Issues in Strongly Correlated Electron Physics: A DMFT ...

Issues in Strongly Correlated Electron Physics: A DMFT ...

Quasi-exactly solvable models in quantum mechanics and Lie algebras S. N. Dolya B. Verkin Institute for Low Temperature Physics and Engineering of the National Academy of Sciences of Ukraine S. N. Dolya JMP, 50 (2009) S. N. Dolya JMP, 49 (2008). S. N. Dolya O. B. Zaslavskii J. Phys. A: Math. Gen. 34 (2001) S. N. Dolya O. B. Zaslavskii J. Phys. A: Math. Gen. 34 (2001) S. N. Dolya O. B. Zaslavskii J. Phys. A: Math. Gen. 33 (2000) Outline 1. QES-extension (A) 2. quadratic QES - Lie algebras 3. physical applications 4. QES-extension (B) 5. cubic QES - Lie algebras sl2(R)Hamiltonians Representatio n: n J 0 x x 2 J x J x 2 x nx Invariant subspace (partial algebraization) Turbiner et al What is being studied? Hamiltonians are formulated in terms of QES Lie algebras. eigenvalues and eigenfunctions when possible. How this is being studied? Invariant subspaces: Vn = span{ f1 ( x), f 2 ( x),..., f n 1 ( x), f n ( x)} Nonlinear QES Lie algebras X

ij s , X c X X c p q pq i j pq X s c pq QESextension: 0. our strategy 1) We find a general form of the operator of the second order P2 for which subspace M2 = span{f1, f2} is preserved. 2) We make extension of the subspace M2 M4 = span{f1, f2, f3, f4} 3) We find a general form of the operator of the second order P4 for which subspace M4 is preserved. 4) we obtain the explicit form of operator P2(N+1) that acts on the elements of the subspace M2(N+1) = {f1,f2,, f2(N+1)} QES-extension: I. M 2 ={ f , f } 0 Select the minimal invariant subspace 0 P2 f 0 = c1 f 0 c2 f 0 Condition for the

subspace M2 P2 f 0 = c3 f 0 c4 f 0 d2 d q ( x) 2 f ( x) p( x) f ( x) r ( x) f ( x) = 0 dx dx f ( x) = f ( x) 0 ; d f ( x) = f ( x) = f ' ( x) dx 0 2 Select the invariant operator d d P2 = a3 ( x) 2 a2 ( x) a0 ( x) dx dx extension for the minimal invariant subspace II. QESextension: M 2 ={ f , f } 0 0 M 4 ={ f , f , f , f } 0 1

0 1 Condition for the subspace M4 P4 f = c01 f c02 f c03 f c04 f ...................................................................... ...................................................................... 0 0 0 1 1 P4 f1 = c11 f 0 c12 f 0 c13 f1 c14 f1 QESextension: Extension for the minimal invariant subspace III. M 2 M 4 ... M 2(N+1) { f 0 , f 0 } { f 0 , f1 , f 0 , f1 } ... { f 0 , f1 ,... f N , f 0 , f1 ,..., f N } Conditions of the QESextension: W f , f , f , f 0 Wronskian matrix 1 0 1

0 1 2 order ( P2 ) order( P4 ) ... order( P2 N 1 ) 2. Order of derivatives hypergeometric function Realization (special functions: hypergeometric, Airy, Bessel ones) d2 d x 2 s dx dx _ f ( x) = 0 F1 ; x s _ f 0 ( x) = 0 F1 ; x s 1 0 1 f 0 ( x ) = 0 M 2 ={ f , f } 0 0 QES-extension: act more M 2 M 4 ... M 2(N+1)

{ f 0 , f 0 } { f 0 , f1 , f 0 , f1 } ... { f 0 , f1 ,... f N , f 0 , f1 ,..., f N } Particular choice of QES extension f n = x f n 1 = x n f 0 , f n = x f n 1 = x n f 0 , 1 N R = span{ f , f ,... f , f , f ,..., f } QES-extension: Example 1 _ f = x 0F1 ; x s 1 _ f = x 0F1 ; x s n n d2 d J = x 2 s 1 , dx dx 1 f J f 1 n n f n

