What is the shape of states of two qubits?

Joint work with P. Pikul, S. Weis, K. Życzkowski, and I.M. Spitkovsky.

The geometry of the set of density operators of size $d$ gets increasingly complex as the dimension grows: see the note about mixed states for details. Conceptually, while the set of states of a single qubit forms a (Bloch) Ball, the set of states of the next nontrivial case, a qutrit ($d=3$), the set is parameterized by eight independent real numbers with additional convoluted polynomial constraints. For two qubits, $d=4$, and any Hermitian operator can be parameterized via 16 real parametrs; for a density operator $\Tr \rho=1$ and consequently $r_{00}=\frac14$ in the following decomposition:

\begin{equation}\begin{aligned} \rho =& \sum_{i,j\in \{0,1,2,3\}} r_{ij} \sigma_i \otimes \sigma_j, \end{aligned} \end{equation}

where $\otimes$ is the Kronecker product and

\begin{equation}\begin{aligned} \sigma_0 &= \II_2,& \sigma_1 = X=\begin{pmatrix}0&1\\1&0\end{pmatrix},\\ \sigma_2&=Y=\begin{pmatrix}0&-i\\i&0\end{pmatrix}, &\sigma_3=Z=\begin{pmatrix}1&0\\0&1\end{pmatrix}. \end{aligned} \end{equation}

We are left with 15-dimensional set, with the remaining constraint that $\rho$ is positive semidefinite. In order to get a better grasp at the geometry of this set, we will resolve to study of its images under linear maps: the joint numerical ranges. The numerical ranges are presented as interactive plots, the flat faces are denoted by their outlines.

Joint numerical ranges

Suppose a sequence of Hermitian operators $X_1, \ldots, X_k$ is fixed. If a state $\rho$ is picked, one could calculate the vector of expectation values:

\begin{equation}\begin{aligned} \rho \mapsto \left(\Tr \rho X_1, \ldots, \Tr \rho X_k\right). \end{aligned} \end{equation}

The joint numerical range $L(X_1,\ldots,X_k)$ is the image of the set of states under the above map: it is the set of all possible vectors of expectation values, for all possible physical states:

\begin{equation}\begin{aligned} L(X_1, \ldots, X_k) := \left\{\left(\Tr \rho X_1, \ldots, \Tr \rho X_k\right) : \Tr \rho=1, \rho=\rho^\dagger, \rho\succeq0 \right\}. \end{aligned} \end{equation}

In the article published in 2018, the classification of possible shapes of $L(X_1,X_2,X_3)$ was developed for the case when $X_i$ are Hermitian matrices of size 3. The classification can be thought as based on the following conceptual points:

  1. Joint numerical ranges are convex and compact (being linear images of convex, compact, and finite-dimensional sets of density operators), and the only nontrivial geometry is contained in their boundary.
  2. Generically (for matrices sampled from probability distributions having support in the entire set of hermitian matrices), the boundary shape is described by a polynomial variety and is smooth: it contains no cusps, segments or flat parts (exposed faces).
  3. In nongeneric cases, nontrivial boundary geometry emerges through their exceptional parts: the nonanalytic features mentioned in the previous point.
  4. Every flat part of the boundary is a nontrivial 1 maximizer of some linear functional $\vec x\mapsto \vec n \cdot \vec x$ over points $\vec x$ in the numerical range $L$.

The existence of such a maximizer is a consequence of the eigenspace of maximal eigenvalue of $\sum_{i=1}^3 n_i X_i$ being doubly degenerate 2. Therefore, any flat part is an image of a qubit state space: is is an ellipse, possibly degenerate to a point.

  1. This insight provides further geometric relations between the flat boundary parts, e.g. two flat parts must share a common point, since two-dimensional spaces (related to preimages of the flat parts) in three-dimensional space must intersect.

Since some time, we are attempting to develop a similarly-structured classification of the 3D numerical ranges of three operators of size $d=4$. This can be understood as a partial classification of the shapes of two qubit states, restricted to 3-dimensional projections of a 15-dimensional set. Several additional complexities arise as compared to the $d=3$ case: the faces can be nonelliptic (corresponding to triple degenerate eigenspaces embedded in the 4-dimensional 2-qubit Hilbert space). The relations between different faces also become nontrivial, since two ellipses may possibly not intersect.

Here we present partial results found by us: examples of nontrivial geometry of such numerical ranges, and some additional observations.

