Projective Ring Line of a Specific Qudit

Introduction

The study of the finite-dimensional Hilbert spaces and their associated generalized Pauli operators has been a forefront issue of the quantum information theory within the past few years. A substantial mathematical insight has been possible thanks to a number of novel graph-combinatorial and algebraic geometrical concepts employed, see, e. g., – and references therein. Among the latter, it is the concept of a projective line defined over a(n associative) ring with unity that acquired a distinguished footing –. In this approach, one simply identifies the points of a projective ring line with the generalized Pauli operators (or the maximum commuting sets of them) pertaining to a given Hilbert space and rephrases their commutation relations in terms of neighbour/distant relations between the points on the line in question. Given this identification, it was possible to “projective-ring-geometrize” any $N$-qubit Hilbert space –, two-qutrits , as well as to get important hints about the smallest composite case, viz. a six-dimensional Hilbert space . A detailed examination of these particular cases led soon to a discovery of a more complex and unifying approach based on group-theoretical considerations . Adopting and properly generalizing the strategy pursued in the last two mentioned papers, we shall demonstrate, on the example of a specific single qudit, that the concept of a projective ring line naturally emerges also in a context slightly different from that introduced and elaborated in –, with the finest traits of the structure of the projective line coming into play.

The Pauli group G𝐺G of a single qudit

The subgroup of the unitary group $\operatorname{U}_{d}$ generated by $X$ and $Z$, known by physicists as the (generalized) Pauli group, will be written as $G$. The operators $X^{0}=:I,X^{1},\ldots,X^{d-1}$ form a cyclic subgroup of $G$ with order $d$; the same properties hold for $Z^{0},Z^{1},\ldots,Z^{d-1}$. Hence

By virtue of (1) and (2), each element of $G$, usually referred to as a (generalized) Pauli operator, can be written in the normal form

It is easy to see that this representation in normal form is unique: From $\omega^{a}X^{b}Z^{c}=\omega^{a^{\prime}}X^{b^{\prime}}Z^{c^{\prime}}$ follows $\omega^{a-a^{\prime}}X^{b-b^{\prime}}Z^{c-c^{\prime}}=I$. As $|0\rangle$ remains fixed under $Z^{c-c^{\prime}}$ we obtain $\omega^{a-a^{\prime}}X^{b-b^{\prime}}|0\rangle=|0\rangle$. This shows $b-b^{\prime}=0$ and $a-a^{\prime}=0$. Thus $Z^{c-c^{\prime}}=I$ which implies $c-c^{\prime}=0$, as required. The uniqueness of the normal form (3) will be crucial for our further exhibition.

We immediately may read off from (2) the following rule for multiplication in $G$, when the factors are given in normal form:

Observe that the product is also in normal form. The term $b^{\prime}c$ in the exponent of $\omega$ on the right hand side shows that $G$ is a non-commutative group. The uniqueness of the normal form implies also that $G$ is a group of order $|G|=d^{3}$.

The commutatorWe shall always be concerned with commutator of operators in the sense of group theory. It must not be confused with the commutator from ring theory which uses addition and multiplication of operators. of two operators $W$ and $W^{\prime}$ is

If $W=\omega^{a}X^{b}Z^{c}$ and $W^{\prime}=\omega^{a^{\prime}}X^{b^{\prime}}Z^{c^{\prime}}$ are given in normal form then it is immediate from (2) that

Recall that two operators commute if, and only if, their commutator (taken in any order) is equal to $I$.

We shall be concerned with two important normal subgroups of $G$:

The centre $Z(G)$ of $G$ is the set of all operators in $G$ which commute with every operator in $G$. An operator $\omega^{a}X^{b}Z^{c}$ given in normal form lies in $Z(G)$ precisely when (5) holds for any choice of $a^{\prime}$, $b^{\prime}$, and $c^{\prime}$. Setting $b^{\prime}:=0$, $c^{\prime}:=1$ we get $b=0$, whereas $b^{\prime}:=1$ and $c^{\prime}:=0$ gives then $c=0$. These necessary conditions are also sufficient, whence

Note that $Z(G)$ is yet another cyclic subgroup of $G$ with order $d$.

The commutator subgroup $[G,G]$ is the smallest subgroup of $G$ which contains all commutators $[W,W^{\prime}]$ with $W,W^{\prime}\in G$. We follow the usual convention to denote the commutator subgroup of $G$ by $G^{\prime}$. From $[Z,X]=\omega I$ follows that all powers of $\omega I$ are elements of $G^{\prime}$. On the other hand (5) shows that there are no other commutators but the powers of $\omega I$. Altogether we obtain

It is easy to see from (5) that each element of $G^{\prime}$ is indeed a commutator, a property which need not be true for the commutator subgroup of an arbitrary group.