J1 fn An = n 2 N 1 dim RN1 = 2 N 1 N = 0,1, 2,... n 0 n N 0 1 n = 0,1,.., N 1, N counter d2 d J =x s 2 N x x 2 dx dx 1 2 2 Bn n An 1 s f n f n 1 =

s n s An 1 f n s 2 Bn 1 f n 1 2n f n n n s f n 1 f n = s f n n n s f n 1 2ns f n 1 Bn = n N N QES-extension: The commutation relations of the operators J1 , J1 = S1 J1 , S1 = 4 J1 J1 2 S1 2 J1 c6 J1 c7 J , S1 = 2 J 1 Casimir operator: 1 2 2J 1 2 FCasimir S1 4 J1 J1 J1 4S1 J1 4 J1 J1 2 N s 4 2 N s 2 J1 J1 2 s 2 s 2 N 6 2 sN J1 s 2 1

c6 2 N s 2 s 2 N c7 s 1 2 2 N s Casimir invariant 2 N R = span{ f , f ,... f , f , f ,..., f } QES-extension: Example 2 N = 0,1, 2,... f = t 1F1 ; t s n n 0 1 dim RN2 = 2 N 1 1 f = t 1F1 ;t s 1 n n 2 d d J2 = t 2 1 s t , dt dt N 0

1 N n = 0,1,.., N 1, N counter 2 d d 2 J2 = t s 2 N t t t N 2 dt dt n A 1 s f B f 2 B / s f f n n

n n 1 n n 1 n J 2 = s n 1 A f B f s 2 B 1 f n n n n 1 n n fn

n f n n s f 1 2 n / s f f n n 1 n n J 2 = n 1 f n n s f

2 ns f f n n 1 n 1 n An = n 2 N Bn = n N QES-extension: The commutation relations of the operators J 2 , J 2 = S 2 J 2 , S2 = 4 J 2 J 2 2 S 2 c5 J 2 c6 J 2 c7 J , S2 = 2 J 2 2 2 J 2 c5 J 2 N s 1 2 FCasimir S 2 4 J 2 J 2 J 2 4S 2 J 2 2c5 J 2 J 2 Casimir J 2 operator: 2 2 N s 4 2 N s 2 J 2 J 2

c5 S 2 2 2 s s 1 N 1 J 2 2 c7 2c5 J 2 s 1 N 3N s N 6 c7 N 2 2 N s s 1 c6 2 N s s 2 2 N c5 2 1 s Casimir invariant 3 N R = span{ f 0 , f1 ,... f N } QES-extension: Example 3 N = 0,1, 2,... dim RN3 = N 1 n f n = 1F1 ;t s n = 0,1,.., N 1, N counter 2 2 d d d d 2 J3 = t 2 s t , J3 = t s N t t t 2 dt dt dt dt J 3 f n = n f n J 3 f n

= s n Cn f n n Bn f n 1 n n s f n 1 n = n Cn = n N 2n Bn = n N QES-extension: The commutation relations of the operators J 3 , J 3 = S3 J 3 , S3 = 4 J 3 J 3 2 S3 c5 J 3 c6 J 3 c7 J , S3 = 2 J 3 c5 s 2 N c6 N s N 2 s c7 s N 2 s 3 2 J 3 c5 J 3 s Two-photon Rabi Hamiltonian bi Hamiltonian describes a two-level system (atom) coupled to a single ode of radiation via dipole interaction. 0 2 2 H R = z a a g a a 2 Two-photon Rabi Hamiltonian 0 2 2 H R = z a a g a a 2

1 0 0 i 0 1 x = , y = , z = ; 0 1 i 0 1 0 L 0 0 4 g E 0 , g , E . 2 0 1 1 = E , L 2 2 L a a g/2 a 2 a2 The two-photon Rabi Hamiltonian L 0 1 1 = E , 2 0 L 2

L11 = 0 L1 = L L E L L E 2 02 x x x x a= ,a = 2 2 The two-photon Rabi Hamiltonian e cx 2 L1e c x 2 = 4 g J 2 2 2 4 g S 2 2 g 2 J 2 4 J 2 02 a0 = 3/4, s = 1/2 (the parameters of subspace 2N ) c= 1 1 g (the parameter of gauge transformation) 2 1 g = E= g 1 g 2 (the parameter of change of variable) 1 2 N 1 1 g 2 (the energy of the Hamiltonian). 2

a0 = 3 4 N 4 N g 2 1 3 5 g 2 /4 t = x 2 The two-photon Rabi Hamiltonian e cx 2 L1e cx2 = = 4 J 3 2 4 g S3 2 g 2 J 3 4 J 3 02 a0 3 N , s = 1/2 (the parameters of subspaces 3N ) 4 2 c= 1 1 g (the parameter of gauge transformation) 2 1 g = E= g 1 g 2 (the parameter of change of variable) 1 N 1 1 g 2 (the energy of the Hamiltonian). 2

a0 = 1 2 N 2 N g 2 1 1 3 g 2 /4 t = x2 Example 2 0 matrix representation 10 0 1 3 6 J , J2 2 1 1 4 0 1 2 3g 9/4 02 3g 2 g 1 L1 2 2 2 g 1 1/4 3 g 4 g 0 condition det(L1) = 0

g 0 = 2 2 4 9 1 4 0 0 1 3 , < 0 < . 8 0 2 2 QES-extension: continuation Example 4 (QES qubic Lie algebra ) VN2 = span{ f 0 , f1 ,... f N , f 0 , f1 ,..., f N } dim VN2 = 2 N 1 n = 0,1,.., N 1, N N = 0,1, 2,... f = 1F1 ; x 1F1 ; x 1 n 1 n n 3 2 d d d 2 2 d

