From stabilizer states to SIC-POVM fiducial states

Abstract

In the stabilizer formalism of quantum computation, the Gottesman–Knill theorem shows that universal fault-tolerant quantum computation requires the resource called magic (nonstabilizerness). Thus stabilizer states serve as “classical states,” and states beyond them are necessary for genuine quantum computation. Characterization, detection, and quantification of magic states are basic issues in this context. In the paradigm of quantum measurement, symmetric informationally complete positive operator valued measures (SIC-POVMs, further abbreviated as SICs) play a prominent role due to their structural symmetry and remarkable features. However, their existence in all dimensions, although strongly supported by extensive theoretical and numerical evidence, remains an elusive open problem (Zauner’s conjecture). A standard method for constructing SICs is via the orbit of the Heisenberg–Weyl group on a fiducial state, and most known SICs arise in this way. A natural question arises regarding the relation between stabilizer states and fiducial states. In this paper, we connect them by showing that they are on two extremes with respect to the $p$-norms of characteristic functions of quantum states. This not only reveals a simple path from stabilizer states to SIC fiducial states, showing quantitatively that they are as far away as possible from each other, but also provides a simple reformulation of Zauner’s conjecture in terms of extremals for the $p$-norms of characteristic functions. A convenient criterion for magic states and some interesting open problems are also presented.

Appendix

Here, we present the detailed proof of the basic properties of $M_p(\rho)$.

Property 1 is evident from the definition by noting that any Clifford operator permutes the Heisenberg–Weyl operators via conjugation, i.e., $VWV^\dagger\in\mathcal P_d$, for all $V\in\mathcal C_d$, $W\in\mathcal P_d$.

Property 2 follows from the classical Minkowski inequality.

Property 3 follows from the abstract Parseval theorem by noting that $\bigl\{\frac{1}{\sqrt d}D_{k,l}\colon k,l\in\mathbb{Z}_d\bigr\}$ constitutes an orthonormal basis for the operator space $L(\mathbb{C}^d)$, which implies that

$$\frac{1}d \sum_{k,l}\lambda_{k,l}^2(\rho)= \operatorname{tr} \rho^2.$$

In particular, for any pure state $\rho=|\psi\rangle\langle \psi |$, we have $\operatorname{tr} \rho^2=1$, which in turn implies Eq. (9).

For property 4, we first note that $|c_{k,l}(\rho)| \le 1$, which follows from the Cauchy–Schwarz inequality as

$$|c_{k,l}(\rho)|^2=| \operatorname{tr} (D_{k,l}\rho)|^2=| \operatorname{tr} (D_{k,l}\sqrt{\rho}\sqrt{\rho}\,)|^2\le \operatorname{tr} \bigl[(D_{k,l}\sqrt{\rho}\,)^\dagger (D_{k,l}\sqrt{\rho}\,)\bigr] \operatorname{tr} (\sqrt{\rho}\sqrt{\rho}\,)=1.$$

Now because $\lambda_{k,l}(\rho)=|c_{k,l}(\rho)|\in[0,1]$ and $p\in[1,2)$, we have

$$ M_p(\rho)=\biggl(\,\sum_{k,l}\lambda_{k,l}^p(\rho)\biggr)^{\!1/p}\ge \biggl(\,\sum_{k,l}\lambda_{k,l}^2(\rho)\biggr)^{\!1/p}=d^{1/p},$$

(A.1)

which implies the above lower bound in inequality (10). In particular, noting orthogonality relation (3), it can be readily verified that any stabilizer state provides the lower bound because any stabilizer state is defined via Eq. (4) in which $\mathcal A$ is a maximal Abelian subgroup of $\mathcal P_d$ containing exactly $d$ elements, i.e., $|\mathcal A|=d$.

Next we show that if the lower bound in inequality (10) is attained by the pure state $\rho=|\psi\rangle\langle\psi|$, then $\rho$ must be a stabilizer state. In fact, suppose the lower bound in inequality (10) is attained by a pure state $\rho$; then the inequality in (A.1) is saturated if and only if $\lambda_{k,l}(\rho)=0$ or 1 (noting that $0\le\lambda_{k,l}(\rho)\le 1)$. By property 3,

$$M_p(\rho)=\biggl(\,\sum_{k,l}\lambda_{k,l}^p\biggr)^{\!1/p}=d^{1/p},$$

we conclude that there are $d$ elements among $\lambda_{k,l}(\rho)$ (including $\lambda_{0,0}(\rho)=1)$ that take the value 1 (we let the index set of such elements be denoted as $L\subset\mathbb{Z}_d\times\mathbb{Z}_d$), and all other take the value 0, that is,

$$\lambda_{k,l}(\rho)=|\langle\psi|D_{k,l}|\psi\rangle |=\begin{cases} 1,& (k,l)\in L, \\ 0,& \text{otherwise}. \end{cases}$$

For any $(k,l)\in L$, define $A_{k,l}=\langle\psi |D_{k,l}|\psi\rangle^{-1}D_{k,l}$; then $\langle\psi |A_{k,l}|\psi\rangle=1$, which implies that ${A_{k,l}|\psi\rangle=|\psi\rangle}$, that is, $|\psi\rangle$ is a common eigenstate of the operators in $\mathcal A=\{A_{k,l}\colon(k,l)\in L\}$ with the common eigenvalue 1. By Eq. (2), we conclude that $A_{k,l}A_{s,t}=cA_{s,t}A_{k,l}$ for some constant $c$ with $|c|=1$. From $A_{k,l}|\psi\rangle=A_{s,t}|\psi\rangle=|\psi\rangle$ for $(k,l),(s,t)\in L$, it follows that $c=1$ when $(k,l),(s,t)\in L$. Consequently, $\mathcal A$ is an Abelian subgroup of the Heisenberg–Weyl group with $d$ elements (a maximal Abelian subgroup), and $|\psi\rangle$ is a common eigenstate of the operators in $\mathcal A$ with the common eigenvalue 1, and therefore $|\psi\rangle$ is a stabilizer state.

