The structure of maps on the space of all quantum pure states that preserve a fixed quantum angle
TTHE STRUCTURE OF MAPS ON THE SPACE OF ALL QUANTUM PURESTATES THAT PRESERVE A FIXED QUANTUM ANGLE
GY ¨ORGY P ´AL GEH´ER AND MICHIYA MORI
Abstract.
Let H be a Hilbert space and P ( H ) be the projective space of all quantum purestates. Wigner’s theorem states that every bijection φ : P ( H ) → P ( H ) that preserves the quantumangle between pure states is automatically induced by either a unitary or an antiunitary operator U : H → H . Uhlhorn’s theorem generalises this result for bijective maps φ that are only assumedto preserve the quantum angle π (orthogonality) in both directions. Recently, two papers, writtenby Li–Plevnik–ˇSemrl and Geh´er, solved the corresponding structural problem for bijections thatpreserve only one fixed quantum angle α in both directions, provided that 0 < α ≤ π holds. In thispaper we solve the remaining structural problem for quantum angles α that satisfy π < α < π ,hence complete a programme started by Uhlhorn. In particular, it turns out that these maps arealways induced by unitary or antiunitary operators, however, our assumption is much weaker thanWigner’s. Introduction
Let H be a complex Hilbert space. For any vector v ∈ H with length one, (cid:107) v (cid:107) = 1, let [ v ]denote the line (one-dimensional subspace) it generates: C · v . From now on whenever we write[ v ] with v ∈ H , it is implicitly assumed that (cid:107) v (cid:107) = 1 holds. Also, given a finite number of vectors v , v , . . . , v n ∈ H with (cid:107) v (cid:107) = (cid:107) v (cid:107) = · · · = (cid:107) v n (cid:107) = 1, the symbol [ v , v , . . . , v n ] stands forthe subspace generated by them. The projective space P ( H ) is the set of all lines in H , that is, P ( H ) = { [ v ] : v ∈ H, (cid:107) v (cid:107) = 1 } . In the mathematical foundations of quantum mechanics a line [ v ]corresponds to a quantum pure state, and P ( H ) to the set of all quantum pure states in a quantumsystem. The so-called quantum angle or Fubini–Study distance between two lines [ u ] , [ v ] ∈ P ( H )is defined by the following formula: (cid:93) ([ u ] , [ v ]) := arccos |(cid:104) u, v (cid:105)| ∈ (cid:104) , π (cid:105) . It is well-known that this is a metric on P ( H ). Moreover, the important quantity called transitionprobability between [ u ] and [ v ] can be expressed as cos (cid:93) ([ u ] , [ v ]), for more details on this see forinstance the introduction of [2].Let us introduce the notation T := { z ∈ C : | z | = 1 } for the complex unit circle. In 1931 Wignerstated the following theorem. Theorem 1.1 (Wigner, [7]) . Let H be a complex Hilbert space with dim H ≥ . Assume that thebijective map φ : P ( H ) → P ( H ) preserves the quantum angle between lines, that is, (cid:93) ( φ ([ u ]) , φ ([ v ])) = (cid:93) ([ u ] , [ v ]) ([ u ] , [ v ] ∈ P ( H )) . (1.1) Then φ is induced by either a unitary or an antiunitary operator U : H → H , namely, we have φ ([ v ]) = [ U v ] ([ v ] ∈ P ( H )) . (1.2) Moreover, two unitary or antiunitary operators U and U induce the same map on P ( H ) if andonly if U = λU holds with some λ ∈ T . Mathematics Subject Classification.
Primary: 47B49, 51A05. Secondary: 47N50.
Key words and phrases.
Projective space, quantum pure state, quantum angle preserving map, Fubini–Studymetric, Wigner symmetry, transition probability preserving map.This research was supported by the London Mathematical Society Grant, Research in Pairs (Scheme 4), ReferenceNo.: 41864.Geh´er was supported by the Leverhulme Trust Early Career Fellowship (ECF-2018-125), and also by the Hun-garian National Research, Development and Innovation Office (Grant no. K115383 and K134944).Mori was supported by Leading Graduate Course for Frontiers of Mathematical Sciences and Physics (FMSP)and JSPS Research Fellowship for Young Scientists (KAKENHI Grant Number 19J14689), MEXT, Japan. a r X i v : . [ m a t h - ph ] F e b GY ¨ORGY P ´AL GEH´ER AND MICHIYA MORI
We note that the reverse direction is trivially true, namely, if φ has the form (1.2), then φ isclearly bijective and (1.1) holds. The real achievement here is that assuming only (1.1) and bijec-tivity already implies the remarkably regular structure (1.2). We call a map a Wigner symmetry ifit possesses the form (1.2). The above theorem became a cornerstone of the mathematical founda-tions of quantum mechanics. One reason being that it plays a crucial role in obtaining the generaltime-dependent Schr¨odinger equation through purely mathematical means. For a nice expositionregarding this we suggest Simon’s paper [5].We note that Wigner himself did not give a mathematically rigorous proof of his statement,indeed, the proof presented in [7] contains gaps. Interestingly enough, it took thirty years for thefirst mathematically rigorous proofs to appear, see [1, 4, 6]. In particular, in [6] Uhlhorn proved amore general version of the above theorem for Hilbert spaces of dimension at least three. Namely,he only assumed the preservation of the quantum logical structure, while Theorem 1.1 assumesthat its complete probabilistic structure is preserved. Still, Uhlhorn’s conclusion is the same asWigner’s, which is a quite remarkable phenomenon.
Theorem 1.2 (Uhlhorn, [6]) . Let H be a complex Hilbert space with dim H ≥ and φ : P ( H ) → P ( H ) be a bijective map preserving orthogonality in both directions, that is, (cid:93) ( φ ([ u ]) , φ ([ v ])) = π ⇐⇒ (cid:93) ([ u ] , [ v ]) = π u ] , [ v ] ∈ P ( H )) . Then φ is a Wigner symmetry. Namely, there exists either a unitary or an antiunitary operator U : H → H such that φ ([ v ]) = [ U v ] ([ v ] ∈ P ( H )) . We note that Uhlhorn’s theorem obviously fails to be true in a two-dimensional Hilbert space,since in that case for every line there exists only one line orthogonal to it. The above two theoremshave been generalised in many ways, more on this can be found in the introduction of [2].In this paper we are interested in the following problem which proposes to generalise Wigner’stheorem along the direction of Uhlhorn.
Problem 1.3.
Fix a quantum angle < α < π . Can we characterise all bijective mappings φ : P ( H ) → P ( H ) that preserve the quantum angle α , that is, (cid:93) ( φ ([ u ]) , φ ([ v ])) = α ⇐⇒ (cid:93) ([ u ] , [ v ]) = α ([ u ] , [ v ] ∈ P ( H ))?We emphasise that, like in Uhlhorn’s theorem, nothing is assumed a priori about other angles,hence (cid:93) ( φ ([ u ]) , φ ([ v ])) (cid:54) = (cid:93) ([ u ] , [ v ]) might happen if (cid:93) ([ u ] , [ v ]) (cid:54) = α . Recently, the papers [2, 3]solved this problem for real Hilbert spaces. However, for complex Hilbert spaces it was onlypartially solved, we state the two relevant theorems below. The first one is the complete solutionfor two-dimensional Hilbert spaces.