J6 = x , J6 = x x N 2 2 x x 3 dx dx dx dx J 6 f n = n f n 1 f n J 6 f n = n 1 Bn f n f n1 n n Bn f n f n 1 n = n; Bn = n N QES-extension: continuation Example 4 (QES qubic Lie algebra ) The commutation relations of the operators J 6 , J 6 = S6 J , S6 = 8 J 6 6 3 6 N J 6 2 c6 J 6 J 6 , S6 = J 6 2 Casimir operator: FCasimir S6 4 J

2 6 J 4 N J c 4 J 2 N J 0 6 c 2 N 2 N 6 4 6 6 2 6 6 Casimir invariant 3 QES-extension: continuation 1) Select the minimal invariant subspace: 2 2 M 3 ={ f ( x ) , f ( x ) f ( x ), f ( x ) } 2) Select the minimal invariant subspace: M 4 ={ f ( x ) g ( x ), f ( x ) g ( x ), f ( x ) g ( x ), f ( x ) g ( x )} Condition for the functions f(x), g(x) L f ( x) L g ( x) 0

2 d d L q( x) 2 p( x) r( x) dx dx QES-extension: continuation Example 5 ( QES Lie algebra 3 N V ) dim VN3 = 3 N 1 N = 0,1, 2,... n = 0,1,.., N 1, N VN3 = span{ f 0 , f1 ,... f N , f 0 , f1 ,..., f N , f 0* , f1* ,..., f N* } 2 f = x f ( x) x 1F1 ; x 1/ 2 1 n n f n = x f ( x) f ( x) x 1F1 ; x 1F1 ; x 1/ 2 3/ 2 n * n f =x n

n 1 2 n 2 f ( x) x n 1 1 1F1 ; x 3/ 2 2 QES-extension: continuation Example 5 ( QES Lie algebra 3 N V ) 3 2 d x d 9N d J 10 = x 2 3 6 N 1 6 x 2 3 x 1 2 x a0 x dx 2 dx 2 dx 4 2 d 3 15 d d 2 d J 10 = x

x 5 2 x 3 x 14 x 2 a1 x a0 x 4 dx dx 4 dx dx 27 6 N 4 , a0 x 2 x 2 N , a1 x 9 N 6 8 x 10 xN 2 x 2 ; 2 J10 n n = C1 n C2 n1 C3 n 1 C4 n 2 C5 n 2 2n 1 4 fn f n , C5 0, C4 0, C4 n 1 * fn 0 n n n 1 2n 1 0 2n 3 4 0 4 ; 2n 1 4 0 QES-extension: continuation Example 6 ( QES Lie algebra

4 dim V N = 4 N 1 4 VN ) N = 0,1, 2,... VN4 = span{ f 0 , f1 ,... f N , f 0 , f1 ,..., f N , n = 0,1,.., N 1, N * 0 * 1 * N , f , f ,..., f , f , f ,..., f } 0 1 N f = x f ( x) g ( x) x 1F1 ; x 1 F1 ; x ; 1/ 2 1/ 2 1 1 n n 1 f n = x x f ( x) g ( x) x 1F1 ; x 1F1 ; x ; 3/ 2 3/ 2 n

n n 1 f = x f ( x) g ( x) x 1F1 ; x 1F1 ; x ; 1/ 2 3/ 2 1 n n f n = x f ( x) g ( x) x 1F1 ; x 1F1 ; x 3/ 2 1/ 2 * n n n QES-extension: continuation Example 6 ( QES Lie algebra 4 N V ) 3 2 d d 9 9 2 2 d J12 = x

x 1 3N 2 N 4 x x xN 3 dx dx 4 2 dx 4 3 2 d d 15 d d 2 2 J12 = x 5 x 3 4 x x 2 2 3 2 x xN 4 dx dx 4 dx dx J12 n n = C1 n C2 n 1 C3 n 1 C4 n 2 C5 n 2 2n 1 4 f n 0 fn , C 0, C 0,

C n n 1 2 n 1 5 4 4 f n* 1 fn 1 0 0 0 2n 1 4 0 0 4 2n 3 4 0 4 0 2n 4

; 3 comparison Angular Momentum L r p p i QES quadratic Lie algebra RN3 J 3 , J 3 , S3 L L FCasimir [ Li , L j ] i ijk Lk X p , X q c ijpq X i X j ... l 0,1,2,.... N = 0,1, 2,.... ml 0, 1, 2,......, l N 0,1, 2,..., dim RN3 = N 1

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