aa r X i v : . [ m a t h - ph ] F e b Non-integer flux — why it does not work.
A. V. Smilga
SUBATECH, Universit´e de Nantes,4 rue Alfred Kastler, BP 20722, Nantes 44307, France ABSTRACT
We consider the Dirac operator on S without one point in the case ofnon-integer magnetic flux. We show that the spectral problem for H = / D can be well defined, if including in the Hilbert space H only nonsingularon S wave functions. However, this Hilbert space is not invariant underthe action of / D — for certain ψ ∈ H , / D ψ does not belong to H anymore.This breaks explicitly the supersymmetry of the spectrum. In the integerflux case, supersymmetry can be restored if extending the Hilbert spaceto include locally regular sections of the corresponding fiber bundle. Fornon-integer fluxes, such an extension is not possible. On leave of absence from ITEP, Moscow, Russia.
Introduction
The famous Atiyah-Singer theorem [1] relates the index of the Dirac operator / D (thedifference n (0) L − n (0) R , n (0) L,R being the number of left-handed (right-handed) zero modesof / D ) on a manifold M to certain topological invariants. The most important of theseinvariants are the Chern classes of the corresponding fiber bundles,Ch( F ) ∝ Tr Z M F ∧ . . . ∧ F . (1)( F = (1 / F µν dx µ ∧ dx ν ). In the simplest nontrivial case of U (1) bundle on S , theindex coincides with the magnetic flux, I = 12 π Z F . (2)When a mathematician talks about a fiber bundle on S , he has in mind a partic-ular construction [2]. The sphere is covered by two overlapping maps with the coor-dinates { x µ } and { x ′ µ } . On each map, the connections A = A µ dx µ and A ′ = A ′ µ dx ′ µ are defined. One requires then that, in the region where the maps overlap, the con-nections A and A ′ are related by a gauge transformation, A ′ µ = Ω − ( x )[ A µ + i∂ µ ]Ω( x ) , (3)where Ω( x ) ∈ U (1) is a well-defined function in the overlap region. With this defini-tion, the magnetic flux (2) is always integer.A convenient explicit realisation of this construction involves two sets of stereo-graphic coordinates { x, y } and { x ′ , y ′ } . One of the maps covers the whole S butthe north pole and another — S but the south pole. The relationship (3) can beestablished on the equator with topology S . The magnetic flux coincides then withthe winding number of the map U (1) → S .One can still pose a naive question. Suppose we consider the Dirac equationon only one such stereographic map, disregarding the singularity at the pole. Themagnetic flux can then acquire an arbitrary value. The question is whether thespectral problem with a fractional flux defined on S \{ one point } is benign and, ifnot, what in particular goes wrong there.The answer is the following. • To begin with, when the north pole of the sphere (corresponding to x = y = ∞ )is disregarded, it seems natural to include into consideration the wave functionsthat are singular at this point and only require for the functions to be squareintegrable. It turns out, however, that, for non-integer fluxes, the Hamiltonian H = / D is not Hermitian in such extended Hilbert space. • We can impose the requirement for the function to be regular at all pointsof S including the north pole. In this case, the Hamiltonian is Hermitian.However, the Dirac operator itself is not well defined in the framework of such1educed Hilbert space H : some nonsingular on S functions become singularwhen acted upon by / D . A mathematician would say that this Hilbert space doesnot constitute a domain of / D . A physicist would notice that certain regular on S functions do not have regular superpartners (obtained by the action of thesupercharges Q, ¯ Q = / D (1 ± σ ) / • The remark above (that supersymmetry is broken if including into