Optimal encoding and decoding of a spin direction

For a system of N spins 1/2 there are quantum states that can encode a direction in an intrinsic way. Information on this direction can later be decoded by means of a quantum measurement. We present here the optimal encoding and decoding procedure using the fidelity as a figure of merit. We compute the maximal fidelity and prove that it is directly related to the largest zeroes of the Legendre and Jacobi polynomials. We show that this maximal fidelity approaches unity quadratically in 1/N. We also discuss this result in terms of the dimension of the encoding Hilbert space.

analysis was presented in [6]. These results have been recently corroborated by numerical analysis [7].
The paper is organized as follows. In section II we introduce our notation and conventions and present a detailed calculation of the maximal fidelity for N = 2. We show that the fidelity obtained by Gisin and Popescu in [5] is optimal (a result also obtained in [8] using different methods). In section III we analyze the more general case of two spin s states. The analysis for any number N of spins is in section IV and our results and discussion are in section V. We conclude with an appendix containing technical details.

II. TWO SPINS
We start by assuming that Alice has two spins in a general eigenstate of n · S (We skip the analysis of the simplest situation in which Alice has only a spin. The reader can find it in [2,6], and our general formulae of section IV can be also particularized to this case). We can think of it as a fixed eigenstate of S z = z · S ( z is the unit vector pointing along the z direction) that Alice has rotated into the direction n = (cos φ sin θ, sin φ sin θ, cos θ). It is convenient to work in the irreducible representations of SU (2). In the present case, 1/2 ⊗ 1/2 = 1 ⊕ 0, the general form of this fixed eigenstate is where, as usual, the normalized states of the basis, |j, m , are labeled by the total spin S and the third component S z : S 2 |j, m = j(j + 1)|j, m and S z |j, m = m|j, m . In the following we stick to the general form (1) to treat all the cases jointly, but one should keep in mind that only combinations with definite S z will be relevant for our analysis. The rotated state U ( n)|A , where U ( n) is the element of the SU (2) group associated to the rotation z → n = R z, is precisely Alice's general eigenstate of n· S. Obviously U ( n) is reducible since it has the form U ( n) = U (1) ( n)⊕U (0) ( n), where U (j) denotes the SU (2) irreducible representation of spin j. Next, Alice sends the rotated state to Bob, who tries to determine n from his measurements. The most general one he can perform is a positive operator valued measurement (POVM). We specify this POVM by giving a set of positive Hermitian operators {O r }, that are a resolution of the identity For each outcome, r, Bob makes a guess, n r , for the direction. As we brought up in the introduction, the quality of the guess is quantified in terms of the fidelity which we can view as a 'score'. To Bob's guess n r , we give the 'score' f = (1 + n · n r )/2. We see that he fidelity f is unity if Bob's guess coincides with Alice's direction and it is zero when they are opposite. Thus, if n is isotropically distributed the average fidelity can be written as where ρ( n) = U ( n)|A A|U † ( n) and dn was defined in the introduction. The evaluation of F can be greatly simplified by exploiting the rotational invariance of the integral (3). If we define R r through the relation and make the change of variables We have where Ω r = U † ( n r )O r U ( n r ).
Notice that in general r Ω r = I. We can regard Ω r as fixed or reference projectors associated to the single direction z. In this sense, they are the counterpart of Alice's fixed state |A . Inserting four times the closure relation k |k k| = I, where k = +, 0, −, s, and {|k } is the basis of the representations 1 ⊕ 0, we obtain Here the indices k, i, j, l, run also over +, 0, −, s; D kj ( n) = [D (1) ⊕ D (0) ] kj ( n) = k|U ( n)|j are the SU (2) rotation matrices in the 1 ⊕ 0 representations, and Now, one can easily evaluate the integrals and obtain the fidelity where the entries marked with * are not relevant for our analysis since we only consider eigenstates of S z for the fixed states |A . These, and the corresponding rotated states U ( n)|A , are the only ones that point along a definite direction in an absolute sense, i.e., even if Alice and Bob do not share a common reference frame. ¿From its definition (10), it follows that ω jj are real nonnegative numbers but ω ij are in general complex numbers for i = j. There are other constrains on ω ij stemming from the condition r O r = I: Because of the Schwarz inequality, we also have |ω 0s | 2 ≤ ω 00 ω ss = ω 00 .
