Equations for the overlaps of a SIC

Throughout fix the integer $d\geq 2$, and let $\omega$ be the primitive $d$-th root of unity $\omega^{:}=e^{{2\pi\over d}i}$. We think of vectors in $\mathbb{C}^{d}$ as periodic signals on the group $\mathbb{Z}_{d}$, and hence index vectors and matrices by elements of $\mathbb{Z}_{d}$. A set of $d^{2}$ unit vectors $(v_{j})$ in $\mathbb{C}^{d}$ (or the lines that they determine) is said to be equiangular if

$$|\langle v_{j},v_{k}\rangle|^{2}={1\over d+1},\qquad j\neq k.$$

(1.1)

In quantum information theory, the corresponding rank one orthogonal projections $(v_{j}v_{j}^{*})$ are said to be a symmetric informationally complete positive operator valued measure, or a SIC for short. The existence of a SIC for every dimension $d$ is known as Zauner’s conjecture (from his 1999 thesis, see [Zau10]), or as the SIC problem.

There are high precision numerical constructions of SICs [RBKSC04], [SG10], [Sco17], and exact SICs in various dimensions [ACFW18], [GS17]. In all of these constructions, the SIC is a Weyl-Heisenberg SIC, i.e., is the orbit $(\rho(g)v)_{g\in G}$ of a fiducial vector $v$ under the unitary irreducible projective representation $\rho:G\to{\cal U}(\mathbb{C}^{\mathbb{Z}_{d}})$ of $G=\mathbb{Z}_{d}\times\mathbb{Z}_{d}$ with Schur multiplier $\alpha$ given by

$$\rho_{jk}=\rho((j,k))=S^{j}\Omega^{k},\qquad\alpha((j_{1},j_{2}),(k_{1},k_{2}))=\omega^{j_{2}k_{1}},$$

(1.2)

where $S$ is the cyclic shift matrix $S_{jk}:=\delta_{j,k+1}$ and $\Omega$ is the diagonal (modulation) matrix $\Omega_{jk}:=\omega^{j}\delta_{jk}$. In this case, the equiangularity condition (1.1) becomes

$$|\langle S^{j}\Omega^{k}v,v\rangle|^{2}={1\over{d+1}},\qquad(j,k)\neq(0,0).$$

(1.3)

In this paper, we consider equations in the variables

$$c_{jk}=\langle S^{j}\Omega^{k}v,v\rangle=\mathop{\rm trace}\nolimits(vv^{*}S^{j}\Omega^{k}),\qquad(j,k)\in\mathbb{Z}_{d}\times\mathbb{Z}_{d},$$

(1.4)

which determine a (Weyl-Heisenberg) SIC. These variables (or scalar multiples of them) are called the overlaps of the SIC. They depend only on the fiducial projector $P=vv^{*}$. The original attempts to find numerical and exact SIC fiducials (using Groebner basis methods) involved polynomial equations in the variables $v_{0},\ldots,v_{d-1}$ and $\overline{v_{0}},\ldots,\overline{v_{d-1}}$, such as the equiangularity condition (1.3), the equations (see [BW07], [Kha08], [ADF14])

$\displaystyle\sum_{r\in\mathbb{Z}_{d}}v_{r}\overline{v}_{r+s}\overline{v}_{r+t}v_{r+s+t}=\begin{cases}0,&s,t\neq 0;\cr{1\over d+1},&s\neq 0,t=0,\quad s=0,t\neq 0;\cr{2\over d+1},&(s,t)=(0,0),\end{cases}$

(1.5)

and the variational characterisation (used for finding numerical SICs)

$${1\over d^{2}}\sum_{(j,k)\in\mathbb{Z}_{d}^{2}}|\langle S^{j}\Omega^{k}v,v\rangle|^{4}={2\over d(d+1)}\|v\|^{4},\qquad\|v\|^{2}=1.$$

(1.6)

More recent exact constructions of SICs [ACFW18] have been in the overlap variables $c_{jk}$ (utilising a natural Galois action on them). Clearly the $c_{jk}$ giving a SIC fiducial projector $vv^{*}$ via (1.4) must satisfy

$$c_{00}=\|v\|^{2}=1,\qquad|c_{jk}|^{2}=|\langle S^{j}\Omega^{k}v,v\rangle|^{2}={1\over d+1},\quad(j,k)\neq(0,0),$$

(1.7)

and also, by the rule $\Omega^{k}S^{j}=\omega^{jk}S^{j}\Omega^{k}$,

$$c_{jk}=\overline{\langle v,S^{j}\Omega^{k}v\rangle}=\overline{\langle\Omega^{-k}S^{-j}v,v\rangle}=\overline{\langle\omega^{jk}S^{-j}\Omega^{-k}v,v\rangle}=\omega^{-jk}\overline{c_{-j,-k}}.$$

(1.8)

These conditions on the overlap variables $c_{jk}$ are not enough to guarantee that they come from a fiducial projector $vv^{*}$ (and hence prove Zauner’s conjecture).

