Quantum Entanglement: An Analysis via the Orthogonality Relation

Abstract

In the literature there has been evidence that a kind of relational structure called a quantum Kripke frame captures the essential characteristics of the orthogonality relation between pure states of quantum systems, and thus is a good qualitative mathematical model of quantum systems. This paper adds another piece of evidence by providing a tensor-product construction of two finite-dimensional quantum Kripke frames. We prove that this construction is exactly the qualitative counterpart of the tensor-product construction of two finite-dimensional Hilbert spaces over the complex numbers, and thus show that composition of quantum systems, especially the phenomenon of quantum entanglement, can be modelled in the framework of quantum Kripke frames. The assumptions used in our construction hint that we need complex numbers in quantum theory. Moreover, for this construction, we give a new and interesting characterization of linear maps of trace 0 in terms of the orthogonality relation.

Appendices

Appendix A

In this appendix, we are going to prove Proposition 2.17 which is crucial to the dimension theory of quantum Kripke frames.

We start from reviewing three results about projective geometry.

The first one is about a way of constructing the linear closure of a set.

Proposition A.1

Let ${\mathfrak {G}} = ( G, {\star } )$ be a projective geometry. For each $A\subseteq G$, define a sequence $\{ A_i \}_{i \in {\mathbb {N}}}$ of subsets of G as follows:

• $A_0=A$;

• $A_{n+1}=\bigcup \{ a\star b\mid a,b\in A_n \}$.

Then ${\mathcal {C}}(A)=\bigcup _{i\in {\mathbb {N}}} A_i$.

Proof

First we prove by induction that $A_i\subseteq {\mathcal {C}}(A)$, for every $i\in {\mathbb {N}}$.

Base Step: $i=0$. By the definition of linear closures $A_0=A\subseteq {\mathcal {C}}(A)$.

Induction Step: $i=n+1$. Let $c\in A_{n+1}$ be arbitrary. By the definition of $A_{n+1}$ there are $a,b\in A_n$ such that $c\in a\star b$. By the induction hypothesis $a,b\in A_n \subseteq {\mathcal {C}}(A)$, so $c\in a\star b\subseteq {\mathcal {C}}(A)$ since ${\mathcal {C}}(A)$ is a subspace.

This finishes the proof by induction. Therefore, $\bigcup _{i\in {\mathbb {N}}} A_i\subseteq {\mathcal {C}}(A)$.

Second we prove that $\bigcup _{i\in {\mathbb {N}}} A_i$ is a subspace including A, and thus ${\mathcal {C}}(A)\subseteq \bigcup _{i\in {\mathbb {N}}} A_i$. By definition $A=A_0\subseteq \bigcup _{i\in {\mathbb {N}}} A_i$. Now let $a,b\in \bigcup _{i\in {\mathbb {N}}} A_i$ be arbitrary. Then there are $n,n'\in {\mathbb {N}}$ such that $a\in A_n$ and $b\in A_{n'}$. Note that by definition $A_i\subseteq A_{i+1}$, for every $i\in {\mathbb {N}}$. Hence $a,b\in A_m$, where $m=max\{ n,n'\}$. Therefore, $a\star b\subseteq A_{m+1}\subseteq \bigcup _{i\in {\mathbb {N}}} A_i$. As a result, $\bigcup _{i\in {\mathbb {N}}} A_i$ is a subspace. $\square$

The second one is a very important and useful result in projective geometry called the projective law.

Theorem A.2

(Corollary 2.4.5 in [38]) Let ${\mathfrak {G}} = ( G, {\star } )$ be a projective geometry. For any non-empty sets $A,B\subseteq G$,

$$\begin{aligned} {\mathcal {C}}(A\cup B)=\bigcup \{ a\star b\mid a\in {\mathcal {C}}(A),\ b\in {\mathcal {C}}(B)\}. \end{aligned}$$

The third one is a corollary of the projective law to be used later.