Theorem 1.4 (Geh´er, [2]) . Let H be a complex Hilbert space with dim H = 2 and fix a number < α < π . Assume that φ : P ( H ) → P ( H ) is a bijective map preserving the quantum angle α inboth directions, that is, (cid:93) ([ u ] , [ v ]) = α ⇐⇒ (cid:93) ( φ ([ u ]) , φ ([ v ])) = α ([ u ] , [ v ] ∈ P ( H )) . Then (i) either φ is a Wigner symmetry, (ii) or α = π , and there exists a Wigner symmetry ψ such that φ ([ v ]) ∈ (cid:110) ψ ([ v ]) , ψ ([ v ]) ⊥ (cid:111) ([ v ] ∈ P ( H )) , (1.3) where ψ ([ v ]) ⊥ denotes the unique line which is orthogonal to ψ ([ v ]) . Moreover, every bi-jective map φ that satisfies (1.3) preserves the angle π . Theorem 1.4 can be proved using the famous Bloch representation and a characterisation ofbijective maps on the unit sphere of a real Hilbert space that preserve a fixed spherical angle (see[2, Theorem 2.1]). The next theorem is the solution for quantum angles at most π . APS THAT PRESERVE A FIXED QUANTUM ANGLE 3
Theorem 1.5 (Geh´er, [2]) . Let H be a complex Hilbert space with dim H ≥ and fix a number < α ≤ π . Assume that φ : P ( H ) → P ( H ) is a bijective map which satisfies (cid:93) ([ u ] , [ v ]) = α ⇐⇒ (cid:93) ( φ ([ u ]) , φ ([ v ])) = α ([ u ] , [ v ] ∈ P ( H )) . Then φ is a Wigner symmetry. In the present paper our goal is to solve Problem 1.3 for the remaining case when dim H ≥ π < α < π . Before we state our main theorem, let us briefly explain the strategy used in [2]to prove Theorem 1.5. For a subset S ⊂ P ( H ), we define its α -set by S (cid:104) α (cid:105) := { [ v ] ∈ P ( H ) : (cid:93) ([ v ] , [ u ]) = α for all [ u ] ∈ S } , and its double- α -set by S (cid:104)(cid:104) α (cid:105)(cid:105) := (cid:16) S (cid:104) α (cid:105) (cid:17) (cid:104) α (cid:105) . The core idea of [2] is to examine the α -sets of pairs of lines. More precisely, it turns out thatif 0 < α < π , then the set { [ v ] , [ v ] } (cid:104) α (cid:105) contains exactly one pair of elements [ w ] , [ w ] with (cid:93) ([ w ] , [ w ]) = α if and only if (cid:93) ([ v ] , [ v ]) = β , where β is explicitly given in terms of α . Hencethe angle β is also preserved by φ . Using this observation it is then possible to construct a sequenceof quantum angles { β n } ∞ n =1 ⊂ (cid:0) , π (cid:1) which are all preserved by φ , moreover, β n (cid:38) n → ∞ .Since small angles are preserved, one can prove that all angles must be preserved. For the case α = π a somewhat modified idea can be applied, which we do not detail here.As was pointed out in [2], the above idea fails to work for quantum angles α > π . The mainresult of this paper is to show that nonetheless the conclusion of Theorem 1.5 holds for all quantumangles. Theorem 1.6.
Let H be a complex Hilbert space with dim H ≥ and fix a number π < α < π .Assume that φ : P ( H ) → P ( H ) is a bijective map which preserves the quantum angle α in bothdirections, namely, it satisfies (cid:93) ([ u ] , [ v ]) = α ⇐⇒ (cid:93) ( φ ([ u ]) , φ ([ v ])) = α ([ u ] , [ v ] ∈ P ( H )) . Then φ is a Wigner symmetry, that is, there exists a unitary or an antiunitary operator U : H → H such that φ ([ v ]) = [ U v ] ([ v ] ∈ P ( H )) . We say that three lines [ v ] , [ v ] , [ v ] are collinear if dim[ v , v , v ] ≤
2. For any (closed) subspace M ⊂ H we may identify the projective space P ( M ) with the subset { [ v ] ∈ P ( H ) : v ∈ M, (cid:107) v (cid:107) =1 } ⊂ P ( H ). If dim M = 2, then we call P ( M ) ( ⊂ P ( H )) a projective line. The following definitionplays a central role in our considerations. Definition 1.7 (Highly- α -symmetric set) . A subset T ⊂ P ( H ) is called highly- α -symmetric if itsatisfies the following three conditions: (i) T = ∞ , (ii) T (cid:104) α (cid:105) = ∞ , (iii) for any subset S ⊂ T with S = 3 , S (cid:104)(cid:104) α (cid:105)(cid:105) = T . We now briefly explain our strategy to prove the above theorem. The aim of the next section isto explore the structure of the α -sets of three collinear lines, and to prove some auxiliary results.Then in sections 3 and 4 we investigate how highly- α -symmetric sets look like when dim H ≥ H = 3, respectively. It turns out that if H has dimension at least four, then a set T is highly- α -symmetric if and only if it is a subset of a projective line with an additional specialstructure, described in Definition 2.6. In case when the dimension of the Hilbert space is three,the aforementioned implication holds only in one direction. In contrast with [2] where α -sets ofpairs of lines were examined, here the core of our method is to explore the shape of double- α -setsof general triples of lines . Using these insights we then prove in Section 5 that all maps φ whichsatisfy our conditions necessarily map projective lines onto projective lines. Finally, an applicationof Theorem 1.4 will complete the proof. GY ¨ORGY P ´AL GEH´ER AND MICHIYA MORI Some preliminary results
From now on H denotes a complex Hilbert space with dim H ≥
3, and α is a fixed angle with π < α < π . We begin with a lemma about some basic properties of α -sets. Lemma 2.1.
We have the following relations: (i) If S ⊂ P ( H ) , then S ⊂ S (cid:104)(cid:104) α (cid:105)(cid:105) . (ii) If S ⊂ S ⊂ P ( H ) , then S (cid:104) α (cid:105) ⊃ S (cid:104) α (cid:105) and S (cid:104)(cid:104) α (cid:105)(cid:105) ⊂ S (cid:104)(cid:104) α (cid:105)(cid:105) . (iii) If S ⊂ P ( H ) , then ( S (cid:104) α (cid:105) ) (cid:104)(cid:104) α (cid:105)(cid:105) = S (cid:104) α (cid:105) . (iv) Every highly- α -symmetric set T satisfies S (cid:104)(cid:104) α (cid:105)(cid:105) = T, S (cid:104) α (cid:105) = T (cid:104) α (cid:105) ( S ⊂ T, S ≥ . Proof.
Points (i)–(ii) are trivial by definition. Point (iii) is an easy application of (i)–(ii), and part(iv) is straightforward from (i)–(iii). (cid:3)
As usual, we say two lines [ u ] , [ v ] ∈ P ( H ) are orthogonal if (cid:93) ([ u ] , [ v ]) = π . We introduce thenotation ⊥ for the orthogonality of vectors and subsets in H , and also for the orthogonality of linesin P ( H ). We continue with two lemmas about the general form of a pair of lines and its α -set. Lemma 2.2.
Let [ v ] , [ v ] ∈ P ( H ) be two different lines. Then there exist an orthonormal system { e , e } ⊂ H and real numbers c ≥ d > , c + d = 1 such that [ v ] = [ ce + ide ] , [ v ] = [ ce − ide ] . Proof.
An application of the famous Bloch representation gives a simple proof. However, in case thereader is not that familiar with it, a more direct proof can be given as follows. Since [ v j ] = [ λv j ]for all λ ∈ T and j = 1 ,
2, without loss of generality we may assume that (cid:104) v , v (cid:105) ≥
0. Hence v + v ⊥ v − v and 0 < (cid:107) v − v (cid:107) ≤ (cid:107) v + v (cid:107) hold. Since (cid:107) v + v (cid:107) + (cid:107) v − v (cid:107) = 4, there existtwo numbers c ≥ d > c + d = 1 and an orthonormal system { e , e } such that v + v = 2 ce and v − v = 2 ide . From here a calculation gives the desired form. (cid:3) We introduce the notation (cid:116) for the disjoint union. We also set a := cos α which we shall usethroughout the paper. Lemma 2.3.
Let { e , e } ⊂ H be an orthonormal system and c ≥ d > with c + d = 1 . Definethe function ρ : [ − θ , θ ] → [0 , , ρ ( θ ) = (cid:114) − (cid:16) ac (cid:17) cos θ − (cid:16) ad (cid:17) sin θ, where • if a ≤ d , then θ = π , • if a > d , then θ is the unique number with < θ < π and (cid:0) ac (cid:1) cos θ + (cid:0) ad (cid:1) sin θ = 1 .Then we have { [ ce + ide ] , [ ce − ide ] } (cid:104) α (cid:105) = (cid:71) (cid:110) A θ : − θ ≤ θ ≤ θ , θ (cid:54) = − π (cid:111) , (2.1) where A θ := (cid:110)(cid:104) ac cos θ · e + ad sin θ · e + h (cid:105) : h ⊥ { e , e } , (cid:107) h (cid:107) = ρ ( θ ) (cid:111) . (2.2) Proof.