considerationonly regular functions) refers both to integer and fractional fluxes. However, ifthe flux is integer, supersymmetry of the Hamiltonian H = / D can be restoredif including into the Hilbert space square integrable functions which behave as ψ ∼ e iφ in the vicinity of the north pole (such that | ψ | = C = 0). They arenothing but the sections of the fiber bundle described above and restricted onthe map not covering the north pole. For noninteger fluxes, this is not possible,which means that supersymmetry is genuinely broken.We guess that the same happens in all other cases when one tries to consider theDirac spectral problem for “crippled” bundles with non-integer Chern classes.Note that we discuss in this paper only compact manifolds. If the manifold isnot compact (for example, R ), the spectrum of / D is continuous [3]. And continu-ous spectrum is not quite the spectrum in a rigid mathematical sense. To pose thequestions about hermiticity, etc., one has to regularize the problem in the infraredrendering the motion finite and the spectrum discrete. One of the ways to performsuch a regularisation is to consider, instead of a non-compact manifold, a manifoldwith a boundary. Then supersymmetry can be preserved by imposing rather compli-cated nonlocal boundary conditions due to Atiyah, Pathody and Singer [4]. We willnot discuss this further. We choose the complex stereographic coordinates w, ¯ w . The metric is ds = 2 dwd ¯ w (1 + ¯ ww ) . (4)It is singular at w = ∞ , but this singularity is integrable. In particular, the volumeof such sphere is V = Z dwd ¯ w (1 + ¯ ww ) = 2 π , (5)corresponding to the radius R = √ /
2. Consider the action [5] S = Z dtd θ (cid:20) − DW ¯ D ¯ W W W ) + G ( ¯ W , W ) (cid:21) , (6)where D = ∂∂θ − i ¯ θ ∂∂t , ¯ D = − ∂∂ ¯ θ + iθ ∂∂t W = w + √ θψ − iθ ¯ θ ˙ w, ¯ W = ¯ w − √ θ ¯ ψ + iθ ¯ θ ˙¯ w (7)are chiral (0 + 1) − dimensional superfields, ¯ DW = D ¯ W = 0, and G ( ¯ W , W ) is anarbitrary real function. After deriving the component Lagrangian, the Hamiltonianand quantizing the latter, we obtain a certain supersymmetric quantum mechanicalspectral problem, which is isomorphic to the Dirac spectral problem. It involves thegauge potentials A w, ¯ w = ( − i∂G, i ¯ ∂G ) and the magnetic field F w ¯ w = − F ¯ ww = 2 i∂ ¯ ∂G .Let us choose G = − q W W ) . (8)Then the magnetic field area density (1 + ¯ ww ) F w ¯ w is constant on S and q is thevalue of the flux (2). The quantum supercharges (isomorphic to / D (1 ± σ ) /
2) are Q = i (1 + ¯ ww ) ψ (cid:20) ∂ − ¯ w (1 − q )2(1 + ¯ ww ) (cid:21) ¯ Q = i (1 + ¯ ww ) ¯ ψ (cid:20) ¯ ∂ − w (1 + q )2(1 + ¯ ww ) (cid:21) . (9)They act upon the two-component wave functionsΨ cov = Ψ F =0 ( ¯ w, w ) + ψ Ψ F =1 ( ¯ w, w ) (10)normalized with the covariant measure µ d ¯ wdw = √ g d ¯ wdw = d ¯ wdw (1 + ¯ ww ) , (11)( F = ψ ¯ ψ = ψ ∂∂ψ is the operator of the fermion charge commuting with the Hamilto-nian.) The operators Q, ¯ Q are (naively) Hermitially conjugate one to another withthe measure (11), ¯ Q = µ − Q † µ . The supercharges (9) are nilpotent and { ¯ Q, Q } isthe Hamiltonian. In the sector F = 0, the latter reads H F =0 = − (1 + ¯ ww ) ∂ ¯ ∂ + κ (1 + ¯ ww )( ¯ w ¯ ∂ − w∂ ) + κ ¯ ww + κ , (12)where κ = 1 − q . (13)It commutes with the angular momentum operator L = ¯ w ¯ ∂ − w∂ . Note that theHamiltonian (12) coincides up to a constant shift with the Hamiltonian for the scalarcharged particle in the field of a monopole with the magnetic charge q ′ = q − H F =1 is similar, one has only to interchange w and ¯ w andinverse the sign of q . 