In Section 2, we define a linear operator $T$, which is an example of the projective Fourier transform, which allows us to reconstruct the fiducial projector as $vv^{*}=Tc$ from a suitable $c=(c_{jk})$. We prove that in addition to (1.7) and (1.8), the simple condition

$$\mathop{\rm trace}\nolimits((Tc)^{4})=1$$

ensures that a $c$ gives a SIC fiducial (Theorem 2.1). We then give some examples, and describe the action of the Clifford group on the SIC fiducials give by overlaps $c$.

In Section 3, we give some interesting properties of $T$, i.e., the projective Fourier transform of $G=\mathbb{Z}_{d}\times\mathbb{Z}_{d}$. In particular, we show that $(\sqrt{d}T)^{6d}=(-1)^{{1\over 2}d(d-1)}I$, and a variant has order $4d$. To our knowledge, this is only the second example of a Fourier transform of finite order, after the (discrete) Fourier transform for a finite abelian group $G=\mathbb{Z}_{d}$ (which satisfies $F^{4}=I$).

In Section 4, we give another system of equations in the overlaps $c$ that determine a SIC. These involve the symbol ($z$–transform) of the rows of $c$. The symbols for $c$ giving a SIC turn out to have interesting Riesz-type factorisation properties. We use these to describe the (sporadic) SICs for $d=3$, which are parametrised by a hypocycloid.

Since the $\rho$ of (1.2) is a unitary irreducible projective representation of dimension $d$, it follows that $(\rho(g))_{g\in G}$ is a tight frame (called a nice error frame with index group $G$ [CW17]) for the $d\times d$ matrices with the Frobenius inner product

$$\langle A,B\rangle:=\mathop{\rm trace}\nolimits(AB^{*})=\sum_{j,k}a_{jk}\overline{b_{jk}},$$

i.e.,

$$A={d\over|G|}\sum_{g\in G}\langle A,\rho(g)\rangle\rho(g),\qquad\forall A\in\mathbb{C}^{d\times d}.$$

(2.9)

In this particular case, $(\rho(g))_{g\in G}=(S^{j}\Omega^{k})$ is an orthogonal basis. Taking $A=vv^{*}$ above gives the following formula for reconstruction from the overlaps $c_{jk}=\langle S^{j}\Omega^{k}v,v\rangle$

$$vv^{*}={d\over d^{2}}\sum_{j,k}\langle vv^{*},S^{-j}\Omega^{-k}\rangle S^{-j}\Omega^{-k}={1\over d}\sum_{j,k}\omega^{jk}c_{jk}S^{-j}\Omega^{-k}={1\over d}\sum_{j,k}c_{jk}(S^{j}\Omega^{k})^{*},$$

since $\omega^{jk}S^{-j}\Omega^{-k}=(S^{j}\Omega^{k})^{*}$, and $(S^{-j}\Omega^{-k})^{*}=\omega^{jk}S^{j}\Omega^{k}$ gives

$$\langle vv^{*},S^{-j}\Omega^{-k}\rangle=\mathop{\rm trace}\nolimits(vv^{*}(S^{-j}\Omega^{-k})^{*})=\mathop{\rm trace}\nolimits(v^{*}\omega^{jk}S^{j}\Omega^{k}v)=\omega^{jk}\langle S^{j}\Omega^{k}v,v\rangle=\omega^{jk}c_{jk}.$$

Motivated by this, we define a linear map $T:\mathbb{C}^{\mathbb{Z}_{d}\times\mathbb{Z}_{d}}\to\mathbb{C}^{\mathbb{Z}_{d}\times\mathbb{Z}_{d}}$ by

$$Tc:={1\over d}\sum_{j,k}c_{jk}(S^{j}\Omega^{k})^{*}={1\over d}\sum_{j,k}\omega^{jk}c_{jk}S^{-j}\Omega^{-k}.$$

(2.10)

