1 Introduction

We clarify and demystify the notion of entanglement by highlighting the observables role and exploring the quantum conditional probability calculus (see also [1]). Observables’ entanglement express the degree of dependence between two quantum observables A and B for a state \(|\psi \rangle .\) Hence, it is meaningless to speak about entanglement without selecting the fixed observables A and B. In this aspect our work is close to the papers [2, 3]; we cite [3]:

“Here we propose that a partitioning of a given Hilbert space is induced by the experimentally accessible observables.... In this sense entanglement is always relative to a particular set of experimental capabilities.”

The distinguishing feature of our work is exploring the calculus of quantum conditional probabilities (see [1] for the first step in this direction).

Observables’ entanglement differs from the standard tensor product non-factorization approach, see, e.g., [4]:

“A state is said to be entangled if it cannot be written as a convex sum of tensor product states.”

In this paper we work only with pure states.

1.1 Entanglement as Hilbert Space Representation of Observables Dependence

Dependence of observables is formalized on the basis of the calculus of quantum conditional probabilities [5,6,7,8]. For dichotomous observables A and B,  a state \(|\psi \rangle\) is AB-entangled iff

$$P(B=+| A=+; \psi ) \not = P(B=+| A=-; \psi ),$$
(1)

where conditional probabilities are defined via the combination of the Born rule and the Lüders projection postulate [9].

There is nothing mystical in this notion of AB-entanglement. The difference between conditional probabilities quantifies the degree of dependence of B-outcome on A-outcomes. This measure of AB-entanglement can be expressed in the Hilbert space terms and it is similar to the concurrence-measure for the usual entanglement.

AB-entanglement is not coupled to the tensor product structure. It can be considered e.g. for a state space \({{{\mathcal {H}}}}\) having non-factorisable dimension, say \(\textrm{dim} \; {{{\mathcal {H}}}}= 5.\) Generally AB-entanglement is introduced for non-commuting operators (incompatible observables). But, in the main part of this paper we work with commuting operators (with dichotomous spectrum). As can be expected, for non-commuting operators AB- and BA-entanglements do not coincide.

The main part of the article is devoted to the Hilbert space characterization of AB-entangled quantum states. The standard entangled states are defined as non-factorisable w.r.t. the tensor product structure on \({{{\mathcal {H}}}}.\) We demonstrate that, for compatible dichotomous observables, AB-entangled states can be characterized as amplitude non-factorisable states. Here amplitudes are defined as norms of the state projections on subspaces \({{{\mathcal {H}}}}_{AB}(\alpha \beta )\) consisting of common eigenvectors of the operators (with eigenvalues \(\alpha , \beta ).\) Such expansion w.r.t. to the joint eigenvectors was actively explored in works of Ozawa, e.g., [10], devoted to a similar problem.

Conditional probability treatment of entanglement can be considered as indirect contribution to critique of nonlocal interpretation of quantum mechanics (QM) and connection of the violation of the Bell inequalities [11,12,13] with spooky action at a distance (see, e.g., [14,15,16,17,18,19,20,21,22,23,24] for analysis of the nonlocality problem within the QM-formalism; see also for probabilistic analysis [25,26,27,28,29,30,31,32,33,34]).

1.2 Observables Entanglement and Correlation

The description of entangled states as amplitude non-factorisable states leads to the following result. For compatible dichotomous observables, a quantum state \(|\psi \rangle\) is AB-entanglement iff correlation of these observables w.r.t. this state is not factorisable, i.e.,

$$\langle AB\rangle _{\psi} \not = \langle A\rangle _{\psi} \langle B\rangle _{\psi} ,$$
(2)

or observables covariance is non-zero:

$$\textrm{cov}\; (A,B; \psi ) \equiv \langle (A - \langle A\rangle _{\psi} ) (B-\langle B\rangle _{\psi} ) \rangle _{\psi} = \langle AB\rangle _{\psi} - \langle A\rangle _{\psi} \langle B\rangle _{\psi} \not =0.$$
(3)

Let \({{{\mathcal {H}}}}={{{\mathcal {H}}}}_{1} \otimes {{{\mathcal {H}}}}_{2}.\) In this case, it is worth to compare AB-entanglement and usual entanglement w.r.t. to the correlation property. If a quantum state \(|\psi \rangle\) is factorisable, i.e., \(|\psi \rangle = |\phi _{1}\rangle \otimes |\phi _{2}\rangle ,\) then, of course, correlation is factorisable for any pair of operators,

$$\langle AB\rangle _{\psi} = \langle A\rangle _{\psi} \langle B\rangle _{\psi} , \quad \text{ i.e. } \; \textrm{cov}\; (A,B; \psi ) =0.$$
(4)

However, the inverse statement is not correct. For some entangled states, the equality (4) holds true. The latter is impossible for AB-entangled states.

1.3 Observables Entanglement vs. Random Variables Dependence in Classical Probability

We recall that in classical probability theory independence of random variables A and B implies correlation factorization

$$\langle AB \rangle _{{{{\mathcal {P}}}}} = E_{{{{\mathcal {P}}}}} [AB]= E_{{{{\mathcal {P}}}}}[A]E_{{{{\mathcal {P}}}}}[B]= \langle A \rangle _{{{{\mathcal {P}}}}} \langle B \rangle _{{{{\mathcal {P}}}}} , \quad \text{ i.e. } \; \textrm{cov}\; (A, B; {{{\mathcal {P}}}}) =0,$$
(5)

where E denotes the mathematical expectation for a classical probability space \({{{\mathcal {P}}}}\) (see “Appendix” on the notion of Kolmogorov probability space [35]).

In textbooks on classical probability theory one proceeds with a fixed probability space \({{{\mathcal {P}}}},\) so dependence of all quantities on \({{{\mathcal {P}}}}\) is not visualized.

For classical dichotomous random variables, independence is equivalent to uncorrelation; hence, for such random variables dependence is equivalent to correlation, i.e., non-factorization of product’s average. We obtain the same characterization of compatible dichotomous AB-entangled quantum observables. For such observables, the notion of observables entanglement is simply the Hilbert space representation of the notion of dependence for classical random variables, \(\textrm{cov}\; (A,B; {{{\mathcal {P}}}}) \not =0.\) Generally observables entanglement extends the notion of dependence of random variables, within the Hilbert space formalism.

In Sect. 7, we study \(A_{u} B_{u}\)-entanglement for families of dichotomous pairwise compatible observables, i.e., represented by commuting operators, \([A_{u}, B_{u}]=0,\) \({{{\mathcal {U}}}}=(u)\) is a family of unitary transformations. Here generally \([A_{u}, A_{v}]\not =0, [B_{u},B_{v}]\not =0.\) And we found an analog of the singlet state \(|\psi _{{{{\mathcal {U}}}}}\rangle ,\) i.e., a state that is \((A_{u},B_{u})\)-entangled for any transformation \(u \in {{{\mathcal {U}}}}.\)

This construction highlights the essence of quantum entanglement, as formalized within quantum probability calculus in complex Hilbert space. For each concrete pair of compatible dichotomous observables, \(A_{u} B_{u}\)-entanglement is just the Hilbert space representation of dependence of classical random variables or equivalently (since this is the dichotomous case) correlation of random variables. So for each pair, there exists a classical probability space \({{{\mathcal {P}}}}_{u}\) such that \(\textrm{cov}\; (A_{u}, B_{u}; {{{\mathcal {P}}}}_{u}) \not =0,\) i.e., correlation of these two observables can be described within classical probability theory. However, generally it is impossible to construct a probability space \({{{\mathcal {P}}}}\) which would describe dependence for all pairs of observables \(A_{u}, B_{u}.\) In contrast to this classical impossibility, quantum probability can describe all these correlations, as correlations w.r.t. the state \(|\psi _{{{{\mathcal {U}}}}}\rangle .\)

This example illustrates the general advantage of quantum entanglement in the description of observables’ correlations. Thus, the nonclassical essence of quantum observables entanglement is the possibility to model correlations for a wider class of observables than in classical probability theory. Hence, our approach to entanglement highlights not the strength of correlations, but the possibility to realize a wide class of correlated observables for the same quantum state (preparation procedure [36]).

