# How to numerically solve the Schrödinger equation for Lennard-Jones potential?

Hi I have a potential like below:

  V[x_]:= 102*(4343/x^12 - 650/x^6) + 33/x^2


Which is a kind of modified Lennard-Jones potential. Schrödinger equation is 1D and time independent. Iam looking for the bound state enegy and wavafunction. How can solve this potential numerically in Mathematica?

• You are going to have to give some more details as to what you want, as well as what have you tried. – KraZug Jul 18 '18 at 10:06
• Which kind of Schrödinger equation are you trying to solve (write it down)? Are you looking for bound states, scattering states, wave packets... how many dimensions? This may be related: Find eigen energies of time-independent Schrödinger equation – Jens Jul 18 '18 at 21:09
• Hi modified the question – Xavier Jul 19 '18 at 16:35
• Check our NDEigensystem, but keep in mind that it returns the smallest eigenvalues (and associated eigenfunctions) by magnitude. Using a shift will likely be very helpful. – chuy Jul 19 '18 at 17:45

Here is a modified version of my answer to How to numerically solve a 1-d time-independent Schrödinger equation? that solves this problem.

I use NDEigensystem after manually shifting the potential so its bottom is at zero energy. This way the eigenvalues are automatically sorted in ascending order. In the output, I undo that shift so the eigenvalues are given on the original energy scale which has its zero at the point where the potential becomes non-binding. You can see that the largest eigenvalue is above zero, so that all the results up to this value represent the complete list of bound states.

In the second half of the code, I plot the eigenfunctions, superimposed on the potential with a baseline corresponding to their eigenvalue.

V[x_] := 102*(4343/x^12 - 650/x^6) + 33/x^2

n = 25;

cutoffDistance = 10;

shift = -FindMinValue[V[x], x];

{ev, ef} =
NDEigensystem[{shift f[x] + V[x] f[x] - 1/2 f''[x],
DirichletCondition[f[x] == 0, True]}, f, {x, 0, cutoffDistance},
n, Method -> {"SpatialDiscretization" -> {"FiniteElement", \
{"MeshOptions" -> {"MaxCellMeasure" -> 0.001}}},
"Eigensystem" -> {"Arnoldi", "MaxIterations" -> 10000}}];
evShifted = ev - shift

(*
==> {-2332.04, -2076.97, -1840.3, -1621.45, -1419.88, -1235.01, \
-1066.23, -912.955, -774.545, -650.364, -539.751, -442.027, -356.495, \
-282.438, -219.119, -165.779, -121.639, -85.8998, -57.7403, -36.3193, \
-20.776, -10.2314, -3.79112, -0.553967, 0.450964}
*)

With[{amplitudes = Table[30, n]},
Show[Plot[
Evaluate[
Table[evShifted[[i]] + amplitudes[[i]] ef[[i]][x], {i, n}]], {x,
0, cutoffDistance},
PlotRange -> {-shift, Max[evShifted] + Max[amplitudes]},
Epilog -> {Gray, Dashed,
Table[Line[{{0, evShifted[[i]]}, {cutoffDistance,
evShifted[[i]]}}], {i, n}]}, AspectRatio -> 3],
Plot[V[x], {x, 0.0001, cutoffDistance},
PlotStyle -> Directive[Dashed, Darker@Red], PlotRange -> All],
ImageSize -> 400
]] To see the plots more clearly, you would have to zoom in on the graphics, or restrict the PlotRange, as I do here (also increased number of PlotPoints):

With[{amplitudes = Table[30, n]},
Show[Plot[
Evaluate[
Table[evShifted[[i]] + amplitudes[[i]] ef[[i]][x], {i, n}]], {x,
0, cutoffDistance}, PlotPoints -> 100,
PlotRange -> {-shift, Max[evShifted] + Max[amplitudes]},
Epilog -> {Gray, Dashed,
Table[Line[{{0, evShifted[[i]]}, {cutoffDistance,
evShifted[[i]]}}], {i, n}]}, AspectRatio -> 3],
Plot[V[x], {x, 0.0001, cutoffDistance},
PlotStyle -> Directive[Dashed, Darker@Red], PlotRange -> All],
PlotRange -> {{1, 4}, {-shift, 0}}, AxesOrigin -> {0, -shift},
ImageSize -> 400
]] • Very nice work (+1). Are the uppermost eigenvalues reliable, or do they shift if you increase cutoffdistance? – Michael Seifert Jul 19 '18 at 18:30
• @MichaelSeifert indeed, choosing cutoffDistance can have an effect on the eigenvalues close to the "ionization threshold" where the waves penetrate more deeply into the classically forbidden region. But I tried changing the value to 20 without any effect, except on the positive eigenvalues. This is expected because they are traveling waves which always reach the boundary, no matter how far away it is. The question explicitly asked for bound states, so I think this approach is fine. – Jens Jul 19 '18 at 18:51
• @MichaelSeifert I think to be completely sure about the near-threshold eigenvalues one could run the eigensolver with a different shift, adjusted so that the most negative eigenvalues are reshuffled to appear later in the list ev. Then reduce the number n. This will focus the solver more narrowly on the eigenvalues you're interested in. But I didn't investigate this further because the question is quite broad. – Jens Jul 19 '18 at 18:57

Here's a shameless plug for something I've been working on.

