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4 A Sample Problem and its Solution

 Besides being able to investigate many solutions of the one-dimensional Schrödinger equation, we can also use the applet as a tool to allow students to get answers to more interesting problems.

A problem that we have given in our first year graduate quantum classes is the polarizability of a single hydrogen atom and the van der Waals interaction strength between two well separated hydrogen atoms.     This requires the students to work out the solution in terms of matrix elements, to calculate those matrix elements numerically using wave functions produced by our applet, and to combine the results to get a value that can be compared to experiments and other calculations.

We also provide a sketch of our solution which students would receive after the completion of the assignment.

4.1 Hydrogen Atom Polarizability Problem

A hydrogen atom is in its ground state. The Hamiltonian of the hydrogen atom in atomic units e = $\hbar$ = m = 1 , and with the center of mass at the origin is

H = - $\displaystyle{\textstyle\frac{1}{2}}$$\displaystyle\nabla^{2}_{}$ - $\displaystyle{\textstyle\frac{1}{r}}$ (16)

When a weak constant electric field along $\hat{z}$ of magnitude E is applied to this hydrogen atom, the perturbing potential is

V = zE, (17)

and a dipole moment is induced proportional to the electric field. The proportionality constant is $\alpha$ , the polarizability.

Use perturbation theory to write an expression for the polarizability. Show that polarizability has units of volume, so the expression in atomic units needs to be multiplied by the Bohr radius cubed to convert to other units.

A direct expansion of the Schrödinger differential equation in powers of E , first done by I. Waller, Zeits. f. Physik 38, 635 (1926) and P.S. Epstein, Phys. Rev. 28, 695 (1926), gives an analytic value of the polarizability of 4.5 a.u.

Physically you should expect the weak electric field applied to the atom to distort the ground state wave function only slightly. However, usual perturbation theory uses the unperturbed atomic p states which have a much larger spatial extent than the 1s ground state. Convince yourself that enforcing the boundary condition that all wave functions go to zero at a radius r = rc will have no effect on the ground state with or without a weak field if rc is large enough that the perturbed wave function has gone substantially to zero. By inspection, the value of the 1s state is less than 10- 4 atomic units at r = 10 a.u. However, the p states are much more excited by being confined to this sphere of radius rc . Because of this, these confined atomic orbitals have been named ``fireballs'' by Otto Sankey who has used them to do electronic structure calculations in solids (see for example, O.F. Sankey and Niklewski, Phys. Rev. B40, 3979 (1989)). Use the Schrödinger applet to calculate numerically the energies and wave functions of the first ten unnormalized p fireball radial functions, and save the results to 10 files. Do the angular integrations analytically, and write a computer program to calculate the matrix elements

$\displaystyle\langle$npz|z|1s$\displaystyle\rangle$ (18)

by doing the radial integrations numerically. Trapezoidal rule integration is adequate. Show that you get the correct polarizability to 3 places (about all that can be expected from our numerics) with only 6 or 7 fireball states.

4.2 Hydrogen Atom van der Waals Interaction

Now assume that you have two well-separated ground-state hydrogen atoms with atomic states that do not overlap. You can assume the protons are infinitely massive with fixed positions in space. The vector from proton 2 to proton 1 defines the z axis, and the separation distance is R . Take the electron coordinates $\vec{r}_{1}^{}$ and $\vec{r}_{2}^{}$ as the vectors pointing from proton 1 to electron 1 and from proton 2 to electron 2 respectively as in figure 1.
Figure 1: Geometry for the van der Waals problem
\rotatebox {270}{\includegraphics[width=3in]{}}\end{figure}

Show that ignoring the coulomb interactions between the particles in different hydrogen atoms gives the unperturbed Hamiltonian

H = - $\displaystyle{\textstyle\frac{1}{2}}$$\displaystyle\nabla_{1}^{2}$ - $\displaystyle{\textstyle\frac{1}{r_1}}$ - $\displaystyle{\textstyle\frac{1}{2}}$$\displaystyle\nabla_{2}^{2}$ - $\displaystyle{\textstyle\frac{1}{r_2}}$ (19)

and if R is large compared to the atomic size, the remaining coulomb interactions can be approximated by the dipole-dipole interaction

V = $\displaystyle{\frac{x_1 x_2 + y_1 y_2 - 2 z_1 z_2}{R^3}}$. (20)

Use your same 10 matrix elements to calculate the van der Waals interaction between two Hydrogen atoms given by the second order perturbed energy. Compare your numerical result with the result of L. Pauling and J.Y. Beach, Phys. Rev. 47, 686 (1935), who calculated using a variational method,

VvdW(r) = - $\displaystyle{\frac{6.499}{R^6}}$, (21)

in atomic units.

