Unitary Spherical Super-Landau Models Andrey Beylin Department of Physics, University of Miami Based on the paper: Unitary Spherical Super-Landau Models arXiv:08064716 A.B., Thomas L. Curtright, Evgeny Ivanov, Luca Mezincescu, Paul K. Townsend

Story of the question Original Model by Landau (1930) – particle on the plane First extension by Haldane (1983) – particle on the sphere with magnetic field generated by a monopole. First supersymmetric extension on the supersphere Superflag model Restoring unitarity for planar super Landau models. Revision of the supersphere and superflag. Further generalizations? C P 1 ! ( j ) » = S U 2 S U ( 2 j 1 ) = [ £ ] 2:10:05 AM

Outline of the Talk Planar cases Supersphere Superflag Construction Fixing negative norms with “metric operator” Hidden N=2 worldline supersymmetry. Supersphere Superflag Fixing negative norms Equivalence between supersphere and superflag models Further generalization? Summary and conclusions. 2:10:05 AM

References Fuzzy $CP(n|m)$ as a quantum superspace arXiv:hep-th/0311159 A super-flag Landau model arXiv:hep-th/0404108 Planar super-Landau models arXiv:hep-th/0510019 E.Ivanov, L.Mezincescu and P.K.Townsend Planar Super-Landau Models Revisited arXiv:hep-th/0612300 T.Curtright, E.Ivanov, L.Mezincescu and P.K.Townsend Supersymmetrizing Landau models arXiv:hep-th/07052249 E.Ivanov Unitary spherical Super-Landau models arXiv:hep-th/08064716 A.B., T.Curtright, E.Ivanov, L.Mezincescu and P.K.Townsend 2:10:05 AM

Model on the superplane We define superplane as a superspace parameterized by complex coordinates . Superplane Lagrangian is Full symmetry group is infinite dimensional, while those inherited from supersphere are super-translations, super-rotations, and phase rotation . The superplane can be viewed as the coset superspace C ( 1 j ) ( z ; ³ ) 1 L = b + f ; 1 L b = j _ z 2 ¡ i N ( ¹ ) ; f ³ + ´ : 1 I U ( 1 j ) I U ( 1 j ) = [ £ Z ] 2:10:05 AM

Model on the planar superflag Ghosts in the quantum theory for the superplane suggest spontaneous breaking of symmetry. Introduce interaction with Goldstino in the Lagrangian Also we can understand this Lagrangian geometrically and write it in terms of invariant forms on the coset superspace I U ( 1 j ) » 1 L = ¡ 1 + ¹ » ¢ j _ z 2 ³ ´ i N M K = I U ( 1 j ) [ £ Z ] 2:10:05 AM

‘Metric operator’ Imagine we have a quantum system and its natural inner product is not positive definite. Let be a complete set of energy eigenvectors with Lets fix our norm with ‘metric operator’ Where j f A i 1 S i g n ( h f A j B ) = ¡ ± 1 h f A j B i ´ G 1 G j f A i ´ = ( ¡ ) g ; y 1 2:10:05 AM

‘Improved’ Hermitian conjugation With respect of new norm we will get Hermitian conjugation which is different from the original one so we get Here we introduce shift operator h f A j O B i = G ­ y ® ¡ 1 ¢ O z ´ G ¡ 1 y = + S S O ´ G ¡ 1 £ y ; ¤ 2:10:05 AM

Shift operators Lemma. Since , the Hamiltonian is Hermitian in both inner products Moreover, if the operator is a constant of motion, then the corresponding shift operator is also a constant of motion. This is the signal that the system may have some `hidden' symmetries. [ G ; H ] = 1 H 1 H = y z . 1 O 1 2:10:05 AM

Superplane Quantum problem for the particle on the superplane is easily factorizable. All states except LLL have a negative norm. It is possible to find ‘metric operator’ Using Lemma we find that shift operators associated with odd generators are new integrals of motion. They satisfy which is just worldline supersymmetry. Infinitesimal transformations generated by are atypical G 1 f S ; z g = 2 N H 1 N = 2 1 ² S + ¹ z 1 ± z = ² _ ³ ; ¡ ¹ 1 2:10:05 AM

Planar superflag(1) Difference form the superplane case is that we have an additional parameter . It is possible to solve this model with factorization method as well, but unitarity properties will depend on . Consider different cases: will return us on the supersphere, after elimination of , which become auxiliary. For situation is similar to the superplane – we can obtain ‘metric operator’ , fix the norm and recover worldline supersymmetry. But there are no supersymmetric ground states, so worldline supersymmetry is spontaneously broken for . M 1 M 1 M = 1 » 1 M < 1 G 1 N = 2 1 M < 1 2:10:05 AM