Notice that by our assumptions we always have c > a . Since 0 < ac ≤ ad , the function θ (cid:55)→ (cid:0) ac (cid:1) cos θ + (cid:0) ad (cid:1) sin θ is positive-valued, monotone non-increasing on [ − π , , π ]. As ac <
1, we have a real number 0 < θ ≤ π with the desired property.Consider an arbitrary line [ v ] ∈ P ( H ). We may take numbers c ≥ c ∈ C and a vector h ⊥ { e , e } such that c + | c | + (cid:107) h (cid:107) = 1 and [ v ] = [ c e + c e + h ]. Then we have [ v ] ∈{ [ ce + ide ] , [ ce − ide ] } (cid:104) α (cid:105) if and only if | c c + ic d | = | c c − ic d | = a. This is equivalent to • either c > c ∈ R and ( c c ) + ( c d ) = a , APS THAT PRESERVE A FIXED QUANTUM ANGLE 5 • or c = 0 and | c | d = a , in which case we may assume without loss of generality that c = ad .Therefore c c = a cos θ and c d = a sin θ for some − π ≤ θ ≤ π , which proves the ⊆ part of (2.1).The ⊇ part of (2.1) and the disjointness are obvious. (cid:3) Note that in case when θ = π , then the set A − π is well defined by (2.2), however, we have A − π = A π . Throughout the paper whenever we use the symbols c and d , it is always assumed that c ≥ d > c + d = 1. Therefore, like in the above proof, the inequality c > a is automaticallysatisfied.Straightforward calculations give the following properties of ρ , which are also illustrated inFigure 1 for the reader’s convenience: • ρ is an even continuous function on [ − θ , θ ], differentiable on ( − θ , θ ), and ρ (cid:48) (0) = 0, • if d < (cid:113) , then ρ is strictly increasing on [ − θ , , θ ], • if d = (cid:113) , then ρ is the constant √ − a function, • ρ ( θ ) = 0 if and only if a ≥ d . π - π (a) When a < d < (cid:113) . Then θ = π and ρ ( θ ) > . π - π (b) When a = d < (cid:113) . Then θ = π , and ρ ( θ ) = 0 . π - π θ - θ (c) When a > d . Then d < (cid:113) , 0 < θ < π and ρ ( θ ) = 0 . π - π (d) When d = (cid:113) . Then θ = π , and ρ is apositive constant function. Figure 1.
Illustration of the function ρ .The following two lemmas give the general form of a collinear triple of lines and its α -set. Lemma 2.4.
Let [ v ] , [ v ] , [ v ] ∈ P ( H ) be three collinear lines that are pairwise different. Thenthere exist an orthonormal system { e , e } ⊂ H , three numbers λ , λ , λ ∈ T , and two real numbers c ≥ d > , c + d = 1 such that [ v j ] = [ ce + λ j de ] ( j = 1 , , . Proof.
An application of the Bloch representation gives a geometric and simple proof. We giveanother more direct proof here. By Lemma 2.2, we can write [ v ] = [ c f + i d f ] and [ v ] = [ c f − i d f ] GY ¨ORGY P ´AL GEH´ER AND MICHIYA MORI where { f , f } is an orthonormal system, c ≥ d > c + d = 1. A straightforward calculationgives that |(cid:104) v , cos tf + sin tf (cid:105)| = |(cid:104) v , cos tf + sin tf (cid:105)| (cid:16) ≤ t ≤ π (cid:17) . We may take numbers c ≥ c ∈ C such that c + | c | = 1 and [ v ] = [ c f + c f ]. Onthe one hand, suppose that c ≥ c . Then | c | ≤ d , |(cid:104) v , f (cid:105)| = |(cid:104) v , f (cid:105)| = c ≤ c = |(cid:104) v , f (cid:105)| and |(cid:104) v , f (cid:105)| = |(cid:104) v , f (cid:105)| = d ≥ | c | = |(cid:104) v , f (cid:105)| . Therefore there exists a 0 ≤ t ≤ π such that with e := cos tf + sin tf we have |(cid:104) v , e (cid:105)| = |(cid:104) v , e (cid:105)| = |(cid:104) v , e (cid:105)| . (2.3)On the other hand, if c < c , then we prove the existence of a line [ e ] with (2.3) in a very similarway.Now, let [ e ] be the unique line which is orthogonal to [ e ] and is contained in the subspace[ v , v ]. Parseval’s formula implies |(cid:104) v , e (cid:105)| = |(cid:104) v , e (cid:105)| = |(cid:104) v , e (cid:105)| . By interchanging the role of e and e if necessary, we may assume c := |(cid:104) v , e (cid:105)| ≥ |(cid:104) v , e (cid:105)| =: d ,which completes the proof. (cid:3) Lemma 2.5.
Let c ≥ d > such that c + d = 1 , λ , λ , λ ∈ T pairwise different, and { e , e } an orthonormal system of H . Set S := { [ ce + λ j de ] : j = 1 , , } ⊂ P ( H ) . (i) If a > d , then S (cid:104) α (cid:105) = (cid:40)(cid:104) ac e + h (cid:105) : h ∈ H, (cid:107) h (cid:107) = (cid:114) − a c , h ⊥ { e , e } (cid:41) . (ii) If a ≤ d , then S (cid:104) α (cid:105) = (cid:40)(cid:104) ac e + h (cid:105) : h ∈ H, (cid:107) h (cid:107) = (cid:114) − a c , h ⊥ { e , e } (cid:41)(cid:71) (cid:40)(cid:104) ad e + h (cid:105) : h ∈ H, (cid:107) h (cid:107) = (cid:114) − a d , h ⊥ { e , e } (cid:41) . Proof.
Note that c > a . Consider an arbitrary line [ v ] ∈ P ( H ). We may take numbers c ≥ c ∈ C and a vector h ⊥ { e , e } such that c + | c | + (cid:107) h (cid:107) = 1 and [ v ] = [ c e + c e + h ]. Thenwe have [ v ] ∈ S (cid:104) α (cid:105) if and only if (cid:12)(cid:12) c c + c λ j d (cid:12)(cid:12) = a ( j = 1 , , . Since the numbers λ j are pairwise different, a simple geometric observation implies that | c c + c λd | = a ( λ ∈ T ) . Thus [ v ] ∈ S (cid:104) α (cid:105) if and only if • either c = 0, c = ac , • or c = 0, | c | = ad .Note that without loss of generality c > (cid:3) We finish this section with an important definition.
Definition 2.6 (Circle) . For any orthonormal system { e , e } ⊂ H and numbers c , d > , c + d =1 , the set of the form { [ c e + λ d e ] : λ ∈ T } is called a circle. APS THAT PRESERVE A FIXED QUANTUM ANGLE 7
Set M := [ e , e ] with the above vectors and consider the Bloch representation of P ( M ) (seefor instance [2]). Remark that a straightforward calculation shows that the image of the circle { [ c e + λ d e ] : λ ∈ T } is an actual circle on the surface S , hence the above choice of the name.Moreover, it is a great (or geodesic) circle if and only if c = d = √ .In the forthcoming two sections we shall explore how the double- α -set of S looks like, and willalso examine highly- α -symmetric sets in detail.3. The structure of highly- α -symmetric sets in the at least four-dimensional case Our goal in this section is to show that highly- α -symmetric sets are exactly circles in P ( H ) ifdim H ≥
4. First, we calculate the double- α -set of S from Lemma 2.5. Lemma 3.1.
Using the notation and assumptions of Lemma 2.5, suppose that dim H ≥ . Thenwe have S (cid:104)(cid:104) α (cid:105)(cid:105) = { [ ce + λde ] : λ ∈ T } . Proof.
Recall that c > a . Define C := (cid:40)(cid:104) ac e + h (cid:105) : h ∈ H, (cid:107) h (cid:107) = (cid:114) − a c , h ⊥ { e , e } (cid:41) . As C ⊆ S (cid:104) α (cid:105) , we have C (cid:104) α (cid:105) ⊇ S (cid:104)(cid:104) α (cid:105)(cid:105) . Consider a line [ v ] = [ c e + c e + k ] with c ≥ c ∈ C , k ∈ H , k ⊥ { e , e } , and c + | c | + (cid:107) k (cid:107) = 1. We have [ v ] ∈ C (cid:104) α (cid:105) if and only if (cid:12)(cid:12)(cid:12) c ac + (cid:104) k, h (cid:105) (cid:12)(cid:12)(cid:12) = a (cid:32) h ∈ H, (cid:107) h (cid:107) = (cid:114) − a c , h ⊥ { e , e } (cid:33) . Notice that the inner product (cid:104) k, h (cid:105) above runs through a closed disk of radius (cid:107) k (cid:107) · (cid:113) − a c onthe complex plane. As c > a , we obtain k = 0 and c = c , hence C (cid:104) α (cid:105) = { [ ce + λde ] : λ ∈ T } . In case of (i) of Lemma 2.5, this completes the proof. On the other hand, in case of (ii) of Lemma2.5, we easily see the reverse inclusion S (cid:104)(cid:104) α (cid:105)(cid:105) ⊇ C (cid:104) α (cid:105) , hence the proof is done. (cid:3) Observe that Lemmas 2.5 and 3.1 imply the following.