1.4 EPR-Argument and Schrödinger’s Viewpoint on Entanglement

Although this article is mainly directed to the mathematical analysis of the probabilistic structure of QM, in the form of complex Hilbert space representation, we would like to make the following foundational remark.

The probabilistic viewpoint on the “EPR-paradox” [37] is presented in Schrödinger’s paper [38, 39] that initiated the modern theory of entanglement. However, this theory ignores the important message of Schrödinger: entanglement characterizes probability update for the outcomes of observable B conditioned on the outcomes of observable A. In the framework of [38, 39] it is meaningless to speak about entanglement without specifying the observables. The state update—the Hilbert space representation of the probability update—encodes the procedure of conditional prediction. For Schrödinger, the quantum formalism is a mathematical machinery for probability prediction and a quantum state is a part of such machinery (cf. with the Växjö interpretation of QM [40, 41] and QBism [42,43,44] as well as with reality without realism interpretation of QM [45, 46]). We can say that Schrödinger interpreted quantum probabilities as conditional probabilities. This viewpoint was latter expressed in the works of Koopman [5], Ballentine [6, 7], Khrennikov [8].

From the probabilistic viewpoint the EPR argument [37] is based outcomes’ prediction with conditional probability one. In the Hilbert space formalism such update is represented with the special class of observables entangled states (see [1] for EPR entangled states; see also Ozawa [47]).

2 Quantum Conditional Probability

All probabilities under consideration depend of quantum state \(|\psi \rangle ,\) i.e., \(P\equiv P_{\psi} ,\) but to make the notation shorter, we omit the state index.

In the quantum formalism an observable A (with a discrete range of values) is represented by a Hermitian operator

$$A= \sum _{\alpha} \alpha E_{A}(\alpha ),$$
(6)

where \(E_{A}(\alpha )\) is projection onto the space \({{{\mathcal {H}}}}_{A}(\alpha )\) of eigenvectors for the eigenvalue \(\alpha .\) For a pure state \(\vert \psi \rangle ,\) the probability to get the outcome \(A=\alpha\) is given by the Born’s rule:

$$P(A=\alpha )= \Vert E_{A}(\alpha ) \psi \Vert ^{2}.$$
(7)

A measurement with the outcome \(A= x\) generates back-action onto system’s state:

$$\vert \psi \rangle \rightarrow \vert \psi \rangle _{\alpha} ^{A}= E_{A}(\alpha )\vert \psi \rangle / \Vert E_{A}(\alpha ) \psi \Vert .$$
(8)

This is the projection postulate in the Lüders form [9]. We remark that von Neumann [48] applied this postulate only to observables with non-degenerate spectra. For observables with degenerate spectra, he considered more general class of back-actions. Nowadays, this class if formalized within theory of quantum instruments [49, 50].

The projection postulate is the mathematical tool for quantum conditioning. Measurements of another observable, say B,  conditioned on the outcome \(A=\alpha\) lead to probability (see, e.g., [8]):

$$P(B= \beta | A= \alpha , \psi )= \Vert E_{B}(\beta ) \psi _{A}^{\alpha }\Vert ^{2}= \frac{ \Vert E_{B}(\beta ) E_{A}(\alpha ) \psi \Vert ^{2}}{\Vert E_{A}(\alpha ) \psi \Vert ^{2}}.$$
(9)

We shall use this definition of conditional probability to define the special form of entanglement.

3 Observables Entanglement

Consider two (generally incompatible) observables A and B represented by Hermitian operators (denoted by the same symbols) with eigenvalues \(X_{a}= \{\alpha _{1},\ldots , \alpha _{m}\}\) and \(X_{b}= \{\beta _{1},\ldots , \beta _{k}\}.\)

Definition 1

In the state \(|\psi \rangle ,\) the outcome \(B=\beta\) depends on the outcomes of A if for at least two values of \(A, \alpha = \alpha _{i}, \alpha _{j},\) the corresponding conditional probabilities don’t coincide,

$$P( B= \beta | A= \alpha _{i}) \not = P( B= \beta | A= \alpha _{j}).$$
(10)

Thus, the probability to get the outcome \(B=\beta\) if the preceding A-measurement had the outcome \(A=\alpha _{i}\) differs from the probability to get the same outcome \(B=\beta\) if the preceding A-measurement had the outcome \(A=\alpha _{j}.\)

The outcome \(B=\beta\) does not depend on the outcomes of A iff

$$P( B= \beta | A= \alpha _{i}) = P( B= \beta | A= \alpha _{j}),\; \text{ for } \text{ all } \text{ pairs } \;\alpha _{i}, \alpha _{j},$$
(11)

i.e. the conditional probability for this outcome is constant w.r.t. the outcomes of A. Denote it \(P( B= \beta | A).\) Does it coincide with unconditional probability \(P( B= \beta )?\) (See Sect. 8).

Definition 2

For observables A and B,  a state \(|\psi \rangle\) is called AB-entangled if all outcomes of B depend on outcomes of A,  i.e. for all \(\beta\) condition (10) holds for some \(\alpha _{i}, \alpha _{j}.\)

We note that this is the purely probabilistic definition, it does not involve the notion of Hilbert space, it can be applied to any statistical physical theory in that system’s state generates conditional probabilities. This definition formalizes dependence of observables. In Sect. 8 we present the detailed comparison with the notion of dependence of random variables in classical probability theory. For classical random variables, AB-entanglement due to Definition 2 is reduced to the well known notion of dependent random variables.

For incompatible observables, AB-entanglement does not imply BA-entanglement. A state \(|\psi \rangle\) is called \(A\leftrightarrow B\)-entangled if it is both AB- and BA-entangled. For compatible observables, AB- and BA-entanglements w.r.t. to some state are equivalent (see Sect. 8).