3cting with the hamiltonian (12) on the wave function Ψ = ¯ w m F ( ¯ ww ) in thesector with a given angular momentum m , we obtain the equation − u (1 + u ) F ′′ ( u ) − ( m + 1)(1 + u ) F ′ ( u ) + κ uF ( u ) + κm (1 + u ) F ( u )= ( λ − κ ) F ( u ) , (14)where u = ¯ ww and λ is the spectral parameter. Introducing the variable z = 1 − u u (15)(it is nothing but cos θ , θ being the polar angle on S ), we derive( z − F ′′ ( z ) + 2( z + m ) F ′ ( z ) + 2 κ ( κ + m )1 + z F = [ λ − κ (1 − κ )] F ( z ) . (16)There are two families of formal solutions: F mn ( z ) = (1 + z ) − κ P m, − m − κn ( z ) [ λ n = n ( n + 1 − κ )] , ˜ F mn ( z ) = (1 + z ) κ + m P m,m +2 κn ( z ) [˜ λ n = ( m + n + 2 κ )( m + n + 1)] , (17)where P α,βn ( z ) are the Jacobi polynomials, P α,βn ( z ) = 12 n n X k =0 (cid:18) n + αk (cid:19) (cid:18) n + βn − k (cid:19) (1 + z ) k ( z − n − k . (18)For α > − , β > −
1, the Jacobi polynomials are mutually orthogonal on the interval z ∈ ( − ,
1) with the weight µ = (1 − z ) α (1 + z ) β .An important observation is that, for integer q , one has not two families of solu-tions, but actually only one such family. Indeed, for integer α, β , the Jacobi polyno-mials satisfy the following interesting relations [6] P − α,βn + α = 2 − α ( z − α n !( n + α + β )!( n + α )!( n + β )! P α,βn ,P α, − βn + β = 2 − β ( z + 1) β n !( n + α + β )!( n + α )!( n + β )! P α,βn . (19)Using the second relation, it is easy to see that, for integer q , the functions F mn ( z )and ˜ F mn ( z ) coincide up to a shift of n and an irrelevant factor. Picking up only thesquare integrable functions, we obtain the so called monopole harmonics [6, 7],Ψ ( q ) F =0 mn = e imφ (1 − z ) | m | / (1 + z ) | m +2 κ | / P | m | , | m +2 κ | n ( z ) ,n = 0 , , . . . ( q > , n = 1 − q, . . . ( q ≤ ,m = − n, . . . , n + q − e iφ = p ¯ w/w (we do not bother about the normalization coefficients). Thespectrum is λ ( q ) F =0 mn = ( | m + κ | + n + κ )( | m + κ | + n + 1 − κ ) . (21)4he eigenfunctions and the eigenvalues in the sector F = 1 are given by the sameexpressions with the change q → − q, m → − m (the eigenfunctions involve, of course,the extra fermion factor ψ ).If choosing q = 1 in the sector F = 0 (or q = − F = 1), the expres-sions are simplified. In this case, κ = 0 and the Hamitonian (12) is reduced to theordinary Laplacian . The eigenfuctions represent then the Gegenbauer polynomials.They are all regular on S . The spectrum is λ = l ( l + 1) (with l = n + | m | ) as itshould be.For other integer q , the situations is somewhat more complicated. There arefunctions with m + 2 κ = 0 and m = 0 which behave as ∼ e imφ in the vicinicy of thenorth pole and are nothing but the sections of the fiber bundle associated with thegauge field. All monopole harmonics (20) are square integrable on S .One can make now two important observations: • For integer q , one can be convinced that the action of the supercharge Q on asquare integrable eigenfunction (20) is also square integrable. The same con-cerns the action of ¯ Q on the square integrable eigenfunctions of H F =1 . In otherwords, the Hilbert space spanned by the functions (20) in the two fermionicsectors lies in the domain of Q and ¯ Q , and hence in the domain of / D . As aresult, the full spectrum of the hamiltonian H = / D is supersymmetric — allexcited states are doubly degenerate. • The Witten index n (0) F =0 − n (0) F =1 of this system (alias, the Atiyah-Singer indexof / D ) is equal to q . Let us see now what happens if q is not integer. The first immediate observation isthat the functions ˜ F mn are not expressed via F mn anymore, and we have a priori twodifferent families of solutions.As an example, consider the case q = 1 /
2. The solutions areΨ (1 / F =0 mn = e imφ (1 − z ) m/ (1 + z ) − / − m/ P m, − m − / n ( z ) , ˜Ψ (1 / F =0 mn = e imφ (1 − z ) m/ (1 + z ) / m/ P m,m +1 / n ( z ) . (22)We are interested, however, only in normalizable solutions. Thus, when m > mn is admissible, while the normalization integral R | Ψ mn | dz diverges ar z = −