Corollary 3.2. If dim H ≥ , then every circle in P ( H ) is highly- α -symmetric. For the remaining part of this section our aim is to prove the reverse.
Lemma 3.3.
Assume that dim H ≥ . Then every highly- α -symmetric set T satisfies one of thefollowing points: (i) either T is contained in a projective line, (ii) or there exists a subspace M with dim M = 3 such that for all [ v ] , [ v ] , [ v ] ∈ T pairwisedifferent elements we have [ v , v , v ] = M .Proof. Suppose that there exist [ u ] , [ u ] , [ u ] ∈ T collinear and pairwise different. Then, byLemmas 2.4 and 3.1, the set T = { [ u ] , [ u ] , [ u ] } (cid:104)(cid:104) α (cid:105)(cid:105) is a circle, hence (i) follows.From now on we assume otherwise. Consider three arbitrary pairwise different lines[ v ] , [ v ] , [ v ] ∈ T . Set M := [ v , v , v ] which is a three-dimensional subspace. Our goal is toprove T ⊂ P ( M ), which will complete the proof. Note that { [ v ] , [ v ] , [ v ] } (cid:104) α (cid:105) = (cid:8) [ u + w ] : u ∈ M, w ⊥ M, (cid:107) u (cid:107) + (cid:107) w (cid:107) = 1 , |(cid:104) u, v (cid:105)| = |(cid:104) u, v (cid:105)| = |(cid:104) u, v (cid:105)| = a (cid:9) . As this set is equal to T (cid:104) α (cid:105) , it is not empty. Let [ x + y ] ∈ P ( H ) be an arbitrary line where x ∈ M , y ⊥ M and (cid:107) x (cid:107) + (cid:107) y (cid:107) = 1. Clearly, we have [ x + y ] ∈ T = { [ v ] , [ v ] , [ v ] } (cid:104)(cid:104) α (cid:105)(cid:105) if and only if |(cid:104) x, u (cid:105) + (cid:104) y, w (cid:105)| = a (3.1) GY ¨ORGY P ´AL GEH´ER AND MICHIYA MORI holds for all u ∈ M, w ⊥ M, (cid:107) u (cid:107) + (cid:107) w (cid:107) = 1 , |(cid:104) u, v (cid:105)| = |(cid:104) u, v (cid:105)| = |(cid:104) u, v (cid:105)| = a . We point out thatthe only restriction on w above, apart from being orthogonal to M , concerns its norm. Therefore,if [ x + y ] ∈ T with x (cid:54) = 0, y (cid:54) = 0, then T contains collinear triples, namely { [ x + λy ] : λ ∈ T } ⊂ T, which is a contradiction.The above observations imply T ⊂ P ( M ) ∪ P ( M ⊥ ), where M ⊥ denotes the largest subspacein H orthogonal to M . On the one hand, if dim H ≥ y ] ∈ T ∩ P ( M ⊥ ), then (3.1) cannothold. Hence in that case indeed T ⊂ P ( M ) follows. On the other hand, if dim H = 4, then T ⊂ P ( M ) ∪ { [ e ] } where e ⊥ { v , v , v } , (cid:107) e (cid:107) = 1. Assume for a moment that [ e ] ∈ T . Then aconsideration of { [ v ] , [ v ] , [ e ] } instead of { [ v ] , [ v ] , [ v ] } gives that T ⊂ P ([ v , v , e ]) ∪ { [ f ] } where f ⊥ { v , v , e } , (cid:107) f (cid:107) = 1. Since v / ∈ [ v , v , e ], we have [ v ] = [ f ]. In such a way we eventuallyobtain that T ⊂ ( P ([ v , v , e ]) ∪ { [ v ] } ) ∩ ( P ([ v , v , e ]) ∪ { [ v ] } ) ∩ ( P ([ v , v , e ]) ∪ { [ v ] } ) ∩ ( P ( M ) ∪ { [ e ] } ) . Hence T = 4, a contradiction. Therefore, [ e ] / ∈ T , and we conclude T ⊂ P ( M ). (cid:3) Lemma 3.4.
Assume that dim H ≥ . Let { e , e , e } ⊂ H be an orthonormal system, c ≥ d > with c + d = 1 , and c , c ∈ C , c > , | c | + | c | + c = 1 . Set [ v ] = [ ce + ide ] , [ v ] = [ ce − ide ] , [ v ] = [ c e + c e + c e ] ∈ P ( H ) , and define the function z : [ − θ , θ ] → C , z ( θ ) = c ac cos θ + c ad sin θ, (3.2) where θ and ρ are as in Lemma 2.3. Then for each − θ ≤ θ ≤ θ we have (cid:0) A θ ∩ { [ v ] } (cid:104) α (cid:105) (cid:1) = ∞ if and only if one of the following possibilities happens: (i) either | z ( θ ) | − c ρ ( θ ) < a < | z ( θ ) | + c ρ ( θ ) , (ii) or z ( θ ) = 0 and ρ ( θ ) = ac ,where A θ is as in (2.2) .Moreover, we have (cid:0) A θ ∩ { [ v ] } (cid:104) α (cid:105) (cid:1) = 1 if and only if (iii) z ( θ ) (cid:54) = 0 , and either a = | z ( θ ) | − c ρ ( θ ) , or a = | z ( θ ) | + c ρ ( θ ) .Proof. An element (cid:2) ac cos θ · e + ad sin θ · e + h (cid:3) of A θ is in { [ v ] } (cid:104) α (cid:105) if and only if | z ( θ ) + c (cid:104) e , h (cid:105)| = a. Notice that if we go through all elements of A θ , then the complex number c (cid:104) e , h (cid:105) goes througha closed (possibly degenerate) disk of radius c ρ ( θ ). This radius is 0 if and only if ρ ( θ ) = 0.Assume that z ( θ ) (cid:54) = 0. Then by some elementary geometric observations we obtain the followingpossibilities: • if | z ( θ ) | − c ρ ( θ ) > a or a > | z ( θ ) | + c ρ ( θ ), then A θ ∩ { [ v ] } (cid:104) α (cid:105) = ∅ , • if | z ( θ ) | − c ρ ( θ ) = a or a = | z ( θ ) | + c ρ ( θ ), then (cid:0) A θ ∩ { [ v ] } (cid:104) α (cid:105) (cid:1) = 1, • if | z ( θ ) | − c ρ ( θ ) < a < | z ( θ ) | + c ρ ( θ ), then (cid:0) A θ ∩ { [ v ] } (cid:104) α (cid:105) (cid:1) = ∞ .In case when z ( θ ) = 0, then we obtain the following possibilities: • if c ρ ( θ ) < a , then A θ ∩ { [ v ] } (cid:104) α (cid:105) = ∅ , • if c ρ ( θ ) ≥ a , then (cid:0) A θ ∩ { [ v ] } (cid:104) α (cid:105) (cid:1) = ∞ .Notice that the case c ρ ( θ ) > a is included in (i) in the statement of the lemma. (cid:3) Notice that (cid:0) A θ ∩ { [ v ] } (cid:104) α (cid:105) (cid:1) is either 0, or 1, or ∞ , provided that dim H ≥
4. Now, we are inthe position to prove the main result of this section.
Lemma 3.5.
