Kochen-Specker sets in four-dimensional spaces

TL;DR

This paper introduces an analytical method to construct an infinite family of Kochen-Specker sets in four-dimensional space, providing a computer-free proof, advancing the understanding of quantum contextuality.

Contribution

It presents the first analytical construction of an infinite family of Kochen-Specker sets in R^4, moving beyond computer-based methods.

Findings

Constructed an infinite family of Kochen-Specker sets in R^4

Provided a short, computer-free proof of the sets' properties

Enhanced understanding of quantum contextuality in four dimensions

Abstract

For the first time we construct an infinite family of Kochen-Specker sets in a space of fixed dimension, namely in R^4. While most of the previous constructions of Kochen-Specker sets have been based on computer search, our construction is analytical and it comes with a short, computer-free proof.

Equations

$$R_{m,s}=\left(\begin{array}[]{cc}\cos\left(\frac{2\pi s}{m}\right)&-\sin\left(\frac{2\pi s}{m}\right)\\ \sin\left(\frac{2\pi s}{m}\right)&\cos\left(\frac{2\pi s}{m}\right)\end{array}\right).$$

$$c^2=-\frac{\cos \left(2\pi \left(\frac{k_p}{p}-\frac{k_q}{q}\right)\right)}{\cos \left(2\pi \left(\frac{k_p}{p}+\frac{k_q}{q}\right)\right)}.$$

$$a=\left(\begin{array}[]{c}(1-c)\cos\frac{2\pi k_{q}}{q}\\ (1-c)\sin\frac{2\pi k_{q}}{q}\\ -(1+c)\sin\frac{2\pi k_{q}}{q}\\ (1-c)\cos\frac{2\pi k_{q}}{q}\end{array}\right),\ \ \ b=\left(\begin{array}[]{c}c+1\\ 0\\ 0\\ c-1\end{array}\right).$$

$$k_x=\{\begin{array}{l}\lceil \frac{x}{4}\rceil \text{mbox}ifx\equiv 3(\mathrm{mod}4)\\ \lfloor \frac{x}{4}\rfloor \text{mbox}ifx\equiv 1(\mathrm{mod}4).\end{array}$$

$$\frac{\pi }{2}<2\pi \left(\frac{k_p}{p}+\frac{k_q}{q}\right)<\frac{3\pi }{2}\text{mbox}and-\frac{\pi }{2}<2\pi \left(\frac{k_p}{p}-\frac{k_q}{q}\right)<\frac{\pi }{2}$$

$$-\frac{\cos \left(2\pi \left(\frac{k_p}{p}-\frac{k_q}{q}\right)\right)}{\cos \left(2\pi \left(\frac{k_p}{p}+\frac{k_q}{q}\right)\right)}>0.$$

$$a=\left(\begin{array}[]{c}1-c\\ 0\end{array}\right)\otimes R_{q}\left(\begin{array}[]{c}1\\ 0\end{array}\right)+\left(\begin{array}[]{c}0\\ 1+c\end{array}\right)\otimes R_{q}\left(\begin{array}[]{c}0\\ 1\end{array}\right)$$

$$b=\left(\begin{array}[]{c}c+1\\ 0\end{array}\right)\otimes\left(\begin{array}[]{c}1\\ 0\end{array}\right)+\left(\begin{array}[]{c}0\\ c-1\end{array}\right)\otimes\left(\begin{array}[]{c}0\\ 1\end{array}\right).$$

$$a^T(R_p^m\otimes R_q^n)b=0.$$

$$a^T(R_p^m\otimes R_q^n)b$$

$$\left(0,\>1+c\right)R_{p}^{m}\left(\begin{array}[]{c}c+1\\ 0\end{array}\right)=\left(1-c,\>0\right)R_{p}^{m}\left(\begin{array}[]{c}0\\ c-1\end{array}\right)=0$$

$$\left(0,\>1\right)R_{q}^{n-1}\left(\begin{array}[]{c}1\\ 0\end{array}\right)=\left(1,\>0\right)R_{q}^{n-1}\left(\begin{array}[]{c}0\\ 1\end{array}\right)=0.$$

