 \\ p_I - p_z &amp; 0 &amp; 0 &amp; p_I + p_z \\ \end{bmatrix}$$
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
• Any 2-qubit state where both qubits are maximally mixed, ρA = ρB = I/2, is bell-diagonal in some local basis. I.e. there exist local unitaries U1, U2 such that U1 ⊗ U2ρABU1 †  ⊗ U2 †  is bell-diagonal.quant-ph/9607007

### Visualization

thumb

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 β1β3 can be plotted in 3D. In these coordinates the Bell states are located at

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

thumb

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 pi to γi is its own inverse.

Category:Quantum States