Disproving hidden variable models with spin magnitude conservation

The squares of the three components of the spin-s operators sum up to $s(s+1)$. However, a similar relation is rarely satisfied by the set of possible spin projections onto mutually orthogonal directions. This has fundamental consequences if one tries to construct a hidden variable (HV) theory describing measurements of spin projections. We propose a test of local HV-models in which spin magnitudes are conserved. These additional constraints imply that the corresponding inequalities are violated within quantum theory by larger classes of correlations than in the case of standard Bell inequalities. We conclude that in any HV-theory pertaining to measurements on a spin one can find situations in which either HV-assignments do not represent a physical reality of a spin vector, but rather provide a deterministic algorithm for prediction of the measurement outcomes, or HV-assignments represent a physical reality, but the spin cannot be considered as a vector of fixed length.

Introduction. Quantum theory allows to probabilistically predict measurement outcomes, however it says nothing about the underlying physical reality of a system that is measured. In the early days of quantum physics it was often assumed that an extension of the theory using some additional hidden variables (HVs) would provide one with mathematical tools to describe the underlying reality. However, this assumption lead to fundamental physical consequences, such as nonlocality or contextuality [1][2][3][4][5][6][7][8][9][10].
In this work we revisit the above problem and show that any HV theory of spin either describes the underlying physical reality, but the magnitude of the spin is not conserved, or the magnitude is conserved, but HVs do not determine the underlying physical reality -they only provide a deterministic algorithm for calculating the outcomes of measurements. This observation is a nonclassical aspect of spin systems that is related, but distinct, from the Bell [1], the Kochen-Specker [4], and the Leggett theorems [5].
We argue that a physical HV model should not only be local, realistic and noncontextual, but should also obey the conservation laws. In case of spin systems the model should conserve angular momentum, i.e., the length of spin vectors should be fixed (see [11][12][13] for another additional constraints on HV models). As a result, our model extends the nonclassical behaviour to a broader class of quantum states. There are spin states that are local and noncontextual, but whose spin vector cannot be described by a magnitude conserving model. Apart from foundational importance, the spin magnitude conserving HV model can have potential applications in various fields of quantum physics. Moreover, we show that the model can be tested in experiments analogous to the ones testing standard locality and noncontextuality. New experiments can be performed and some old ones can be re-examined.
Let us first recall some basic facts. Consider a spin s particle whose spin state is represented by a vector or a density matrix in a (2s + 1)-dimensional Hilbert space. The number s is either an integer or a half-integer. The average values of spin coordinates can be determined with a help of the three spin operatorsŜ x ,Ŝ y , andŜ z [19]. The spectrum of each operator is: s, s − 1, . . . , −s. These eigenvalues correspond to possible spin projections onto a given axis. Interestingly,Ŝ 2 x +Ŝ 2 y +Ŝ 2 z = s(s + 1)1 1, where1 1 is the identity operator. The physical meaning of the above formula is that the length of the spin vector is s(s + 1) and that within quantum formalism this property is conserved because it does not depend on a spin state. In addition, the relation for the sum of squares does not depend on the choice of directions x, y, and z. What matters, is that the directions are mutually orthogonal.
However, imagine that there is a HV theory that assigns a well defined value to each spin operator. More precisely, each spin operatorŜ n , where n denotes the direction in space, is preassigned one of 2s + 1 eigenvalues, say v(Ŝ n ) = s−1. The Kochen-Specker theorem [4] states that for s ≥ 1 such assignment is not possible for some properly chosen sets of directions, but it is in principle possible to do it for three mutually orthogonal directions x, y and z for any s.
Here, we focus on the following problem. We know that it is possible to deterministically assign eigenvalues to three spin operators for mutually orthogonal axes. This results in a HV description of spin vector s = (v(Ŝ x ), v(Ŝ y ), v(Ŝ z )). However, is it possible to satisfy The above question has a positive answer for some values of s, but before we provide a general answer, let us consider three examples. For s = 1/2 the HV model leads to s = (± 1 2 , ± 1 2 , ± 1 2 ), hence s · s = 3/4 = s(s + 1). Therefore, for s = 1/2 a HV model naturally conserves the spin magnitude.
The smallest s for which all HV assignments do not satisfy s · s = s(s + 1) is s = 3/2. This time the eigenvalues of a spin operator are ± 3 2 and ± 1 2 , whereas s(s + 1) = 15 4 . All HV assignments can be split into four classes: . They lead to the following values of s · s: 27 4 , 19 4 , 11 4 and 3 4 , respectively. None of them is equal to 15 4 , therefore the physical interpretation of the HV model is either that s represents the physical reality but the spin magnitude is not conserved, or that s does not represent the physical reality but provides a deterministic method for calculating outcomes of measurements.