$$a^T(R_p^m\otimes R_q^n)b=(1-c^2)(R_p^m{)}_{1,1}(R_q^{n-1}{)}_{1,1}+(c^2-1)(R_p^m{)}_{2,2}(R_q^{n-1}{)}_{2,2}=0$$

$$a^T{M}^{1-i}{M}^{i-r}b=a^T{M}^{1-r}b=a^T(I\otimes R_q^2)b.$$

$$a^T(R_p^{-1}\otimes R_q^{-1})a$$

$$(2c^2+2)(R_p{)}_{1,1}(R_q{)}_{1,1}+(2-2c^2)(R_p{)}_{1,2}(R_q{)}_{1,2}.$$

$$c^2$$

$$b^T(R_p^{-1}\otimes R_q)b$$

$$(M^{i}a{)}_2$$

$$=$$

$$(M^{i}a{)}_3$$

$$=$$

$$(R_p^m{)}_{1,1}(R_q^{n-1}{)}_{1,2}=(R_p^m{)}_{2,1}(R_q^{n-1}{)}_{2,2}=0.$$

$$Z=\left(\begin{array}[]{cc}0&-1\\ 1&0\end{array}\right).$$

$$M^{i}a=\left(\begin{array}[]{c}\pm(1-c)\\ 0\\ 0\\ \pm(1+c)\end{array}\right)$$

$$\{\{a,a+1\}:a\in {\mathbb{Z}}_n\}\cup \{\{a,a+k\}:a\in {\mathbb{Z}}_n\}.$$

Full text

1 Introduction

The Kochen-Specker theorem (KS theorem) is an important result in quantum mechanics [4]. It demonstrates the contextuality of quantum mechanics, which is one of its properties that may become crucial in quantum information theory [3]. In this paper we focus on proofs of the KS theorem that are given by showing that, for $n\geq 3$, there does not exist a function $f:{\mathbb{C}}^{n}\rightarrow\{0,1\}$ such that for every orthogonal basis $B$ of ${\mathbb{C}}^{n}$ there exists exactly one vector $x\in B$ such that $f(x)=1$ (where ${\mathbb{C}}^{n}$ denotes the $n$-dimensional vector space over the field of complex numbers). This particular approach has been used in many publications, see for example [1, 5, 8] and many references cited therein. Definition 1.1 given below formalizes one common way of constructing such proofs, using a simple parity argument. Therefore the structures that satisfy Definition 1.1 are sometimes referred to as parity proofs of KS theorem.

Definition 1.1**.**

We say that $({\mathcal{V}},{\mathcal{B}})$ is a Kochen-Specker pair in ${\mathbb{C}}^{n}$ if it meets the following conditions:

• (1)

${\mathcal{V}}$ is a finite set of vectors in ${\mathbb{C}}^{n}$.

• (2)

${\mathcal{B}}=(B_{1},\ldots,B_{k})$ where $k$ is odd, and for all $i=1,\ldots,k$ we have that $B_{i}$ is an orthogonal basis of ${\mathbb{C}}^{n}$ and $B_{i}\subset{\mathcal{V}}$.

• (3)

For each $v\in{\mathcal{V}}$ the number of $i$ such that $v\in B_{i}$ is even.

It is quite common in the literature [2, 5, 8] to refer to a KS pair as a KS set, and we will do so sometimes in this paper.

An extensive and recent summary of known examples of KS sets in low dimensions is presented in [7]. It turns out that until recent time, the vast majority of known examples have been found by computer search, however without much insight in the sets generated [7]. More recently, first computer-free constructions have appeared that relate KS sets to some other mathematical structures such as Hadamard matrices [6]. In this paper we continue this trend by giving a simple, computer-free construction of an infinite family of KS sets in ${\mathbb{R}}^{4}$. This is the first time that an infinite family of KS sets in a space of fixed dimension is found. Moreover, four is the smallest possible dimension in which KS sets described in Definition 1.1 can exist, since the definition clearly requires the dimension to be even, and the KS theorem only holds in dimension at least 3.

