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

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

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? – David Bar Moshe 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.) – Paul B. Slater Nov 24 '19 at 12:17

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! – Sanchayan Dutta 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.) – Paul B. Slater Nov 24 '19 at 14:53
• @PaulB.Slater updated the answer – Danylo Y 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. – Paul B. Slater 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. – Paul B. Slater Nov 25 '19 at 12:51

As indicated by Danylo in his anwser, eq. (32) in arXiv: 1103.2030 presents the sixteen vectors ("ignoring overall phases and normalisation") $$$$\left( \begin{array}{cccc} x & 1 & 1 & 1 \\ x & 1 & -1 & -1 \\ x & -1 & 1 & -1 \\ x & -1 & -1 & 1 \\ i & x & 1 & -i \\ i & x & -1 & i \\ -i & x & 1 & i \\ -i & x & -1 & -i \\ i & i & x & -1 \\ i & -i & x & 1 \\ -i & i & x & 1 \\ -i & -i & x & -1 \\ i & 1 & -i & x \\ i & -1 & i & x \\ -i & 1 & i & x \\ -i & -1 & -i & x \\ \end{array} \right),$$$$ where $$$$x=\sqrt{2+\sqrt{5}}.$$$$ 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 ($$d=4$$) $$$$|\langle\psi_i||\psi_j\rangle|^2 = \frac{d \delta_{ij}+1}{d+1}, i,j=1,2,\ldots, d^2 ,$$$$ given in eq. (7) in arXiv: 1805.03955, giving us the explicit set requested in the question.