Corollary A.3

Let ${\mathfrak {G}} = ( G, {\star } )$ be a projective geometry. For any $a\in G$ and $A\subseteq G$, ${\mathcal {C}}(\{ a\}\cup {\mathcal {C}} (A))={\mathcal {C}}(\{ a\} \cup A)$.

Proof

By definition ${\mathcal {C}}(A)$ is a subspace, and thus ${\mathcal {C}}({\mathcal {C}}(A)) = {\mathcal {C}}(A)$. Using the projective law,

$$\begin{aligned} {\mathcal {C}}(\{ a\} \cup A)&= \bigcup \{ c\star d\mid c\in {\mathcal {C}}(\{ a\}),\ d\in {\mathcal {C}}(A) \} \\&= \bigcup \{ c\star d\mid c\in {\mathcal {C}}(\{ a\}),\ d\in {\mathcal {C}}({\mathcal {C}}(A)) \} \\&= {\mathcal {C}}( \{ a \} \cup {\mathcal {C}}(A)) \end{aligned}$$

$\square$

Now we are going to prove that in a quantum Kripke frame, for any $n\in {\mathbb {N}}$ and $s_1,...,s_n\in \Sigma$, ${\mathcal {C}}(\{ s_1,...,s_n\})={\sim }{\sim }\{ s_1,...,s_n\}$. Please remind that, according to Theorem 2.16, in a quantum Kripke frame ${\mathfrak {F}} = (\Sigma , {\rightarrow })$, for any $s, t \in \Sigma$, ${\sim }{\sim }\{ s, t \}$ is the line $s \star t$ in the projective geometry corresponding to ${\mathfrak {F}}$.

Lemma A.4

In a quantum Kripke frame ${\mathfrak {F}} = ( \Sigma , {\rightarrow })$, if $P\subseteq \Sigma$ is closed, it is a subspace of ${\textbf{G}} ({\mathfrak {F}})$.

Proof

Assume that $s,t\in P$. Then $\{ s,t\}\subseteq P$. Applying 2 of Lemma 2.15 twice, one can obtain ${\sim }{\sim }\{ s,t\}\subseteq {\sim }{\sim }P$. Since P is closed, ${\sim }{\sim }\{ s,t\}\subseteq {\sim }{\sim }P=P$. $\square$

Lemma A.5

In a quantum Kripke frame ${\mathfrak {F}} = ( \Sigma , {\rightarrow })$, ${\sim }Q={\sim }{\mathcal {C}}(Q)$, for every $Q\subseteq \Sigma$.

Proof

By definition $Q\subseteq {\mathcal {C}}(Q)$, so ${\sim }{\mathcal {C}}(Q)\subseteq {\sim }Q$ by 2 of Lemma 2.15. It remains to show that ${\sim }Q\subseteq {\sim }{\mathcal {C}}(Q)$.

We define a sequence of sets $\{ Q_i \}_{ i \in {\mathbb {N}}}$ in the same way as in Proposition A.1. Then by the proposition ${\mathcal {C}}(Q)=\bigcup _{i\in {\mathbb {N}}} Q_i$. It is easy to see from the definition that ${\sim }{\mathcal {C}}(Q)={\sim }\bigcup _{i\in {\mathbb {N}}} Q_i=\bigcap _{i\in {\mathbb {N}}} {\sim }Q_i$. We prove ${\sim }Q \subseteq \bigcap _{i\in {\mathbb {N}}} {\sim }Q_i = {\sim }{\mathcal {C}}(Q)$ by showing that ${\sim }Q\subseteq {\sim }Q_i$, for every $i\in {\mathbb {N}}$. Use induction on i.

Base Step: $i=0$. ${\sim }Q\subseteq {\sim }Q={\sim }Q_0$ obviously holds.