1. When m <
0, it may seem at first that there areno normalizable solutions whatsoever due to the divergence at z = 1. Well, thisdivergence is there for n < | m | . But for larger n , one can use the first relation inEq.(19) and express P −| m | , ±| m |∓ / n ( z ) ∝ (1 − z ) | m | P | m | , ±| m |∓ / n −| m | ( z ) . (23) A mathematician would remark that, in this case, the twisted Dirac complex is equivalent tothe untwisted Dolbeault ( q = 1) or anti-Dolbeault ( q = −
1) complex. See e.g. [5] for more detaileddiscussion. z = 1 (that corresponds to w = ¯ w = 0) disappears and wehave only to take care about the divergence at z = − w = ¯ w = ∞ ). For m ≤ − mn and only the family Ψ mn is admissible. All thesenormalized solutions are regular on S .The values m = 0 , − two normal-izable families: Ψ n , ˜Ψ n = (1 + z ) ∓ / P , ∓ / n ( z ) , Ψ − ,n , ˜Ψ − ,n = e − iφ (1 + z ) ± / (1 − z ) / P , ± / n − ( z ) (24)(we used the property (23) in the second line). Half of these states are singular on S , while another half are not.We can see now that the Hamiltonian (12) is not Hermitian in the Hilbert spaceincluding all the functions in (24). Indeed, consider the sector m = 0 and restrictourselves with the case n = 0. There are two normalizable states, | i = (1 + z ) / with λ = 12 , | i = (1 + z ) − / with λ = 0 . (25)The states | , i have different eigenvalues, but are not orthogonal to each other, h | i 6 = 0. This means that the hermiticity of (12) is lost. One can also see that, ifcomparing the matrix elements h | H | i and h | H | i , they differ by an integral of atotal derivative h | H | i − h | H | i ∼ Z ∂ (cid:20) w ww (cid:21) dwd ¯ w ∼ Z d x ∂ i h x i r i = 0 . (28)This analysis can be generalized for other values of q . In Fig. 1, we plotted thepower γ in the asymptotic behavior of the wave functions at infinity , Ψ ( q ) mn ∼ | w | − γ . A similar phenomenon can be observed in other settings. Consider for example the covariantLaplacian on S . Its explicit expression in stereographic coordinates is H = −△ S = − f ∂ i f ∂ i (26)with f = 1 + r /
4. (The metric is then ds = d~x /f , the radius of such 3-sphere being R = 1.)This operator has some set of regular on S eigenfunctions, in particular the eigenfunction Ψ = 1with λ = 0. One can try to consider, however, the operator (26) on S \{ pole } and include into theconsideration also singular at infinity functions with the only requirement for them to be squareintegrable. There is one such singular square integrable eigenfunction of H : Ψ = √ f with theeigenvalue λ = − /
4. (Recall that the normalization integral includes the measure µ = √ g = 1 /f .)As the eigenvalues are different and h | i 6 = 0, hermiticity in the extended Hilbert space that includesΨ is lost. One can also compare h | H | i and h | H | i and find out that h | H | i − h | H | i ∼ Z d x ∂ i h x i r i = 0 . (27) m=−2 m=−1 m=0 m=1 m=1m=0m=0 γ m=−1 Figure 1: The eigenstates of (12) for different q . Solid lines — regular functions.Dashed lines — singular but square integrable functions. q γ m=2m=1m=2m=1 m=3 m=2 m=3 Figure 2: The same in the sector F = 1.7ote that it is not the energy that is plotted, such that a crossing of the lines inFig. 1 does not generically mean a degeneracy of the levels. However, at γ = 0, eachcrossing involves a single tower of states (there is also the quantum number n thatmarks the levels of these towers) rather than two different towers.Positive γ correspond to nonsingular at the north pole functions. The functionswith − < γ < γ ≤ − m = 0 , − q = 1 /
2, but for any q ∈ (0 , q , the two extra towers are present in a different pair of sectors.For example, if q = 2 .