Assume that dim H ≥ . Then a set T ⊂ P ( H ) is highly- α -symmetric if and only ifit is a circle.Proof. Corollary 3.2 gives one direction. To prove the reverse implication, assume that T is highly- α -symmetric. By Lemma 3.3 and the assumption T = ∞ , we may take a pair of different elements[ v ] , [ v ] ∈ T such that they are not orthogonal. Then, by Lemma 2.2, we have [ v ] = [ ce + ide ]and [ v ] = [ ce − ide ] for some orthonormal system { e , e } ⊂ H and real numbers c > d > c + d = 1. Define A θ , ρ and θ as in Lemma 2.3. Consider an arbitrary third element [ u ] ∈ APS THAT PRESERVE A FIXED QUANTUM ANGLE 9 T \ { [ v ] , [ v ] } . If [ u ] sits on the projective line spanned by [ v ] and [ v ], then by Lemmas 2.4 and3.1, the set T = { [ v ] , [ v ] , [ u ] } (cid:104)(cid:104) α (cid:105)(cid:105) is a circle.From now on we assume that T ∩ P ([ v , v ]) = { [ v ] , [ v ] } . By Lemma 3.3, there exists a unitvector e ⊥ { e , e } such that T ⊂ P ([ e , e , e ]). Consider two arbitrary (not necessarily different)lines [ v ] , [ (cid:98) v ] ∈ T \ { [ v ] , [ v ] } . We may take numbers c , c ∈ C , c > | c | + | c | + c = 1, (cid:98) c , (cid:98) c ∈ C , (cid:98) c > | (cid:98) c | + | (cid:98) c | + (cid:98) c = 1 such that[ v ] = [ c e + c e + c e ] and [ (cid:98) v ] = [ (cid:98) c e + (cid:98) c e + (cid:98) c e ] . By (iv) of Lemma 2.1, we have { [ v ] , [ v ] , [ v ] } (cid:104) α (cid:105) = T (cid:104) α (cid:105) = { [ v ] , [ v ] , [ (cid:98) v ] } (cid:104) α (cid:105) . By Lemma 2.3, this implies A θ ∩ { [ v ] } (cid:104) α (cid:105) = A θ ∩ { [ (cid:98) v ] } (cid:104) α (cid:105) (3.3)for all − θ ≤ θ ≤ θ . We define the functions z and (cid:98) z by (3.2) and (cid:98) z : [ − θ , θ ] → C , (cid:98) z ( θ ) = (cid:98) c ac cos θ + (cid:98) c ad sin θ. Clearly, (3.3) is equivalent to the following for all − θ ≤ θ ≤ θ : a = | z ( θ ) + c (cid:104) e , h (cid:105)| ⇐⇒ a = | (cid:98) z ( θ ) + (cid:98) c (cid:104) e , h (cid:105)| ( h ⊥ { e , e } , (cid:107) h (cid:107) = ρ ( θ )) . (3.4)Assume for a moment that [ e ] ∈ T . Substitute [ v ] = [ e ]. Then for all − θ ≤ θ ≤ θ we have a = |(cid:104) e , h (cid:105)| ⇐⇒ a = | (cid:98) z ( θ ) + (cid:98) c (cid:104) e , h (cid:105)| ( h ⊥ { e , e } , (cid:107) h (cid:107) = ρ ( θ )) . Since T (cid:104) α (cid:105) = ∞ , there exists at least one pair ( θ, h ) which solves both equations above. Notethat (cid:104) e , h (cid:105) (cid:54) = 0, and that ( θ, λh ) also solves the first, hence the second, equation for all λ ∈ T . Bya simple geometric consideration one sees that this can happen only if (cid:98) c = 1. Therefore [ (cid:98) v ] = [ e ],which further implies the contradiction T = { [ v ] , [ v ] , [ e ] } . Hence we obtain [ e ] / ∈ T .Therefore, neither z nor (cid:98) z is the constant zero function. In particular, since their images arecontained in (possibly degenerate) ellipses, they have at most two zeros. We distinguish two cases. Case 1. When for every θ ∈ [ − θ , θ ] we have (cid:0) A θ ∩ { [ v ] } (cid:104) α (cid:105) (cid:1) ≤ . Define the set F := (cid:110) θ ∈ [ − θ , θ ] : (cid:16) A θ ∩ { [ v ] } (cid:104) α (cid:105) (cid:17) = 1 (cid:111) . Since T (cid:104) α (cid:105) = ∞ , we obtain F = ∞ . By (iii) of Lemma 3.4, we infer that (cid:12)(cid:12) | z ( θ ) | − a (cid:12)(cid:12) = c ρ ( θ ) ( θ ∈ F ) . (3.5)We claim that (3.5) implies that | z ( θ ) | is constant on [ − θ , θ ]. In order to see this, we take thesquare of both sides in (3.5), rearrange the equation, and take squares again: (cid:0) | z ( θ ) | + a − c ρ ( θ ) (cid:1) = (2 a | z ( θ ) | ) ( θ ∈ F ) . (3.6)Notice that ρ ( θ ) and | z ( θ ) | are complex linear combinations of cos θ, sin θ and cos θ sin θ . Hencethey, and in particular the right-hand side of (3.6), are complex linear combinations of 1 , cos(2 θ )and sin(2 θ ). The left-hand side of (3.6) can be written in the form( a + b cos(2 θ ) + c sin(2 θ )) = a + b cos (2 θ ) + c sin (2 θ ) + 2 ab cos(2 θ ) + 2 ac sin(2 θ ) + 2 bc cos(2 θ ) sin(2 θ )with some complex numbers a , b , c . Note that this expression is a complex linear combination of1 , cos(2 θ ) , sin(2 θ ) , cos(4 θ ) and sin(4 θ ). Since both sides of (3.6) are trigonometric polynomials andthey coincide on the infinite set F ⊂ (cid:2) − π , π (cid:3) , they must coincide on the whole real line. Hencethe coefficients on both sides with respect to 1 , cos(2 θ ) , sin(2 θ ) , cos(4 θ ) and sin(4 θ ) have to be thesame. Since it is zero for sin(4 θ ), we obtain that b = 0 or c = 0. Assume we have b = 0, then theleft-hand side of (3.6) is a + c sin (2 θ ) + 2 ac sin(2 θ ) = a + c − c θ ) + 2 ac sin(2 θ ) . But since the coefficient of cos(4 θ ) is also zero, we obtain that c = 0. Therefore | z ( θ ) | is indeed a(non-zero) constant function. Similarly, we obtain the same conclusion for the c = 0 case. Using this information in (3.5) we obtain that ρ ( θ ) is constant on F , hence on [ − θ , θ ]. Therefore weinfer c = d = √ , which contradicts our assumption c > d , so the present case cannot happen. Case 2. When there exists a (cid:101) θ ∈ [ − θ , θ ] such that (cid:0) A (cid:101) θ ∩ { [ v ] } (cid:104) α (cid:105) (cid:1) = ∞ holds. We claimthat there is a non-degenerate interval J ⊆ [ − θ , θ ] such that (i) from Lemma 3.4 holds for all θ ∈ J . If (cid:101) θ satisfies (i), then this is clear from the continuity of z and ρ . Suppose (cid:101) θ (cid:54) = 0 and itsatisfies (ii), namely, z ( (cid:101) θ ) = 0 and ρ ( (cid:101) θ ) = ac . In this case if we move θ a little bit away from (cid:101) θ but closer towards 0, then (as c > d >
0) both | z ( θ ) | and ρ ( θ ) increase continuously. Hence weget the desired interval. Finally, assume that (cid:101) θ = 0 and it satisfies (ii), namely, z (0) = 0 and ρ (0) = ac . Consequently, c = 0, and since z is not constant zero, c (cid:54) = 0. We only have to observethat | z ( θ ) | = ad | c sin θ | is differentiable from the right at 0, and that this half-sided derivative is | c | ad >
0. Since ρ (cid:48) (0) = 0, we get the same conclusion by elementary calculus.Now, for all θ ∈ J there exists a non-degenerate arc C θ in the complex plane such that a = | z ( θ ) + c (cid:104) e , h (cid:105)| ⇐⇒ a = | (cid:98) z ( θ ) + (cid:98) c (cid:104) e , h (cid:105)|⇐⇒ (cid:104) e , h (cid:105) ∈ C θ ( h ⊥ { e , e } , (cid:107) h (cid:107) = ρ ( θ )) . As the radii of the circles containing the arcs z ( θ ) + c C θ and (cid:98) z ( θ ) + (cid:98) c C θ are both equal to a , weobtain (cid:98) c = c . A consideration of their centres also gives z ( θ ) = (cid:98) z ( θ ) ( θ ∈ J ). Since both z and (cid:98) z are trigonometric polynomials, their coincidence on the interval J implies c = (cid:98) c , c = (cid:98) c , andhence [ (cid:98) v ] = [ v ]. So this second case cannot happen either. The proof is done. (cid:3) As it turns out the above lemma fails in three dimensions. The aim of the next section is toexplore what can be said about highly- α -symmetric sets in that case.At this point the reader has the option to proceed with Section 5 and read the proof of Theorem1.6 in the case when dim H ≥ The structure of highly- α -symmetric sets in the three-dimensional case We start with a simple statement.