Let us discuss the implications of these equations for different values of m.
The fixed state |A for m = 1 is simply |A = |1, 1 , i.e., A + = 1 and A 0 = A − = A s = 0. In this case the fidelity is given by the element W ++ of (12), where the second condition in (13) has been used. The maximal value, which we denote by F + , is then This value occurs for The case m = −1, for which |A = |1, −1 , is completely analogous with the index substitution + ↔ −. The maximal value of the fidelity is also F − = 3/4.

Case m = 0
For m = 0 one has |A = A 0 |1, 0 + A s |0, 0 , with |A 0 | 2 + |A s | 2 = 1. The maximal fidelity is the largest eigenvalue of the 2 × 2 submatrix of (12) corresponding to the m = 0 subspace: It reaches its maximal value, F 0 , for Substituting back into (18) we obtain [5] The corresponding eigenvector is where the phase is the unconstrained parameter δ = arg ω s0 . Notice that the family of states (21) contains entangled as well as unentangled states. With the choice e iδ = ±1 one obtains the product states | ↑↓ , | ↓↑ ; precisely those considered by Gisin and Popescu [5], which led them to the conclusion that anti-parallel spins are better than parallel spins for encoding a direction. ¿From this analysis one can also obtain important information about the optimal POVM. Taking into account that one can always take the projectors O r to be one-dimensional [9], we can write Bob's reference projectors Ω r as where |Ψ r are normalized states and c r are positive numbers. The values of ω ij (see Eq. 10) endow the information about the components of |Ψ r in the spherical basis (8). To be specific, consider states with m = 0. The maximal fidelity condition (19) implies that the states |Ψ r must have also m = 0, hence |Ψ r = α r |1, 0 + β r |0, 0 . This result is, to some extent, what one expects: In order for a POVM to be optimal, the measurement must project on states as similar as possible to the signal state. Further, the Schwarz inequality (14) becomes equality if and only if α r /β r = λ for all r. If this is the case, the fidelity can reach the maximal value F 0 . Then, imposing the POVM conditions (13) it is straightforward to verify that all |Ψ r must coincide with a single state, which we denote by |B , The relative weights of the |1, 0 and |0, 0 components, √ 3 : 1, are easily understood as being the square root of the dimension of the Hilbert spaces corresponding to j = 1 and j = 0. We therefore see that optimal POVMs can be obtained by rotating the single reference state |B . The weights c r are free parameters except for the constrain Because the Hilbert space has dimension four, a POVM (optimal or not) must consist of at least four projectors. Let us show that indeed an optimal POVM with this minimal number of projectors exists. Since the number of projectors in the POVM equals the dimension of the Hilbert space, we are actually dealing with a von Neumann measurement, i.e., Hence, Ψ r |Ψ r = 1 ⇒ c r = 1 for the four values of r, which is, of course, consistent with (24). Inverting (7) and taking into account (22), we see that the four unit vectors n r have to be chosen so that By symmetry, they should correspond to the vertices of a tetrahedron inscribed in a unit sphere, i.e., n r = (cos φ r sin θ r , sin φ r sin θ r , cos θ r ) with It is easy to verify that with this choice condition (26) is fulfilled and the maximal fidelity (20) is attained. One can check that the four projectors (26) are equal to those already considered by Gisin and Popescu in [5]. Our aim here was just to present a motivated explanation for their choice of POVM. Finite optimal POVMs for N > 2 are less straightforward to obtain. However, the results of [3,4], which enables us to construct finite POVMs for code states with maximal m, |N/2, N/2 = | ↑↑ N ) . . . ↑ , can also be used here for other values of m. We will comment on this issue in our last section.
