Quantum states: an analysis via the orthogonality relation

Abstract

From the Hilbert space formalism we note that five simple conditions are satisfied by the orthogonality relation between the (pure) states of a quantum system. We argue, by proving a mathematical theorem, that they capture the essentials of this relation. Based on this, we investigate the rationale behind these conditions in the form of six physical hypotheses. Along the way, we reveal an implicit theoretical assumption in theories of physics and prove a theorem which formalizes the idea that the Superposition Principle makes quantum physics different from classical physics. The work follows the paradigm of mathematical foundations of quantum theory, which I will argue by methodological reflection that it exemplifies a formal approach to analysing concepts in theories.

Appendix

In this appendix, I will develop the technicalities used in the proof of Proposition 4.5 in Sect. 4.3.

We fix a Kripke frame $\mathfrak {F} = ( \Sigma , {\rightarrow } )$ satisfying Reflexivity and Symmetry throughout this appendix for convenience.

Note that Items 1 to 5 in Lemma 2.8 only need Reflexivity and Symmetry in their proofs, so they hold in $\mathfrak {F}$.

1.1 Subframe

In this subsection, we introduce and discuss the notion of a subframe.

Definition A.1

A subframe of $\mathfrak {F}$ is an ordered pair $(P , {\rightarrow _P})$ such that

• 1. $P \subseteq \Sigma$ is non-empty and closed in $\mathfrak {F}$;

• 2. ${\rightarrow _P}$ is the restriction of ${\rightarrow }$ to P, i.e. ${\rightarrow _P} = {\rightarrow } \cap (P \times P)$.

Remark A.2

Let $(P , {\rightarrow _P})$ be a subframe of $\mathfrak {F}$.

• 1. For any $s,t \in P$, $s \rightarrow t$ if and only if $s \rightarrow _P t$.

• 2. For any $A \subseteq P$ and $s,t \in P$, s and t are indistinguishable with respect to A in $(P , {\rightarrow _P})$ if and only if they are indistinguishable with respect to A in $\mathfrak {F}$.

• 3. $(P , {\rightarrow _P})$ is a Kripke frame satisfying Reflexivity and Symmetry, so the results in this appendix apply to it as well.

Given this remark, in the following, I denote the restriction of ${\rightarrow }$ in $\mathfrak {F}$ to P by ${\rightarrow }$ instead of ${\rightarrow _P}$. I also use the same symbol ${\approx _A}$ indexed by $A \subseteq P$ to denote the indistinguishability relation with respect to A in $(P , {\rightarrow })$ and that in $\mathfrak {F}$. On the contrary, I denote the orthocomplement of $A \subseteq P$ in $(P , {\rightarrow })$ by ${\sim }_P A$, because the orthocomplements in the subframe are different from those in $\mathfrak {F}$.

I start from some simple facts about the orthocomplements and the closed subsets in a subframe.

Lemma A.3

Let $(P , {\rightarrow })$ be a subframe of $\mathfrak {F}$.

• 1. ${\sim }_P A = {\sim }A \cap P$, for every $A \subseteq P$.

• 2. For each $A \subseteq P$, if ${\sim }_P {\sim }_P A = A$, then ${\sim }{\sim }A = A$.

• 3. If $\mathfrak {F}$ satisfies Representation, then, for each $A \subseteq P$, ${\sim }{\sim }A = A$ implies that ${\sim }_P {\sim }_P A = A$.

Proof

For Item 1: Easy verification.

For Item 2: Assume that ${\sim }_P {\sim }_P A = A$. By Lemma 2.8$A \subseteq {\sim }{\sim }A$. It remains to show that ${\sim }{\sim }A \subseteq A$. Assume that $s \not \in A$. By the assumption $s \not \in {\sim }_P {\sim }_P A$. Then there is a $t \in {\sim }_P A$ such that $s \rightarrow t$. By 1 $t \in {\sim }A \cap P$. Hence $t \in {\sim }A$ is such that $s \rightarrow t$, and thus $s \not \in {\sim }{\sim }A$.