Lemma 4.1.
The α -set S (cid:104) α (cid:105) of any subset S ⊂ P ( H ) is closed. In particular, every highly- α -symmetric set T is compact, hence they contain at least one element that is not an isolated pointof T . The proof is straightforward, hence it is omitted. We now prove the three-dimensional versionof Lemma 3.1.
Lemma 4.2.
Using the notation and assumptions of Lemma 2.5, suppose that dim H = 3 andthat e ⊥ { e , e } is a unit vector. Then we have the following possibilities: (i) if either c √ c ≥ a > d , or ( a, c, d ) = (cid:16) √ , (cid:113) , √ (cid:17) , or ( a, c, d ) = (cid:16) √ , √ , √ (cid:17) , then wehave S (cid:104)(cid:104) α (cid:105)(cid:105) = (cid:110) [ ce + λde ] : λ ∈ T (cid:111) (cid:71) (cid:115) − a − a c e + λ a (cid:113) − a c e : λ ∈ T , (4.1)(ii) otherwise we have S (cid:104)(cid:104) α (cid:105)(cid:105) = { [ ce + λde ] : λ ∈ T } . (4.2)Note that if ( a, c, d ) = (cid:16) √ , (cid:113) , √ (cid:17) , then (4.1) becomes S (cid:104)(cid:104) α (cid:105)(cid:105) = (cid:40)(cid:34)(cid:114) e + λ (cid:114) e (cid:35) : λ ∈ T (cid:41) (cid:71) (cid:40)(cid:34)(cid:114) e + λ (cid:114) e (cid:35) : λ ∈ T (cid:41) , (4.3)and if c √ c = a , then (4.1) is S (cid:104)(cid:104) α (cid:105)(cid:105) = (cid:110) [ ce + λde ] : λ ∈ T (cid:111) (cid:71) { [ e ] } . APS THAT PRESERVE A FIXED QUANTUM ANGLE 11
In particular, if ( a, c, d ) = (cid:16) √ , √ , √ (cid:17) , then (4.1) takes the form S (cid:104)(cid:104) α (cid:105)(cid:105) = (cid:40)(cid:34)(cid:114) e + λ (cid:114) e (cid:35) : λ ∈ T (cid:41) (cid:71) { [ e ] } . (4.4) Proof of Lemma 4.2.
As in the proof of Lemma 3.1, we set C := (cid:40)(cid:34) ac e + λ (cid:114) − a c e (cid:35) : λ ∈ T (cid:41) . Consider a line [ v ] = [ c e + c e + c e ] with c ≥ c , c ∈ C and c + | c | + | c | = 1. Weobtain that [ v ] ∈ C (cid:104) α (cid:105) if and only if (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) c ac + c λ (cid:114) − a c (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = a ( λ ∈ T ) . As c > a , this is equivalent to • either c = c , | c | = d and c = 0, • or c = 0, | c | = (cid:114) − a − a c and | c | = a (cid:113) − a c .Note that a > − a c holds if and only if a > c √ c . Therefore we obtain the following twopossibilities: • if a > c √ c , then C (cid:104) α (cid:105) = { [ ce + λde ] : λ ∈ T } , • if a ≤ c √ c , then C (cid:104) α (cid:105) = { [ ce + λde ] : λ ∈ T } (cid:71) (cid:115) − a − a c e + λ a (cid:113) − a c e : λ ∈ T . In particular, this completes the case when a > d , since then S (cid:104) α (cid:105) = C (cid:104) α (cid:105) holds.In what follows, we shall handle the cases a < d and a = d separately. Set D := (cid:40)(cid:34) ad e + λ (cid:114) − a d e (cid:35) : λ ∈ T (cid:41) . Suppose that a < d . Then similarly as for C (cid:104) α (cid:105) (where c > a was automatic), we obtain thefollowing: • if a > d √ d , then D (cid:104) α (cid:105) = { [ ce + λde ] : λ ∈ T } , • if a ≤ d √ d , then D (cid:104) α (cid:105) = { [ ce + λde ] : λ ∈ T } (cid:71) (cid:115) − a − a d e + λ a (cid:113) − a d e : λ ∈ T . Recall that S (cid:104) α (cid:105) = C ∪ D . Hence we observe that { [ ce + λde ] : λ ∈ T } ⊆ S (cid:104)(cid:104) α (cid:105)(cid:105) = C (cid:104) α (cid:105) ∩ D (cid:104) α (cid:105) ⊆ { [ ce + λde ] : λ ∈ T } (cid:116) { [ e ] } . Therefore, after some easy calculations we obtain the following, which completes the a < d case: • if ( a, c, d ) = (cid:16) √ , √ , √ (cid:17) , then we have (4.1) and (4.4), • otherwise, we have (4.2). Finally, let us assume that a = d . In this case D = { [ e ] } , hence S (cid:104)(cid:104) α (cid:105)(cid:105) = C (cid:104) α (cid:105) ∩ { [ e ] } (cid:104) α (cid:105) = C (cid:104) α (cid:105) ∩ { [ de + x ] : x ⊥ e , (cid:107) x (cid:107) = c } . If we also have a = d > c √ c , then this clearly gives (4.2). Otherwise, S (cid:104)(cid:104) α (cid:105)(cid:105) = [ ce + λde ] , (cid:115) − d − d c e + λ d (cid:113) − d c e : λ ∈ T (cid:92) (cid:110) [ de + x ] : x ⊥ e , (cid:107) x (cid:107) = c (cid:111) . This gives (4.2), unless (cid:114) − d − d c = d , which happens if and only if ( a, c, d ) = (cid:16) √ , (cid:113) , √ (cid:17) . Inthis latter case we obtain (4.1) and (4.3), which completes the proof. (cid:3) Assume that the assumption of (ii) in Lemma 4.2 holds. Then by Lemma 2.5 the circle in (4.2)is highly- α -symmetric. In the next lemma we investigate the other case. Lemma 4.3.
Assume that dim H = 3 and that the assumptions of (i) in Lemma 4.2 hold. Thenthe set S (cid:104)(cid:104) α (cid:105)(cid:105) in (4.1) is not highly- α -symmetric.Proof. Our strategy is to find four lines [ u ] , [ u ] , [ u ] ∈ S (cid:104)(cid:104) α (cid:105)(cid:105) and [ w ] ∈ P ( H ) such that[ w ] ∈ { [ u ] , [ u ] , [ u ] } (cid:104) α (cid:105) \ S (cid:104) α (cid:105) (4.5)which, by (iii)–(iv) of Lemma 2.1, will prove our statement. Let 0 < t < π and consider the unitvector w := ac cos t · e + ac sin t · e + (cid:114) − a c e , Note that (cid:104) w, e j (cid:105) (cid:54) = 0 ( j = 1 , , w ] / ∈ S (cid:104) α (cid:105) . Using elementarycalculus, it is easy to see that for small enough t > < c (cid:16) ac cos t (cid:17) − d (cid:16) ac sin t (cid:17) < a < c (cid:16) ac cos t (cid:17) + d (cid:16) ac sin t (cid:17) . (4.6)Hence there exists a number λ ∈ T \ { , − } with[ u ] := [ ce + λde ] , [ u ] := [ ce + λde ] ∈ { [ w ] } (cid:104) α (cid:105) ∩ S (cid:104)(cid:104) α (cid:105)(cid:105) . In a similar way, we obtain the following for small enough t > < a − (cid:115) − a − a c (cid:16) ac sin t (cid:17) ≤ a ≤ a + (cid:115) − a − a c (cid:16) ac sin t (cid:17) . However, unlike in (4.6), here we have equations if c √ c = a . Therefore, we conclude the existenceof a number µ ∈ T such that[ u ] := (cid:115) − a − a c e + µ a (cid:113) − a c e ∈ { [ w ] } (cid:104) α (cid:105) ∩ S (cid:104)(cid:104) α (cid:105)(cid:105) . The relation (4.5) follows and the proof is complete. (cid:3)
We continue with the analogue of Lemma 3.5 in three dimensions. It basically says that highly- α -symmetric sets are exactly the circles with certain diameters. The lemma also implies someestimations for the diameter. Lemma 4.4.