This can be viewed as the $\alpha$-Fourier transform of [Wal20] (for a Schur multiplier $\alpha$) which is a map $F_{\alpha}:\mathbb{C}^{G}\to\oplus_{\rho}\mathbb{C}^{d_{\rho}\times d_{\rho}}$, where $\rho$ counts over the irreducible projective representations of $G$ with multiplier $\alpha$ (and dimension $d_{\rho}$). Here $G=\mathbb{Z}_{d}\times\mathbb{Z}_{d}$ has just one such representation, the $\rho$ of (1.2), and $F_{\alpha}$ of $\nu=c\in\mathbb{C}^{G}=\mathbb{C}^{\mathbb{Z}_{d}\times\mathbb{Z}_{d}}$ at the unitary representation $\rho$ is

$$(F_{\alpha}\nu)_{\rho}=\sum_{g\in G}\nu(g)\rho(g)^{*}=\sum_{j,k}c_{jk}(S^{j}\Omega^{k})^{*}=d(Tc).$$

Thus $T$ is the projective Fourier transform for the group $G=\mathbb{Z}_{d}\times\mathbb{Z}_{d}$. For this particular group, we can view the image of a vector in $\mathbb{C}^{G}$ as being in $\mathbb{C}^{G}=\mathbb{C}^{\mathbb{Z}_{d}\times\mathbb{Z}_{d}}$, and as a result it is natural to consider powers of the Fourier transform. The only other case that we know of where this can be done is for the ordinary representations of a finite abelian group (where the representations give the character group $\hat{G}$, which can be identified with $G$). In this case the (discrete) Fourier transform has order $4$.

We now use $T$ to characterise those vectors (matrices) $c\in\mathbb{C}^{G}=\mathbb{C}^{\mathbb{Z}_{d}\times\mathbb{Z}_{d}}$ which give a fiducial projector $vv^{*}=Tc$.

The so called $2$–trace $\sum_{j\neq k}\lambda_{j}\lambda_{k}$ of $Tc$ is equal to $\{(\mathop{\rm trace}\nolimits(Tc))^{2}-\mathop{\rm trace}\nolimits((Tc)^{2})\}/2$. Since $Tc$ is Hermitian, $\mathop{\rm trace}\nolimits((Tc)^{2})=\langle Tc,(Tc)^{*}\rangle=\langle Tc,Tc\rangle$, and by the orthogonality of the $\rho_{jk}=S^{j}\Omega^{k}$, we calculate

$$\langle Tc,Tc\rangle={1\over d^{2}}\sum_{j,k}|c_{j,k}|^{2}\langle\rho_{jk}^{*},\rho_{jk}^{*}\rangle={1\over d}\sum_{j,k}|c_{jk}|^{2}.$$

Now by (ii) and (iii),

$$\mathop{\rm trace}\nolimits((Tc)^{2})=\langle Tc,Tc\rangle={1\over d}\sum_{j,k}|c_{jk}|^{2}={1\over d}\Bigl{(}1+(d^{2}-1){1\over d+1}\Bigr{)}=1,$$

and so $\sum_{j\neq k}\lambda_{j}\lambda_{k}=\{(\mathop{\rm trace}\nolimits(Tc))^{2}-\mathop{\rm trace}\nolimits((Tc)^{2})\}/2=(1-1)/2=0$.       

(iv)^′ The characteristic polynomial of $Tc$ has the form $P_{Tc}(\lambda)=\lambda^{d}-\lambda^{d-1}$

(iv)^′′ $\mathop{\rm trace}\nolimits((Tc)^{j})=1$, $j=1,2,\ldots$

(iv)^′′ $\Longrightarrow$ (iv) $\Longrightarrow$ $Tc$ has eigenvalues $1,0,\ldots,0$ $\iff$ (iv)^′ $\Longrightarrow$ (iv)^′′.

The condition (i) allows further variables $c_{jk}$ to be eliminated. When $d$ is odd, half of the $(d^{2}-1)$ variables $c_{jk}$, $(j,k)\neq(0,0)$, can be eliminated. For $d$ even, half of the $d^{2}-4$ variables $c_{jk}$, $(j,k)\not\in\{0,{d\over 2}\}^{2}$, can be eliminated, and

$$c_{{d\over 2},0}=\overline{c_{{d\over 2},0}},\qquad c_{0,{d\over 2}}=\overline{c_{0,{d\over 2}}},\qquad c_{{d\over 2},{d\over 2}}=(-1)^{d\over 2}\overline{c_{{d\over 2},{d\over 2}}},$$

(2.11)

so that $c_{0,{d\over 2}},c_{0,{d\over 2}}\in\mathbb{R}$, and $c_{{d\over 2},{d\over 2}}$ is in $\mathbb{R}$ for ${d\over 2}$ even, and is in $i\mathbb{R}$ for ${d\over 2}$ odd.