Magnitude conservation for arbitrary s. We look for a solution to the following equation: where we used a simplified notation s . The numbers s, s i (i = x, y, z) are either all integers or half-integers. We are going to show that: for the half-integer case Eq. (1) has a solution iff 2s = 1( mod 4), i.e., s = 1 2 , 5 2 , 9 2 , . . .; for the integer case Eq. (1) has a solution iff s cannot be written in either of the forms: where i and j are non-negative integers. The first few integer values of s for which there is no solution are s = 12, 15, 19, 44, 51, . . . The above can be proven by using the Legendre's threesquare theorem of 1798. It tells that a non-negative integer n can be written as a sum of three squares of integers iff n is not of the form where a and b are non-negative integers.
In the former case, the RHS can be written as a sum of three squares of integers. What we should prove is that all three numbers are squares of odd integers. It is clear that the square of an odd integer is odd and the square of an even integer is even. Next, the sum of squares of three odd integers is always 3 (mod 8). This is a straightforward consequence of the fact that (2k+1) 2 = 1 + 8k(k + 1)/2 and k(k + 1)/2 is an integer if k is an integer. The sum of squares of two even numbers and an odd number (the only other way to get an odd RHS) can not be 3 (mod 8), because the square of an even number is either 0 (mod 8) or 4 (mod 8), so this way the LHS is either 1 (mod 8) or 5 (mod 8). This concludes the proof for the half-integer case.
Integer case. How can s(s+1) take the form in Eq. (6)? Since s(s + 1) is even, a in Eq. (6) cannot be 0.
First let s be even. We can assume s = 4 a q, where q is a positive integer. This is because the even factor 4 a in the product s(s+1) must be contained in s. Then, s(s+1) = 4 a (4 a q 2 + q). Hence (4 a q 2 + q) = 8b + 7, which implies that q must be odd. We write 4 a q 2 + q = 7 (mod 8). From previous proof we know that q 2 = 1 (mod 8) which implies 4 a + q = 7 (mod 8). If a = 1 the above condition means q = 3 (mod 8), while if a > 1 the above condition means q = 7 (mod 8). The former case corresponds to (2), while the latter one to (3). Now let s be odd. In this case, s + 1 = 4 a q, where q is an odd integer as in the previous case. Then s(s + 1) = (4 a q −1)4 a q = 4 a (4 a q 2 −q). This is of the form Eq. (6) iff 4 a − q = 7 (mod 8) (see argument above). If a = 1, then the above condition means q = 5 (mod 8), while if a > 1 the above condition means q = 1 (mod 8). The former case corresponds to (4), while the latter one to (5).
Refutation of the HV model. We see that the spin magnitude conserving HV model can be in principle constructed for half of the half-integer cases and most of the integer ones. For the values of s for which it cannot be constructed one can speak of a phenomenon that is analogous to the state-independent contextuality [14,15], namely there is no model for any state. For the remaining cases one needs to look for less straightforward methods to refute this model.
Firstly, note that for most s values, even if there is a model, only a limited number of possible spin projections can be measured if the spin magnitude is to be conserved. The remaining projections can never be measured according to the model, so if they are measured, the model is contradicted. For example, for s = 2 we have s(s + 1) = 6. The only HV assignments allowed by our model are of the form (±2, ±1, ±1) and all permutations thereof. Nevertheless, it is in principle possible to measure that a projection of the spin onto some axis is zero. This value does not occur in any allowable HV assignment, therefore by measuring projection zero one automatically refutes the model. However, there exist states, such as |s = 2, s z = 1 , for which the probability of measuring a projection zero along the directions x, y or z is zero. This is because this state is orthogonal to |s = 2, s x = 0 , |s = 2, s y = 0 , and |s = 2, s z = 0 . Therefore, the above example can be considered as a state-dependent scenario (with respect to measurements along the x, y, and z axes).
Bell-like scenario. In general, the state-dependent HV scenarios are tested with the help of Bell-like inequalities [17]. Therefore, our next goal is to find such an inequality to test our spin magnitude conserving model.
Since we are interested in the magnitudes of spin vectors, we should focus on spin measurements along three mutually orthogonal directionsŜ x ,Ŝ y , andŜ z . Although these measurements cannot be performed jointly, due to a lack of commutation, they are deeply related because of the physical properties of the model studied by us. However, these three measurements are not enough to construct any meaningful Bell-like inequality. We need more measurements. A natural choice, the one we follow, is to consider a bipartite scenario with a spatially separated pair of spins, which we label A and B, and a set of six measurements:Ŝ where i, j = x, y, z.
The general form of the correlation inequality for two spin-s particles and the six spin operators is of the form where c kl are real coefficients and β is the classical bound derived from our model. The above can be rewritten as where ) is the vector of operators and Before we derive an inequality for general s, let us consider few examples for s = 1 and s = 2.
Example 1. Consider two spin-1 particles and the following inequality One can easily show by considering all possible HV assignments, which in our model are of the form s X = (±1, ±1, 0), including all possible permutations, that the classical HV bound taking into account the conservation of the spin vector magnitude is β = −2.
On the other hand, the classical HV bound which does not take into account the spin magnitude and allows for assignments such as (±1, ±1, ±1), or (±1, 0, 0), leads to the boundβ = −3.
