Is there a rank-1 POVM on qubits with $4$ outcomes that is not extremal but such that its resulting shape on the Bloch sphere has a non-zero volume?

Question

Let us consider a rank-1 POVM acting on qubits with $4$ outcomes (that is, all its elements are rank-1). Furthermore, let us assume that this POVM is unbiased, meaning that $\mathrm{Tr}\left[M_i\right]=\frac12$ for all $i\in[4]$. In particular, it is possible to write this POVM as $\left\{\frac12\left|\psi_i\middle\rangle\!\middle\langle\psi_i\right|\right\}$, and thus to represent such a POVM as a list of $4$ quantum states.

One particular example of such a POVM is the SIC-POVM. If we plot the states of this POVM on the Bloch sphere and "triangulize" it, this forms a tetrahedron:

It is well-known that SIC-POVMs are extremal, meaning that their elements are $\mathbb{R}$-linearly independent. On the other hand, plotting the POVM corresponding to $|+\rangle,|-\rangle,|+\mathrm{i}\rangle,|-\mathrm{i}\rangle$, this gives:

This POVM isn't extremal. Coincidently, the shape is forms is just a plane: it has no volume. It's not too hard to show that if it is the case, then the POVM won't be extremal. But is the converse true? Is it possible to find a non-extremal, unbiased, rank-1 POVM such that the resulting shape on the Bloch sphere has a non-zero volume?

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Without loss of generality, we can assume that $\left|\psi_0\right\rangle=|0\rangle$, that $\left|\psi_1\right\rangle=\sqrt{\alpha_1}|0\rangle+\sqrt{1-\alpha_1}|1\rangle$, that $\left|\psi_2\right\rangle=\sqrt{\alpha_2}|0\rangle+\sqrt{1-\alpha_2}\mathrm{e}^{\mathrm{i}\theta_2}|1\rangle$ and that $\left|\psi_3\right\rangle=\sqrt{\alpha_3}|0\rangle+\sqrt{1-\alpha_3}\mathrm{e}^{\mathrm{i}\theta_3}|1\rangle$, with $0<\alpha_i<1$. If I'm not mistaken, the matrices will be $\mathbb{R}$-linearly independent if $\cos\left(\theta_2\right)\neq2\sqrt{\frac{\alpha_1\alpha_2}{\left(1-\alpha_1\right)\left(1-\alpha_2\right)}}$, but this doesn't seem to characterize a plane (and I'm not even sure my computations are correct).

Answer 1

No.

Let $\mathcal{H}_2$ denote the real vector space of $2\times 2$ Hermitian matrices, and define a linear map $\phi:\mathcal{H}_2\rightarrow \mathbb{R}^4$ as follows. $$ \phi(H) = \bigl(\operatorname{Tr}(H),\operatorname{Tr}(\sigma_x H),\operatorname{Tr}(\sigma_y H),\operatorname{Tr}(\sigma_z H)\bigr) $$ This is a linear bijection, and we can use it to translate density matrices to Bloch ball points. Specifically, if we write $$ \phi(\rho) = (1,x,y,z) $$ for $x,y,z\in\mathbb{R}$, then the point representing $\rho$ within the Bloch ball has Cartesian coordinates $(x,y,z).$

Now choose any four $2\times 2$ density matrices $\rho_1,\rho_2,\rho_3,\rho_4$, and write $$ \phi(\rho_k) = (1,x_k,y_k,z_k) $$ for $k = 1,2,3,4.$ Because $\phi$ is a linear bijection, the set $\{\rho_1,\ldots,\rho_4\}$ is linearly independent if and only if $$ \bigl\{\phi(\rho_1),\ldots,\phi(\rho_4)\bigr\} = \bigl\{ (1,x_1,y_1,z_1), \ldots, (1,x_4,y_4,z_4)\bigr\} $$ is linearly independent. Using a well-known fact from convexity theory, this set is linearly independent if and only if the set $$ \bigl\{ (x_1,y_1,z_1), \ldots, (x_4,y_4,z_4)\bigr\} $$ is affinely independent. This is equivalent to the convex hull of these four points having nonzero volume.

Finally, by a well-known fact due to Parthasarathy, which the question alludes to, a POVM $\bigl\{\frac{1}{2}\vert\psi_1\rangle\langle\psi_1\vert, \ldots, \frac{1}{2}\vert\psi_4\rangle\langle\psi_4\vert\bigr\}$ is extremal if and only if it is linearly independent as a set of Hermitian matrices. Thus, by the equivalences above, it is extremal if and only if the convex hull of the four corresponding points on the Bloch sphere has nonzero volume.

Answer 2

Any unbiased, rank-1 POVM with $4$ outcomes on qubits is defined by a single quantum state $|\psi\rangle$ and a unitary $U$, its elements being $U|\psi\rangle$, $UX|\psi\rangle$, $UY|\psi\rangle$ and $UZ|\psi\rangle$. Since rotating by a unitary changes neither the extremality of the POVM nor the volume of the shape it defines on the Bloch sphere, we are free to choose $U=I$ here.