7, these are the sectors m = 1 , F = 1.The structure of the levels shown in Fig. 2 is the same as in the sector F = 0 upto the shift m → m + 2. This means that, e.g. for q = 1 /
2, we have two extratroublesome normalizable singular towers of states in the sectors m = 1 ,
2, etc.Let us try now to redefine a problem and exclude the singular functions fromconsideration. Note that the functions (24) either grow or vanish at infinity, thesection-of-a-bundle behavior ∼ e imφ characteristic for monopole harmonics with in-teger q is not possible here. Thus, our Hilbert space includes only regular on S functions. The Hamiltonian becomes in this case Hermitian in the both fermionicsectors. The trouble strikes back, however, when we try to act on the states inthis reduced Hilbert space by the supercharges. It particular, let us act by the super-charge Q on the states with F = 0. Let first q = 1 / n = 0)nonsingular statesΨ F =000 = 1(1 + ¯ ww ) / , Ψ F =0 − , = w (1 + ¯ ww ) / (29)We derive Q Ψ F =000 = Ψ F =110 = ψ ¯ w (1 + ¯ ww ) / , Q Ψ F =0 − , = Ψ F =100 = ψ ww ) / (30)The second function in Eq.(30) is regular at w = ∞ , whereas the first one is normal-izable but singular !Generically, one can observe that the lines with positive slope in Fig. 1 are shiftedupwards upon the action of Q (such that the parameter γ is increased, increasingthe convergence of the normalization integral.) On the other hand, the lines withnegative slope are shifted downwards. This means that some nonsingular functionsbecome singular (and roughly a half of singular nonrenormalizable functions becomenormalizable). And that means that the supersymmetry of the spectrum is lost.Some eigenfunctions of the Hamiltonian do not have superpartners.The fact that the action of the supercharge can make a singular function out ofa nonsingular one is rather natural, bearing in mind the generic structure of (9) at Thus, the title of this paper implying that noninteger flux does not work is not quite exact. Tosome extent, it works . This concerns especially the nonsupersymmetric problem of a scalar chargedparticle in the magnetic monopole field. It is benign even if the monopole charge is not integer. w , Q ∼ ¯ ww ( ∂ + 1 /w ) ∼ ¯ w . A special explanation may rather be required whythis does not happen for integer q . Well, the reason is that the problematic functionΨ with positive γ and negative slope giving singular Q Ψ fuses at integer q with apositive slope function for which Q Ψ is not singular.A similar phenomenon (the loss of apparent supersymmetry due to the fact thatthe superpartners of some states lie outside the Hilbert space and thus do not belongto the spectrum of the Hamiltonian) is known to show up for some other systems.The simplest example [8] is probably Witten’s supersymmetric Hamiltonian H = p W ′ ( x )] W ′′ ( x )2 [ ¯ ψ, ψ ] (31)with the superpotential W ′ = − ωx + 1 /x . In the bosonic sector F = 0, the Hamilto-nian is H B = p ω x − ω ( x ) ∝ exp {− ω x / } has the negative energy E = − ω , whichobvously is not consistent with supersymmetry. The reason for the trouble is thatthe function Ψ does not belong to the