How is a two qubit mixed state represented in the form of Bloch vector?



This is impossible in general, because the space of two-qubit mixed states is significantly more complicated than the Bloch sphere.


The space $D_N$ of all mixed states of an $N$-level quantum system has $N^2-1$ real dimensions. For a qubit $D_2$ has only three dimensions and can be readily visualized as a subset of $\mathbb{R}^3$. However, for two qubits $N=2^2 = 4$ and $D_4$ is $15$-dimensional. Consequently, there is no visualization aid analogous to the Bloch sphere that allows us to imagine the full space all at once.

Algebraic representation

Nevertheless, we can use other techniques to study $D_4$. Algebraically, the Bloch sphere can be thought of as a map that takes a state $\rho_1 \in D_2$ to $x_{\rho_1} \in \mathbb{R}^3$ such that

$$ \rho_1 = \frac{I + x_{\rho_1} \cdot \sigma_1}{2}\tag1 $$

where $\sigma_1 = (\sigma_x, \sigma_y, \sigma_z)^T$ is a $3$-vector of Pauli operators. This representation is unique because $\{I, \sigma_x, \sigma_y, \sigma_z\}$ is a basis of the real vector space of Hermitian operators in two dimensions. Similarly, we can map every two-qubit state $\rho_2\in D_4$ to a $x_{\rho_2} \in \mathbb{R}^{15}$ such that

$$ \rho_2 = \frac{I + \sqrt{3}\,x_{\rho_2} \cdot \sigma_2}{2}\tag2 $$


$$ \sigma_2 = \begin{pmatrix} I \otimes \sigma_x\\ I \otimes \sigma_y\\ I \otimes \sigma_z\\ \sigma_x \otimes I\\ \sigma_x \otimes \sigma_x\\ \sigma_x \otimes \sigma_y\\ \sigma_x \otimes \sigma_z\\ \sigma_y \otimes I\\ \sigma_y \otimes \sigma_x\\ \sigma_y \otimes \sigma_y\\ \sigma_y \otimes \sigma_z\\ \sigma_z \otimes I\\ \sigma_z \otimes \sigma_x\\ \sigma_z \otimes \sigma_y\\ \sigma_z \otimes \sigma_z\\ \end{pmatrix} $$

is a $15$-vector of non-identity two-qubit Pauli operators and the scaling factor $\sqrt{3}$ ensures that $\|x\|_2 \le 1$ for convenience. As before, $x_{\rho_2}$ is unique for a given state $\rho_2$.

Back to geometry

An immediate consequence of $(2)$ that distinguishes the shape of $D_4$ from $D_2$ is that $D_4$ is not symmetric with respect to the maximally mixed state. This is a first sign that the shape of $D_4$ is more complicated than a sphere.

Even though $D_4$ cannot be visualized all at once, we can visualize various cross-sections through $D_4$. Some $3$-dimensional cross sections turn out to be balls, some are tetrahedrons and others have more complicated shapes.

See the paper "An elementary introduction to the geometry of quantum states with pictures" or the book "Geometry of Quantum States: An Introduction to Quantum Entanglement" by Ingemar Bengtsson and Karol Życzkowski for more details.

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    $\begingroup$ I would also suggest the review paper published by Ingemar Bengtsson and Karol Życzkowski "An Introduction to Quantum Entanglement: a Geometric Approach" arxiv.org/pdf/quant-ph/0606228.pdf $\endgroup$ – KAJ226 Jan 27 at 19:42
  • $\begingroup$ In the equation mentioned above, 𝑥𝜌2∈ℝ15, should the radius be equal to the Euclidean distance in 15 dimensional space? I mean say (𝑥𝜌2 ~ r1, r2, ....r15) then the radius is (r1^2 + r2^2 + ... + r15^2)^(1/2)? Is my understanding correct? $\endgroup$ – Ganesh M Feb 9 at 12:42
  • $\begingroup$ Remember that the 15-dimensional space of mixed two-qubit states is not a sphere, so the set does not have a radius in the sense that sphere does. However, the set is contained in a sphere of radius $1$. And yes, this radius is computed using the usual Euclidean distance formula $r = \|x_{\rho_2}\|_2 = \sqrt{x_1^2 + \dots + x_{15}^2}$. $\endgroup$ – Adam Zalcman Feb 9 at 15:51

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