Problem A (variational calculation)Solution program (instructor's Fortran version) [var.f90] First, let's look at how the energies of the first four states converge as a function of the number of basis states, N=Nx*Ny, first using Nx=Ny: Nx Ny Nx*Ny E0 E1 E2 E3 ------------------------------------------------------------------------ 5 5 25 0.01455951 0.02205329 0.04358191 0.06356153 10 10 100 0.01420417 0.02106122 0.04320811 0.06354212 15 15 225 0.01396114 0.02087201 0.04298447 0.06353202 20 20 400 0.01392182 0.02071306 0.04294863 0.06352973 25 25 625 0.01389049 0.02068858 0.04291843 0.06352809 30 30 900 0.01388495 0.02067539 0.04291384 0.06352769 35 35 1225 0.01388330 0.02067261 0.04291210 0.06352759 40 40 1600 0.01388304 0.02067227 0.04291190 0.06352755 45 45 2025 0.01388291 0.02067210 0.04291175 0.06352754 50 50 2500 0.01388262 0.02067129 0.04291153 0.06352751 55 55 3025 0.01388170 0.02067057 0.04291075 0.06352745 60 60 3600 0.01388129 0.02066912 0.04291042 0.06352741 Since the system is rectangular, and longer in the y-direction, one might expect the convergence to be faster for rectangular lattices with Ny > Nx. Let's look at results for Ny=2*Nx: Nx Ny Nx*Ny E0 E1 E2 E3 ------------------------------------------------------------------------ 5 10 50 0.01422518 0.02110244 0.04322777 0.06355619 10 20 200 0.01392777 0.02072316 0.04295374 0.06353090 15 30 450 0.01388708 0.02067886 0.04291568 0.06352840 20 40 800 0.01388417 0.02067411 0.04291288 0.06352777 25 50 1200 0.01388315 0.02067215 0.04291198 0.06352767 30 60 1800 0.01388166 0.02066972 0.04291074 0.06352749 35 70 2400 0.01388077 0.02066861 0.04290998 0.06352742 40 80 3200 0.01388065 0.02066840 0.04290986 0.06352739 Indeed, for large N the energy here is lower (compare, e.g., the results for 45*45 and 30*60). For completeness, let's also look at the case Ny=Nx/2: Nx Ny Nx*Ny E0 E1 E2 E3 ------------------------------------------------------------------------ 10 5 50 0.01454808 0.02203783 0.04357100 0.06355273 20 10 200 0.01420087 0.02105476 0.04320518 0.06354146 30 15 450 0.01395945 0.02086954 0.04298302 0.06353145 40 20 800 0.01392084 0.02071138 0.04294779 0.06352954 50 25 1200 0.01388998 0.02068776 0.04291799 0.06352793 60 30 1800 0.01388459 0.02067480 0.04291353 0.06352762 70 35 2400 0.01388310 0.02067227 0.04291193 0.06352753 80 40 3200 0.01388288 0.02067201 0.04291177 0.06352752 Here the convergence is definitely worse than in the two preceding cases. The fiure below shows the data for large N. Regardless of the ratio Ny/Nx, the energy should converge to the same value when N goes to infinity (which seems very plausible here). Note, however, how un-smooth the convergence is. Obviously, not only the shape of the "box" matters in how the energies converge as a function of N and Ny/Nx, but also the internal potential structure. It may not always be easy to see what would be the best ratio to use (and here we have only given a couple of examples, without exploring this in detail).
![]() Now let's look at the wave functions. These were calculated using Nx=Ny=40:
![]() ![]() Since the square is graphed we don't see the phase. Looking at the wave function itself, the ground state is (as expected) found to be symmetric and the first excited state is anti-symmetric.