domain of the supercharge Q ∝ p + iW ′ , Q Ψbeing not square integrable. To make the spectrum supersymmetric, we should inthis case restrict the Hilbert space and consider only the functions that vanish at theorigin, Ψ(0) = 0. With this restriction, the ground state ˜Ψ ( x ) = x Ψ ( x ) has zeroenergy.On the other hand, for a magnetic field with noninteger flux, there is no way tomake the Pauli Hamiltonian supersymmetric. We have just shown that the Hilbertspace of regular functions on S does not constitute the domain of Q , but this istrue also for any restricted or enhanced Hilbert space. If we start, for example, fromthe space of square integrable functions, we can easily see that the action of thesupercharge on the singular square integrable functions marked with dashed lines inFigs. 1,2 would produce the functions for which the normalization integral diverges.The last comment we want to make is the following. We have seen that theHamiltonian (12) is not Hermitian in the Hilbert space of square integrable func-tions. However, the spectrum of this Hamiltonian is real and it is known that suchHamiltonians belong to the class of quasi-Hermitian (or crypto-Hermitian [9]) Hamil-tonians that can be rendered real, if redefining the inner product in a special way[10]. It is also possible to do for the Hamiltonian (12)) (and also for the more simpleHamiltonian (26)), if defining the inner product in a usual way h | i = R Ψ ∗ Ψ µ d x for nonsingular on S functions, but postulating that extra singular normalizablestates are orthogonal to nonsingular ones. No doubt, such an inner product is veryunnatural, but it can in principle be chosen and then the Hamiltonians (12) and (26)would be Hermitian. I am indebted to E. Ivanov and A. Wipf for useful discussions.9 eferences [1] M.F. Atiyah and I.M. Singer, Annals Math. (1968) 484,546; (1971)119,139.[2] see e.g. T. Eguchi, P.B. Gilkey, and A.J. Hanson, Phys. Repts. (1980) 213.[3] Y. Aharonov and A. Casher, Phys. Rev. A (1979) 2461; R. Musto, L.O’Raifertaigh, and A. Wipf, Phys. Lett. B (1986) 433.[4] M.F. Atiyah, V.K. Patodi and I.M. Singer, Math. Proc. Camb. Phil. Soc. (1975) 43; M. Ninomiya and C.I. Tan, Nucl. Phys. B245 (1985) 199; H. Roemerand P.B. Schroer, Phys. Lett.
B21 (1977) 182; P. Forgacs, L. O’Raifertaigh, andA. Wipf, Nucl. Phys.
B293 (1987) 559.[5] E.A. Ivanov and A.V. Smilga, arXiv: 1012.2069 [hep-th].[6] T.T. Wu and C.N. Yang, Nucl. Phys.
B107 (1976) 365.[7] S.Kim and C. Lee, Ann. Phys. (2002) 390 [hep-th/0112120].[8] M.A. Shifman, A.V. Smilga, and A.I. Vainshtein, Nucl. Phys.
B299 (1988) 79.[9] J. Feinberg and A. Zee, Phys.Rev.
E59
Methods of Matrix Algebra , Sect. IV.10. Academic Press, New York,1965; J.B. Bronzan, J.A. Shapiro, and R.L. Sugar, Phys. Rev.
D14 (1976) 618;D. Amati, M. Le Bellac, M. Ciafaloni, and G. Marchesini, Nucl. Phys.
B112 (1976) 107; E. Calicetti, S. Graffi, and M. Maioli, Comm. Math. Phys. (1980)51; F.G. Scholtz, H.B. Geyer, and F.J.W. Hahne, Ann. of Phys. (NY) (1992) 74; C.M. Bender and S. Boettcher, Phys. Rev. Lett. (1998) 5243;A.Mostafazadeh, J. Math. Phys. (2002) 2814, 3944; C.M. Bender, Rept.Prog. Phys.70