Induction Step: $i=n+1$. Let $s\in {\sim }Q$ and $t\in Q_{n+1}$ be arbitrary. By definition there are $u,v\in Q_n$ such that $t\in {\sim }{\sim }\{ u, v \}$. By IH $s\in {\sim }Q\subseteq {\sim }Q_n$. Hence $s\not \rightarrow u$ and $s\not \rightarrow v$, i.e. $s\in {\sim }\{ u,v\}$. Since $t\in {\sim }{\sim }\{ u, v \}$, $s\not \rightarrow t$. Since t is arbitrary, $s\in {\sim }Q_{n+1}$. Therefore, ${\sim }Q\subseteq {\sim }Q_{n+1}$. $\square$

The following lemma suggests a way to get a bigger closed set from a smaller one using linear closures. It is a special case of Proposition 14.2.5 in [38]. Since a direct proof is not long, we give it here to avoid introducing general terminologies.

Lemma A.6

In a quantum Kripke frame ${\mathfrak {F}} = ( \Sigma , {\rightarrow })$, let $s\in \Sigma$ and $P\subseteq \Sigma$ be closed. Then ${\mathcal {C}}(\{ s\} \cup P)$ is closed.

Proof

If $s \in P$, then ${\mathcal {C}} (\{ s \} \cup P) = {\mathcal {C}} (P) = P$ is closed by Lemma A.4 and the definition of subspaces. It remains to show the case when $s \not \in P$. Since P is closed, $s \not \in {\sim }{\sim }P$, so there is a $u \in {\sim }P$ such that $s \rightarrow u$.

By Lemma 2.15${\mathcal {C}}(\{ s\} \cup P) \subseteq {\sim }{\sim }{\mathcal {C}}(\{ s\} \cup P)$. It remains to show that ${\sim }{\sim }{\mathcal {C}}(\{ s\} \cup P) \subseteq {\mathcal {C}}(\{ s\} \cup P)$. By Lemma A.5 it suffices to show that ${\sim }{\sim }(\{ s\} \cup P) \subseteq {\mathcal {C}}(\{ s\} \cup P)$.

Let $w \in {\sim }{\sim }(\{ s\} \cup P)$ be arbitrary. If $w = s$, then $w \in {\mathcal {C}}(\{ s\} \cup P )$; so it remains to deal with the case when $w \ne s$. By Lemma 4.11 in [41] there is a $v \in {\sim }{\sim }\{ w, s \}$ such that $u \not \rightarrow v$. Since $s \rightarrow u$ and $u \not \rightarrow v$, $s \ne v$. Hence by Lemma 4.12 in [41] $w \in {\sim }{\sim }\{ s, v \}$. To show that $w \in {\mathcal {C}} ( \{ s \} \cup P )$ and thus finish the proof, it remains to show that $v \in P = {\sim }{\sim }P$.

Let $x \in {\sim }P$ be arbitrary. When $x = u$, then $v \not \rightarrow u$ follows from the construction of v. When $x \ne u$, by Lemma 4.11 in [41] there is a $y \in {\sim }{\sim }\{ x, u \}$ such that $y \not \rightarrow s$. Since $x, u \in {\sim }P$, $y \in {\sim }{\sim }\{ x, u \} \subseteq {\sim }{\sim }{\sim }P = {\sim }P$. Together with $y \not \rightarrow s$, we have $y \in {\sim }\{ s \} \cap {\sim }P = {\sim }( \{ s \} \cup P )$. Hence $w \not \rightarrow y$. Since $v \in {\sim }{\sim }\{ w, s \}$, $v \not \rightarrow y$. Since $s \rightarrow u$ and $y \not \rightarrow s$, $y \ne u$. Hence by Lemma 4.12 in [41] $x \in {\sim }{\sim }\{ y, u \}$. Since $v \not \rightarrow u$ and $v \not \rightarrow y$, $v \in {\sim }\{ u, y \}$, so $x \not \rightarrow v$. Since x is arbitrary, $v \in {\sim }{\sim }P = P$. $\square$

Finally we are ready to prove Proposition 2.17.

Proposition A.7

(Proposition 2.17) In a quantum Kripke frame ${\mathfrak {F}} = ( \Sigma , {\rightarrow })$, for any $n\in {\mathbb {N}}$ and $s_1,...,s_n\in \Sigma$, ${\mathcal {C}}(\{ s_1,...,s_n\})={\sim }{\sim }\{ s_1,...,s_n\}$.