Assume that dim H = 3 . Then for any set T ⊂ P ( H ) and orthonormal system { e , e } ⊂ H the following hold: (i) If T is highly- α -symmetric, then it is a circle. (ii) If a (cid:54) = √ , c ≥ d > a , c + d = 1 , then the circle { [ ce + λde ] : λ ∈ T } is highly- α -symmetric. APS THAT PRESERVE A FIXED QUANTUM ANGLE 13 (iii) If a = √ , c > d > a , c + d = 1 , then the circle { [ ce + λde ] : λ ∈ T } is highly- α -symmetric. (iv) If < d < min (cid:110) a, (cid:113) − a − a (cid:111) , c + d = 1 , then the circle { [ ce + λde ] : λ ∈ T } is nothighly- α -symmetric.Proof. Parts (ii)–(iv) easily follow from Lemmas 2.5 and 4.2. For (iv) we additionally note that d < (cid:113) − a − a implies a < c √ c .In order to prove (i), assume that T is highly- α -symmetric. Suppose that there are three differentelements [ v ] , [ v ] , [ v ] ∈ T which sit on the same projective line. Then by Lemmas 2.4, 4.2 and4.3, the set T = { [ v ] , [ v ] , [ v ] } (cid:104)(cid:104) α (cid:105)(cid:105) is a circle.From now on, we shall assume that no three different elements of T are collinear. Our aim is toobtain a contradiction. By Lemma 4.1, we may take a line [ v ] ∈ T that is not isolated in T . Takeanother line [ v ] ∈ T \ { [ v ] } . They can be written as [ v ] = [ ce + ide ] and [ v ] = [ ce − ide ]with some orthonormal system { e , e } ⊂ H and real numbers c ≥ d > c + d = 1. Let e ⊥ { e , e } be a unit vector. In what follows, we use the same symbols as in the proof ofLemma 3.5. Namely, we consider two arbitrary lines [ v ] , [ (cid:98) v ] ∈ T \ { [ v ] , [ v ] } which may bewritten as [ v ] = [ c e + c e + c e ] and [ (cid:98) v ] = [ (cid:98) c e + (cid:98) c e + (cid:98) c e ], where c , c , (cid:98) c , (cid:98) c ∈ C , c > (cid:98) c > | c | + | c | + c = | (cid:98) c | + | (cid:98) c | + (cid:98) c = 1. By (iv) of Lemma 2.1, we have { [ v ] , [ v ] , [ v ] } (cid:104) α (cid:105) = T (cid:104) α (cid:105) = { [ v ] , [ v ] , [ (cid:98) v ] } (cid:104) α (cid:105) . By Lemma 2.3 this implies A θ ∩ { [ v ] } (cid:104) α (cid:105) = A θ ∩ { [ (cid:98) v ] } (cid:104) α (cid:105) for all − θ ≤ θ ≤ θ , where A θ := (cid:110)(cid:104) ac cos θ · e + ad sin θ · e + λρ ( θ ) e (cid:105) : λ ∈ T (cid:111) . In particular, observe that the cardinality c ( θ ) := (cid:16) A θ ∩ { [ v ] } (cid:104) α (cid:105) (cid:17) does not depend on the specific choice of [ v ] ∈ T \ { [ v ] , [ v ] } . We have the following for all − θ ≤ θ ≤ θ : a = | z ( θ ) + c ρ ( θ ) λ | ⇐⇒ a = | (cid:98) z ( θ ) + (cid:98) c ρ ( θ ) λ | ( λ ∈ T ) . (4.7)Exactly the same argument as in the proof of Lemma 3.5 right after (3.4) shows that [ e ] / ∈ T ,hence neither z nor (cid:98) z is the constant zero function. We distinguish two cases. Case 1. When for all θ ∈ [ − θ , θ ] we have c ( θ ) < ∞ . As can be seen by a simple geometricconsideration, in this case for all − θ < θ < θ (which implies ρ ( θ ) >
0) both equations of (4.7)have at most two solutions. In particular, there is no solution if z ( θ ) = 0. Let F be the set ofthose θ ∈ ( − θ , θ ) for which there is at least one solution λ . Note that F = ∞ , as T (cid:104) α (cid:105) = ∞ .For all θ ∈ F let λ ( θ ) and λ ( θ ) denote the two solutions, which might coincide for some θ . Byelementary geometry, one sees that z ( θ ) and (cid:98) z ( θ ) are real linearly dependent for all θ ∈ F . Indeed,we can easily see the following: if λ ( θ ) = λ ( θ ), then both { z ( θ ) , λ ( θ ) } and { (cid:98) z ( θ ) , λ ( θ ) } are reallinearly dependent; if λ ( θ ) (cid:54) = λ ( θ ), then both z ( θ ) and (cid:98) z ( θ ) are orthogonal to λ ( θ ) − λ ( θ ) inthe complex plane. Hence for all θ ∈ F a (cid:61) (cid:16) z ( θ ) (cid:98) z ( θ ) (cid:17) = (cid:61) (cid:16) c (cid:98) c (cid:17) c cos θ + (cid:61) (cid:16) c (cid:98) c (cid:17) d sin θ + (cid:61) (cid:16) c (cid:98) c + c (cid:98) c (cid:17) cd sin θ cos θ. (4.8)Note that a trigonometric polynomial has infinitely many zeros on a compact interval if and onlyif it is the constant zero function on R . Therefore, the right-hand side of (4.8) is zero for all real θ .By substituting θ = 0 , π , arccos c , we obtain that each of the following is a real linearly dependentsystem in C : { c , (cid:98) c } , { c , (cid:98) c } , { c + c , (cid:98) c + (cid:98) c } . (4.9)Assume for a moment that c and c are real linearly independent complex numbers. Then (4.9)implies (cid:98) c = qc and (cid:98) c = qc with some 0 (cid:54) = q ∈ R . Notice that this forces [ (cid:98) v ] to lie on the projective line spanned by [ e ] and (cid:104) c | c | + | c | e + c | c | + | c | e (cid:105) , hence the contradiction T ≤ c and c are real linearly dependent, hence |(cid:104) v , v (cid:105)| = | cc − idc | = (cid:112) c · | c | + d · | c | ≤ c. Since [ v ] ∈ T \ { [ v ] , [ v ] } was arbitrary, we obtain thatinf { (cid:93) ([ v ] , [ u ]) : [ u ] ∈ T \ { [ v ] }} > . This contradicts our assumption that [ v ] is not an isolated point of T , so this case cannot happen. Case 2. When there exists a (cid:101) θ ∈ [ − θ , θ ] such that c ( (cid:101) θ ) = ∞ holds. In this case ρ ( (cid:101) θ ) > λ ∈ T . Therefore, (cid:98) z ( (cid:101) θ ) = (cid:98) c ac cos (cid:101) θ + (cid:98) c ad sin (cid:101) θ = 0 , z ( (cid:101) θ ) = c ac cos (cid:101) θ + c ad sin (cid:101) θ = 0 (4.10)and c = (cid:98) c = aρ ( θ ) . After some easy calculation we infer from (4.10) that (0 , (cid:54) = ( (cid:98) c , (cid:98) c ) = µ ( c , c )holds with some µ ∈ T . Therefore, [ (cid:98) v ] must lie on the projective line spanned by [ e ] and (cid:104) c | c | + | c | e + c | c | + | c | e (cid:105) . However, since [ (cid:98) v ] ∈ T \ { [ v ] , [ v ] } was arbitrary, this implies thecontradiction T ≤
4. So this case cannot happen either, the proof is done. (cid:3) Proof of the main theorem
This section is devoted to the final step of the proof of our main result.
Proof of Theorem 1.6.