KS sets are key tools for proving some fundamental results in quantum theory and they also have various applications in quantum information processing, see [2, 5] and the references therein.

2 The new construction

For integers $m>0$ and $s\geq 0$ we define the matrix

[TABLE]

By $\otimes$ we denote the Kronecker product of matrices. We now state our main result.

Theorem 2.1**.**

Let $p,q\geq 3$ be relatively prime odd integers, let $k_{p},k_{q}$ be integers relative prime to $p$ and $q$ respectively, and let $c$ be a non-zero real number such that

[TABLE]

Let $a,b$ be vectors in ${\mathbb{R}}^{4}$ defined by

[TABLE]

Let $r$ be the unique integer in the interval $(0,pq)$ such that $r\equiv 1\pmod{p}$, $r\equiv-1\pmod{q}$. Let $M=R_{p,k_{p}}\otimes R_{q,k_{q}}$. We have:

(i) For $0\leq i<pq$ the set $B_{i}=\{M^{i}a,M^{i-1}a,M^{i}b,M^{i-r}b\}$ is an orthogonal basis of ${\mathbb{R}}^{4}$.

(ii) Let ${\mathcal{V}}=\{M^{i}a\>:\>0\leq i<pq\}\cup\{M^{i}b\>:\>0\leq i<pq\}$ and let ${\mathcal{B}}=\{B_{i}\>:\>0\leq i<pq\}$. Then $({\mathcal{V}},{\mathcal{B}})$ is a Kochen-Specker pair.

Proof.

(i) The fact that $B_{0}$ is an orthogonal basis of ${\mathbb{R}}^{4}$ is proved in Lemma 2.3 below. Since $M$ is orthogonal, it follows that $B_{i}$ is an orthogonal basis of ${\mathbb{R}}^{4}$ for each $i$, since $B_{i}$ is the image of $B_{0}$ under $M^{i}$.

(ii) Let the indices of $B$ be taken modulo $pq$. For each $i$, the vector $M^{i}a$ belongs to two bases, namely $B_{i}$ and $B_{i+1}$, and the vector $M^{i}b$ belongs to two bases, namely $B_{i}$ and $B_{i+r}$. In Proposition 2.4 below we show that the vectors in the set ${\mathcal{V}}$ are pairwise linearly independent. Hence each element of ${\mathcal{V}}$ belongs to exactly two bases in ${\mathcal{B}}$, and all conditions for Kochen-Specker pair are satisfied. ∎

The condition that $p$ and $q$ are relatively prime is used to guarantee the existence of the integer $r$ with the given properties, using the Chinese remainder theorem.

We note that for given $p$ and $q$ it is always possible to choose $k_{p}$ and $k_{q}$ such that $c$ is a non-zero real number. There are many possible choices; one of them is

[TABLE]

Then it is easy to show that

[TABLE]

hence

[TABLE]

We will now work towards proving that $B_{0}$ is an orthogonal basis. For simplicity we will write just $R_{p}$ and $R_{q}$ instead of $R_{p,k_{p}}$ and $R_{q,k_{q}}$, respectively. To reduce the use of brackets in upcoming calculations we will assume throughout that the ordinary matrix product has a higher precedence among algebraic operations than the Kronecker product. For example, the notation $X\otimes YZ$ denotes $X\otimes(YZ)$.

We note that the vectors $a$ and $b$ defined in Theorem 2.1 can be written as

[TABLE]

and

[TABLE]

Lemma 2.2**.**

Assume that $m=0$ or $n=1$. Then

[TABLE]

Proof.

Since the matrices $R_{p}$ and $R_{q}$ are orthogonal, we have

[TABLE]

After expanding the right-hand side into four terms and simplifying each of them as $(T\otimes U)(V\otimes W)(X\otimes Y)=TVX\otimes UWY$ we get

[TABLE]

If $m=0$ then

[TABLE]

while if $n=1$ then

[TABLE]

Hence if $m=0$ or $n=1$ then

[TABLE]

because $(R_{s}^{u})_{1,1}=(R_{s}^{u})_{2,2}$ for all $s,u$. ∎

Lemma 2.3**.**

The set $B_{0}$ defined in Theorem 2.1 is an orthogonal basis of ${\mathbb{R}}^{4}$.

