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Added a sample output for the small tJ ham with 4 sites and 1 hole.
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examples/tJ_4.out
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examples/tJ_4.out
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1-D t-J Eigenproblem, n=12
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start: 0 end: 12
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HAMILTONIEN t-J
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Le nombre de trou est : 1
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Famille 1 : F
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LECTURE DES ATOMES, DES LIAISONS, DES INTEGRALES
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================ CLUSTER 1 ==================
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Liaisons entre les atomes 3
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Les atomes 1 et 2 forment la liaison 1 qui est de type 1
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Les atomes 2 et 3 forment la liaison 2 qui est de type 1
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Les atomes 3 et 4 forment la liaison 3 qui est de type 1
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=============================================
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Le nombre total d atomes est 4
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=============================================
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Nombre de J differents 3
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type de liaison 1 1
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type de liaison 2 1
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type de liaison 3 1
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type de J 1.0000000000000000
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type de J -1.0000000000000000
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type de J 0.0000000000000000
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Parametres : Jz 1 = 1.0000000000000000
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Parametres : Jxy 1 = 1.0000000000000000
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Parametre : t 1 = -1.0000000000000000
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Parametres : Jz 2 = 1.0000000000000000
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Parametres : Jxy 2 = 1.0000000000000000
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Parametre : t 2 = -1.0000000000000000
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Parametres : Jz 3 = 1.0000000000000000
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Parametres : Jxy 3 = 1.0000000000000000
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Parametre : t 3 = -1.0000000000000000
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coucoudslect3
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coucou
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Parametres pour le t-J
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xj1 = 1.0000000000000000E-002
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xj2 = 0.0000000000000000
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xt1 = -0.20000000000000001
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xt2 = 0.0000000000000000
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xv1 = 0.0000000000000000
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xv2 = 0.0000000000000000
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xv3 = 0.0000000000000000
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xbj = 0.0000000000000000
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xbt = 0.0000000000000000
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xeneparJ = 0.0000000000000000
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xeneperpJ = 0.0000000000000000
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xeneparT = 0.0000000000000000
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xeneperpT = 0.0000000000000000
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xenediagJ = 0.0000000000000000
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xenediagT = 0.0000000000000000
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xspar = -0.0000000000000000
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xsperp = -0.0000000000000000
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Le systeme comporte 0 plaquettes.
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Spin total 0
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Nombre de vecteurs demande 8
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Nombre maximal d iterations de Davidson 280
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Vecteur calcule le plus bas 1
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Variable Nes4 (vecteurs d essai) 0
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Nombre de determinants en donnees 100
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Variable Ysuiv (suivre le vecteur initial) F
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Seuil au dela duquel seront ecrits les vecteurs 9.9999999392252903E-009
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Option d ecriture des determinants sur FIL2 T
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=======nombre de centres de spin alpha===== 2
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nt2= 3
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nt1 4
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nt2= 3 nbeta= 1
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Time used to build the matrix: 0.007272
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time = 0.007272 mpiid = 0
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Time used to assemble the matrix: 0.000024
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Time used: 0.000336
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Number of iterations of the method: 1
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Solution method: krylovschur
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Number of converged eigenvalues: 2
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Stopping condition: tol=1e-09, maxit=10000000
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k ||Ax-kx||/||kx|| <S>
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----------------- ----------------- ------------------
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-3.855773 6.4802e-16 0.500000 0.000000 0.000000 1.000000
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-3.414214 1.59311e-15 0.500000 0.000000 0.000000 1.000000
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