After dwelling on minimal POVMs, it is convenient to consider also the other end of the spectrum: POVMs with infinitely many outcomes or continuous POVMs [10]. They will be used in the general analysis in the sections below, where they will prove very efficient. Recall that for any finite measurement on isotropic distributions it is always possible to find a continuous POVM that gives the same fidelity [3]. Therefore, restricting ourselves to this type of measurements do not imply any loss of generality. We illustrate this point for N = 2 and m = 0 to introduce the notation that will be used in the following sections.
We have seen that the matrix elements ω ij contain all the information required for computing the fidelity, independently of any particular choice of POVM. Any measurement for which ω ij satisfy the condition (17) for m = 1 or (19) for m = 0 is surely optimal. A continuous POVM is just a particularly simple and useful realization. It amounts to taking the index r to be continuous, i.e., where the subindex B in the invariant measure refers to Bob (measuring device). Substituting (22) into (10) one obtains in the continuous version where |B is the normalized state (23) and c( n B ) is a continuous positive weight, which plays the role of c r and according to (24) must satisfy We now show that in fact c( n B ) is a constant and, hence, equal to 4. Condition (26) reads which is equivalent to Using the well known orthogonality relation of the matrix representations of SU (2) [11], one obtains which is just the total dimension (3 + 1) of the Hilbert space to which the state (23) belongs. Therefore, the projectors O( n B ) = c U ( n B )|B B|U † ( n B ) in (31) describe an optimal continuous POVM. They are obtained from the fixed state (23) in a manner analogous to the construction of the minimal POVM in (26) and (27), excepting the constant factor c required by the normalization of the matrix representations of SU (2).
To complete the analysis of N = 2, we calculate the maximal fidelity for a given (non-optimal) fixed state |A with m = 0. Without any loss of generality it can be written as where we have used the same phase convention as in (21). From (12), and the constrains (13) and (14), it is straightforward to see that the maximal value of the fidelity is To attain this value, Bob must perform an optimal POVM, characterized by (23). He may use, for instance, the minimal one (Eqs. 26-27), or the continuous one, O( n B ). This result shows that for any fixed state (35) with 1/2 < |A 0 | < √ 3/2 the fidelity is higher than that of the parallel case (i.e., m = ±1) for which F = F ± = 3/4.

III. TWO SPINS S
Imagine now that instead of two spin-1/2 Alice can use two equal arbitrary spins s 1 = s 2 = s to encode the directions. This can be seen as a generalization of the simple case studied in the preceding section. However the most important feature of this analysis, as it will be shown in section IV, is that it provides the solution of our original problem, namely, that of obtaining the maximal fidelity when Alice has N spin-1/2 particles at her disposal.