We remark that condition (10) characterizing AB-entanglement should be completed by the conditional probability existence constraints:

$$P(A= \alpha )\not = 0, \quad \alpha \in X_{a} .$$
(12)

In the Hilbert space formalism the basic inequality (10) determining outcome dependence can be written as

$$\frac{ \Vert E_{B}(\beta ) E_{A}(\alpha _{i}) \psi \Vert ^{2}}{\Vert E_{A}(\alpha _{i}) \psi \Vert ^{2}}\not = \frac{ \Vert E_{B}(\beta ) E_{A}(\alpha _{j}) \psi \Vert ^{2}}{\Vert E_{A}(\alpha _{j}) \psi \Vert ^{2}}.$$
(13)

It implies that

$$|| E_{B}(\beta ) E_{A}(\alpha _{i}) \psi ||\; || E_{A}(\alpha _{j}) \psi ||\not = || E_{B}(\beta ) E_{A}(\alpha _{j}) \psi ||\; || E_{A}(\alpha _{i}) \psi ||.$$
(14)

Proposition 1

Inequality (14) is necessary and sufficient condition for dependence of the outcome \(B=\beta\) on the outcomes \(A=\alpha _{i}, \alpha _{j}\).

Proof

(1) Inequality (10) implies inequality (14).

(2) Now let inequality (14) hold for some \(\beta\) and \(\alpha _{i}, \alpha _{j}.\)

Generally we have

$$P(A=\alpha _{i})= ||E_{A}(\alpha _{i}) \psi ||^{2} =\left\| \sum _{\beta} E_{B}(\beta )E_{A}(\alpha _{i}) \psi \right\| ^{2} = \sum _{\beta} || E_{B}(\beta )E_{A}(\alpha _{i}) \psi ||^{2}.$$

Hence, if \(P(A=\alpha _{i})=0,\) then, for any \(\beta ,\) \(|| E_{B}(\beta )E_{A}(\alpha _{i}) \psi ||^{2}=0.\) Hence, inequality (14) is violated. Thus, inequality (14) implies that \(P(A=\alpha _{i})\not =0\) and \(P(A=\alpha _{j})\not =0;\) conditional probabilities are well defined. Hence, inequality (14) can be written in the form (10). So, the outcome \(B=\beta\) depends on the outcomes \(A=\alpha _{i}, \alpha _{j}.\) \(\square\)

Proposition 1a

Let observable A be dichotomous. Then inequalities

$$|| E_{B}(\beta ) E_{A}(+) \psi || \;||E_{A}(-) || \not = || E_{B}(\beta ) E_{A}(-) \psi || \; ||E_{A}(+)||, \quad \beta \in X_{b},$$
(15)

are the necessary and sufficient conditions for AB-entanglement.

Proposition 2

AB-entanglement is invariant under unitary transformations.

Proof

Let U be a unitary operator and let \(A_{U}=UAU^{\star} , B_{U}=UBU^{\star} , \vert \psi _{U} \rangle = U \vert \psi \rangle .\) Then \(|| E_{B_{U}}(\beta ) E_{A_{U}}(\alpha ) \psi _{U} \rangle = || E_{B}(\beta ) E_{A}(\alpha ) \psi ||\) and \(|| E_{A_{U}}(\alpha ) \psi _{U}||= || E_{A}(\alpha _{i}) \psi ||.\) \(\square\)

3.1 Measure of Observables’ Entanglement

Now we introduce a quantitative measure of AB-entanglement.

Definition 3

For quantum observables A and B,  AB-concurrence in the state \(|\psi \rangle\) is defined as

$${{{\mathcal {M}}}}_{AB}(\psi )= \sum _{\beta} \sum _{\alpha \not =\alpha ^{\prime} } |P( B= \beta | A= \alpha ) - P( B= \beta | A= \alpha ^{\prime} )|$$
(16)

(see Sect. 6.1 on comparison with the standard concurrence measure of entanglement). The crucial issue is that AB-concurrence depends on a pair of observables. The maximally AB-conditional probability entangled states are considered in Sect. 3.4, these are the EPR-states invented in [1].

In operator terms, AB-concurrence in state \(|\psi \rangle\) is given by

$${{{\mathcal {M}}}}_{AB}(\psi )= \sum _{\beta} \sum _{\alpha \not =\alpha ^{\prime} } \Big | \; \frac{||E_{B}(\beta ) E_{A}(\alpha ) \psi ||^{2}}{||E_{A}(\alpha ) \psi ||^{2}}- \frac{||E_{B}(\beta ) E_{A}(\alpha ^{\prime} ) \psi ||^{2}}{||E_{A}(\alpha ^{\prime} )\psi ||^{2}} \; \Big |.$$
(17)

We also define re-normalized AB-concurrence as

$${{{\mathcal {C}}}}_{AB}(\psi )= \sum _{\beta} \sum _{\alpha \not =\alpha ^{\prime} } \Big | \; ||E_{B}(\beta ) E_{A}(\alpha ) \psi ||\; ||E_{A}(\alpha ^{\prime} ) \psi || - ||E_{B}(\beta ) E_{A}(\alpha ^{\prime} ) \psi || \; ||E_{A}(\alpha )\psi || \; \Big |.$$
(18)

3.2 Dichotomous Observables: Equivalence of Outcomes’ Dependence and Entanglement

Proposition 3

For dichotomous observables A and B,  dependencies of the values \(B=-\) and \(B=+\) on the outcomes of A are equivalent. Thus, each dependence is equivalent to AB-entanglement.

Proof

In the state \(|\psi \rangle\) the value \(B=-\) depends on the outcomes of A if

$$P( B= -| A= +) \not = P( B= - | A= -).$$
(19)

This automatically implies that even the value \(B=+\) depends on the outcomes of A, 

$$\begin{aligned} P( B= +| A= +)= & {} 1- P( B= -| A= +) \\&\not =&1- P( B= -| A= -) = P( B= + | A= -), \end{aligned}$$

i.e., the state \(|\psi \rangle\) is AB-entangled. \(\square\)

Proposition 3a

For dichotomous observables A and B,  AB-entanglement is characterized by the single constraint:

$$|| E_{B}(+) E_{A}(+) \psi || \;||E_{A}(-) || \not = || E_{B}(+) E_{A}(-) \psi || \; ||E_{A}(+)||.$$
(20)

A state \(|\psi \rangle\) is AB-disentangled, iff

$$|| E_{B}(+) E_{A}(+) \psi || \;||E_{A}(-) || = || E_{B}(+) E_{A}(-) \psi || \; ||E_{A}(+)||.$$
(21)

3.3 EPR-Entanglement

In article [1] we invented the notion of entanglement which is also coupled to the probability update via conditional measurement, so called EPR-entanglement. Consider two observables, generally incompatible, represented by operators A and B.

Definition 1a

In the state \(|\psi \rangle ,\) the observables A and B are perfectly conditionally correlated (PCC) for the values \((A=\alpha , B=\beta )\) if the conditional probability to get the outcome \(B=\beta\) if the preceding A-measurement had the outcome \(A=\alpha\) equals to 1,

$$P(B= \beta | A= \alpha )= \frac{ \Vert E_{B}(\beta ) E_{A}(\alpha ) \vert \psi \rangle \Vert ^{2}}{\Vert E_{A}(\alpha ) \vert \psi \rangle \Vert ^{2}} = 1.$$
(22)

The order of observations is important even for compatible observables and to be more precise we have to speak about AB-PCC. More generally consider observables with values \((\alpha _{i})\) and \((\beta _{i})\) and some set \(\Gamma\) of pairs \((\alpha _{i}, \beta _{j}).\)

Definition 2a

(EPR entanglement) If, for all pairs from the set \(\Gamma ,\) \(\vert \psi \rangle\) is PCC-state, then such state is called EPR entangled w.r.t. \(\Gamma .\)

We are interested in sets \(\Gamma\) such that each of \(\alpha\) and \(\beta\) values appears in the pairs once and only once. We call such EPR entanglement complete.