The group generated by $S$ and $\Omega$ is called the Heisenberg group, and its normaliser in the unitary matrices is the Clifford group. Indeed, if a $a\in{\operatorname{C}}(d)$, then

$$a\rho_{jk}a^{-1}=z_{a}(j,k)\rho_{\psi_{a}(j,k)},\qquad\forall(j,k)\in\mathbb{Z}_{d}^{2},$$

where $\psi_{a}$ is matrix multiplication by an element of $\mathop{\it SL}\nolimits_{2}(\mathbb{Z}_{d})$. The Clifford group ${\operatorname{C}}(d)$ maps SIC fiducials to SIC fiducials, via the action

$$a\cdot(vv^{*}):=(av)(av)^{*}=a(vv^{*})a^{-1},\qquad a\in{\operatorname{C}}(d).$$

The induced action on the overlaps of the fiducial is given by

	$\displaystyle(a\cdot c)_{jk}$	$\displaystyle=\mathop{\rm trace}\nolimits(a(vv^{*})a^{-1}S^{j}\Omega^{k})=\langle a^{-1}S^{j}\Omega^{k}av,v\rangle=\langle z_{a^{-1}}(j,k)\rho_{\psi_{a^{-1}}(j,k)}v,v\rangle$
		$\displaystyle=z_{a^{-1}}(j,k)c_{\psi_{a^{-1}}(j,k)}.$

In [BW19], it is shown that the Clifford group is generated by the scalar matrices, $S$, $\Omega$, the Fourier transform $F$ and the Zauner matrix $Z$, where

$$F_{jk}:={1\over\sqrt{d}}\omega^{jk},\qquad Z_{jk}:=\zeta^{d-1}\mu^{j(j+d)},\quad\mu:=e^{{2\pi\over 2d}i},\ \zeta:=e^{{2\pi\over 24}i}.$$

For these (see [Wal18])

$$(S^{a}\Omega^{b}\cdot c)_{jk}=\omega^{ak-bj}c_{jk},\qquad(F\cdot c)_{jk}=\omega^{-jk}c_{k,-j},\qquad(Z\cdot c)_{jk}=\mu^{j(j+d-2k)}c_{k-j,-j}.$$

When a (Weyl-Heisenberg) SIC fiducial $vv^{*}$ is known, there is always appears to be one which is given by an eigenvector $v$ of $Z$ (indeed these are often searched for directly). Correspondingly, the overlaps satisfy $Z\cdot c=c$, i.e., the equations

$$\mu^{j(j+d-2k)}c_{k-j,-j}=c_{jk},$$

which allows a further reduction of the variables $c_{jk}$.

Here we consider some properties of $T:\mathbb{C}^{\mathbb{Z}_{d}\times\mathbb{Z}_{d}}\to\mathbb{C}^{\mathbb{Z}_{d}\times\mathbb{Z}_{d}}$ given by (2.10), i.e., the projective Fourier transform of $G=\mathbb{Z}_{d}\times\mathbb{Z}_{d}$. It follows from the Plancherel formula for projective representations [Wal20], or (2.9) that $\sqrt{d}T$ is unitary. Indeed, (2.9) can be written as $I=T\Lambda$, where $\Lambda:\mathbb{C}^{G}\to\mathbb{C}^{G}:A\mapsto(\langle A,\rho(g)\rangle)_{g\in G}$ satisfies

$$\langle A,A\rangle={1\over d}\sum_{g\in G}|\langle A,\rho(g)\rangle|^{2}={1\over d}\langle\Lambda A,\Lambda A\rangle,$$

so that ${1\over\sqrt{d}}\Lambda$ is unitary, and hence $\sqrt{d}T$ is unitary.

We now show that $\sqrt{d}T$ has finite order ($6d$ or $12d$), i.e., the projective Fourier transform for $\rho$ of (1.2) has finite order (Theorem 3.1). To do this, we need a technical lemma (Lemma 3.1), based on the Zauner matrix $Z$ (of order $3$), which can be factored

$$Z=\zeta^{d-1}RF,\quad\zeta:=e^{2\pi i\over 24},\qquad(R)_{jk}=\mu^{j(j+d)}\delta_{jk},\quad\mu:=e^{2\pi i\over 2d},$$

where $F$ is the Fourier matrix, and $R$ is diagonal. The strong form of Zauner’s conjecture is that there is a SIC fiducial which is an eigenvector of $Z$, for every dimension $d$.