According to the Clebsch-Gordan decomposition, a normalized eigenvector of the total spin in the z-direction with eigenvalue m A can be written as where J = 2s. The state |A and its components A j should carry the label m A to denote the different eigenvalues of S z , however, we will drop it to simplify the notation. A general eigenstate of n · S has the form U The POVM projectors can be constructed from a fixed state |B of the form Note that |B is an eigenvector of S z with eigenvalue m B , though we also drop the label m B here. The absolute value of the coefficients B j and the positive weight c are determined by the completeness relation dn B O( n B ) = I, which using (33) leads to the normalization condition and a value for c given by Notice that the factor 2j + 1 in (40) is just the dimension of the Hilbert space of the irreducible representation j of SU (2), and c is the dimension of the total Hilbert space. Thus, (39) is the straight generalization of the states (23). The fidelity can be written as where The integral in (42) can be easily computed by noticing that cos θ = D (1) 00 ( n). Using again the orthogonality relations (33) we have were j 1 m 1 ; j 2 m 2 |j 3 m 3 are the Clebsch-Gordan coefficients of j 1 ⊗ j 2 → j 3 . The fidelity can be cast as where the last term is zero for m A < m B and the coefficients µ j and ν j are The phases δ j in (47) are arbitrary. They are just the generalization of the single free phase of (23). Here we have δ j = arg(B * j B j−1 ). The maximal fidelity is achieved by choosing δ j equal to the phases of the signal state |A : We see now that all terms in (45) are explicitly positive with the exception of the last one, which necessarily vanishes for optimal states |A , i.e., A j = 0 for j < m. Gathering all these results, we obtain for the fidelity: Here A t = (|A J |, |A J−1 |, |A J−2 |, . . .) is the transpose of A, and M is a real matrix of tridiagonal form with and The largest eigenvalue, x l , of M determines the maximal fidelity through the relation To find x l , we set up a recursion relation for the characteristic polynomial of M: with the starting values Q −1 (x) = 0 and Q 0 (x) = 1. Eq. 54 resembles the recursion relation of orthogonal polynomials, but at first sight the solution does not seems straightforward at all. We thus work out in detail the simplest case for which m A = m B = 0. For this particular instance (54) reads where we have used the definitions (46), (47) and (52). We can rewrite (55) as It is now apparent that the terms inside the square brackets can be absorbed into a redefinition of the characteristic polynomial through a x-independent change of normalization, namely, This leads us to the recursion relation of the Legendre polynomials: Working along the same lines, it is easy to convince oneself that the general solution of (54) is, up to a normalization factor, the Jacobi polynomial P a,b l (x) [12]: where and m is defined in (43). Note that m can be written simply as m = (a + b)/2. Note also that P 0,0 l is the Legendre polynomial P l . ¿From the result (A12) in Appendix A it turns out that the maximal value of the fidelity (53) is attained for m A = m B = 0, i.e., precisely the particular case of Legendre polynomials discussed above. Thus, from (53) we have where x a,b n stands for the largest zero of P a,b n (x). The fact that m A = m B = 0 implies maximal fidelity can be translated into physical terms by saying that Alice's states and Bob's projectors must effectively span the largest possible Hilbert space. For a fixed choice of m A , not necessarily optimal, the best m B is that for which the Hilbert spaces spanned by U ( n)|A and U ( n B )|B coincide, i.e., m A = m B = m. In this case, the maximal value of the fidelity is given by (53), with x l = x 0,2m J+1−m , i.e., F = (1 + x 0,2m J+1−m )/2 < F max . One reaches the same conclusion if m B is fixed and m A can be adjusted for best results (see discussion in appendix A after Eq. A12).

IV. GENERAL CASE: N SPINS
We now show that the solution we have obtained in the preceding section is in fact of general validity. Recall that in our original problem Alice has N spins. Let us suppose that N is even (N odd will be considered below). As usual, Alice constructs her states by rotating a fixed eigenstate of S z . In terms of the irreducible representations of SU (2), such state can be written as: The main difference with the previous example of two equal spins s is that for j < N/2 the irreducible representations U (j) appear more than once in the Clebsch-Gordan decomposition of (1/2) ⊗N . Hence, we label the different occurrences with the index α, which we can view as a new quantum number required to break the degeneracy of Alice's system of spins under global rotations. Similarly, the expression for Bob's fixed state |B is However, it is known that equivalent matrix representations are not orthogonal under the group integration, i.e., for α = β one has in general The fixed projectors in the square brackets will be judiciously chosen to eliminate the off-diagonal terms coming from the mixing of equivalent representations in the closure relation. The projector O( n B ) are explicitly of rank higher than one. However, recalling [9], we can view the right hand side of (66) as defining a sum of rank one projectors The two points of view are equivalent if the averaged fidelity is used as a figure of merit. In a suggestive compact notation we can write where |B ≡ and Next, we introduce a set of orthonormal vectors {b α j }; and define the vectors B α j as With the above definitions one can easily see that dn B O( n B ) = I and, hence, the set of projectors (66) defines a POVM.