For Item 3: Assume that $\mathfrak {F}$ satisfies Representation and ${\sim }{\sim }A = A$. By 1 ${\sim }_P {\sim }_P A = {\sim }({\sim }A \cap P) \cap P$, so it amounts to show that ${\sim }({\sim }A \cap P) \cap P = A$.

On the one hand, since $A \subseteq P$ and $A = {\sim }{\sim }A \subseteq {\sim }( {\sim }A \cap P )$ by Lemma 2.8, $A \subseteq {\sim }({\sim }A \cap P) \cap P$.

On the other hand, let $s \in {\sim }({\sim }A \cap P) \cap P$ be arbitrary. Then $s \in P$ and $s \in {\sim }({\sim }A \cap P)$. To show that $s \in A$, by the assumption it suffices to show that $s \in {\sim }{\sim }A$. Hence we let $t \in {\sim }A$ be arbitrary and try to show that $s \not \rightarrow t$. Consider two cases.

• Case 1: $t \in {\sim }P$.

  Since $s \in P$, it follows that $s \not \rightarrow t$.

• Case 2: $t \not \in {\sim }P$.

  By Representation there is a $t_\parallel \in P$ such that $t \approx _P t_\parallel$. Since $A \subseteq P$ and $t \in {\sim }A$, $t_\parallel \in {\sim }A$. Hence $t_\parallel \in P \cap {\sim }A$. Since $s \in {\sim }({\sim }A \cap P)$, $t_\parallel \not \rightarrow s$. Since $s \in P$, by $t \approx _P t_\parallel$ we have $t \not \rightarrow s$, so $s \not \rightarrow t$.

$\square$

Proposition A.4

Let $(P , {\rightarrow })$ be a subframe of $\mathfrak {F}$. If $\mathfrak {F}$ satisfies both Separation and Representation, then so is $(P , {\rightarrow })$.

Proof

For Representation, let $A \subseteq P$ and $s \in P$ be such that $A = {\sim }_P {\sim }_P A$ and $s\not \in {\sim }_P A$. By Lemma A.3$A = {\sim }{\sim }A$. Moreover, since $s \not \in {\sim }_P A$ and $s \in P$, by Lemma A.3$s \not \in {\sim }A$. By Representation there is an $s' \in A$ such that $s \approx _A s'$.

For Separation, assume that $s,t \in P$ are such that $s \ne t$. Since $\mathfrak {F}$ satisfies Separation, there is a $w \in \Sigma$ such that $w \rightarrow s$ and $w \not \rightarrow t$. Since $w \rightarrow s$, $w \not \in {\sim }P$. By Representation there is a $w' \in P$ such that $w \approx _P w'$. Hence $w' \in P$ is such that $w' \rightarrow s$ and $w' \not \rightarrow t$. $\square$

1.2 Maximal Opposite Family

Here we prove some useful results about maximal opposite families.

Corollary A.5

Let $\{ P_i \mid i \in I \}$ be a maximal opposite family in $\mathfrak {F}$. $\emptyset = \bigcap \{ {\sim }P_i \mid i \in I \}$.

Proof

If I is a singleton, the result holds by Lemma 2.8 and Remark 3.2. In the following we focus on the case when I has at least two elements.

Take an $i^* \in I$. By Lemma 3.4$P_{i^*} = \bigcap \big \{ {\sim }P_i \mid i \in I \setminus \{ i^* \} \big \}$. Hence $\bigcap \{ {\sim }P_i \mid i \in I \} = {\sim }P_{i^*} \cap \bigcap \big \{ {\sim }P_i \mid i \in I \setminus \{ i^* \} \big \} = {\sim }P_{i^*} \cap P_{i^*} = \emptyset$. $\square$

Lemma A.6

If $\mathfrak {F}$ satisfies Separation and Representation, then, for any closed $P \varsubsetneqq \Sigma$ and $s \not \in {\sim }P$, there are $s_\parallel \in P$ and $s_\bot \in {\sim }P$ such that $s_\parallel \in {\sim }{\sim }\{ s , s_\bot \}$.