Proof

We prove by induction that, for every $n\in {\mathbb {N}}$, ${\mathcal {C}}(\{ s_1,...,s_n \})$ is closed.

Base Step: $n=0$. By convention $\{ s_1,...,s_n \} = \emptyset$, so ${\mathcal {C}} (\emptyset ) = \emptyset$ is closed by Lemma 2.15.

Induction Step: $n=k+1$. By the induction hypothesis ${\mathcal {C}}(\{ s_1,...,s_k\})$ is closed. Then by Lemma A.6${\mathcal {C}}(\{ s_{k+1}\} \cup {\mathcal {C}}(\{ s_1,...,s_k\}))$ is closed. By Corollary A.3

$$\begin{aligned} {\mathcal {C}}(\{ s_{k+1}\} \cup {\mathcal {C}}(\{ s_1,...,s_k\}))={\mathcal {C}}(\{ s_{k+1}\} \cup \{ s_1,...,s_k\})={\mathcal {C}}(\{ s_1,...,s_k,s_{k+1}\}) \end{aligned}$$

Hence ${\mathcal {C}}(\{ s_1,...,s_k,s_{k+1}\})$ is closed. This finishes the proof by induction.

Now by Lemma A.5${\sim }{\sim }\{ s_1,...,s_n\}={\sim }{\sim }{\mathcal {C}} (\{ s_1,...,s_n\} )$. As ${\mathcal {C}} (\{ s_1,...,s_n\} )$ is closed, ${\sim }{\sim }\{ s_1,...,s_n\}={\mathcal {C}} (\{ s_1,...,s_n\} )$. $\square$

Appendix B

In this appendix, we list the results in projective geometry in the literature which are used in this paper. The following results with the notations are from [38], and the footnotes are added by me.

Proposition 2.3.3 (Page 34) For every subset A of a projective geometry G there exists a smallest subspace ${\mathcal {C}} (A)$ containing A and the so obtained operator ${\mathcal {C}}$ is a closure operator on G.

Proposition 3.1.13 (Page 59) Let G be a projective geometry and ${\mathcal {C}}$ the closure operator associated to it. Then the set G together with the closure operator ${\mathcal {C}}$ is a geometry.¹³

Lemma 6.3.4 (Page 137) Let $V_1$, $V_2$ be vector spaces over $K_1$, $K_2$ respectively, $f: V_1 \rightarrow V_2$ and $h: V_1 \rightarrow V_2$ two additive maps¹⁴ such that $h(x) \in K_2 \cdot f(x)$ for all $x \in V_1$. We suppose that $f(V_1)$ contains at least two linearly independent vectors. Then there exists a unique $\mu \in K_2$ such that $h(x) = \mu \cdot f(x)$ for all $x \in V_1$.

Theorem 10.3.1 (Page 243) For a partial map $g: G \dashrightarrow G'$ between arguesian geometries the following conditions are equivalent:

• (1) if homogeneous coordinates $u: {\mathcal {P}} V \rightarrow G$ and $u': {\mathcal {P}} V' \rightarrow G'$ are given, there exists a semilinear map $f: V \rightarrow V'$ such that $g = u' \circ {\mathcal {P}} f \circ u^ {-1}$,

• (2) g is described by a semilinear map in some homogeneous coordinates,

• (3) g is the composite of two non-degenerate morphisms,

• (4) g is the composite of finitely many non-degenerate morphisms.

Definition 10.3.2 (Page 243) A morphism $g: G \dashrightarrow G'$ between arguesian geometries is called arguesian if it satisfies the equivalent conditions of the preceding theorem.

Proposition 11.2.5 (Page 259) For any vector space V one has a natural isomorphism $\Theta _V: {\mathcal {P}} (V^*) \rightarrow ({\mathcal {P}} V)^*$.¹⁵ It is induced by the map $\Omega : (V^*)^\bullet \rightarrow ({\mathcal {P}}V)^*, \varphi \mapsto {\mathcal {P}}(\text {ker}\ \varphi )$.¹⁶

Definition 13.4.1(Page 310) A dualized (projective) geometry is a projective geometry G together with a subspace $\Gamma \subseteq G^*$ of the dual geometry satisfying $\bigcap \Gamma = \emptyset$.