The fidelity can be read off from (45) and it is given by where the phases have been chosen so that ν j , A α j and B α j are real. In general b α j ∈ R k , where k must be greater or equal than the highest degeneracy of the irreducible representations in the Clebsch-Gordan series of (1/2) ⊗N , since otherwise (70) could not be fulfilled. Equation 72 suggests the definition which enables us to write Using Schwarz inequality we have The right hand side is exactly the fidelity (45) of the preceding section with the substitution This equation shows that the existence of several equivalent representations in the Clebsch-Gordan decomposition of Alice's Hilbert space cannot be used to increase the value of the fidelity already obtained in section III. The equality holds when all vectors A j are parallel, in which case we recover (45). The square root on the right hand side of (76) plays the role of an effective component of |A on the Hilbert space of a single irreducible representation j. The specific ways |A projects on each one of the equivalent representations are of no relevance, providedÃ j do not change. As far as the fidelity is concerned, all them are equivalent to taking a state |Ã that belongs to N/2 ⊕ (N/2 − 1) ⊕ (N/2 − 2) ⊕ · · · (no duplications), with the corresponding components given byÃ j .
As we have just seen, the maximal fidelity can be achieved from a code state containing only one of each irreducible representations. This type of states are formally the same as those considered in the simplified example of two equal spins s 1 = s 2 = s studied in section III, for which s ⊗ s = J ⊕ (J − 1) ⊕ · · · ⊕ 0, with J = 2s = N/2. The problem of an even number of spins is thus completely solved: according to (53) the maximal fidelity is given by where x 0,0 N/2+1 is the largest zero of the (Legendre) polynomial P N/2+1 (x) = P 0,0 N/2+1 (x). For an odd number of spins we can proceed as in section III but considering now states with two different spins: s 1 = s, s 2 = s + 1/2. The corresponding Clebsch-Gordan decomposition is also non-degenerate: s ⊗ (s + 1/2) = J ⊕ (J − 1) ⊕ · · · ⊕ 1/2, with J = 2s + 1/2 = N/2. The results from (37) to (54) are still valid (for the value of J we have just specified). The optimal values of m A and m B are again the minimal ones: m A = m B = 1/2. The maximal fidelity is where x 0,1 N/2+1/2 stands for the largest zero of the Jacobi polynomial P 0,1 N/2+1/2 (x). This completes the solution of the general problem.
It is physically obvious that the larger the number of spins Alice can use the better she should be able to encode n. One thus expects that the maximal fidelity should increase monotonously with N . It is interesting to obtain this result from the properties of the zeroes of the Jacobi polynomials. For an even number of spins, N = 2n − 2, the corresponding zero is x 0,0 n , whereas for N + 1 it is x 0,1 n , and x 0,1 n−1 for N − 1. Proving that F N −1 < F N < F N +1 amounts to showing that  but this is just a particular case of (A9) for a = 0 and b = 1.
Not only the optimal strategy Alice can devise with N spins leads to a fidelity larger than F N −1 . She can also use non-optimal ones and still exceed F N −1 . E.g., for N = 4, the choice m A = m B = 1, which is non-optimal, gives a fidelity F = (10 + √ 10)/15 > (6 + √ 6)/10 = F 3 . This is also a trivial consequence of (A9) as in this case one has x 0,2 2 > x 0,1 2 . In physical terms, this is telling us that the dimension of the Hilbert space spanned by U ( n)|A and U ( n B )|B when N = 4 and m A = m B = 1 (including equivalent spin representations only once) is still larger than the maximal available dimension for N = 3.

