Lnear sstem of the Schrödnger equaton Notes on Quantum Mechancs htt://quantum.bu.edu/notes/quantummechancs/lnearsstems.df Last udated Wednesda, October 9, 003 :0:08 Corght 003 Dan Dll (dan@bu.edu) Deartment of Chemstr, Boston Unverst, Boston MA 05 In numercal alcatons of quantum mechancs n chemstr, a owerful method of solvng the Schrödnger equaton, H Y a = E a Y a s to exress the unnown wavefuncton, Y a, n terms of nown bass functons,, as Y a = c a. The result s a lnear sstem of equatons for the exanson coeffcents, c a. Here s what the lnear sstem loos le for an exanson n terms of three bass functons. where H - E a H H 3 H H - E a H 3 H 3 H 3 H 33 - E a z c a c a c 3 a z = 0 H = HxL * H HxL x. For ths examle, we have three equatons n four unnowns: the three exanson coeffcents, c a, and the energ, E a. The wa to roceed s to frst determne the values of the energ b requrng that the matrx H - E a H H 3 H H - E a H 3 H 3 H 3 H 33 - E a does not have an nverse, for f t dd have an nverse, then the soluton would be H - E a H H 3 H H - E a H 3 H 3 H 3 H 33 - E a z z - H - E a H H 3 H H - E a H 3 H 3 H 3 H 33 - E a z c a c a c 3 a z = c a c a c 3 a z = 0, that s, all of the exanson coeffcents would be equal to 0 and so the wavefuncton, Y a, would vansh!
Lnear sstem of the Schrödnger equaton Now, the nverse of a matrx s roortonal to the recrocal of ts determnant, and so we can ensure that a matrx does not have an nverse b arrangng for ts determnant to be equal to 0. We do ths b fndng those values of E a at whch the determnant vanshes. For an n µ n matrx, the determnant s an n-th order olnomal n E Ça and so there wll be as man values of the energ at whch the determnant vanshes as there are bass functons. Once the values of the energ, E a, have been determned, we are left wth a homogeneous sstem of equatons for the unnown exanson coeffcents, c a. The wa to solve ths sstem of equatons s to set one of the exanson coeffcents equal to, set asde the corresondng equaton of the lnear sstem, and then solve the remanng set of equatons. For the 3 µ 3 examle, settng c a =, the remanng equatons to solve are H - E a H 3 z c a z =- H z. H 3 H 33 - E a c 3 a H 3 We can solve ths to for the remanng coeffcents, c a and c 3 a. The result s Y a = N H + c a + 3 a c 3 a L where the normalzaton constant, N, corrects for the fact that we have set c a =. The last ste s to determne the value of N usng» Y a» x = N À + c a + 3 a c 3 a À x = N H + c a + c 3 a L, where n the second equalt we have assumed the orthonormalt of the bass functons,. The fnal result s that Y a = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ è!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ H + c a + 3 a c 3 a L + c a + c 3 a s the normalzed wavefuncton corresondng to the energ E a. à Partcle on a bum rng To see how ths method of solvng the Schrödnger equaton wors, let's consder a artcle movng on a bum rng. A artcle on a smooth rng has onl netc energ, snce the otental energ s zero on the rng. We can mae the rng "bum" b addng the otental energ V HfL = Ñ ÅÅÅÅÅÅÅÅ I sn HfL. Here s a lot of ths otental energ, n unts Ñ ê H IL. Corght 003 Dan Dll (dan@bu.edu). All rghts reserved
