Solutions for homework assignment #5

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).

The ground state energy versus the number of states (N=Nx*Ny) for different ratios of Nx and Ny.

Now let's look at the wave functions. These were calculated using Nx=Ny=40:

The squared ground state (left) and first excited state (right) wave functions.

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.

The squared second (left) and third excited state (right) wave functions.

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.