Non-elliptical faces

As mentioned previously, there is an entire new category of flat faces: the nonelliptical ones. Remember that the elliptical shape of flat faces for $d=3$ comes from the fact that they are images of qubit state space embedded in the qutrit states. The ellipse is, in a way, just a squished Bloch ball. Similarly, flat nonelliptical faces for $d=3$ are images of qutrit state space. And shapes of those are known. In the following examples we present the possible shapes found through numerics and educated guesses; they are grouped by the total number of nonelliptical faces.

Due to an argument similar to the qutrit case 3, there exist additional geometric relations between the flat faces. If a nonelliptic face is present, any other flat part must share at least a single point with it, reducing the possible geometries. If there exist three nonelliptic faces, there must exist a corner point, and consequently the operators $X_1, X_2, X_3$ decompose in some basis to

\begin{equation} X_i = \begin{pmatrix} X'_i & 0\\0&x_i\end{pmatrix}, \end{equation}

where $X'_i$ is a 3-by-3 Hermitian matrix and $x_i$ is a real number. Thus, the numerical range $L(X_1, X_2, X_3)$ must decompose into a convex hull of $L(X'_1,X'_2,X'_3) \cup \{(x_1, x_2, x_3)\}$, and its geometry is determined completely by the already known classification of qutrit numerical ranges. Here we discuss the nontrivial examples, for which no such reduction takes place.

One nonelliptical face

\begin{equation} (X_1,X_2,X_3)=\left(\left( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & i \\ 0 & 1 & -i & 0 \\ \end{array} \right)\left( \begin{array}{cccc} -1 & i & 0 & i \\ -i & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ -i & 0 & 1 & 0 \\ \end{array} \right)\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 \\ \end{array} \right)\right) \end{equation}

Two nonelliptical faces, zero ellipses

A first example in which the face-face intersection is visible. As mentioned previously, if in the joint numerical range of matrices of size $d$ a flat face is present, it is an image of a set of states of lower dimensionality $ d'< d $ embedded in the qudit state space. Consequently, if two faces are present, corresponding to $d'_1$ and $d'_2$, they must intersect nonempty if $d'_1+d'_2 > d$, and the interesection corresponds to another subset with dimension at least $ d'_1 +d'_2 -d $. Here, if an nonelliptical face is present, it is an image of an embedded qutrit, and intersection of two nonelliptical faces must be a segment (unless the joint numerical range is degenerate), and any pair of nonelliptical and elliptical face must intersect at a point.

\begin{equation} (X_1,X_2,X_3)=\left(\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right),F X_1 F^\dagger,\left( \begin{array}{cccc} 0 & \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{2} & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \frac{1}{2} & 0 & 0 & 0 \\ \end{array} \right)\right), \end{equation} where $F$ is the Fourier matrix: \begin{equation} F=\frac12\left( \begin{array}{cccc} 1 & 1 & 1 & 1 \\ 1 & i & -1 & -i \\ 1 & -1 & 1 & -1 \\ 1 & -i & -1 & i \\ \end{array} \right). \end{equation}

Two nonelliptical faces, one ellipse

The ellipse must intersect with the other two faces.

\begin{equation} (X_1,X_2,X_3)=\left(\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right),\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right),F \left( \begin{array}{cccc} -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) F^\dagger\right). \end{equation}

Two nonelliptical faces, disconnected ellipses

Similarly, both of the ellipses must intersect with both nonelliptical faces. Here, the ellipses do not share a point:

\begin{equation} (X_1,X_2,X_3)=\left( \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right),F X_1 F^\dagger,\left( \begin{array}{cccc} 0 & 0 & 0 & \frac{1}{2} \\ 0 & 0 & \frac{1}{2} & 0 \\ 0 & \frac{1}{2} & 0 & 0 \\ \frac{1}{2} & 0 & 0 & 0 \\ \end{array} \right) \right). \end{equation}

… and here, they do:

\begin{equation} (X_1,X_2,X_3)=\left( \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right),F X_1 F^\dagger,\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & \frac{1}{2} & 0 \\ 0 & \frac{1}{2} & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right) \right). \end{equation}

Two nonelliptical faces, three disconnected ellipses

\begin{equation} (X_1,X_2,X_3)=\left(\left( \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right),\left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right),\frac12 F \left( \begin{array}{cccc} -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right) F^\dagger \right). \end{equation}

Examples dual to transverse spectrahedral classification

Examples here are based on the classification of transverse quartic spectrahedra: for a given spectrahedron, a linear transformation of matrices can be made to transform it into a form dual to a numerical range. These transformations are rather complex, and while the end matrices can be written in a closed form, it's not particularly simple. So, please find the numerical approximation to the defining matrices in the CSV file: each block of four lines is one of the matrices. This example is dual to the $\rho=\sigma=10$ transverse spectrahedron example; thus, it has 10 exposed faces (which must be ellipses) in the boundary.