Let M be an arbitrary two-dimensional subspace of H . In what followswe shall prove that there exists another two-dimensional subspace N such that φ maps P ( M )onto P ( N ). Then a straightforward application of Theorem 1.4 gives that the restriction φ | P ( M ) preserves every quantum angle, which in turn completes the proof.Fix c := (cid:113) , if dim H ≥ a (cid:54) = (cid:113) (cid:113) , if dim H = 3 and a = (cid:113) and d := (cid:112) − c . By Lemmas 3.5 and 4.4, every circle of the form C ([ e ] , [ e ]) := { [ c e + λd e ] : λ ∈ T } , where { e , e } is an orthonormal system, is highly- α -symmetric. We obviously have P ( M ) = (cid:91) (cid:8) C ([ e ] , [ e ]) : { e , e } is an orthonormal basis of M (cid:9) . It is apparent from Definition 1.7 and the properties of φ , that φ and φ − map highly- α -symmetricsets onto highly- α -symmetric sets. In particular, φ ( P ( M )) is a union of circles of the form D ([ e ] , [ e ]) := φ (cid:0) C ([ e ] , [ e ]) (cid:1) .Observe that if C ([ e ] , [ e ]) ∩ C ([ f ] , [ f ])) ≥ { e , e } and { f , f } of M , then D ([ e ] , [ e ]) and D ([ f ] , [ f ]) are contained in the same projective line.Indeed, there exist two different lines [ u ] , [ u ] ∈ P ( M ) such that { [ u ] , [ u ] } ⊆ C ([ e ] , [ e ]) ∩ C ([ f ] , [ f ]). Set [ v ] := φ ([ u ]) and [ v ] := φ ([ u ]). Since [ v ] , [ v ] ∈ D ([ e ] , [ e ]) ∩ D ([ f ] , [ f ]), weconclude D ([ e ] , [ e ]) ∪ D ([ f ] , [ f ]) ⊆ P ([ v , v ]). Note that P ([ v , v ]) is equal to the projectiveline generated by φ ([ c e + d e ]) and φ ([ c e − d e ]).From here we distinguish between two cases. Case 1. When dim H ≥ or a (cid:54) = (cid:113) holds. Then c = d = √ , and it is rather straight-forward to see from the Bloch representation that C ([ e ] , [ e ]) ∩ C ([ f ] , [ f ])) ≥ { e , e } and { f , f } of M . Indeed, the Bloch representations of thesecircles are great circles on S . However, let us give here a more direct proof of the inequality C ([ e ] , [ e ]) ∩ C ([ f ] , [ f ])) ≥
2. If { [ e ] , [ e ] } = { [ f ] , [ f ] } , then this is obvious, so from nowon we assume otherwise. There exist numbers a , b > a + b = 1, µ ∈ T such that f may APS THAT PRESERVE A FIXED QUANTUM ANGLE 15 be assumed to have the form a e + µ b e . Consequently, f may be assumed to have the form b e − µ a e . Then (cid:20) √ e + iµ √ e (cid:21) = (cid:20) a − i b √ e + µ b + i a √ e (cid:21) = (cid:20) √ f − i √ f (cid:21) ∈ C ([ e ] , [ e ]) ∩ C ([ f ] , [ f ]) , and similarly (cid:20) √ e − iµ √ e (cid:21) = (cid:20) a + i b √ e + µ b − i a √ e (cid:21) = (cid:20) √ f + i √ f (cid:21) ∈ C ([ e ] , [ e ]) ∩ C ([ f ] , [ f ]) . Fix an orthonormal basis { e , e } of M . We obtain φ ( P ( M )) = φ (cid:16)(cid:91) (cid:8) C ([ f ] , [ f ]) : { f , f } is an orthonormal basis of M (cid:9)(cid:17) = (cid:91) (cid:8) D ([ f ] , [ f ]) : { f , f } is an orthonormal basis of M (cid:9) ⊆ P ( N ) , where P ( N ) is the projective line generated by φ ([ c e + d e ]) and φ ([ c e − d e ]). However, bythe very same reasons, the inverse φ − maps P ( N ) into some projective line P ( L ). Since we have P ( M ) ⊆ φ − ( P ( N )), we infer P ( M ) = φ − ( P ( N )), which in turn completes the proof of this case. Case 2. When dim H = 3 and a = (cid:113) are satisfied. Then c = (cid:113) and d = (cid:113) . One easilysees that it suffices to show the following: for any two orthonormal bases { e , e } and { f , f } of M , there exists a third orthonormal basis { g , g } of M such that C ([ e ] , [ e ]) ∩ C ([ g ] , [ g ])) ≥ , C ([ g ] , [ g ]) ∩ C ([ f ] , [ f ])) ≥ . (5.1)Again, one way to verify this is by utilising the Bloch representation, however, let us show itdirectly here. If [ e ] = [ f ] and [ e ] = [ f ], then this is obvious, so from now on we assumeotherwise. Then there are numbers 0 ≤ a < , < b ≤ a + b = 1, µ ∈ T such that f and f may be assumed to have the forms a e + µ b e and b e − µ a e , respectively. Note that C ([ f ] , [ f ]) = (cid:40)(cid:34)(cid:114)
712 ( a e + µ b e ) + λ (cid:114)
512 ( b e − µ a e ) (cid:35) : λ ∈ T (cid:41) = (cid:40)(cid:34)(cid:32)(cid:114) a + λ (cid:114) b (cid:33) e + µ (cid:32)(cid:114) b − λ (cid:114) a (cid:33) e (cid:35) : λ ∈ T (cid:41) . As b >
0, we obtain that C ([ e ] , [ e ]) ∩ C ([ f ] , [ f ])) ≥ λ ∈ T \ {− , } such that (cid:12)(cid:12)(cid:12)(cid:113) a + λ (cid:113) b (cid:12)(cid:12)(cid:12) = (cid:113) . The latter equationis equivalent to (cid:12)(cid:12)(cid:12) a + λ (cid:113) b (cid:12)(cid:12)(cid:12) = 1. Since (cid:12)(cid:12)(cid:12) a − (cid:113) b (cid:12)(cid:12)(cid:12) ≤ max (cid:110) a , (cid:113) (cid:111) <
1, the inequality C ([ e ] , [ e ]) ∩ C ([ f ] , [ f ])) ≥ a + (cid:113) b >
1. A simple calculation givesthat this is further equivalent to < a < a = |(cid:104) e , f (cid:105)| . Therefore, if we have < |(cid:104) e , f (cid:105)| <
1, then (5.1) holds with g = e and g = e . On the other hand, if 0 ≤ |(cid:104) e , f (cid:105)| = a ≤ , then choose g := √ e + µ √ e and g := √ e − µ √ e . We have |(cid:104) g , e (cid:105)| = √ > and |(cid:104) g , f (cid:105)| = 1 √ a + 1 √ b ≥ √ √ > . This completes the proof. (cid:3)
We close our paper with mentioning that even though H was assumed to be a Hilbert space,our method clearly works for general complex inner product spaces as well. In that case, the onlychange we have to make in the statement of Theorem 1.6 is to replace “unitary or an antiunitaryoperator” with “bijective linear or conjugate-linear isometry”, since the former term is usuallyused only for Hilbert spaces. References [1] V. Bargmann, Note on Wigner’s theorem on symmetry operations,
J. Math. Phys. (1964), 862–868.[2] G.P. Geh´er, Symmetries of Projective Spaces and Spheres, Int. Math. Res. Not. IMRN (2020), 2205–2240.[3] C.-K. Li, L. Plevnik, and P. ˇSemrl, Preservers of matrix pairs with a fixed inner product value,
Oper. Matrices (2012), 433–464.[4] J.S. Lomont, and P. Mendelson, The Wigner unitary-antiunitary theorem, Ann. Math. (1963), 548–559.[5] B. Simon, Quantum dynamics: from automorphism to Hamiltonian, Studies in Mathematical Physics, Essays inhonor of Valentine Bargmann , eds. E.H. Lieb, B. Simon, A.S. Wightman, Princeton Series in Physics, PrincetonUniversity Press, Princeton, 327–349, 1976.[6] U. Uhlhorn, Representation of symmetry transformations in quantum mechanics,
Ark. Fysik (1963), 307–340.[7] E.P. Wigner, Gruppentheorie und ihre Anwendung auf die Quantenmechanik der Atomspektrum , Fredrik Viewegund Sohn, 1931.
Gy¨orgy P´al Geh´er, Department of Mathematics and Statistics, University of Reading, Whiteknights,P.O. Box 220, Reading RG6 6AX, United Kingdom
Email address : [email protected] or [email protected]
Michiya Mori, Graduate School of Mathematical Sciences, The University of Tokyo, Komaba,Tokyo, 153-8914, Japan
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