Proof.

The matrix $M$ is orthogonal, hence $M^{j}=(M^{-j})^{T}$ for all $j$. Note that $M^{-r}=R_{p}^{-1}\otimes R_{q}$ by the properties of $r$. The dot product of $b$ and $M^{-1}a$ can be written as $a^{T}Mb$. Furthermore the dot product of $M^{i-1}a$ and $M^{i-r}b$ can be written as

[TABLE]

Therefore we need to prove the following six equalities, which we write in a uniform way that allows us to treat four of the six cases simultaneously.

1. (i)

$a^{T}\left(I\otimes I\right)b=0$ 2. (ii)

$a^{T}\left(R_{p}^{-1}\otimes R_{q}^{-1}\right)a=0$ 3. (iii)

$b^{T}\left(R_{p}^{-1}\otimes R_{q}\right)b=0$ 4. (iv)

$a^{T}\left(R_{p}^{-1}\otimes R_{q}\right)b=0$ 5. (v)

$a^{T}\left(R_{p}\otimes R_{q}\right)b=0$ 6. (vi)

$a^{T}\left(I\otimes R_{q}^{2}\right)b=0$.

Equalities (i), (iv), (v) and (vi) follow from Lemma 2.2.

In case (ii) we get

[TABLE]

Since $(R_{s})_{1,1}=(R_{s})_{2,2}$ and $(R_{s})_{1,2}=-(R_{s})_{2,1}$ for each $s$, the last expression simplifies to

[TABLE]

This is equal to 0 exactly when

[TABLE]

In case (iii) we get

[TABLE]

which is equal to (12). Hence assuming that $c^{2}$ equals the expression (13), which is a necessary and sufficient condition for equality (ii) to hold, also implies that equality (iii) holds.

This completes the proof of Lemma 2.3 and hence also the proof of Theorem 2.1. ∎

Next we show that the $2pq$ vectors used in our construction are pairwise linearly independent. Strictly speaking this is not required for the proof of the Kochen-Specker pair property, however it may be of interest for example in the physical implementations of our construction.

Proposition 2.4**.**

Let $M,a,b$ be as in Theorem 2.1. The $2pq$ vectors$M^{i}a\ (0\leq i<pq)$ and $M^{i}b\ (0\leq i<pq)$ are pairwise linearly independent.

Proof.

For a positive integer $s$ let $\zeta_{s}=e^{2\pi\sqrt{-1}/s}$ denote the primitive $s$-th root of unity in ${\mathbb{C}}$. The eigenvalues of $R_{p,k_{p}}^{s}\otimes R_{q,k_{q}}^{t}$ are $\zeta_{p}^{\pm sk_{p}}\zeta_{q}^{\pm tk_{q}}$ with all four combinations of $\pm$ signs in the exponents. Since $\gcd(p,q)=\gcd(p,k_{p})=\gcd(q,k_{q})=1$ and $p,q$ are odd, it follows that the eigenvalues $\zeta_{p}^{\pm sk_{p}}\zeta_{q}^{\pm tk_{q}}$ are not real unless $s=t=0$. It follows that $M^{i}$ does not have a real eigenvector for $0<i<pq$, hence $M^{i}a$ and $M^{j}a$ are linearly independent whenever$i\not\equiv j\pmod{pq}$, and likewise $M^{i}b$ and $M^{j}b$ are linearly independent whenever $i\not\equiv j\pmod{pq}$.