V. DISCUSSION AND OUTLOOK
In this paper we have addressed the problem of optimizing strategies for encoding and decoding directions on the quantum states of a system of N spins. We have restricted ourselves to states that point along a definite direction in an intrinsic way, namely, to eigenstates of n · S. This case is of great interest since no prior knowledge of any sender's (Alice's) reference state or frame by the recipient (Bob) is required at all for a viable transfer of the information. We have optimized both Alice's states and Bob's measurements. Our results are summarized in (77) and (78), where we give the maximal averaged fidelities F N . Interestingly enough, these results can be written in terms of the largest zeroes of the Jacobi polynomial, which are known to play an important role in angular momentum theory and are intimately related to the matrix representations of SU (2). The states that lead to the maximal fidelities are among those that have the smallest (non-negative) value of n · S, namely, m = 0 for N even and m = 1/2 for N odd, but still span the largest Hilbert space under rotations.
We display the values of the maximal fidelity for N up to seven in table I for illustrational purposes. It shows, e.g., that the optimal encoding with three spins (m = 1/2) gives F 3 = (6 + √ 6)/10 ∼ 0.845, which is already larger than the corresponding maximal value for four parallel spins (m = 2): F = 5/6 ∼ .833 [2]. This illustrates a general feature: the optimal strategies discussed here lead to fidelities that increase with N much faster than that of sending parallel spins. In fact, Eq. A13 shows that F N approaches unity quadratically in the number of spins, namely where ξ ∼ 2.4 is the first zero of the Bessel function J 0 (x). In contrast, if parallel spins are used the maximal fidelity approaches unity only linearly, F ∼ 1 − 1/N . This can be understood in terms of the dimension d of the Hilbert space used effectively in each case, which is a direct sum of the Hilbert spaces of the irreducible representations of SU (2) involved. Here effectively means 'nonredundantly', thus equivalent representations count only once. Encoding with N parallel spins uses only the Hilbert space of the representation N/2, whose dimension is d = N + 1, whereas our optimal strategy uses a much larger Hilbert space, with d = (N/2 + 1) 2 for N even and d = (N/2 + 1) 2 − 1/4 for N odd; in both cases d ∼ N 2 . We are led to the conclusion that the fidelity as a function of d tends to unity as where a is of order one and depends on the particular strategy. Improvements on the approach discussed in this paper can only come from encoding and decoding procedures that make extensive use of the available Hilbert space, namely, strategies that use the redundant equivalent representations. In [6] we presented a strategy for which the maximal fidelity approaches unity exponentially in the number of spins, i.e., F ∼ 1 − 2 −N . We argue there that this encoding is likely to lead to the maximal fidelity one can possibly achieve with N spins, since it makes effective use of the whole Hilbert space of the system, for which d = 2 N (thus, Eq. 81 also holds in this case). The corresponding encoding process, however, involves complicated unitary operations and, moreover, it seems to require that Alice and Bob share a common reference frame [13].
We have obtained our general results using continuous POVMs, but finite ones can also be designed. For N parallel spins (m A = m B = N/2), a general recipe for finite optimal POVMs exists [3], and minimal versions for up to N = 7 can be found in [4]. The unit vectors n r associated to the outcomes of these POVMs are the vertices of certain polyhedra inscribed in the unit sphere. For N ≤ 7 we have explicitly verified that these very same polyhedra can be use to design finite optimal POVMs for any value of m A = m B ≤ N/2. Moreover, the minimal POVMs of [4] remain minimal for the states considered here. We have discussed this issue in detail for N = 2 in section II. For N = 3 the polyhedron corresponding to the minimal POVM is the octahedron [4]. One can easily verify that O r = U ( n r )|B B|U † ( n r ), where |B is given in (39) with m B = 1/2, 3/2, fulfill the completeness condition (2). We hence believe that the discretization of a continuous POVM is a geometrical problem, i.e., it seems to be independent of the states |B .