For example, for two dichotomous observables with \(\alpha , \beta =\pm 1,\) we consider, e.g., the set of the pairs \((A=+, B=-), (A=-, B=+),\) in short, EPR \(A=-B\) entanglement, or the pairs \((A=+, B=+), (A=-, B=-),\) EPR \(A=B\) entanglement. Consider such EPR-entanglements.

Let us start with \(A=-B\) entanglement, i.e., \(P(B=-|A=+)=1\) and \(P(B=+|A=-)=1.\) Thus, \(P(B=+|A=+)=0\) and \(P(B=-|A=-)=0,\) and \(P(B=-|A=+)=1 \not = P(B=-|A=-)=0\) and \(P(B=+|A=-)=1 \not = P(B=+|A=+)=0.\) In this case EPR-entangled state is automatically AB-entangled in the sense of the present paper.

In the same way \(A=B\) entanglement, i.e., \(P(B=+|A=+)=1\) and \(P(B=-|A=-)=1.\) Thus, \(P(B=-|A=+)=0\) and \(P(B=+|A=-)=0,\) and \(P(B=+|A=+)=1 \not = P(B=+|A=-)=0\) and \(P(B=-|A=-)=1 \not = P(B= -|A=+)=0.\) And again EPR-entangled state is automatically AB-entangled.

Thus, EPR AB-entanglement is just the very special case of AB-entanglement.

3.4 Maximally AB-Entangled States

For dichotomous observables the measure of AB-entanglement (16) has the form

$$\begin{aligned} {{{\mathcal {M}}}}_{AB}(\psi )= & {} |P( B=+| A= -) - P( B= +| A= +)|\\{} & {} \quad + |P( B=-| A= -) - P( B= -| A= +)|, \end{aligned}$$

hence it can written as

$${{{\mathcal {M}}}}_{AB}(\psi ) = 2|P( B=+| A= -) - P( B= +| A= +)|.$$
(23)

From this formula, we immediately obtain the following characterization of maximally AB-entangled states:

Proposition 4

The AB-concurrence \({{{\mathcal {M}}}}_{AB}\) approaches its maximal value, \({{{\mathcal {M}}}}_{AB}(\psi )=2,\) if and only if \(|\psi \rangle\) is EPR AB-entangled.

4 Joint Eigenvectors’ Superposition

Our aim is find interrelation between AB-entanglement and the standard one. The latter is considered in the case of compatible observables.Footnote 1

Now we assume, except Sect. 8 that observables are dichotomous and compatible, i.e., represented by commuting Hermitian operators A and B, \([A,B]=0,\) with eigenvalues \(\alpha , \beta = \pm 1.\)

The state space can be represented as the direct sum of subspaces for the joint eigenvectors, \(A \vert \phi _{\alpha \beta }\rangle =\alpha |\phi _{\alpha \beta }\rangle , B \vert \phi _{\alpha \beta }\rangle = \beta |\phi _{\alpha \beta }\rangle ,\alpha , \beta = \pm ,\)

$${{{\mathcal {H}}}} = {{{\mathcal {H}}}}_{ab}(++) \oplus {{{\mathcal {H}}}}_{ab}(+-)\oplus {{{\mathcal {H}}}}_{ab}(-+) \oplus {{{\mathcal {H}}}}_{ab}(--).$$
(24)

Any (normalized) vector can be represented as superposition

$$\vert \psi \rangle = \vert \psi _{++} \rangle +\vert \psi _{+-}\rangle + \vert \psi _{-+}\rangle + \vert \psi _{--}\rangle ,$$
(25)

where \(\vert \psi _{\alpha \beta }\rangle \in {{{\mathcal {H}}}}_{ab}(\alpha \beta )\) and

$$1= \Vert \psi \Vert ^{2} = \Vert \psi _{++} \Vert ^{2} + \Vert \psi _{+-}\Vert ^{2} + \Vert \psi _{-+} \Vert ^{2} + \Vert \psi _{--} \Vert ^{2}.$$
(26)

To be able to determine the A-conditional probabilities, we have to assume that

$$\vert \psi _{++}\rangle + \vert \psi _{+-}\rangle \not =0, \; \vert \psi _{-+}\rangle + \vert \psi _{--}\rangle \not =0.$$
(27)

This is the constraint for operating with AB-entanglement.

We calculate the quantities for Formula (15)

$$\begin{aligned}{} & {} \langle \psi |E_{A}(+)| \psi \rangle = \Vert \psi _{++} \Vert ^{2} + \Vert \psi _{+-} \Vert ^{2}, \quad \langle \psi |E_{A}(-)| \psi \rangle = \Vert \psi _{-+}\Vert ^{2} + \Vert \psi _{--} \Vert ^{2},\\{} & {} \langle \psi |E_{B}(-) E_{A}(+)| \psi \rangle = \Vert \vert \psi _{+-}\rangle \Vert ^{2}, \quad \langle \psi |E_{B}(-) E_{A}(-)| \psi \rangle = \Vert \vert \psi _{--}\rangle \Vert ^{2}. \end{aligned}$$

Inequality (14) has the form:

$$\begin{aligned}{} & {} \Vert \psi _{+-}\Vert ^{2} \Big (\Vert \psi _{-+}\Vert ^{2} + \Vert \psi _{--}\Vert ^{2}\Big ) \\{} & {} \quad \not = \Vert \psi _{--}\Vert ^{2} \Big ( \Vert \psi _{++}\Vert ^{2} + \Vert \psi _{+-}\Vert ^{2}\Big ). \end{aligned}$$

And it is reduced to the inequality

$$\Vert \psi _{+-} \Vert ^{2} \Vert \psi _{-+} \Vert ^{2} \not = \Vert \psi _{--}\Vert ^{2} \Vert \psi _{++}\Vert ^{2}.$$
(28)

By Proposition 1 this is the necessary and sufficient condition for AB-entanglement. We note that the constraint (27) that is needed to be able to define quantum conditional probabilities is automatically encoded in (27).

5 Tensor Product Case

Now let us equip the state space with a tensor product structure \({{{\mathcal {H}}}}= {{{\mathcal {H}}}}_{1} \otimes {{{\mathcal {H}}}}_{2}\) and represent the operators A and B as tensor products, i.e., \(A= a \otimes I\) and \(B= I \otimes b,\) where \(a: {{{\mathcal {H}}}}_{1} \rightarrow {{{\mathcal {H}}}}_{1}\) and \(b: {{{\mathcal {H}}}}_{2} \rightarrow {{{\mathcal {H}}}}_{2}.\) We remark that \({{{\mathcal {H}}}}_{1}= {{{\mathcal {H}}}}_{1}(+) \oplus {{{\mathcal {H}}}}_{1} (-)\) and \({{{\mathcal {H}}}}_{2}= {{{\mathcal {H}}}}_{2}(+) \oplus {{{\mathcal {H}}}}_{2}(-).\) Each \(|\phi _{A} \rangle \in {{{\mathcal {H}}}}_{1}\) and \(|\phi _{B} \rangle \in {{{\mathcal {H}}}}_{2}\) can be represented as superpositions: \(|\phi _{A} \rangle = |\phi _{a,-} \rangle + |\phi _{a.+} \rangle\) and \(|\phi _{B} \rangle = |\phi _{b,-} \rangle + |\phi _{b.+} \rangle .\) Hence, for a factorisable state \(|\psi \rangle = |\phi _{A} \rangle |\phi _{B} \rangle ,\) superposition (25) has the form:

$$|\psi \rangle = |\phi _{a,+} \rangle |\phi _{b,+} \rangle + |\phi _{a,+} \rangle |\phi _{b,-} \rangle + |\phi _{a,-} \rangle |\phi _{b,+} \rangle + |\phi _{a,-} \rangle |\phi _{b,-} \rangle .$$
(29)

