# The Number Behind the Simplest SIC–POVM

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## Abstract

The simple concept of a SIC poses a very deep problem in algebraic number theory, as soon as the dimension of Hilbert space exceeds three. A detailed description of the simplest possible example is given.

## Keywords

Equiangular lines SIC-POVMs Number theory## 1 Introduction to the SIC Existence Problem

Physicists rarely care about numbers; that is to say, they rarely care about the *nature* of numbers. Indeed, they routinely and unquestioningly rely on the real number system. Still, doubts are sometimes expressed. Thus, Schrödinger referred to quantum mechanics as a “makeshift”, because it seemingly does not at all challenge the notion of the continuum [1]. Be that as it may, there is an easily formulated quantum mechanical question forcing us to come directly to grips with a major unsolved problem concerning numbers.

*d*dimensions, can one find \(d^2\) vectors \(|\psi _I\rangle \) such that

Numbers can be added and multiplied to form new numbers. Indeed they form sets, known as *fields*, that are closed under addition, subtraction, multiplication, and division. A standard example is the field of rational numbers \(\mathbf{Q}\), which is generated from the integers by applying the field operations. Every field having an infinite number of elements contains \(\mathbf{Q}\) as a subfield. Examples include the real number field \(\mathbf{R}\) (against which Schrödinger expressed his reservations) and the complex number field \(\mathbf{C}\). The ancient Greeks (in particular, Eudoxos) gave a definition of the real numbers, but they also payed particular attention to a smaller extension of \(\mathbf{Q}\), consisting of numbers that can be geometrically constructed using ruler and compass. Algebraically this corresponds to the requirement that all the numbers that occur are given in terms of nested square roots of rational numbers. This number field turned out to be too small for some purposes, for instance one needs the cube root \(\root 3 \of {2}\) in order to duplicate the cube. More generally we can consider fields built using nested *radicals*, including cube roots, quartic roots, and so on. During the Italian Renaissance there was a race to express the solutions of polynomial equations in terms of radicals, but it was eventually shown by Abel and Galois that this is not possible in general. The lesson learned from this brief excursion into history is that the choice of the number field depends on the particular task one is facing.

The question is: what number field is needed to construct SICs? A conjectural, but precise and highly remarkable, answer is now available [5]. Our purpose here is to describe this answer using the simplest non-trivial example, where almost all the calculations can be done with pencil and paper.

## 2 Introducing the Weyl–Heisenberg Group

If \(d = 2\) the SIC existence question is trivial. Moreover, the vectors being sought are then always fully determined by a complete set of mutually unbiased bases, in a geometrically natural way. Although it is considerably harder to see (one will have to read several papers in order to patch a proof together [6, 7, 8, 9]), this statement holds also when \(d = 3\). Numerically, SICs have been found in every dimension where they have been looked for (this includes all dimensions \(d \le 121\), and a few more [10, 11]), but no existence proof has been found, and beyond three dimensions it is very hard to see an underlying pattern in the solutions. Yet closer inspection reveals that there is a pattern also when \(d \ge 4\). This pattern, so far as it is understood, resides in number theoretical properties of exact solutions for SICs [5, 12], and the simplest non–trivial example occurs when \(d = 4\).

To see how exact solutions can be obtained, we first take note of the fact that all known SICs (with one exception) form orbits under the Weyl–Heisenberg group, a discrete group first brought into quantum mechanics by Weyl [13]. For \(d \le 3\) it has been proved that every SIC arises in this way [7, 8], and for all prime *d* that the Weyl–Heisenberg group is the only possible group [14]. (The exception is generated, when \(d = 8\), by another group. Like so many exceptional structures it is related to octonions [15], and here we will assume that it can be left aside as a curiousity.)

*d*, and the phase factor \(\omega \) is a primitive

*d*th root of unity. It is convenient to introduce yet another phase factor \(\tau = - e^{i\pi /d}\) [16], and to define the displacement operators

*algebraic numbers*, that is by roots of polynomials with integer coefficients. It is the precise nature of these numbers that is so surprising.

*u*. Once this number is known, the entire SIC can be reconstructed from Eq. (5). For all \(d > 4\) there are several independent numbers.

*u*.

## 3 The Number Field of the Example

Our first concern is to determine the smallest number field to which the number *u* belongs. Call it \(\mathbf{Q}(u)\), since it is an extension of the rational numbers. We will freely use the fact that the number field we are looking for is a subfield of the complex numbers. This gives the enterprise a concrete flavour, and simplifies some statements compared to those found in textbooks.