![]() ![]() From these graphs we can see that the n=1 and n=3 states look very much like bonding states of a diatomic molecule (large probability between the attractive centers). The n=2 state looks like and anti-bonding state. The n=4 state doesn't really look like a state of a diatomic molecule, because the effects of the finite size of the "box" have become important here, but still the state has some anti-bonding features (low probability in the region between the attractive centers). |
Problem B (discretized real-space / Lanczos calculation)Solution program [lanc.f90] In the discretized real-space calculation, we check the convergence of the first four energies as a function of the discretization delta and the number of Lanczos iterations: delta=0.2 (25*50 lattice elements). ================================================================= iterations E0 E1 E2 E3 ----------------------------------------------------------------- 50 0.01424877 0.03982384 0.07577415 0.11434867 100 0.01104897 0.01692824 0.04023400 0.05891621 150 0.01104894 0.01692816 0.04023066 0.05850876 200 0.01104894 0.01692816 0.04023066 0.05850875 300 0.01104894 0.01104894 0.01692816 0.01692816 400 0.01104894 0.01104894 0.01105785 0.01692816 After 200 Lanczos iterations (basis states) the four lowest energies appear to be well converged to 8 decimal places. After that, we can see the successive appearance of multiple copies of the same state energi), due to non-othogonality problems. These problems could be easily overcome by the re-orthogonalization procedure discussed in class. Next, we successively double the number of lattice elements in each direction, to investigate the convergence of the energies as a function of the discretization cell-size delta. delta=0.1 (50*100 lattice elements). ================================================================= iterations E0 E1 E2 E3 ----------------------------------------------------------------- 100 0.01359333 0.04270708 0.08373779 0.14099236 200 0.01324518 0.01992994 0.04191934 0.06348454 300 0.01324517 0.01992912 0.04191724 0.06106467 400 0.01324517 0.01992912 0.04191724 0.06106464 500 0.01324517 0.01324522 0.01992912 0.02006401 600 0.01324517 0.01324517 0.01992912 0.01992912 delta=0.05 (100*200 lattice elements). ================================================================= iterations E0 E1 E2 E3 ----------------------------------------------------------------- 100 0.06298295 0.12106840 0.25709737 0.48510783 200 0.01721634 0.04861115 0.07595067 0.13176396 300 0.01391936 0.02144963 0.04370488 0.07304734 400 0.01356656 0.02030848 0.04243093 0.06330082 500 0.01356638 0.02030761 0.04242152 0.06229634 600 0.01356638 0.02030761 0.04242152 0.06229564 700 0.01356638 0.02030761 0.04242152 0.06229563 800 0.01356638 0.02030761 0.04242152 0.06229563 delta=0.025 (200*400 lattice elements). ================================================================= iterations E0 E1 E2 E3 ----------------------------------------------------------------- 100 0.05420624 0.41064007 0.92209001 1.72127298 200 0.02016086 0.08355610 0.27699021 0.46687546 300 0.01728184 0.05908541 0.11035425 0.22896869 400 0.01459613 0.02183848 0.06553600 0.13353513 500 0.01376280 0.02055673 0.06244828 0.07478966 600 0.01372735 0.02049812 0.05139688 0.06462293 700 0.01372423 0.02049013 0.04317888 0.06311201 800 0.01372417 0.02048990 0.04267267 0.06291673 900 0.01372417 0.02048990 0.04266768 0.06291183 1000 0.01372417 0.02048990 0.04266759 0.06291163 1100 0.01372417 0.02048990 0.04266759 0.06291158 1200 0.01372417 0.02048990 0.04266759 0.06291157 Note that the details of the convergence depend on the initial wave function, which here was generated at random in all cases. Your initial wave function will be different and you should see different numbers. The over-all rate of convergence should be similar, however, and the final converged result should be the samae. Note also that the non-orthogonality problems become less severa with increasing number of cells. Let's now check the consistency of the ground state energy with the result of the variational calculation. The next plot shows the Lanczos results for E0 and E1 versus delta (circles), compared with the variational result (squares at delta=0). The trend versus delta is clearly consistent with the same energies in the limit of delta=0. Note that the leading delta dependence is linear, not quadratic (as is the case for a purely kinetic hamiltonian). Evidently the discretization error is on order in delta worse when there is a potential energy present. ![]() The wave functions look almost identical to those in the variational calculation and are not graphed here. |