Lnear sstem of the Schrödnger equaton 3 VHfL 0.8 0.6 0. 0. ÅÅÅÅÅ 3 ÅÅÅÅÅÅÅÅÅ f Potental energ, n unts Ñ ê H IL, of a bum rng. The otental energ has mnma along the x drectons and maxma along the drectons. To construct the lnear sstem of the Schrödnger equaton for a artcle movng on a rng n ths otental energ, we need the ntegrals V mm' = F m HfL * ÅÅÅÅÅÅÅÅ Ñ 0 I sn HfL z F m' HfL x. The value of such ntegrals, n unts Ñ ê H IL, s / for m = m', - ê for m = m', and 0 otherwse. Ths means that the ntegrals of the hamltonan, H mm' = Ñ ÅÅÅÅÅÅÅÅ I ÅÅÅÅÅÅÅÅÅÅÅÅ f + sn HfL z - of a artcle on ths bum rng for bass functons wth m = 0, and, are, n unts Ñ ê H IL, m' = 0 m' = m' = m = 0 0 m = 9 0 m = 0 0 3 Use the values of the ntegrals V mm' to show these values are correct. If we exress the wavefuncton of the artcle on the bum rng as the exanson Y a HfL =F 0 HfL c 0 a +F HfL c a +F HfL c a, then the normalzed wavefunctons, Y a HfL, and corresondng energes, E a, n unts Ñ ê H IL, are α Ψ α E α 0.3989. φ.5 0.3987 + 0.07893. φ 0.836 3 0.07893 + 0.3987. φ.5556 Normalzed wavefunctons, Y a HfL, and corresondng energes, E a, n unts Ñ ê H IL, for a artcle n a rng n a otental energ Ñ ê H IL sn HfL, usng a exanson n rng bass functons, F m HfL, for m = 0, and. Corght 003 Dan Dll (dan@bu.edu). All rghts reserved
Lnear sstem of the Schrödnger equaton Show that these results are correct. Here s the comoston of each wavefuncton, Y a HfL, n terms of the ercentage contrbuton of the three rng bass functons, F 0 HfL, F HfL and F HfL. Φ 0 Φ Φ Ψ 0. 0. 00. Ψ 99.6 0.386 0. Ψ 3 0.386 99.6 0. Percentage comoston of each bum rng wavefuncton, Y a HfL, n terms of the fve rng bass states, F m HfL, for m = 0, and. Show that these results are correct. Fnall, here are the robablt denstes corresondng to the three wavefunctons, Y HfL, Y HfL and Y 3 HfL.»Y» 0.7 0.6 0.5 ÅÅÅÅÅ 3 ÅÅÅÅÅÅÅÅÅ f Probablt denstes corresondng to the three wavefunctons Y HfL (horzontal lne), Y HfL (oscllatng thn gre lne) and Y 3 HfL (oscllatng thc gre lne). Show that these results are correct. The horzontal lne s for energ E =.5 Ñ ê H IL; ths energ s ust the sum of the netc energ and the otental energ exectaton values for m =. The oscllatng thn gre lne s for energ E = 0.8 Ñ ê H IL. Ths energ s slghtl less than the sum of the netc energ and the otental energ exectaton values for m = 0. The decrease s due to the robablt denst beng slghtl larger n the otental valles (along x) and slghtl smaller at the otental eas (along ). The oscllatng thc gre lne s for energ E =.56 Ñ ê H IL; ths energ s slghtl more than the sum of the netc energ and the otental energ exectaton values for m =. The ncrease s due to the robablt denst beng slghtl larger at the otental eas (along ) and slghtl smaller n the otental valles (along x). à Imrovng the energes and egenvalues In ths method of solvng the Schrödnger equaton, as addtonal bass functons are added, the more accurate the energes and wavefunctons become. To llustrate ths, we can resolve the artcle on a bum rng usng a fve-member bass consstng of the rng functons F m HfL wth m = 0,,, 3 and. Corght 003 Dan Dll (dan@bu.edu). All rghts reserved
Lnear sstem of the Schrödnger equaton 5 The ntegrals of the hamltonan, H mm' = Ñ ÅÅÅÅÅÅÅÅ I ÅÅÅÅÅÅÅÅÅÅÅÅ f + sn HfL z - of a artcle on ths bum rng for bass functons wth m = 0,,, 3 and, are, n unts Ñ ê H IL, m' = 0 m' = m' = m' = m' = 3 m = 0 0 0 0 m = 9 0 0 m = 0 33 0 0 m = 0 0 0 3 m = 3 0 0 0 9 The normalzed wavefunctons, Y a HfL, and corresondng energes, E a, n unts Ñ êh IL, are α Ψ α E α 0.39878. φ + 0.087 3. φ.9 0.087. φ + 0.39878 3. φ 9.5078 3 0.000978 0.00830859. φ + 0.398856. φ 6.505 0.0859 + 0.398083. φ + 0.00830057. φ.5037 5 0.3987 + 0.083. φ + 0.00038739. φ 0.8 Normalzed wavefunctons, Y a HfL, and corresondng energes, E a, n unts Ñ ê H IL, for a artcle n a rng n a otental energ Ñ ê H IL sn HfL, usng a exanson n rng bass functons, F m HfL, for m = 0,,, 3 and. Here s the comoston of each wavefuncton, Y a HfL, n terms of the ercentage contrbuton of the fve rng bass functons. Φ 0 Φ Φ Φ Φ 3 Ψ 0 0 0 99.9 0.09737 Ψ 0 0 0 0.09737 99.9 Ψ 3 0.0000058 0.0337 99.96 0 0 Ψ 0.3869 99.57 0.039 0 0 Ψ 5 99.6 0.3869 0.000096 0 0 Percentage comoston of each bum rng wavefuncton, Y a HfL, n terms of the fve rng bass states, F m HfL, for m = 0,,, 3 and. Fnall, here are the robablt denstes corresondng to the fve wavefunctons, Y a HfL. 0.8 0.7 0.6 0.5 ÅÅÅÅÅ 3 ÅÅÅÅÅÅÅÅÅ Probablt denstes corresondng to the fve wavefunctons Y a HfL. At f =ê, the orderng of the curves and redomnant rng bass functon contrbuton, from bottom to to, are Ha, ml = H5, 0L (blac curve), Ha, ml = H, L (red curve), Ha, ml = H3, L (green curve), Ha, ml = H, 3L (ellow curve), and Ha, ml = H, L (blue curve). Corght 003 Dan Dll (dan@bu.edu). All rghts reserved