Hence,

$$\Vert \psi _{+-} \Vert ^{2} \Vert \psi _{-+} \Vert ^{2} = \Vert \psi _{--}\Vert ^{2} \Vert \psi _{++}\Vert ^{2}.$$
(30)

So, we proved the following simple, but important theorem:

Theorem 1

Let \({{{\mathcal {H}}}}= {{{\mathcal {H}}}}_{1} \otimes {{{\mathcal {H}}}}_{2}\) and \(A= a \otimes I, B= I \otimes b,\) where operators a, b have dichotomous spectra. Then any AB-entangled state \(|\psi \rangle\) is also entangled w.r.t. the tensor product structure.

Hence, AB-entangled states form the subclass \({{{\mathcal {E}}}}_{AB}\) of the class of all entangled states \({{{\mathcal {E}}}}\) for the tensor product decomposition \({{{\mathcal {H}}}}= {{{\mathcal {H}}}}_{1} \otimes {{{\mathcal {H}}}}_{2}.\) We shall see that \({{{\mathcal {E}}}}_{AB}\) is a proper subclass of \({{{\mathcal {E}}}}.\)

6 AB-Entanglement as Amplitude Factorization and Observables’ Correlation

The standard entanglement is characterized via the condition of state non-factorization w.r.t. the tensor product structure. We want to find an analogous condition for AB-entanglement. We start with studying the case of four dimensional state space.

6.1 Four Dimensional State Space

Consider now two commuting operators A and B acting in four dimensional space \({{{\mathcal {H}}}}\) having dichotomous spectra \(\alpha , \beta = \pm 1\) and common eigenvectors for all possible combinations of \(\alpha , \beta ,\) \(A \vert \alpha \beta \rangle = \alpha \vert \alpha \beta \rangle , B \vert \alpha \beta \rangle = \beta \vert \alpha \beta \rangle ,\) i.e., all joint eigensubspaces are one dimensional. In this case \({{{\mathcal {H}}}}\) and the operators can be represented in the tensor form, \({{{\mathcal {H}}}}= {{{\mathcal {H}}}}_{1} \otimes {{{\mathcal {H}}}}_{2}\) (two qubit state space) and \(A= a \otimes I, B=I \otimes b;\) we can use the standard notion of entanglement and compare it with AB-entanglement.

Any state can be represented as superposition

$$\vert \psi \rangle = c_{++}\vert ++ \rangle + c_{+-} \vert +-\rangle + c_{-+}\vert -+\rangle + c_{--}\vert --\rangle ,$$
(31)

where \(|c_{++}|^{2} + |c_{+-}|^{2} + |c_{-+}|^{2} + |c_{--}|^{2} =1.\)

This state is factorisable if and only if the coefficients satisfy the following equality:

$$c_{+-} c_{-+} = c_{--} c_{++}.$$
(32)

One of the measures of standard entanglement is concurrence and, for \(\vert \psi \rangle\), concurrence is given by

$${{{\mathcal {C}}}}(\psi )= |c_{+-} c_{-+} - c_{--} c_{++} |.$$
(33)

So. a state is entangled iff \({{{\mathcal {C}}}}(\psi )\not = 0.\) Re-normalized AB-concurrence (18) is written as

$${{{\mathcal {C}}}}_{AB}(\psi )= |\;|c_{+-} c_{-+}| - |c_{--} c_{++}|\; |.$$
(34)

So a state is AB entangled iff \({{{\mathcal {C}}}}_{AB}(\psi )\not = 0.\)

Definition 4

A state \(|\psi \rangle\) is called AB amplitude-factorisable w.r.t. the pair of compatible observables A and B with dichotomous spectrum if the absolute values of the coordinates in the joint eigenvectors basis are factorisable, i.e., for (31),

$$|c_{\alpha \beta }|= \lambda _{\alpha} \; \mu _{\beta} , \quad \text{ where } \; \alpha , \beta = \pm ,$$
(35)

where \(\lambda _{\alpha} , \;\mu _{\beta} \ge 0\) and

$$\lambda _{+}^{2} + \lambda _{-}^{2} =1, \quad \mu _{+}^{2} + \mu _{-}^{2}= 1.$$
(36)

The state factorization implies the amplitude factorization, but not vice verse: condition (35) does not take into account the presence of phases, but the latter can generate the standard entanglement even for amplitude-factorisable states.

Let

$$\begin{aligned} |\psi \rangle= & {} (\lambda _{+} |+ \rangle + e^{i \theta _{1}} \lambda _{-} |- \rangle ) \otimes (\mu _{+} |+ \rangle + e^{i \theta _{2}} \mu _{-} |- \rangle ) \\= & {} \lambda _{+} \mu _{+} |++ \rangle + e^{i \theta _{2}} \lambda _{+}\mu _{-} |+ -\rangle + e^{i \theta _{1}} \lambda _{-} \mu _{+} |-+ \rangle + e^{i (\theta _{1}+\theta _{2})} \lambda _{-} \mu _{-} |-- \rangle . \end{aligned}$$

Now consider the state

$$|\psi \rangle =|++ \rangle + |+ -\rangle - |-+ \rangle +|-- \rangle ,$$

it is entangled in ordinary sense, although it is amplitude factorisable.

The violation of relation (28) has the form of the equality:

$$|c_{+-}|^{2} |c_{-+}|^{2} = |c_{--}|^{2} |c_{++}|^{2}$$
(37)

Theorem 2

Let \(\textrm{dim}\; {{{\mathcal {H}}}}= 4.\) For a pair of compatible dichotomous observables A and B,  a state \(|\psi \rangle\) is AB-entangled if and only if it is amplitude non-factorisable.