Now, we proceed to establish the right-hand inequality in (10). By writing

$$M_p(\rho)=\biggl(\,\sum_{k,l}\lambda_{k,l}^p(\rho)\biggr)^{\!1/p}=\biggl(1+\sum_{k,l\neq 0}\lambda_{k,l}^p(\rho)\biggr)^{\!1/p}$$

and invoking the Hölder inequality for $p\in[1,2)$, we have

$$\begin{aligned} \, \sum_{k,l\neq 0}\lambda_{k,l}^p(\rho)&\le \biggl(\,\sum_{k,l\neq 0}(\lambda_{k,l}^p(\rho))^{2/p}\biggr)^{\!p/2}\biggl(\,\sum_{k,l\neq 0}1\biggr)^{\!1-p/2}= \\ &=\biggl(\,\sum_{k,l}\lambda_{k,l}^2(\rho)-1\biggr)^{\!p/2}(d^2-1)^{1-p/2}= \\ &=(d-1)^{p/2}(d^2-1)^{1-p/2} = (d-1)(d+1)^{1-p/2}, \end{aligned}$$

which implies the upper bound in inequality (10). The upper bound can be attained by any fiducial state (assuming its existence), which can be verified as follows. Suppose that $|f\rangle$ is a SIC fiducial state and let $|f_{k,l}\rangle=D_{k,l}|f\rangle$; then $|f_{0,0}\rangle=|f\rangle$, and

$$\biggl\{E_{k,l}=\frac{1}d |f_{k,l}\rangle\langle f_{k,l}|\colon k,l\in\mathbb{Z}_d\biggr\}$$

constitutes a SIC. In particular,

$$|\langle f|f_{k,l}\rangle|^2=\frac{1}{d+1},\qquad(k,l)\neq(0,0).$$

It follows that

$$\begin{aligned} \, M_p(|f\rangle\langle f|)&=\biggl(\,\sum_{k,l}| \operatorname{tr} (D_{k,l} |f\rangle\langle f|)|^p\biggr)^{\!1/p}= \biggl(1+\sum_{k,l\neq 0}|\langle f|f_{k,l}\rangle|^p\biggr)^{\!1/p}= \\ &=\biggl(1+(d^2-1)\biggr(\frac{1}{d+1}\biggr)^{\!p/2\,}\biggr)^{\!1/p}=(1+(d-1)(d+1)^{1-p/2})^{1/p}, \end{aligned}$$

which is precisely the upper bound in inequality (10).

Conversely, if the upper bound in inequality (10) is attained by the state $\rho=|\psi\rangle\langle \psi|$, then by the equality condition of the Hölder inequality, we conclude that all $\lambda_{k,l}(\rho)=|c_{k,l}(\rho)|=| \operatorname{tr} (D_{k,l}\rho)|$ are equal for $(k,l)\neq(0,0)$. By property 3,

$$\sum_{k,l\neq 0}\lambda_{k,l}^2(\rho)+1=d,$$

it follows that

$$(d^2-1)| \operatorname{tr} (D_{k,l}\rho)|^2=|\langle \psi|D_{k,l}|\psi\rangle|^2=d-1,\qquad(k,l)\neq (0,0)$$

which in turn implies that

$$|\langle\psi|D_{k,l}|\psi\rangle|^2=\frac{1}{d+1},\qquad (k,l)\neq (0,0).$$

Furthermore, by Eq. (2), the above equation implies that

$$|\langle\psi_{k,l}|\psi_{s,t}\rangle|^2=\frac{1}{d+1},\qquad (k,l)\neq (s,t),$$

where $|\psi_{k,l}\rangle=D_{k,l}|\psi\rangle$. Consequently, $|\psi\rangle$ is a fiducial state.

For property 5, noting that $\lambda_{k,l}(\rho)\in[0,1]$ and $p>2$, we have

$$M_p(\rho)=\biggl(\,\sum_{k,l}\lambda_{k,l}^p(\rho)\biggr)^{\!1/p}\le\biggl(\,\sum_{k,l}\lambda_{k,l}^2(\rho)\biggr)^{\!1/p}=d^{1/p},$$

which implies the upper bound in inequality (11). That this upper bound is attained if and only if $\rho$ is a stabilizer state can be established similarly to the lower bound case in property 4.

When $p>2$, by the (reverse) Hölder inequality, we have

$$\sum_{k,l\neq 0}\lambda_{k,l}^p(\rho)\ge \biggl(\,\sum_{k,l\neq 0}(\lambda_{k,l}^p(\rho))^{2/p}\biggr)^{\!p/2}\biggl(\,\sum_{k,l\neq 0}1\biggr)^{\!1-p/2}= (d-1)(d+1)^{1-p/2},$$

which implies the lower bound in inequality (11). That this lower bound can be attained if and only if $\rho$ is a fiducial state follows from the same argument as in property 4.