Without loss of generality, $|\psi\rangle$ can be written as $|\psi\rangle=\sqrt{p}|0\rangle+\sqrt{1-p}\mathrm{e}^{\mathrm{i}\theta}|1\rangle$ with $p\in[0, 1]$ and $\theta\in[0,2\pi)$. We can decompose each element of this POVM in the following basis of the $2\times2$ Hermitian matrices: $$\left\{|0\rangle\!\langle0|,|0\rangle\!\langle1|+|1\rangle\!\langle0|,\mathrm{i}|0\rangle\!\langle1|-\mathrm{i}|1\rangle\!\langle0|,|1\rangle\!\langle1|\right\}\,.$$ Now, we know that a rank-1 POVM is extremal if and only if its elements are linearly independent. This is equivalent to saying that the matrix of the coefficients of the POVM's elements must be invertible. This matrix is the following: $$ \begin{pmatrix} p&\sqrt{p(1-p)}\cos(\theta)&-\sqrt{p(1-p)}\sin(\theta)&1-p\\ 1-p&\sqrt{p(1-p)}\cos(\theta)&\sqrt{p(1-p)}\sin(\theta)&p\\ 1-p&-\sqrt{p(1-p)}\cos(\theta)&-\sqrt{p(1-p)}\sin(\theta)&p\\ p&-\sqrt{p(1-p)}\cos(\theta)&\sqrt{p(1-p)}\sin(\theta)&1-p \end{pmatrix} $$ It's possible to show that the determinant of this matrix is equal to $\sin(2\theta)p(1-p)(1-2p)$. From there, we can deduce that this POVM won't be extremal if and only if: $$p=0\lor p=1\lor p=\frac12\lor\theta=0\mod\frac\pi2\,.$$ This is equivalent to saying that $|\psi\rangle$ belongs to either the $XY$, $ZX$ or $YZ$ plane of the Bloch sphere. Note that in this case, all the elements of the POVM are going to stay on the same plane.

Thus, by rotating the POVM with an arbitrary $U$, a rank-1, unbiased non-extremal POVM on qubits with $4$ outcomes has all its elements lying on a plane on the Bloch sphere. We thus have equivalence between being extremal and defining a shape with positive volume in this case.

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To see that any 4 element rank-1 POVM has the form $$\{U|\psi\rangle, UX|\psi\rangle, UY|\psi\rangle, UZ|\psi\rangle \}$$ consider the following.

For any state $|\phi\rangle$ the coordinates of the corresponding point ${\bf v} = (v_x,v_y,v_z)$ on the Bloch sphere can be written as $$ \begin{align} v_x = \langle \phi | X |\phi\rangle = Tr(X|\phi\rangle\langle\phi |),\\ v_y = \langle \phi | Y |\phi\rangle = Tr(Y|\phi\rangle\langle\phi |), \\ v_z = \langle \phi | Z |\phi\rangle = Tr(Z|\phi\rangle\langle\phi |). \end{align} $$

Let $|\phi_1\rangle, |\phi_2\rangle, |\phi_3\rangle, |\phi_4\rangle$ form a POVM and ${\bf v_1}, {\bf v_2}, {\bf v_3}, {\bf v_4}$ are their corresponding Bloch vectors. From the POVM defining condition one can easily deduce that $$ {\bf v_1 + v_2 + v_3 + v_4 = 0}. $$

In particular, this means that $$ \frac{1}{2}({\bf v_1 + v_2}) = -\frac{1}{2}({\bf v_3 + v_4}), $$ that is, the vectors from the origin to the middles of opposite edges are collinear, have the same length but opposite directions. Thus, opposite edges must have the same length too (since ${\bf v_i}$ all have the same length) and $$ {\bf v_1 \cdot v_2} = {\bf v_3 \cdot v_4}, $$ where $(\cdot)$ denotes the inner product.

Since it's true for any pair of opposite edges one can deduce that vectors that join middles of opposite edges are mutually orthogonal. Thus we can rotate the whole configuration so that they will match the axes. Now it's easy to see that ${\bf v_i}$ (after the rotation) are just reflections of each other around the axes.

Answer 3

If we take a general qubit state $|\psi\rangle$ and a general gate $U$ then $$\{U|\psi\rangle, UX|\psi\rangle, UY|\psi\rangle, UZ|\psi\rangle \}$$ form a POVM. In fact, any other 4 element POVM has this form (I can't give you a reference, but I'm sure it's a known fact).

It's easy to see how it looks on the Bloch sphere. Take a random point, then rotate it by 180 degrees over x,y,z axes to obtain three additional points. Then rotate everything however you wish.

In the resulting tetrahedron the three segments that join middles of the opposite edges will be mutually orthogonal and intersect in the center of the sphere. The square that corresponds to $|+\rangle,|-\rangle,|+\mathrm{i}\rangle,|-\mathrm{i}\rangle$ is a degenerate case, where one of the three segments has 0 length.

It's not hard to see that the resulting tetrahedron lies in a plane if and only if the starting point was in one of the coordinate planes. Then it follows that the 4-gon must be a rectangle in a great circle.

Now we can just check all such rectangles. Moreover, without loss of generality we can assume it lies in the zy plane, i.e. $U=I$, $|\psi\rangle = \sqrt{\alpha}|0\rangle+\sqrt{1-\alpha}|1\rangle$. My calculations show they are not extremal for any $\alpha$.

EDIT

I just realized you're also asking about the existence of non extremal yet not plane configurations (there's something wrong in the body of your question). Can't say for sure, but this should give a way to approach the problem.