Proof

As we have seen, AB-disentangled is characterized by the equality (37). We prove that this equality is equivalent to the amplitude factorization. It is clear that amplitude factorization implies (37). Now let this equality holds true. We set (under the condition that the denominators are not equal to zero):

$$\lambda _{+}=\frac{|c_{++}|}{\sqrt{|c_{++}|^{2} + |c_{-+}|^{2}}},\quad \lambda _{-}=\frac{|c_{-+}|}{\sqrt{|c_{++}|^{2} + |c_{-+}|^{2}}},$$
(38)
$$\mu _{+}= \frac{|c_{++}|}{\sqrt{|c_{++}|^{2} + |c_{+-}|^{2}}},\quad \mu _{-}=\frac{|c_{+-}|}{\sqrt{|c_{++}|^{2} + |c_{+-}|^{2}}}.$$
(39)

It is easy to check that these quantities factorize the modules of the coordinates \(c_{\alpha , \beta }.\) For example,

$$\lambda _{+} \mu _{-} = |c_{++}| |c_{+-}|/ \sqrt{(|c_{++}|^{2} + |c_{-+}|^{2})(|c_{++}|^{2} + |c_{+-}|^{2})}.$$

So, the square of the denominator is given by

$$|c_{++}|^{4} + |c_{++}|^{2}|c_{+-}|^{2} + |c_{++}|^{2} |c_{-+}|^{2} + |c_{+-}|^{2} |c_{-+}|^{2}.$$

On the other hand, the AB-disentanglement equation gives us

$$|c_{++}|^{2}( 1- |c_{++}|^{2} - |c_{+-}|^{2} - |c_{-+}|^{2}) = |c_{+-}|^{2} |c_{-+}|^{2}.$$

Hence, denominator equals to \(|c_{++}|\) and \(\lambda _{+} \mu _{-} = |c_{+-}|.\)

Now consider the case in that one of denominators in (38), (39) equals to zero, say \(c_{++}=0, c_{-+}=0.\) In this case amplitude factorization can be shown directly. We just set \(\lambda _{-} = |c_{--}|, \lambda _{+} = |c_{+-}|, \mu _{-}=1, \mu _{+}=0.\)

Since some states belonging to \({{{\mathcal {E}}}}\) can be amplitude factorisable, embedding \({{{\mathcal {E}}}}_{AB} \rightarrow {{{\mathcal {E}}}}\) is not surjection.

Gnerally for a state \(|\psi \rangle ,\) we set \(\langle A \rangle _{\psi} = \langle \psi | A |\psi \rangle , \langle B \rangle _{\psi} = \langle \psi | B |\psi \rangle , \langle A B \rangle _{\psi} = \langle \psi | AB |\psi \rangle .\) Covariance of quantum observables A and B is defined as

$$\textrm{cov}(A,B)= \langle (A - \langle A \rangle _{\psi} )(B- \langle B \rangle _{\psi} ) \rangle _{\psi} = \langle AB \rangle _{\psi} - \langle A \rangle _{\psi} \langle B \rangle _{\psi} .$$
(40)

Observables are called AB-uncorrelated in state \(|\psi \rangle\) if \(\textrm{cov}(A,B) = 0\) or correlation factorization condition holds:

$$\langle A B \rangle _{\psi} = \langle A \rangle _{\psi} \langle B \rangle _{\psi} .$$
(41)

Otherwise they are called correlated. That is, if \(\textrm{cov}(A,B)\not =0.\) \(\square\)

Theorem 3

A pair of compatible dichotomous observables A and B are AB-entangled in a state \(|\psi \rangle\) iff they are correlated: \(\textrm{cov}(A,B)\not =0,\) i.e.,

$$\langle A B\rangle _{\psi} \not = \langle A \rangle _{\psi} \langle B\rangle _{\psi} .$$
(42)

Proof

Set \(a=|c_{++}|^{2}, b= |c_{--}|^{2}, c= |c_{+-}|^{2}, d= |c_{-+}|^{2}.\) Set \(x= a+b, y= c+d.\) We remark that \(x+y=1.\) The latter implies that \(x^{2} - y^{2}= x-y.\) Hence,

$$\langle A B\rangle _{\psi} = a+b - c - d= x-y= x^{2} - y^{2} = a^{2}+b^{2}- c^{2}-d^{2} + 2ab - 2cd.$$

Then

$$\begin{aligned}{} & {} \langle A \rangle _{\psi} \langle B\rangle _{\psi} = ((a+c) -(d+b))((a+d) - (b+c))=a^{2}+b^{2}- c^{2}-d^{2} - 2ab + 2cd\\{} & {} \quad = x^{2} - y^{2}+ 4ab -4 cd. \end{aligned}$$

Hence, the correlation \(\langle A B\rangle _{\psi}\) is factorisable if and only if \(ab=cd,\) i.e., the amplitude factorization condition (37) is satisfied. \(\square\)

This proposition supports our invention of entanglement as dependence of observables. Disentanglement corresponds to the classical probabilistic situation of independence of observables given by random variables A and B. For such random variables, independence implies \(\langle A B\rangle _{P}= \langle A \rangle _{P} \langle B\rangle _{P},\) where P is the probability measure (it plays the role of a quantum state). For dichotomous random variables independence is equivalent uncorrelation (see Sect. 8).

It is easy to show that in the quantum theory factorization of correlations does not imply the standard disentanglement, because the amplitude factorization condition can hold for an entangled state. In contrast, by Theorem 3AB-disentanglement is equivalent to correlation factorization for these observables.

6.2 State Space of Arbitrary Dimension

Now we turn to Sect. 4. Let in superposition (25) all vectors \(\vert \psi _{\alpha \beta } \rangle \not =0.\) In particular, this presumes that all subspaces \({{{\mathcal {H}}}}_{AB}(\alpha \beta )\) are nontrivial.

We rewrite this superposition as (31), where

$$c_{\alpha \beta } = ||\psi _{\alpha \beta }||$$
(43)

and \(\vert \alpha \beta \rangle = \vert \psi _{\alpha \beta } \rangle / ||\psi _{\alpha \beta }||.\) Hence, we again can restrict consideration to the four dimensional case. Instead of the expansion of a state in the basis of joint eigenvectors of the operators A and B,  we consider its expansion in its projections on joint eigensubspaces \({{{\mathcal {H}}}}_{AB} (\alpha \beta ), \alpha \beta = \pm 1.\) We remark that here we do not consider any tensor product structure.

The violation of AB-entanglement condition (28) has the form of equality:

$$\Vert \psi _{+-} \Vert ^{2} \Vert \psi _{-+} \Vert ^{2} = \Vert \psi _{--}\Vert ^{2} \Vert \psi _{++}\Vert ^{2}.$$
(44)

Definition 4a

A state \(|\psi \rangle\) is called AB amplitude-factorisable if

$$||\psi _{\alpha \beta }||= \lambda _{\alpha} \mu _{\beta} , \quad \alpha , \beta = \pm ,$$
(45)

where \(\lambda _{\alpha} , \mu _{\beta} \ge 0\) and

$$\lambda _{+}^{2} + \lambda _{-}^{2} =1, \quad \mu _{+}^{2} + \mu _{-}^{2}= 1.$$
(46)

Theorem 2a

For compatible dichotomous observables A and B,  a state \(|\psi \rangle\) is AB-entangled if and only if it is amplitude non-factorisable.

Proof

AB-disentangled is characterized by the equality (44). This equality is equivalent to the amplitude factorization. The proof is in the line with the proof for the four dimensional case. It is evident that amplitude factorization implies equality (44) Now let the latter hold true. We set

$$\lambda _{+}=\frac{||\psi _{++}||}{\sqrt{||\psi _{++}||^{2}+ ||\psi _{-+}||^{2}}},\quad \lambda _{-}=\frac{||\psi _{-+}||}{\sqrt{||\psi _{++}||^{2} + ||\psi _{-+}||^{2}}},$$
(47)
$$\mu _{+}= \frac{||\psi _{++}||}{\sqrt{||\psi _{++}||^{2} + ||\psi _{+-}||^{2}}}, \quad \mu _{-}=\frac{||\psi _{-+}||}{\sqrt{||\psi _{++}||^{2} + ||\psi _{+-}||^{2}}}.$$
(48)

It is easy to check that these quantities factorize the modules of the coordinates \(c_{\alpha , \beta }.\) The cases of zero denominators are analyzed similarly to the four dimensional case. \(\square\)

As in the the four dimensional case, we obtain characterization of AB-entanglement as observables’ correlation:

Theorem 3 is valid in the general case.

