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I am trying to calculate the energy of a quantum mechanical rotor by plugging a potential function predicted from DFT calculations in this paper: DOI:10.1039/C8CC02650E

The potential can be approximated by:

U[x_] := (-V Sin[3 x ]^4 + V) Cos[3 x ]^2 + (V/2 Cos[6 x ] + V/2) Sin[3 x ]^2;

Where V is the height of the potential, and x is the angle of rotation of the H3 group of atoms.

It is also reported that for this -CH3 the Schrodinger equation can be written simply as:

$\qquad -B\frac{d^2\psi}{dx^2}+U(x)~\psi=Ei~\psi$

where $B = 0.665 meV$

I have been trying to use the method published here:

$\qquad$ Find eigen energies of time-independent Schrödinger equation

I modified the potential and pfun1, as well as the limits on x. But I think that I should change the boundary conditions on $\psi$? 

V = 13.17;
U[x_] := (-V Sin[3 x ]^4 + V)*Cos[3 x ]^2 + (V/2 Cos[6 x ] + V/2)*
Sin[3 x ]^2;

pfun1 = ParametricNDSolveValue[{-B ψ''[x] + U[x] ψ[x] == 
Ei ψ[x], ψ[0.] == 0., ψ[1.] == 
1.}, ψ, {x, -2 π, 2 π}, {Ei}]

I'm also not sure about the value of x to supply to FindRoot.

Plot[pfun1[Ei][?], {Ei, 0, 10}, ImageSize -> Medium]
val1 = Map[FindRoot[pfun1[Ei][?], {Ei, 0}] &, {1, 2, 3}]
Jason B.
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KTO
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  • You'll want to use something like {eV, eF} = NDEigensystem[{-B \[Psi]''[x] + U[x] \[Psi][x], \[Psi][0] == \[Psi][2 \[Pi]]}, \[Psi], {x, 0, 2 \[Pi]}, 3] where the last number is the number of eigenvalues eV and eigenfunctions eF you wish to calculate. The point is that the variable x is an angle and therefore you need periodic boundary conditions at x = 0, 2 Pi. The answer you were using was written before the new NDEigensystem had been introduced... – Jens Sep 05 '18 at 17:04
  • Perhaps this can come handy: Find the Spectrum of a Schrödinger Operator – Vitaliy Kaurov Sep 05 '18 at 22:25
  • Thank you @Jens - most useful! For some reason it refuses to plot the eigenfunctions using Plot - just gives a blank graph. – KTO Sep 18 '18 at 17:24
  • Thanks @VitaliyKaurov - very nice example indeed! – KTO Sep 18 '18 at 17:25

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