Finally, one can find that in quantum theory the spin magnitude conserving bound can be violated down to 5616. This means that in the corresponding scenario quantum theory cannot be described by a model in which spin magnitudes are conserved, but can be described, in principle, by the model in which the magnitudes are not conserved. Interestingly, the quantum state giving the maximal violation β Q above is not a maximally entangled two-qutrit state. It is a partially entangled state with two equal Schmidt coefficients, and the third Schmidt coefficient being nonzero (a similar situation occurs in the case of the well-known CGLMP inequality [16]).
Example 2. Consider another inequality and two spin-1 particles This time β = −4,β = −7 and β Q = − √ 17 = −4.1231. Example 3. Finally, consider two spin-2 particles and the following inequality For this example β = −20,β = −34 and β Q = −20.1897. Note that there is a geometric interpretation of the inequalities above. Namely, the set of local distributions with conserved spin magnitude form a polytope P in the (9-dimensional) space of correlations. This polytope P is completely characterized by a finite number of vertices, each of them corresponding to a possible deterministic HV assignment. In the HV model, any correlation can be expressed as a convex combination of deterministic HV assigments, hence it is a point lying inside the polytope P. Conversely, any point outside the polytope P can be detected with inequalities which define the facets of the polytope P. Each of the three examples above are such kind of inequalities, which detect quantum correlations not reproducible with HV models.
Inequality for general s. The inequalities in the above examples seem not to be related and one may get an impression that for each s one needs to derive an independent inequality, which would make the problem much more complicated. However, we show that a universal approach is possible and one can derive an inequality which can be violated for any s. The main idea behind the derivation is to use the symmetry of the two spin-s singlet state and to assume that the matrix C is an orthogonal rotation matrix, i.e., the columns and the rows of C are normalized and orthogonal and its determinant is one.
Singlet state of two spin-s particles. Consider the following state This is a singlet state of two spin-s particles. It is maximally entangled and the corresponding total angular momentum is zero. As a result, the state is invariant under rotations generated by e i n·S (A) θ ⊗ e i n·S (B) θ . Next, let us consider correlations The above can be evaluated using the formula for square pyramidal numbers n k=1 k 2 = n(n + 1)(2n + 1) 6 .
We get The rotational symmetry implies Therefore j=x,y,z However, the above can be rewritten as where C is an orthogonal matrix whose entries (c xx , c xy , . . . ) are given by The other way around. For a pair of arbitrary s-spin particles the operator where C is a rotation matrix, has an eigenvalue −s(s+ 1) with the corresponding eigenvector |φ . In fact, this is the smallest eigenvalue of the operator (25), because it has the same spectrum as S (A) · S (B) , for which −s(s + 1) is the smallest eigenvalue. This stems from and since (S (A) ) 2 = (S (B) ) 2 = s(s + 1) one gets Classical bound. We know that if C is a rotation matrix, then in quantum theory the following expression has a lower bound β Q = −s(s + 1). In particular, there exists a state |φ which corresponds to an eigenvalue −s(s + 1). The goal is to show that the classical bound β (taking into account the conservation of spin vector magnitude) obeys β > −s(s + 1).
In case of our HV model the above expression takes form where the average is taken with respect to some probability distribution over vectors s (X) )) whose entries are integers or half-integers. Moreover, S (X) · S (X) = s(s + 1), due to the fact that the HV model conserves the spin magnitude.
The lower bound is determined by two vectors a and b which minimize the expression a · C · b, i.e., Because C is a rotation matrix one can write C · b = b ′ , where b ′ · b ′ = s(s + 1). It is in principle possible that b ′ = −a, which would lead to β = −s(s + 1). However, we need to remember that the vector a is made of integers or half-integers and therefore only a subset of orthogonal matrices of measure zero gives b ′ that is also made of integers or half-integers. In particular, for a rotation matrix C whose entries are irrational numbers one always gets β > −s(s + 1).  Example of an inequality for a general s. Consider the following rotation matrix This matrix leads to a Bell-like inequality that tests the spin magnitude conserving HV model for two spin-s particles. In Table I there are some examples of quantum values and classical bounds. Note that the inequality generated by (31) can be regarded as an entanglement witness, similarly as in [18], where C plays a role of a scalar product metric. If S (A) · C · S (B) ρ < β sep , the state ρ is entangled. The threshold β sep = min ρ prod S (A) · C · S (B) ρ prod is a minimum over all product states. For C given by Eq. (31), β sep is simply equal to s 2 , which is less than β for all values of s mentioned in Table I. For all such cases, the violation of the inequality implies entanglement.
Conclusions. We have shown that if, except for the assumptions of locality and realism, one also requires that spin magnitudes are conserved in a Bell-type experiment, it leads to a stronger version of Bells theorem. The corresponding inequalities are violated by a wider class of states than in the case of standard Bell inequalities. We conclude that one can find situations in which HV models are in conflict with the conservation of angular momen-tum law and cannot be represented by states of a physical spin.