6 Lnear sstem of the Schrödnger equaton The frst thng to note s that, as n the three-member bass, the denstes (a = and 5) that are reduced at the otental bums and enhanced at the otental valles corresond to lower energes than the sum of the netc energ and the otental energ exectaton values for domnate value of m for the gven Y a, and the denstes (a =, 3 and ) that are enhanced at the otental bums and reduced at the otental valles corresond to hgher energes than the sum of the netc energ and the otental energ exectaton values for domnate value of m for the gven Y a. The second thng to note s that whereas n the three-member bass the denst of the level wth m = remaned constant and ts energ dd not change from the sum of the netc energ and the otental energ exectaton values for m =, now the denst s shfted awa from the bums nto the valles, and the energ s corresondngl lowered. Ths s balanced b a shft of the redomnantl m = 3 denst awa from the valles and onto the bums, and t corresondng ncreased energ, relatve to the sum of the netc energ and the otental energ exectaton values for m = 3. à Comarng the three- and fve-member bass results It s ver nstructve to comare the energ changes that result usng the three-member bass wth those that result usng the fve member bass. Here s a tabulaton of the three-member bass results, m E m E m +V m E α E α HE m V m L 0 0 0.5 0.8 0.056.5.5 0.056 6.5.5 0. 3 9 Rng functon netc energes, E m, rng functon netc energes lus otental bum otental energes, V m, energes E a, and the change n energ, E a - HE m - V m L, for a artcle on a rng n a otental energ sn HfL, usng a exanson n rng bass functons, F m HfL, for m = 0, and. All energes are n unts Ñ ê H IL. and here s a tabulaton of the fve-member bass results, m E m E m +V m E α E α HE m V m L 0 0 0.5 0.8 0.056.5.5 0.00 6 6.5 6.5 0.005.5.9 0.0078 3 9 9.5 9.5 0.0078 Rng functon netc energes, E m, rng functon netc energes lus otental bum otental energes, V m, energes E a, and the change n energ, E a - HE m - V m L, for a artcle on a rng n a otental energ sn HfL, usng a exanson n rng bass functons, F m HfL, for m = 0,,, 3 and. All energes are n unts Ñ ê H IL. The e thng to note n these results and ther comarson s that the energes, E a, for the fve-member bass are lower than those for the three member bass. Ths s an examle of a general result nown as the searaton theorem. The theorem has two arts: Frst, unless we use all of the bass functons of a comlete set, the resultng energes, E a, wll be hgher than the exact energes; second, the more bass functons we nclude, the closer wll be the energes, E a, to the exact energes, that s, enlargng a lnear sstem alwas lowers energes, E a. Ths roert of lnear sstems of the Schrödnger equaton means that we can get good aroxmatons to exact energes of at least the lowest several levels wthout havng to usng a huge number of bass functons. The reason s that addng more and more bass functons results n smaller and smaller mrovement to the energes of the lowest levels. It s for ths reason that ths method of worng wth the Schrödnger equaton s so owerful and so wdesread n alcatons of quantum mechancs to chemcal sstems. Corght 003 Dan Dll (dan@bu.edu). All rghts reserved