The ring associated with G𝐺G

A symplectic module associated with G𝐺G and the commutation algebra of Pauli operators

As $(G,\cdot)$ is a non-commutative group, it cannot be isomorphic to the additive group of any module. Recall that the factor group of $G$ by any normal subgroup is commutative if, and only if, this normal subgroup contains the commutator subgroup $G^{\prime}$. This means that the “largest” commutative group we can obtain from $G$ by factorisation is the factor group

Taking into account our normal form (3) and the description of $G^{\prime}$ in (6), the group $G/G^{\prime}$ comprises all cosets

Each element of $G/G^{\prime}$ can be written in a unique way in this normal form. As a by-product of this uniqueness, we learn from (7) that the factor group $G/G^{\prime}$ has order $d^{2}$. Multiplication in $G/G^{\prime}$ is governed by the formula

Recall that our main goal is to describe whether or not two operators of $G$ commute. Since $G/G^{\prime}$ is a commutative group, any information of this kind is eliminated by our passage from $G$ to the factor group $G/G^{\prime}$. This is why in the following construction we use not only the group $G/G^{\prime}$, but also the group $G$ and the commutator subgroup $G^{\prime}$:

Let $G^{\prime}X^{b}Z^{c}$ and $G^{\prime}X^{b^{\prime}}Z^{c^{\prime}}$ be elements of $G/G^{\prime}$ in normal form. We associate with them the commutator

This assignment uses the group $G$. It is independent of the choice of representatives from the cosets $G^{\prime}X^{b}Z^{c}$ and $G^{\prime}X^{b^{\prime}}Z^{c^{\prime}}$, since $a$ and $a^{\prime}$ do not appear on the right hand side of (5).

Summing up, we see that the set of operators in $G$ which commute with a fixed operator $\omega^{a}X^{b}Z^{c}$ corresponds to the perpendicular set (shortly the perp-set) of $(b,c)$, viz.

The perp-set of $(b,c)$ is closed under addition and multiplication by ring elements. Also, it is non empty, since

We are now in a position to state a first, preliminary result about perp-sets.

By (10), a necessary and sufficient condition for $(u,v)$ to lie in $(b,c)^{\perp}$ reads

This is a set of $18-6=12$ vectors, because $(2,0)$, $(4,0)$ and $(0,0)$ are vectors which belong to all three points.

A particular case: d𝑑d is square-free

While Theorem 1 describes the perp-set of any admissible vector, the result for non-admissible vectors is unsatisfactory. The aim of this section is to improve the results of Theorem 1 under the additional hypothesis that the number $d$ is square-free. Throughout this section we adopt the assumption that

of $r$ finite fields. Let us recall how this isomorphism arises: We consider the ring elements

Since each ideal $J^{(k)}$ is isomorphic to a field, the last equation is equivalent to

We are now in a position to state our main result. Note that the set-theoretic union of points gives a set of vectors.

The vector $(b,c)$ is contained in precisely

The set-theoretic union of these points equals the perpendicular set of the vector $(b,c)$.

The perpendicular set of the vector $(b,c)$ satisfies

We obtain $u^{(j)}\neq 0$ from (18), whence $({b^{\prime}}{}^{(j)},{c^{\prime}}{}^{(j)})$ is one of the $p_{j}-1$ distinct multiples of $(b^{(j)},c^{(j)})$ by a non-zero factor in $J^{(j)}$. Next, (19) implies $u^{(k)}=0$, whence $({b^{\prime}}{}^{(k)},{c^{\prime}}{}^{(k)})$ is one of the $p_{k}^{2}-1$ non-zero pairs with entries from $J^{(k)}$. These necessary conditions are also sufficient so that we obtain

As this is a determinant over the field $J^{(j)}$, and because the second row is non-zero, we can define $v^{(j)}\in J^{(j)}$ via $(x^{(j)},y^{(j)})=v^{(j)}(b^{(j)},c^{(j)})$. If $k\in K$ and $(x^{(k)},y^{(k)})\neq(0,0)$ we set $v^{(k)}:=1^{(k)}$, otherwise we let $v^{(k)}:=0$.

Ad (c): By (b), it suffices to count the number of vectors $(b^{\prime\prime},c^{\prime\prime})$ which are a multiple of an admissible vector as described in part (a) of the present proof. For each $j\notin K$ the pair $(b^{\prime\prime}{}^{(j)},c^{\prime\prime}{}^{(j)})$ can be chosen as any of the $p_{j}$ multiples of $(b^{(j)},c^{(j)})$ by a factor in $J^{(j)}$, whereas for each $k\in K$ the pair $(b^{\prime\prime}{}^{(k)},c^{\prime\prime}{}^{(k)})$ can be chosen arbitrarily in $p_{k}^{2}$ ways. Hence there are

With the settings and notations of Theorem 2 the number of operators in the generalized Pauli group $G$ which commute with the operator $\omega^{a}X^{b}Z^{c}\in G$ equals the value given by (17) multiplied by d.

Conclusion

Acknowledgements

This work was supported by the Science and Technology Assistance Agency under the contract $\#$ APVT–51–012704, the VEGA grant agency projects $\#$ 2/6070/26 and $\#$ 7012 and by the $\langle$Action Austria–Slovakia$\rangle$ project $\#$ 58s2 “Finite Geometries Behind Hilbert Spaces.”

References