The exposed faces were determined numerically, by sampling points4 $\vec f \in \partial L(F_1, F_2, F_3)$, determining the corresponding normal vectors (which can be expressed as a semidefinite optimization), and finally grouping the normal vectors by their $N$-digit decimal approximations (so the vectors $\vec n$ and $\vec n'$ are grouped if and only if the first $N$ digits of $n_1$ are equal to first $N$ digits of $n_1'$ etc.). Then, the groups are analyzed. Usually, most groups emerging from this numerical procedure are singletons (i.e. most normal vectors are far apart), there are some groups of $2-10$ elements, and 10 groups containing a significant fraction of the sampled points. So, let's see the numbers! Here, $50000$ points from $\partial L$ were samples, the corresponding normal vectors were determined, and subsequently grouped into groups of matching first $N=3$ digits5, the most populous groups have the following number of elements

\begin{equation} ({\bf6202,3293,2091,1246,1199,860,645,486,292,284},7,6,6,6,6,5,5,5,5,5,\ldots). \end{equation}

This striking falloff in group population after first $10$ elements more or less justifies this numerical procedure, but it is always possible that it is just a numerical coincidence. However, let's just take these 10 normal vectors (or rather, the averages of the determined groups), and try to numerically determine the faces. They are clearly visibly, and indeed very fragile to any changes to normal vectors: a relative perturbation of order $10^{-3}$ destroys the numerically determined faces.

With matrices defined floating point numbers, there is really no alternative to the numerical face determination: tne exact degeneracy of extremal eigenspaces (needed for the existence for a flat part) does not even make sense in this context. So, a numerical approximation of the face being a maximizer of $\vec x \mapsto \vec n \cdot \vec x$ is calculated by sampling points from normals $\vec n'$ such that $\lvert \vec n-\vec n'\rvert<\varepsilon$, and here $\varepsilon\approx 10^{-6}$.

We reserve the numerical approximation for this case only; it is necessary due to the complexity of finding strictly defined three 4-by-4 matrices with the right properties to produce 10 ellipses in the boundary of the numerical range. However, for lower number of ellipses the strict defining matrices can be found.

Four ellipses

For the following matrices, the connectivity structure of ellipses is isomorphic to the cycle graph $C_4$:

\begin{equation} (X_1,X_2,X_3)=\left( \left( \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right), \left( \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ \end{array} \right), \left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ \end{array} \right) \right). \end{equation}

Five ellipses

For the following matrices, the connectivity structure of ellipses is isomorphic to the wheel graph $W_5$:

\begin{equation} (X_1,X_2,X_3)=\left(\left( \begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right), \left( \begin{array}{cccc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \end{array} \right), \left( \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ \end{array} \right) \right). \end{equation}

Asymmetric dice composed of 6 ellipses

The ellipses in the following example are completely disconnected. This is not the case in general; there exist similar examples for which the connectivity graph of the flat parts is not empty.

\begin{equation} (X_1,X_2,X_3)=\left(\frac{1}{\sqrt2}\left( \begin{array}{cccc} 1 & 1 & 0 & 0 \\ 1 & -1 & 0 & 0 \\ 0 & 0 & -1 & 1 \\ 0 & 0 & 1 & 1 \\ \end{array} \right), \left( \begin{array}{cccc} 0 & -1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ \end{array} \right), \left( \begin{array}{cccc} 0 & 0 & 0 & -1 \\ 0 & 0 & i & 0 \\ 0 & -i & 0 & 0 \\ -1 & 0 & 0 & 0 \\ \end{array} \right) \right). \end{equation}

  1. That is: it contains more than a single point. ↩

  2. If the eigenspace was triple degenerate, the operator $\sum_{i=1}^3 n_i X_i$ is proportional to identity and the problem has a reduced effective dimension. The classification of this case is known.  ↩

  3. A qutrit and qubit state space embedded in a two-qubit space must share a common point, and thus any pair of elliptic and nonelliptic face must share at least a single common point. Similarly, intersection of any pair of qutrit space in a two-qubit space must be an embedded qubit state space; consequently, any pair of two nonelliptic flat faces must share at least a segment. ↩

  4. Such that $\vec f=\vec d t$, where $\vec d$ is sampled uniformly from the 2-sphere. ↩

  5. For instance, the vectors $(0.{\bf12}3,0.{\bf45}6,0.{\bf78}9)$ and $(0.{\bf12}0,0.{\bf45}9,0.{\bf78}5)$ would be placed in a single group for $N=2$: they are identical after rounding to the first $N=2$ digits. ↩