Proof

Assume that $P \varsubsetneqq \Sigma$ is closed and $s \not \in {\sim }P$. Note that ${\sim }P \ne \emptyset$ for $P \ne \Sigma$. If $s \in P$, then by just letting $s_\parallel = s$ and picking an arbitrary $s_\bot \in {\sim }P$ we will have $s_\parallel \in {\sim }{\sim }\{ s , s_\bot \}$. In the following, we focus on the case when $s \not \in P$.

Since P is closed, $s \not \in {\sim }{\sim }P$. By Representation there is an $s_\bot \in {\sim }P$ such that $s \approx _{{\sim }P} s_\bot$. Since $s \not \in {\sim }P$, $s \ne s_\bot$, so $s \not \in {\sim }{\sim }\{ s_\bot \}$ by Separation and Lemma 2.8. By Representation there is an $s_\parallel \in {\sim }\{ s_\bot \}$ such that $s \approx _{{\sim }\{ s_\bot \}} s_\parallel$. It follows that $s_\parallel \in {\sim }{\sim }\{ s , s_\bot \}$.

It remains to show that $s_\parallel \in P$. Suppose (towards a contradiction) that $s_\parallel \not \in P$. Then $s_\parallel \not \in {\sim }{\sim }P$. By Representation there is an $s'_\parallel \in {\sim }P$ such that $s_\parallel \approx _{{\sim }P} s'_\parallel$. One the one hand, since $s'_\parallel \rightarrow s'_\parallel$, $s_\parallel \rightarrow s'_\parallel$. On the other hand, since $s_\parallel \not \rightarrow s_\bot$ and $s_\bot \in {\sim }P$, $s'_\parallel \not \rightarrow s_\bot$. Since $s \approx _{{\sim }P} s_\bot$ and $s'_\parallel \in {\sim }P$, $s \not \rightarrow s'_\parallel$. Therefore, $s'_\parallel \in {\sim }\{ s , s_\bot \}$. Since $s_\parallel \in {\sim }{\sim }\{ s , s_\bot \}$, $s_\parallel \not \rightarrow s'_\parallel$. As a result, we have got a contradiction. $\square$

Finally, we prove a lemma for building maximal opposite families used in the proof of Proposition 4.5.

Lemma A.7

Suppose that $\mathfrak {F}$ satisfies Separation and Representation. Let I and J be two disjoint, non-empty sets, $\{ P_i \mid i \in I \}$ a maximal opposite family in $\mathfrak {F}$, $i^* \in I$, $\{ Q_j \mid j \in J \}$ a maximal opposite family in the subframe $( P_{i^*} , {\rightarrow } )$ of $\mathfrak {F}$ and $K = ( I \setminus \{ i^* \} ) \cup J$. The following element in $\wp (\wp (\Sigma ) \setminus \{ \emptyset \} )$ is a maximal opposite family in $\mathfrak {F}$: for each $k \in K$,

$$\begin{aligned} O_k = \left\{ \begin{array}{ll} P_k , &{} \text{ if } k \in I \setminus \{ i^* \} \\ Q_k , &{} \text{ if } k \in J \end{array} \right. \end{aligned}$$

Proof

The case when $I = \{ i^* \}$ follows from Remark 3.2. It remains to consider the case when I is not a singleton.

In this case, it can be verified that $\{ O_k \mid k \in K \}$ is an opposite family.