We can do arbitrary potentials in up to about 4 dimensions (depending on the quality of your eigensolver) with a technique called discrete variable representation, more commonly known as DVR.

It effectively sets up the Hamiltonian in a grid-localized basis, allowing for convenient discrete treatments of problems over grids by diagonalizing the representation of the potential.

I've got a small framework for this setup in one of my packages. The easiest installation is via paclet server. Here's how it works.

<< ChemToolsDVR


### Create a DVR Object:

There are many different ways to set up a DVR suitable for different potentials and systems, so I classify these by the grid they operate on or the basis the DVR originates from. You've got decay at both ends, so we'll use a standard Cartesian grid in 1D:

dvr = ChemDVRObject["Cartesian1DDVR"];


### Get the Energies:

There are lots of little options that we can set, but for now we'll just supply the "PotentialFunction", the "Range" the discretization should operate over, and the number of "Points" in the discretization.

dvr[
"Energies",
"Points" -> 250,
"Range" -> {1, 10},
"PotentialFunction" -> (N[102*(4343/#^12 - 650/#^6) + 33/#^2] &)
] // AbsoluteTiming

{0.370715, {-2332.04, -2076.97, -1840.3, -1621.45, -1419.88, \
-1235.01, -1066.23, -912.947, -774.521, -650.323, -539.709, -442.013, \
-356.537, -282.549, -219.291, -165.99, -121.862, -86.1105, -57.9222, \
-36.4642, -20.8822, -10.3017, -3.83119, -0.570611, 0.444762}}


If we increase the "Points" we get a correspondingly more refined answer

### Wavefunctions and Plots

We can also get the energies and wavefunctions themselves (this calls Eigensystem instead of Eigenvalues and can be less efficient on large systems):

{ev, ef} =
dvr[
"Wavefunctions",
"Points" -> 250,
"Range" -> {1, 3.5},
"PotentialFunction" -> (N[102*(4343/#^12 - 650/#^6) + 33/#^2] &)
]; // AbsoluteTiming

{0.330197, Null}


And finally here's a plot of them (note that I used a ton more "Points" just to improve the smoothing on the wavefunctions):

dvr[
"Points" -> 650,
"Range" -> {1, 10},
"PotentialFunction" -> (N[102*(4343/#^12 - 650/#^6) + 33/#^2] &),
(* Plotting Options *)
"PlotDisplayMode" -> Show,
"WavefunctionSelection" -> 10,
"WavefunctionScaling" -> Scaled[.001],
"WavefunctionShifting" -> "Energy",
"WavefunctionClipping" -> None,
PlotRange -> {{1, 3}, {-2500, 0}}
] • It works! (+1). The eigenvalues in with your choice of 250 points don't quite agree with what I got (for the highest energies). But when I repeat your code with 650 points your results seem to agree with what I got. I like the documentation Wiki at the bottom of the Github page you linked. Are your grids equally spaced finite-difference grids? – Jens Jul 19 '18 at 21:44
• @Jens Yeah that's to be expected. 250 is a pretty rough discretization (although good enough for most things). Bad scaling on the technique, though, so it can really only go up to ~3D problems and maybe some 4D if you have lots of RAM. I think FEA may have a similar issue, though. – b3m2a1 Jul 19 '18 at 21:47
• Very nice! I had a brief look at the lecture you linked. What, in your opinion, are the main advantages of the DVR over other methods? – user21 Jul 20 '18 at 4:57
• @user21 the primary advantage is the ease with which any potential may be treated. Matrix representations of coordinate operators in the DVR basis are diagonal, and most potentials in chemistry share this. The kinetic energy can then be treated once and never again. I don’t really know enough to compare it to FEM, but I have read that chemists ran into relative difficulty in applying FEA to chemical problems. Why that is I unfortunately don’t know. One final nice DVR property is that no integrals need to be computed, which is different from many basis set treatments of physical problems. – b3m2a1 Jul 20 '18 at 7:54
• @b3m2a1, thanks for the info. Should you accidentally come across an example that it supposed to not work well with FEM, I'd appreciate if you could forward that. Thanks. Good luck with you package! – user21 Jul 20 '18 at 14:37