The $(j,k)$-block $A_{jk}$ of the (block) matrix representation $[\sqrt{d}T]$ of $\sqrt{d}T$ is given by

	$\displaystyle A_{jk}v$	$\displaystyle=\hbox{$j$-th column of $\sqrt{d}T([0\ldots 0,v,0\ldots 0])$ \quad($v$ is the $k$-th column)}$
		$\displaystyle={1\over\sqrt{d}}\sum_{a,b}[0\ldots 0,v,0\ldots 0]_{ab}(S^{a}\Omega^{b})^{*}e_{j}={1\over\sqrt{d}}\sum_{a}v_{a}\Omega^{-k}S^{-a}e_{j}$
		$\displaystyle={1\over\sqrt{d}}\sum_{a}(\Omega^{-k}P_{-1}S^{-j}e_{a})v_{a}={1\over\sqrt{d}}(\Omega^{-k}P_{-1}S^{-j})v,$

so that

$$A_{jk}={1\over\sqrt{d}}\Omega^{-k}P_{-1}S^{-j},\qquad(A_{jk})_{ab}={1\over\sqrt{d}}\omega^{-ak}\delta_{a,j-b}.$$

The $(j,k)$-block $B_{jk}$ of $[\sqrt{d}T]^{2}$ is given by

	$\displaystyle(B_{jk})_{ab}$	$\displaystyle=\bigl{(}\sum_{r}A_{jr}A_{rk}\bigl{)}_{ab}=\sum_{r}\sum_{t}(A_{jr})_{at}(A_{rk})_{tb}={1\over d}\sum_{r}\sum_{t}\omega^{-ar}\delta_{a,j-t}\omega^{-tk}\delta_{t,r-b}$
		$\displaystyle={1\over d}\omega^{-a(j-a+b)}\omega^{-(j-a)k}={1\over d}\omega^{a^{2}-aj+ak-jk-ab}.$

The $(j,k)$-block $C_{jk}$ of $[\sqrt{d}T]^{3}$ is given by

	$\displaystyle(C_{jk})_{ab}$	$\displaystyle=\bigl{(}\sum_{r}B_{jr}A_{rk}\bigl{)}_{ab}=\sum_{r}\sum_{t}(B_{jr})_{at}(A_{rk})_{tb}=\sum_{r}\sum_{t}(B_{jr})_{at}(A_{rk})_{tb}$
		$\displaystyle={1\over d\sqrt{d}}\sum_{r}\sum_{t}\omega^{a^{2}-aj+ar-jr-at}\omega^{-tk}\delta_{t,r-b}={1\over d\sqrt{d}}\sum_{r}\omega^{a^{2}-aj+ar-jr-a(r-b)}\omega^{-(r-b)k}$
		$\displaystyle={1\over d\sqrt{d}}\omega^{a^{2}-aj+ab+bk}\sum_{r}\omega^{-r(j+k)}={1\over\sqrt{d}}\omega^{a^{2}-aj+ab+bk}\delta_{j,-k}=(R^{2}\Omega^{-j}F\Omega^{k})_{ab}\delta_{j,-k},$

so that

$$C_{jk}=\begin{cases}0,&k\neq-j;\cr R^{2}\Omega^{-j}F\Omega^{-j},&k=-j.\end{cases}$$

It therefore follows, that $[\sqrt{d}T]^{6}$ is block diagonal, with diagonal blocks

$$Q_{jj}=C_{j,-j}C_{-j,j}=(R^{2}\Omega^{-j}F\Omega^{-j})(R^{2}\Omega^{j}F\Omega^{j})=\Omega^{-j}(R^{2}F)^{2}\Omega^{j}.$$

Thus $[\sqrt{d}T]^{6d}$ is block diagonal, and, by Lemma 3.1, its diagonal blocks simplify to

$$\Omega^{-j}(R^{2}F)^{2d}\Omega^{j}=\Omega^{-j}(-1)^{{1\over 2}d(d-1)}I\Omega^{j}=(-1)^{{1\over 2}d(d-1)}I,$$

i.e., $[(\sqrt{d}T)^{6d}]=[(-1)^{{1\over 2}d(d-1)}I]$.