Example 13.4.2 (Page 310) Let V be a dualized vector space, i.e. a vector space with a vector subspace $V' \subseteq V^*$ of the algebraic dual which is separating. This means that for every $x \ne 0$ there exists $l \in V'$ such that $l (x) = 1$. Then ${\mathcal {P}} (V)$ together with the subspace $\Theta _V ({\mathcal {P}}(V'))$ (cf. 11.2.5) is a dualized geometry.

Proposition 13.5.3 (Page 316) Let $V_1$ and $V_2$ be dualized vector spaces. Then for every non-degenerate continuous homomorphism $g: {\mathcal {P}} (V_1) \dashrightarrow {\mathcal {P}} (V_2)$ there exists a continuous quasilinear map $f: V_1 \rightarrow V_2$ such that $g = {\mathcal {P}} f$.

Proposition 14.1.4 (Page 324) Let G be an orthogeometry. Then G together with the set $\Gamma := \{a^\bot \mid a \in G \}$ is a dualized geometry.

Definition 14.4.1 (Page 334) An adjunction between two orthogeometries $G_1$ and $G_2$ consists of two partial maps $g_1: G_1 \dashrightarrow G_2$ and $g_2: G_2 \dashrightarrow G_1$ satisfying

• 1) $\text {Ker}\ g_1 = (\text {Im}\ g_2)^\bot$,

• 2) $\text {Ker}\ g_2 = (\text {Im}\ g_1)^\bot$,

• 3) for all $a \in \text {Dom}\ g_1$ and $b \in \text {Dom}\ g_2$ one has $g_1 a \perp b$ iff $a \perp g_2 b$.

Proposition 14.4.4 (Page 335) For a partial map $g_1: G_1 \dashrightarrow G_2$ between orthogeometries the following conditions are equivalent:

• (1) there exists a partial map $g_2: G_2 \dashrightarrow G_1$ such that $(g_1, g_2)$ is an adjunction,

• (2) $g_1$ is a continuous homomorphism.

Lemma 14.4.9 (Page 337) A quasilinear map $f: V_1 \rightarrow V_2$ is continuous if and only if there exists a quasilinear map $f^\circ : V_2 \rightarrow V_1$ such that $\Phi _2 (fx,y) = \tau (\Phi _1 (x,f^\circ y))$ for all $x \in V_1$ and $y \in V_2$. The map $f^\circ$ is unique and called the adjoint of f.

The following results with the notations are from [39].

Definition (Page 62) Let ${\textbf{P}}$ be a projective space.¹⁷ We say that in ${\textbf{P}}$ the theorem of Pappus holds if any two intersecting lines g and h with $g \ne h$ satisfy the following condition. If $A_1, A_2, A_3$ are distinct points on g and $B_1, B_2, B_3$ are distinct points on h all different from $g \cap h$ then the points

$$\begin{aligned} Q_{12}:= A_1 B_2 \cap B_1 A_2,\ Q_{23}:= A_2 B_3 \cap B_2 A_3 \text{ and } Q_{31}:= A_3 B_1 \cap B_3 A_1 \end{aligned}$$

lie on a common line.

Theorem 2.2.2 (Page 62) Let V be a vector space over a division ring F. Then the theorem of Pappus holds in ${\textbf{P}} (V)$ if and only if F is commutative (in other words, if F is a field).

Theorem 2.2.3 Hessenberg’s Theorem (Page 65) Let ${\textbf{P}}$ be an arbitrary projective space. If the theorem of Pappus holds in ${\textbf{P}}$ then the theorem of Desargues is also true in ${\textbf{P}}$.

Theorem 3.6.7 (Page 132) The projective collineations of ${\textbf{P}} (V)$ are precisely the products of central collineations.