*u*, we immediately conclude that \(-u\), 1 /

*u*, and \(-1/u\) also belong to the field. So does the number

*x*is more manageable than

*u*itself, which is related to the fact that it is symmetric under exchanges \(u \leftrightarrow 1/u\). Going on in this way, we notice that

*u*can be obtained from them, which means that the field can be equivalently written as \(\mathbf{Q}(u) = \mathbf{Q}(\sqrt{5}, \sqrt{2}, i\sqrt{\sqrt{5}+1})\). Indeed the latter three generators of the field were used in the early references [10, 12]. However, the field is not yet large enough to contain the number

*i*, or the number \(\tau = - (1+i)/\sqrt{2}\) that appears in the reconstruction formula (5). We definitely want to extend our number field so that the number \(\tau \) is included. And we want to do so in a principled way. This means that Galois theory must come into play.

*minimal polynomial*, with coefficients among the integers, of the algebraic number

*u*. Minimal polynomials of degree

*n*always have

*n*distinct roots, otherwise they would not be minimal (that is, have the lowest possible degree). Mathematica gives the minimal polynomial for

*u*after only a moment’s hesitation. It is of degree 8:

*u*of a degree not exceeding seven. In general the dimension of a field, such as \(\mathbf{Q}(u)\), considered as a vector space over its ground field, in this case \(\mathbf{Q}\), is called its

*degree*. So the conclusion so far is that

*Galois group*of the field; when Galois first studied it he regarded it as the group that permutes the roots of the minimal polynomial. The order of the Galois group equals the degree of the extension.

The leading coefficient of our minimal polynomial equals 1, and it is a palindromic polynomial, in an obvious sense. The first property implies that *u* is an *algebraic integer* (by definition), and the second property implies that 1 / *u* is another root of same polynomial. It follows that both *u* and 1 / *u* are algebraic integers. Therefore (again by definition) *u* is an algebraic *unit*. Another peculiarity of our polynomial is that only even powers appear in it, meaning that the phase factors \(-u\) and \(-1/u\) are roots of the polynomial too.

To complete the list of phase factors appearing in Eq. (7) we also need the number \(-1\). It is a root of the polynomial \(p_0(t) = t+1\), whose leading coefficient is again 1. Therefore \(-1\) is an algebraic integer too, and because \(p_0(t)\) is palindromic it is an algebraic unit as well. In this context we call it a ‘baby unit’. Every phase factor in Eq. (7) is an algebraic unit.

*d*[5]. This granted, it is easy to work out the minimal polynomial \(p_1(t)\) by hand. First we find the minimal polynomial for the number \(x = u + 1/u\), namely

*x*is an algebraic integer, but it is not a unit. Now we use the fact that whenever the minimal polynomial \(p_{2n}\) of an algebraic number

*z*is a palindromic polynomial of degree 2

*n*, it can be obtained as

*normal*extension of \(\mathbf{Q}\) allowing us to

*split*the polynomial, that is to say we need to include all its roots in a field which will be larger than just \(\mathbf{Q}(u)\). Because the polynomial is palindromic and depends only on even powers we know that a full factorization must take the form

*r*. (It is a unit by construction.) Writing this out leads to a second degree polynomial equation for \(r^2\) with coefficients in the field \(\mathbf{Q}(u)\), which we can solve. But we can also simply guess the solution. From Eq. (10) it is evident that \(\sqrt{5}\) will be left invariant when we permute the roots that we have so far. This suggests that it should go to \(-\sqrt{5}\) under the exchange \(u \leftrightarrow r\). If so,

*r*can be obtained by changing the sign in front of \(\sqrt{5}\) in the expression for

*u*. If we perform the replacements

*r*by changing the sign in front of the second term. A direct calculation confirms that

*r*is a root of \(p_1\).

*r*is obviously a root of

*r*, with coefficients in \(\mathbf{Q}(u)\), is

*u*, since the polynomial admits eight roots over \(\mathbf{F}_2\). The degrees of these extensions are

The order of the Galois group equals the degree of the extension, and this group permutes the roots of the polynomials that were used to define the field. The story becomes particularly simple when the extension is normal. If all the roots of the degree 8 polynomial had been in the field obtained by adjoining one of its roots, the extension would have been already normal, and the Galois group would have had order 8. On the other hand we might have found only one root in the first step, and would then have been left with an irreducible polynomial of order 7. The second extension would then have had degree \(8\cdot 7\) over the field of rationals. In the worst case scenario an extension of degree 8! would be needed to split the polynomial, and the Galois group would then be the symmetric group \(S_8\). Our Galois group must have the comparatively modest order 16.