7 Analog of Singlet State for Observables Entanglement

Consider a family of pairs of operators \(( A_{u}, B_{u}),\) where u is some parameter. Can one find a quantum state \(|\psi \rangle\) that is \(A_{u} B_{u}\)-entangled for all these pairs?

We assume that, for any u,  \([ A_{u}, B_{u}] =0,\) but it may be that, for some pairs u, v,  \([ A_{u}, A_{v}] \not =0\) or (and) \([ B_{u}, B_{v}] \not =0.\)

Let us consider the tensor product case \({{{\mathcal {H}}}}= H \otimes H,\) where \(\textrm{dim} \; H=2,\) and two operators in H, a, b,  with the eigenbases \((f_{+}, f_{-})\) and \((g_{+}, g_{-}),\) where \(a f_{\pm }= \pm f_{\pm }, b g_{\pm }= \pm g_{\pm }.\) Set \(A= a \otimes I, B= I \otimes b.\)

Take now any unitary transformation u in H and consider operators of the form

$$a_{u} = u a u^{\star} ,\quad b_{u} = u b u^{\star} ,$$
(49)

which are diagonal w.r.t. to the bases \((f_{u,+}, f_{u,-})\) and \((g_{u,+}, g_{u,-}),\) obtained with the unitary transformation u from the bases \((f_{+}, f_{-})\) and \((g_{+}, g_{-}).\)

Then consider state \(\vert \psi \rangle\) having the form

$$\vert \psi \rangle = c_{+-} \vert f_{+} g_{-}\rangle + c_{-+} \vert f_{-} g_{+}\rangle ,\quad c_{-+}, c_{-+} \not =0.$$
(50)

This state is AB-entangled. And it preserves its form under the unitary transformation \(U= u\otimes u\) if and only if \(c_{+-}= - c_{-+},\) i.e.,

$$\vert \psi \rangle = (\vert f_{+} g_{-}\rangle - \vert f_{-} g_{+}\rangle )/ \sqrt{2}= (\vert f_{u,+} g_{u,-}\rangle - \vert f_{u,-} g_{u,+}\rangle )/ \sqrt{2}.$$
(51)

Hence, this state is AB-entangled both for the pair A, B and the pair \(A_{u}= a_{u}\otimes I , B_{u}= I \otimes b_{u}\) for any unitary transformation \(u: H \rightarrow H.\)

In particular, let us consider the standard singlet state

$$\vert \psi _{\textrm{singlet}} \rangle = (\vert f_{+} f_{-}\rangle - \vert f_{-} f_{+}\rangle )/\sqrt{2}.$$
(52)

This state is \(AA^{\prime}\)-entangled for the operators \(A=a \otimes I\) and \(A^{\prime} =I \otimes a.\) And it is also \(A_{u} A_{u}^{\prime}\)-entangled for the operators \(A_{u}=a_{u} \otimes I\) and \(A_{u}^{\prime} =I \otimes a_{u}.\)

The singlet state is entangled (in the standard non-factorization sense) independently from the selection of bases, i.e., the operators A and B. However, this is not the case of the operator based entanglement. It is easy to find u such that \(\vert \psi _{\textrm{singlet}} \rangle\) is \(AA_{u}^{\prime}\)-disentangled, i.e., amplitude factorisable (cf. discussion in Sect. 6.1). Let

$$\vert f_{-}\rangle = u_{-+} |f_{u,+}\rangle + u_{--} |f_{u,-}\rangle , \quad \vert f_{+}\rangle = u_{++} |f_{u,+}\rangle + u_{+-} |f_{u,-}\rangle ,$$

where \(u_{-+} {\bar{u}}_{++} + u_{--} {\bar{u}}_{+-}=0,\; |u_{-+}|^{2} + |u_{--}|^{2} =1, \; |u_{++}|^{2} + |u_{+-}|^{2} =1.\) Then

$$\vert \psi _{\textrm{singlet}} \rangle = (u_{-+} |f_{+} f_{u,+}\rangle + u_{--} |f_{+} f_{u,-}\rangle + u_{++} |f_{-} f_{u,+}\rangle - u_{+-} |f_{-} f_{u,-}\rangle ) /\sqrt{2}.$$

Now select \(u_{-+} = u_{+-}= u_{--}= - u_{++}= 1/\sqrt{2}.\) Then

$$\vert \psi _{\textrm{singlet}} \rangle = (|f_{+} f_{u,+}\rangle + |f_{+} f_{u,-}\rangle + |f_{-} f_{u,+}\rangle - |f_{-} f_{u,-}\rangle ) /2.$$

This state is amplitude factorisable (but not tensor product factorisable) and hence \(AA_{u}^{\prime}\)-disentangled.

Starting with the conditional probabilistic definition of AB-entanglement, we analyzed its representation in the complex Hilbert space formalism and its interrelation (in the tensor product case) with the standard entanglement.

For compatible dichotomous observables, AB-entanglement is equivalent to amplitude non-factorization, where amplitudes are state’s projections on joint eigensubspaces of the operators. This characterization leads to the equivalence of AB-entanglement to correlation of observables. With such construction, we can proceed in an arbitrary Hilbert space, i.e., without referring to the tensor product structure.

8 AB-Entanglement vs. Dependence of Classical Random Variables

We recall that in classical probability theory discrete random variables A and B are independent iff their joint probability distribution (JPD) is factorisable, i.e., for all \(\alpha , \beta ,\)

$$P(A= \alpha , B= \beta )= P(A= \alpha ) P(B= \beta ).$$
(53)

This definition is equivalent to reduction of conditional probabilities to “absolute probabilities”:

$$P(B= \beta |A= \alpha )= P(B= \beta ),\quad \text{ if }\; P(A= \alpha )\not =0,$$
(54)
$$P(A= \alpha |B= \beta )= P(A= \alpha ), \quad \text{ if }\; P(B= \beta )\not =0.$$
(55)

Thus, for non-zero probabilities \(P(A= \alpha )\) and \(P(B= \beta )\) independence—probability factorization (53), is equivalent to conditional independence (54), (55).