Planar superflag(2) In case we have states with negative norm only for . All the lowest levels have positive norm. This implies that ‘metric operator’ have to be level dependent. While it is possible to find so all the states have positive norm, we will not have worldline supersymmetry in this case. For this case anticommutator of shift operators will be M > 1 l > 2 M 1 ( l < 2 M ) 1 G 1 f S ; z g = 2 N j H s u y ( M > ) 1 2:10:05 AM

Supersphere (1) Riemann supersphere is a complex Kähler supermanifold. Its geometry is determined by Kähler potential, metric, connections The classical Lagrangian of the superspherical Landau model is C P ( 1 j ) » = S U 2 K = l o g ¡ 1 + z ¹ ³ ¢ ; B A @ i d Z ´ ( ) L = _ Z A ¹ B g + N ³ ´ 1 2:10:05 AM

Supersphere (2) After quantization we get this form of the Hamiltonian The energy levels of the Landau model of a supersphere could be found exactly using a factorization method. Energy eigenvalues are Wave functions H = ¡ ( 1 ) a + b g A ¹ B r N ; @ K E ` = ( + 2 N ) 1 ª ( N ) ` = r + 1 A ¢ 2 ¡ © : ; ¹ B 2:10:05 AM

Supersphere (3) The Hilbert space has a natural -invariant norm Norm of the eigenvectors on the component level S U ( 2 j 1 ) j ª 2 = R d ¹ e ¡ K ¤ ; z @ ³ : 1 j ª ( N ) ` 2 = C Z d z ¹ 1 + · ¡ A Ã Â ¢ F ¸ : 2:10:05 AM

Superflag(1) is a complex Kähler supermanifold, which we call superflag, with coordinates In contrast with the supersphere, Lagrangian for the Landau model on the superflag built with the invariant form, not the Kähler potential of the superflag. We have The superflag Lagrangian now will be S U ( 2 j 1 ) = [ £ ] Z M = ( z ; ³ » ) ¹ 1 ! + = K ¡ 1 2 n _ z £ » ¹ ³ ¤ o ; ¢ ( ) : L = j ! + 2 h _ Z M ( N A B ) c : i 1 2:10:05 AM

Superflag(2) Resulting Hamiltonian for the quantum theory For an integer the Hamiltonian may be diagonalized with energy levels and eigenfunctions given by H N = ¡ K 2 1 ³ r ( ) z » ´ ¹ ; A @ i : $2N$ E N = ` ( 2 + 1 ) ; : © ( ` ) = D 2 N + 1 ¢ a n z ; ³ » > ´ r ¡ : 2:10:05 AM

Superflag(3) Invariant inner product on the superflag is given by Final result for Hamiltonian eigenfunctions in component fields after all integrations and removing derivatives h ¨ j ª i = R d z ¹ @ ³ » K ¡ 2 ¤ : 1 j ª ( ` ) N 2 = C Z d z ¹ 1 + £ ½ M ¡ Ã A F Â ¾ : 2:10:05 AM

Unitary norm As for planar cases we want to find ‘metric operator’, but we have to restrict ourselves with . We have two cases Here are generators of the . M < 1 2 M < ¡ N 1 G a n = 1 2 N + ` h J ( F ¡ M ) i : ¡ 2 N 1 < M . ~ G a n = 1 ¡ 8 ( F 2 M N ) + : ( F ; J 3 § ¦ Q ) 1 S U ( 2 j 1 ) 2:10:05 AM

Hidden symmetries There are hidden worldline supersymmetry for planar superflag at . For superflag model in the case we does not have worldline supersymmetry. But if Lagrangian possess ‘enlarged’ symmetry group This symmetry algebra is a subalgebra of the enveloping algebra of . For LLL is an exception, it doesn’t have symmetry. M · 1 ¡ 2 N 1 < M · M 6 = 1 f J § ; 3 Z ~ ¦ Q · G g 1 s u ( 2 j ) 1 s u ( 2 j 1 ) M = 1 S U ( 2 j ) 1 2:10:05 AM

Generalized superplane Superplane model have atypical transformation with respect to hidden worldline supersymmetry. Consider two chiral superfields and in . Superplane Landau model can be recovered from Our idea is to generalize this Lagrangian We can solve it for N = 2 1 © 1 ª 1 N = 2 ; d 1 S = ¡ R d t 2 µ © ¹ + ª ½ £ D ¤ 1 S = ¡ R d t 2 µ © K ( ; ¹ ) + V ª ½ £ D ¤ : 1 K ( © ; ¹ ) = l o g ¡ 1 + ¢ V 2 : 2:10:05 AM

Summary and conclusion Superflag Landau model with charges is quantum equivalent to the superspherical Landau model with charge . All superflag Landau models admit positive norm. In general, this norm have to be a ‘dynamical’ combination of the original and ‘alternative’ norm involving a ‘metric operator’. Redefinition of Hermitian conjugation give rise to `shift‘ operators which turn to be new ‘hidden’ symmetry generators. This generators form a closed subset for 2 N = ¡ 1 ; M 2 N 1 ¡ 2 N < M 1 2:10:05 AM