*G*can be written down immediately. With an eye on the polynomial \(p_2(t)\), we find them to be

*H*. We need an additional element with the property that \(g_4(u) = r\). To see what it does to

*r*we note that

*G*is to observe that the abelian group

*H*is a normal subgroup since

*soluble*. A group

*G*is called soluble if it admits a sequence of normal subgroups \(H_k\) so that

The generators of the Galois group, and how they act

| | \(\sqrt{5}\) | \(\sqrt{2}\) | \(i\sqrt{1+\sqrt{5}}\) | | \(\tau \) | |
---|---|---|---|---|---|---|---|

\(g_1\) | 1 / | | \(\sqrt{5}\) | \(\sqrt{2}\) | \(-i\sqrt{1+\sqrt{5}}\) | \(-i\) | \(1/\tau \) |

\(g_2\) | \(-u\) | \(-r\) | \(\sqrt{5}\) | \(-\sqrt{2}\) | \(-i\sqrt{1+\sqrt{5}}\) | | \(-\tau \) |

\(g_3\) | | 1 / | \(\sqrt{5}\) | \(\sqrt{2}\) | \(i\sqrt{1+\sqrt{5}}\) | \(-i\) | \(1/\tau \) |

\(g_4\) | | | \(-\sqrt{5}\) | \(-\sqrt{2}\) | \(\sqrt{\sqrt{5}-1}\) | | \(-\tau \) |

*H*. Hence the abelian group

*H*is the Galois group of the field considered as an extension of \(\mathbf{Q}(\sqrt{5})\). This makes the field an

*abelian extension*of the

*real quadratic field*\(\mathbf{Q}(\sqrt{5})\). The field \(\mathbf{E}_\mathbf{R} = \mathbf{Q}(\sqrt{5}, \sqrt{2})\) is totally real, in the sense that every embedding of this field into \(\mathbf{C}\) is real. This field can be extended to either \(\mathbf{Q}(u)\), which contains the overlap phases, or to \(\mathbf{Q}(r)\), and the two of them are related by the automorphism \(g_4\) when regarded as subfields of \(\mathbf{Q}( u,r)\), which is the field needed to construct the SIC projectors. Judging from the way the various patterns recur in higher dimensions [5, 12], the correct way of looking at the fields we have encountered is as given in Fig. 1.

## 4 Some Further Results I was Told About

We have reached the conclusion that the SIC phases are units in an interesting field. Now the set of units in a given field form a multiplicative *unit group*, and it is natural to ask how the SIC phases are positioned within that group.

*torsion*subgroup, and evidently consists of the units \(\pm 1\). A possible set of generators of the infinite factors is [22]

Remarkably, \(u_{(3)}\) is equal to the SIC phase *u*. Hence the SIC phase has a very special position inside the unit group. Unfortunately, for the fields arising from SICs, the unit group is known only in a handful of cases [5]. The size of the calculation grows quickly with the field—in fact calculating unit groups is one of those problems for which the best known algorithm demands a quantum computer [23].

## 5 How the Example Generalizes

*d*(or 2

*d*if

*d*is even), or extensions thereof. These words carry deep meaning for algebraic number theorists. The most familiar example of a conductor is the integer

*n*in the phase factor \(e^{2\pi i/n}\), when the rational field is extended to the

*cyclotomic field*\(\mathbf{Q}(e^{2\pi i/n})\). Cyclotomic fields house the most general abelian extensions of \(\mathbf{Q}\), and their conductors tell us how they fit together. Considering abelian extensions of the imaginary quadratic fields \(\mathbf{Q}(\sqrt{-D})\) one is led to replace the exponential function with special functions defined on suitable elliptic curves. It seems that, if we want to deal with abelian extensions of the real quadratic field \(\mathbf{Q}(\sqrt{D})\), SICs provide very valuable insights. Some of the details, and how they relate to Kronecker’s

*Jugendtraum*and to Hilbert’s unsolved 12th problem [24], are described in a contribution to this issue by Appleby et al. [25].

Coming back to the question whether SICs exist in all dimensions, if we knew the field and its unit group, and if we had enough information about the position of the SIC phases within the unit group, the question might not look so formidable anymore.

Finally, the reader may well ask for the physical significance of all this. The answer is not known. But the idea that quantum theory unquestioningly accepts the continuum has been effectively contradicted: on the contrary, elementary quantum theory seems to know about some of the deepest discrete structures hidden inside the continuum. This suggests that any challenge to the continuum has to go through quantum mechanics, not around it.

## Notes

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