In quantum theory AB-disentanglement means that there exists an outcome \(B=\beta\) such that

$$P(B= \beta |A= \alpha )= P(B= \beta |A= \alpha ^{\prime} )$$
(56)

for all \(\alpha , \alpha ^{\prime}\) (in the case of \(P(A= \alpha )\not =0, \alpha \in X_{a}).\)

In classical probability, for random variables A and B,  equality (56) is equivalent to equality (54)—for the concrete outcome \(B=\beta\) (for non-zero probabilities for A-outcomes). We prove this to illustrate the role of one of the basic laws of classical probability theory—the formula of total probability (FTP):

$$P(B= \beta )= \sum _{\alpha} P(B= \beta |A= \alpha )P(A= \alpha ).$$
(57)

Denote the quantity determined by (56) as \(P(B= \beta |A).\) We want to show that it coincides with \(P(B= \beta ).\) We put \(P(B= \beta |A)\) in (57):

$$P(B= \beta )= P(B= \beta |A)) \sum _{\alpha} P(A= \alpha )= P(B= \beta |A)).$$
(58)

Thus, we obtain (54) and hence (55) as well as (53)—for the concrete value \(\beta .\)

Generally in quantum probability theory FTP is violated, classical FTP is additively perturbed by the interference term:

$$\begin{aligned} P(B= \beta )= & {} \sum _{\alpha , \alpha ^{\prime} } \langle \psi | E_{A}(\alpha ) E_{B}(\beta ) E_{A}(\alpha ^{\prime} ) | \psi \rangle \\= & {} \sum _{\alpha} P(B= \beta |A= \alpha )P(A= \alpha ) + \delta _{AB}(\beta ), \end{aligned}$$
(59)

where

$$\delta _{AB}(\beta )=\sum _{\alpha \not =\alpha ^{\prime} } \langle \psi | E_{A}(\alpha ) E_{B}(\beta ) E_{A}(\alpha ^{\prime} ) | \psi \rangle .$$
(60)

In RHS, the first summand corresponds to classical FTP and the second one is the interference term.Footnote 2

By trying to perform the above calculations we obtain

$$P(B= \beta ) - P(B= \beta |A)) = \delta _{AB}(\beta ).$$
(61)

Thus, generally the notion of AB-disentanglement differs from the classical notion of independence, formulated as (54). What is about the basic definition of classical independence via probability factorization, see (53)?

In quantum probability sequential JPD is defined as

$$P_{AB}(A=\alpha , B= \beta ) \equiv P(A=\alpha ) P(B=\beta |A=\alpha ).$$
(62)

In operator terms

$$P_{AB}(A=\alpha , B= \beta ) = \langle \psi | E_{A}(\alpha ) E_{B}(\beta ) E_{A}(\alpha )| \psi \rangle = ||E_{B}(\beta ) E_{A}(\alpha ) \psi ||^{2} .$$
(63)

For non-commuting observables, generally

$$P_{AB}(A=\alpha , B= \beta )= ||E_{B}(\beta ) E_{A}(\alpha ) \psi ||^{2} \not =$$
(64)
$$||E_{A}(\alpha ) E_{B}(\beta ) \psi ||^{2} =P_{BA}( B= \beta , A=\alpha ).$$

Definition 5

Observables A and B are called AB-independent w.r.t. state \(|\psi \rangle ,\) if their sequential JPD is factorisable, i.e., for all pairs \(\alpha , \beta ,\)

$$P_{AB}(A= \alpha , B= \beta )= P(A= \alpha ) P(B= \beta ).$$
(65)

Otherwise observables are called AB-dependent in \(|\psi \rangle .\) Since JPD is the sequential JPD, AB- and BA-(in)dependence are not equivalent. (For commuting observables, AB- and BA-(in)dependence are equivalent.)

It is easy to check that two sorts of AB-independence, i.e., via (54), (55) and (65) are equivalent even for quantum probability (for non-zero probabilities of outcomes). They can be used as another approach to the notion of AB-conditional probability (dis)entanglement.

The most interesting for application is entanglement for compatible observables, \([A, B]=0.\) Here quantum FTP (59) is reduced to classical FTP (58), since the interference term (60) equals to zero.

Proposition 5

For compatible observables A and B and state \(|\psi \rangle ,\) the notions of AB-entanglement and AB-dependence are equivalent.

Proof

(1) Let quantum observables be dependent. Then there exists outcomes \(\beta , \alpha\) such that

$$P_{AB}(A= \alpha , B= \beta ) \not = P(A= \alpha ) P(B= \beta ).$$
(66)

In particular. this implies that \(P(A= \alpha )\not =0, P(B= \beta )\not =0,\) since for commuting operators \(P_{AB}(A= \alpha , B= \beta ) = ||E_{B}(\beta ) E_{A}(\alpha ) \psi ||^{2} \le || E_{A}(\alpha ) \psi ||^{2}= P(A= \alpha )\) and in the same way \(P_{AB}(A= \alpha , B= \beta ) \le P(B= \beta ).\) Thus, we have

$$P(B= \beta |A= \alpha ) \not = P(B= \beta ).$$
(67)

Now suppose that \(|\psi \rangle\) is not AB-entangled. Then there exists an outcome \(\beta\) such that the equality \(P(B= \beta |A= \alpha )= P(B= \beta |A= \alpha ^{\prime} )\) holds for all pairs \(\alpha , \alpha ^{\prime} .\) Denote the quantity determined by this constraint by \(P(B= \beta |A).\) Now we apply FTP, and obtain that \(P(B= \beta )= P(B= \beta |A).\) But this contradicts (67).

(2) Let quantum observables be AB-entangled. Then, for each \(\beta ,\) \(P(B= \beta |A= \alpha )\not = P(B= \beta |A= \alpha ^{\prime} )\) for some \(\alpha , \alpha ^{\prime} .\) Suppose that quantum observables are independent, i.e., \(P_{AB}(A= \gamma , B= \beta ) = P(A= \gamma ) P(B= \beta )\) for all \(\gamma , \beta .\) This implies that \(P(B= \beta |A= \gamma ) = P(B= \beta ),\) and hence \(P(B= \beta |A= \alpha )= P(B= \beta |A= \alpha ^{\prime} )\) for \(\alpha , \alpha ^{\prime}\) considered above. \(\square\)

We remark that compatible observables A and B can represented by random variables \(\xi _{A}\) and \(\xi _{B}\) on classical probability space. In this case dependence of observables corresponds to dependence of random variables. By Proposition 5AB-entanglement also corresponds to dependence of random variables. We formulate this coupling as

Proposition 6

For compatible observables, AB-entanglement can be treated as the complex Hilbert space representation of dependence of classical random variables.

Now we compare the notion of uncorrelated classical random variables and quantum observables in the dichotomous case, i.e., \(A, B= \pm 1.\) We recall the following result.

Proposition

Dichotomous random variables are independent iff they are uncorrelated.

This is classical probabilistic counterpart part of Theorem 3.

9 Concluding Remarks

In this article we extended the conditional probability approach to entanglement initiated in [1] for probability one predictions. This approach can be considered as the probabilistic formalization of the original Schrödinger’s viewpoint on entanglement [38, 39]. We recall that he considered entanglement as apart of complex Hilbert space machinery for prediction of outcomes of observables.Footnote 3 And he treated the EPR argument [37] in this way. Hence, for Schrödinger, entanglement is related to a pair of observables A and B,  i.e., a state is entangled w.r.t. to a selected pair, this is AB-entanglement. This original Schrödinger’s treatment of entanglement as observables entanglement was deformed into state-entanglement and highlighting the role of the tensor product structure (with just a few exceptions, see [2, 3]).

In this article entanglement is defined as dependence of observables A and B. This viewpoint on entanglement was reflected in Schrödinger’s “Beschränkung"’ by using for it the meaning “restriction”. Of course, it would be better if Schrödinger would from the very beginning use the word “dependence” and associate dependence of classical random variables with AB-entanglement, as is done in this article.