It remains to consider the possibility that $M^{i}a$ and $b$ are linearly dependent for some $i$. Then $(M^{i}a)_{2}=(M^{i}a)_{3}=0$. Let $M^{-i}=R_{p}^{m}\otimes R_{q}^{n}$. Explicit calculations give

[TABLE]

Assume that the right-hand sides of (24) and (30) both equal zero, and subtract the latter from the former. This gives $(R_{p}^{m})_{1,1}(R_{q}^{n-1})_{1,2}=-(R_{p}^{m})_{2,1}(R_{q}^{n-1})_{2,2}$. After plugging this into (24) and performing some elementary manipulations we deduce that $c=0$ or

[TABLE]

Since we assume a choice of $k_{p},k_{q}$ such that $c\neq 0$, we must have (31). Denote

[TABLE]

A quick argument shows that (31) implies that either both $R_{p}^{m}$ and $R_{q}^{n-1}$ are equal to $Z$ or $-Z$, or they are both equal to $I$ or $-I$. The first case is not possible, since $p$ and $q$ are odd. So assume that both $R_{p}^{m}$ and $R_{q}^{n-1}$ are equal to $I$ or $-I$. Then a calculation similar to those above shows that

[TABLE]

where signs are to be taken consistently. In comparison with equation (1) we see that if $M^{i}a$ and $b$ are linearly dependent, then $c\neq\pm 1$, and $\frac{1-c}{c+1}=\frac{1+c}{c-1}$. The last equation has no real solution, and this completes the proof. ∎

For integers $n$ and $k$ such that $1<k<n-1$ let ${\rm CR}(n,k)$, the chordal ring with parameters $n$ and $k$, be the graph with vertex set ${\mathbb{Z}}_{n}$ (integers modulo $n$) and with edge set

[TABLE]

That is, ${\rm CR}(n,k)$ is the $n$-cycle (“ring”) to which all chords connecting vertices at distance $k$ were added.

Let $G$ be the graph whose vertices are the $2pq$ vectors $M^{i}a\ (0\leq i<pq)$ and $M^{i}b\ (0\leq i<pq)$ forming our KS set (Theorem 2.1) and two vertices are adjacent if and only if they both belong to the basis $B_{i}$ for some $i$. Then it is easily seen from the construction of the bases $B_{i}$ in Theorem 2.1(i) that $G$ is the line graph of the chordal ring ${\rm CR}(pq,r)$. It is of special interest in quantum information theory to know that the graph $G$ is vertex transitive, which is the final statement of this paper.

Proposition 2.5**.**

Let $p$, $q$ and $r$ be as in Theorem 2.1. The graph ${\rm CR}(pq,r)$ is edge transitive.

Proof.

Consider any two edges of ${\rm CR}(pq,r)$. If they are both “ring” edges or they are both chords, then there is a cyclic shift of the vertices of ${\rm CR}(pq,r)$ which is an automorphism of ${\rm CR}(pq,r)$ and it maps one of the edges to the other edge.

We are left with the case when one of the edges is a “ring” edge and the other edge is a chord. Without loss of generality assume that the edges are $\{0,1\}$ and $\{0,r\}$. By the assumption of Theorem 2.1 we have $r\equiv 1\pmod{p}$ and $r\equiv-1\pmod{q}$. Therefore $r$ is relatively prime to $pq$ and $r^{2}\equiv 1\pmod{pq}$ and $r\not\equiv\pm 1\pmod{pq}$. Therefore the mapping $\varphi:a\mapsto ra$ is a bijection from ${\mathbb{Z}}_{pq}$ to ${\mathbb{Z}}_{pq}$, the vertex set of ${\rm CR}(pq,r)$. Its action on ring edges is $\varphi(\{a,a+1\})=\{ra,ra+r\}$ hence each ring edge is mapped to a chord. Its action on chords is $\varphi(\{a,a+r\})=\{ra,ra+r^{2}\}=\{ra,ra+1\}$ hence each chord is mapped to a ring edge. Therefore $\varphi$ is an automorphism of ${\rm CR}(pq,r)$. Since $\varphi(\{0,1\})=\{0,r\}$, it follows that ${\rm CR}(pq,r)$ is edge transitive. ∎

Acknowledgement

Research of both authors was supported in part by the Natural Sciences and Engineering Research Council of Canada (NSERC).