# Give an explicit example of a $d = 4$ SIC-POVM

For $$q=e^{2 \pi i/3}$$, the set of $$d^2$$ vectors ($$d=3$$) $$\begin{equation} \left( \begin{array}{ccc} 0 & 1 & -1 \\ 0 & 1 & -q \\ 0 & 1 & -q^2 \\ -1 & 0 & 1 \\ -q & 0 & 1 \\ -q^2 & 0 & 1 \\ 1 & -1 & 0 \\ 1 & -q & 0 \\ 1 & -q^2 & 0 \\ \end{array} \right) \end{equation}$$

forms a SIC-POVM (symmetric, informationally complete, positive operator-valued measure), as noted in eq. (1) of https://arxiv.org/abs/1109.6514 .

I would similarly like to have a 16-vector counterpart for $$d=4$$ (to use for entanglement detection--per https://arxiv.org/abs/1805.03955). (There is clearly a huge amount of interesting related literature on such topics, but an attempt of mine to find an explicit d=4 counterpart somewhere within it has so far not been successful.)

• Is the SIC defined in proposition 3.4 of arxiv.org/abs/1410.5862v2 appropriate for your purpose? Nov 24 '19 at 7:40
• Thanks for this comment, DBM! Prop. 3.4 would certainly seem to be appropriate. But then the question for me becomes that of giving an explicit representation of the Weyl-Heisenberg group $W \times H$. (I was hoping--admittedly, lazily--to have the requested set of 16 vectors in the indicated form above without at this point having to immediately tackle the underlying clearly sophisticated math in its several details.) Nov 24 '19 at 12:17

As indicated by Danylo in his anwser, eq. (32) in arXiv: 1103.2030 presents the sixteen vectors ("ignoring overall phases and normalisation") $$\begin{equation} \left( \begin{array}{cccc} x & 1 & 1 & 1 \\ x & 1 & -1 & -1 \\ x & -1 & 1 & -1 \\ x & -1 & -1 & 1 \\ i & x & 1 & -i \\ i & x & -1 & i \\ -i & x & 1 & i \\ -i & x & -1 & -i \\ i & i & x & -1 \\ i & -i & x & 1 \\ -i & i & x & 1 \\ -i & -i & x & -1 \\ i & 1 & -i & x \\ i & -1 & i & x \\ -i & 1 & i & x \\ -i & -1 & -i & x \\ \end{array} \right), \end{equation}$$ where $$\begin{equation} x=\sqrt{2+\sqrt{5}}. \end{equation}$$ These form "a set of 16 SIC-vectors covariant under the Heisenberg group".

To now normalize the vectors we need to multiply them by $$\frac{1}{\sqrt{5+\sqrt{5}}}$$. The resultant sixteen vectors $$|\psi_i\rangle$$ satisfy the desired relation ($$d=4$$) $$\begin{equation} |\langle\psi_i||\psi_j\rangle|^2 = \frac{d \delta_{ij}+1}{d+1}, i,j=1,2,\ldots, d^2 , \end{equation}$$ given in eq. (7) in arXiv: 1805.03955, giving us the explicit set requested in the question.

You can find it here Symmetric Informationally Complete Quantum Measurements or here SIC-POVMs: A new computer study, in the appendix B.

Update

Given a single fiducial vector $$v = (a_1,a_2,a_3,a_4)^T \in \mathbb{C}^4$$ it's pretty easy to write down all SIC-POVM vectors. They are just $$C^kS^lv$$ for $$k,l \in \{0..3\}$$, where $$C$$ and $$S$$ are clock and shift matrices given by $$C = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & i & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -i \\ \end{pmatrix}, ~~~ S = \begin{pmatrix} 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ \end{pmatrix}.$$

So $$C^kS^lv = (a_{1-l},i^ka_{2-l},(-1)^ka_{3-l},(-i)^ka_{4-l})^T$$, where $$a_0 = a_4$$, $$a_{-1} = a_3$$, etc.
Note that a phase of any SIC-POVM vector doesn't matter.

Update 2

A simple formula for all 16 vectors can be found here, eq (32)
The non-normalized fiducial vector is just $$\left( \begin{array}{c} \sqrt{2+\sqrt{5}} \\ 1 \\ 1 \\ 1 \end{array} \right)$$ but we must change the matrices $$C$$ and $$S$$ to correctly generate 16 vectors: $$C = e^{i\pi/4}\begin{pmatrix} 0 & 1 & 0 & 0 \\ -i & 0 & 0 & 0 \\ 0 & 0 & 0 & -i \\ 0 & 0 & -1 & 0 \\ \end{pmatrix}, ~~~ S = e^{i\pi/4}\begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -i & 0 & 0 & 0 \\ 0 & i & 0 & 0 \\ \end{pmatrix}.$$

• Hi. Please note that link-only or reference-only answers are not considered to be real answers. If and when you get time, consider elaborating on how the material in the references address the OP's question. Thanks! Nov 24 '19 at 14:08
• Well, I certainly appreciate this answer--but I was still hoping for a fully explicit set of sixteen vectors, very much in the direct manner of the d = 3 example in the question. This is all wonderful, fascinating literature--but requires some mastery/confidence in the underlying concepts (fiducial, Weyl-Heisenberg, frame theory, Clifford group,...) to finally obtain the quite explicit (computationally usable) set which I am seeking. (I had considered in lieu of this question, simply posing it, via email, to I. Bengtsson, whom I strongly suspect could readily extend his d = 3 example.) Nov 24 '19 at 14:53
• @PaulB.Slater updated the answer Nov 24 '19 at 22:03
• The updated answer of Danylo Y would seem to suffice for the indicated purpose of an explicit construction of the desired set of sixteen 4-vectors.. (Required fiducial 4-vectors are given by eqs. (28) and (29) in the first reference [Symmetric Informationally...] of the answer.) I will undertake such a construction--and intend to post the result as a second answer. Nov 25 '19 at 10:44
• Well, getting a little frustrated with the indicated strategy in previous comment. It turns out that I can normalize the candidate sixteen vectors, but the squares of the absolute values of their inner products for different vectors--per eq. (7) in arxiv.org/pdf/1805.03955.pdf--are certainly not equal to $\frac{1}{5}$, nor to any particular constant value. So, the desired FULLY EXPLICIT set of sixteen vectors still seems elusive. Nov 25 '19 at 12:51