V = 8.0000 0 0 0 0 0 7.8013 0 0 0 0 0 7.6050 0 0 0 0 0 7.4113 0 0 0 0 0 7.2200The upper left corner of the kinetic energy matrix should look like this:
T = 400.00000 -200.00000 -0.00000 -0.00000 -0.00000 -200.00000 400.00000 -200.00000 -0.00000 -0.00000 -0.00000 -200.00000 400.00000 -200.00000 -0.00000 -0.00000 -0.00000 -200.00000 400.00000 -200.00000 -0.00000 -0.00000 -0.00000 -200.00000 400.00000and the first 5 energies in this approximation are
0.49992 1.49962 2.49913 3.49930The harmonic oscillator functions should look something like
To make the plot look better, I adapted the sign of the wave functions such the the value for grid point number 5 is positive. Also, I divided the eigenvectors by the square-root of the step size, to normalize them.
With a grid from \(r=1\) to \(r=4\) in steps of 0.05 \(a_0\) I get these energies for \(J=0\), (in cm-1):
e = 2036.26 5956.58 9666.43 13173.30The exact \(v=0, J=0\) result with the formula from Wikipedia is 2046.64 cm-1, and for \(v=3, J=0\) it is 13380.69 cm-1.
The vibrational term values in cm-1:
G(v)=we*(v+0.5)-wexe*(v+0.5)^2+weye*(v+0.5)^3 computed exact diff _____________________________________________ we 4126.75 4138.32 -11.5659 wexe 104.641 89.88 14.7606 weye 0.265076 0 0.2651Vibrational energies in cm-1 for J=0, 1, 2, 3
v\J 0 1 2 3 ________________________________________ 0 2036.26 2077.48 2159.88 2283.35 1 5956.58 5996.44 6076.13 6195.52 2 9666.43 9704.94 9781.93 9897.27 3 13173.30 13210.47 13284.76 13396.07
Note that these results are not necessarily converged to all digits shown. If you use the same grid as mentioned above, you expect these results though.
HTML5-validator Last updated: 29-Mar-2020, by Gerrit C. Groenenboom.