Before showing maximality, I claim that ${\sim }P_{i^*} = \bigcap \{ {\sim }Q_j \mid j \in J \}$: For one direction, for each $j \in J$, $Q_j \subseteq P_{i^*}$, so ${\sim }P_{i^*} \subseteq {\sim }Q_j$. Hence ${\sim }P_{i^*} \subseteq \bigcap \{ {\sim }Q_j \mid j \in J \}$. For the other direction, suppose (towards a contradiction) that $\bigcap \{ {\sim }Q_j \mid j \in J \} \not \subseteq {\sim }P_{i^*}$. On the one hand, there is an $s \in \bigcap \{ {\sim }Q_j \mid j \in J \}$ but $s \not \in {\sim }P_{i^*}$. By Lemma A.6 there are $s_\parallel \in P_{i^*}$ and $s_\bot \in {\sim }P_{i^*}$ such that $s_\parallel \in {\sim }{\sim }\{ s , s_\bot \}$. Since $s \in \bigcap \{ {\sim }Q_j \mid j \in J \}$ and $s_\bot \in {\sim }P_{i^*} \subseteq \bigcap \{ {\sim }Q_j \mid j \in J \}$, $s_\parallel \in \bigcap \{ {\sim }Q_j \mid j \in J \}$. Thus $s_\parallel \in P_{i^*} \cap \bigcap \{ {\sim }Q_j \mid j \in J \}$. On the other hand, since $\{ Q_j \mid j \in J \}$ is a maximal opposite family in the subframe $( P_{i^*} , {\rightarrow } )$ of $\mathfrak {F}$, by Lemma A.3 and Corollary A.5$\emptyset = \bigcap \{ {\sim }_{P_{i^*}} Q_j \mid j \in J \} = \bigcap \{ P_{i^*} \cap {\sim }Q_j \mid j \in J \} = P_{i^*} \cap \bigcap \{ {\sim }Q_j \mid j \in J \}$, which contradicts that $s_\parallel \in P_{i^*} \cap \bigcap \{ {\sim }Q_j \mid j \in J \}$. As a result, ${\sim }P_{i^*} = \bigcap \{ {\sim }Q_j \mid j \in J \}$.

Now we prove maximality. By Lemma 3.4 it suffices to show that, for each $k^* \in K$, $O_{k^*} = \bigcap \big \{ {\sim }O_k \mid k \in K \setminus \{ k^* \} \big \}$. We consider 2 cases.

• Case 1: $k^* \in I \setminus \{ i^* \}$.

  $$\begin{aligned} O_{k^*} = P_{k^*}&= \bigcap \{ {\sim }P_i \mid i \in I \setminus \{ k^* \} \} \qquad \qquad \qquad (\text {Lemma } 3.4) \\&= {\sim }P_{i^*} \cap \bigcap \{ {\sim }P_i \mid i \in I \setminus \{ k^* , i^* \} \} \\&= \bigcap \{ {\sim }Q_j \mid j \in J \} \cap \bigcap \{ {\sim }P_i \mid i \in I \setminus \{ k^* , i^* \} \}\qquad \qquad \qquad \text {(the claim)} \\&= \bigcap \Big \{ {\sim }O_k \mid k \in K \setminus \{ k^* \} \Big \} \end{aligned}$$

• Case 2: $k^* \in J$.

  $$\begin{aligned} O_{k^*} = Q_{k^*}&= \bigcap \{ {\sim }_{P_{i^*}} Q_j \mid j \in J \setminus \{ k^* \} \} \qquad \qquad \qquad (\text {Lemma } 3.4) \\&= \bigcap \{ {\sim }Q_j \cap P_{i^*} \mid j \in J \setminus \{ k^* \} \} \qquad \qquad \qquad (\text {Lemma } A.3) \\&= P_{i^*} \cap \bigcap \{ {\sim }Q_j \mid j \in J \setminus \{ k^* \} \} \\&= \bigcap \{ {\sim }P_i \mid i \in I \setminus \{ i^* \} \} \cap \bigcap \{ {\sim }Q_j \mid j \in J \setminus \{ k^* \} \}\quad (\text {Lemma }3.4) \\&= \bigcap \Big \{ {\sim }O_k \mid k \in K \setminus \{ k^* \} \Big \} \end{aligned}$$

$\square$