The optimal states, |A , can be easily computed from the matrix M in (50), as they are the eigenvectors corresponding to the maximal eigenvalue. Recall that for N = 2 one obtains the one-parameter family of states (21) which includes the product states | ↑↓ , | ↓↑ . For N > 2, product states of the type | ↑↓↑↑↓ · · · do not seem to be optimal. Consider, e.g., N = 4. The optimal eigenvector of M is which is clearly not a product state of the individual spins for any choice of the phases [14]. One could argue that this solution is not entirely general because the Clebsch-Gordan series of (1/2) ⊗4 contains the representation 1 three times and 0 twice, whereas in (82) they appear only once. However, any optimal state has the same 'effective' components A j (see eqs. 75-76), which can be read off from (82): Note now that any product state with m = 0 (two spins up and two spins down), e.g., | ↑↑↓↓ , | ↑↓↓↑ , has an 'effective' Clebsch-Gordan decomposition given byÃ 2 =Ã 1 =Ã 0 = 1/ √ 3, which do not coincide with (83). Therefore, these product states cannot be optimal. Nevertheless, they lead to a maximal fidelity F = (15 + 5 √ 2 + 2 √ 5)/30 ≈ 0.885, which is remarkably close to F 4 ≈ 0.887. This is likely to be the case for arbitrary N . These issues are currently under investigation.
It also follows the differentiation formula The normalization is chosen so that the coefficient A n of the highest power of P a,b n (x) = A n x n + B n x n−1 + · · · is A n = Γ(2n + a + b + 1) 2 n n!Γ(n + a + b + 1) .
Let us recall some basic facts about the zeros of orthogonal polynomials: i) any orthogonal polynomial, P n , of order n has n real simple zeros. For Jacobi polynomials these zeros lie in the interval (−1, 1); ii) the zeros of P n and P n+1 are interlaced; iii) for x greater than the largest zero, the polynomial is a monotonously increasing function (if the polynomial is normalized as in Eq. A5, where A n > 0). In particular, P n (x) must be positive in this region . Now we can prove the results needed in the text. As in there, we denote by x a,b n the largest zero of the polynomial P a,b n (x). Let us start by showing that x a+1,b+1 From property iii above it follows that the left hand side of (A4) is manifestly positive for x > x a,b n . Hence, so it is the right hand side. We conclude that x a+1,b+1 n−1 cannot belong to this region and (A8) follows Next, we prove the inequality We evaluate (A6) at x = x a,b n and use properties ii (⇒ x a,b n−1 < x a,b n ) and iii, which imply that P a,b n−1 (x a,b n ) > 0, to obtain that P a,b−1 n (x a,b n ) > 0. We repeat the process for x = x a,b n−1 and conclude that P a,b−1 n (x a,b n−1 ) < 0. Hence P a,b−1 n has a zero in the interval (x a,b n−1 , x a,b n ). This is necessarily the largest zero x a,b−1 n since, according to (A6) and properties ii and iii, P a,b−1 n (x) > 0 for x > x a,b n . Thus (A9) follows The inequality x a,b+1 n < x a,b−1 n+1 (A10) can be proven as follows. Evaluate (A7) at x = x a,b+1 n so that the left hand side of this equation is zero. The second inequality in (A9) and property iii imply that P a,b n (x a,b+1 n ) > 0. Hence the first term on the right hand side of (A7) must be negative, i.e., P a,b−1 n+1 (x a,b+1 n ) < 0, and (A10) follows immediately, since otherwise property iii would not hold for P a,b−1 n+1 For two given integers l, m consider now the following set of zeros We want to prove that max C l m = x 0,2m l−m .
According to (A8), decreasing m ′′ by one leads us to a larger zero. The maximum is then in the subset {x 0,2m ′ l−m ′ : m ≤ m ′ ≤ l}. The inequality (A10) now implies (A12) Finally, we give the large n (asymptotic) behavior of x a,b n [12]: where ξ a is the first zero of the Bessel function J a (x). For a = 0, which is relevant for our discussion in section V, we also give the subleading term: where ξ 0 = ξ = 2.405. (A15)