4.3 Solution for Hydrogen Polarizabilty and van der Waals Strength

The perturbed wave function is

|$\displaystyle\psi$$\displaystyle\rangle$ = |1s$\displaystyle\rangle$ + $\displaystyle\sum_{n \ne 1s}^{}$$\displaystyle{\frac{\vert n \rangle \langle n\vert V\vert 1s\rangle}{E_{1s}-E_n}}$ (22)

and the dipole moment is

- $\displaystyle{\frac{\langle \psi \vert z \vert\psi\rangle}{\langle \psi \vert\psi\rangle}}$. (23)

Keeping terms linear in V gives the linear response result

$\displaystyle\alpha$ = - 2$\displaystyle\sum_{n \ne 1s}^{}$$\displaystyle{\frac{\vert\langle n\vert z\vert 1s\rangle\vert^2}{E_{1s}-E_n}}$. (24)

The z operator only couples the 1s state to p states. If the usual H atom states are used, the continuum states also must be included by taking the appropriate limit of the sum to integrate over these continuum states.

The angular integrations can be done directly, using for example,

z = r$\displaystyle\sqrt{\frac{4 \pi}{3}}$Y10($\displaystyle\theta$,$\displaystyle\phi$) (25)

to give

$\displaystyle\langle$npz|z|1s$\displaystyle\rangle$ = $\displaystyle{\textstyle\frac{1}{\sqrt{3}}}$$\displaystyle\int_{0}^{\infty}$dr r 3R1s(r)Rnp(r) (26)

The fireballs radial functions were calculated using the applet with the input given by the H-atom 2p state example input except we change Xmax to 10.0 and increment nodes from 0 to 9. The output was saved into the files 2p.out, 3p.out, ... 11p.out by using the Write button. The ground state energy is -0.5 a.u., and the following C code was used to calculate the matrix elements.
#include <stdio.h>
#include <math.h>
#define NBUF 1024

   char s[NBUF];
   double r,r2,psi,psi0,anorm0,anorm,dip,e;
   int i,j,c;
   anorm = anorm0 = dip = 0.0;
   i = j = 0;
   while ((c = getchar()) != EOF) {
      if (c == '\n') {
         i = 0;
         if (j++ == 6) sscanf(s,"# %lg",&e);
         if (s[0] == '#') continue;
 read r and psi, calculate the normalizations and the dipole transition
 matrix elements to the 1s ground state using trapezoidal rule
         sscanf(s,"%lg %lg",&r,&psi);
         psi0 = exp(-r);
         r2 = r*r;
         anorm0 += r2*psi0*psi0;
         anorm += psi*psi;
         dip += r2*psi*psi0;
      } else {
         s[i++] = (char) c;
   dip /= sqrt(anorm0*anorm*3.0);
   printf("%lg %lg\n",dip,e);

The matrix elements and energy denominators were combined in the following C code (which also calculates the van der Waals strength, see below):

#include <stdio.h>
#define N 10

   double dipmat[N],e[N],d,vdw;
   int i,j;
 read in matrix elements and energies
   for (i=0;i<N;i++) scanf("%lg %lg\n",dipmat+i,e+i);
 calculate polarizability
   printf("State       Summed Polarizability\n");
   d = 0.0;
   for (i=0;i<N;i++) {
      d += 2.*dipmat[i]*dipmat[i]/(0.5+e[i]);
      printf("%d              %4.4lg\n",i,d);
 calculate van der Waals strength
   vdw = 0.0;
   for (i=0;i<N;i++) {
      for (j=0;j<N;j++) {
         vdw += 6.*dipmat[i]*dipmat[i]*dipmat[j]*dipmat[j]/(1.0+e[i]+e[j]);
   printf("Van der Waals Strength = %4.4lg\n",vdw);
and combining these together using the unix script (dip is the matrix element executable, and polar is the polarizability and van der Waals strength executable).
(for i in 2p.out 3p.out 4p.out 5p.out 6p.out 7p.out 8p.out 9p.out 10p.out\
   dip <$i 
   done) | polar

gives the result

State       Summed Polarizability
0              3.391
1              4.243
2              4.442
3              4.484
4              4.494
5              4.497
6              4.498
7              4.498
8              4.498
9              4.498
Van der Waals Strength = 6.497
which demonstrates that we only miss the correct result by .002 . Interested students can play around to see what needs to be done to improve this result.

The van der Waals interaction goes through in a similar way. Taylor series expanding the coulomb interactions gives the dipole-dipole term at lowest order. The second order energy is then given by

$\displaystyle\Delta$E = $\displaystyle{\textstyle\frac{1}{R^6}}$$\displaystyle\sum_{n,m \ne 1s,1s}^{}$$\displaystyle{\frac{\vert\langle n m \vert
x_1 x_2 + y_1 y_2 - 2 z_1 z_2 \vert 1s 1s\rangle\vert^2 }{2E_{1s}-E_n-E_m}}$   
  = $\displaystyle{\textstyle\frac{6}{R^6}}$$\displaystyle\sum_{n,m \ne 1s,1s}^{}$$\displaystyle{\frac{\vert\langle n m \vert
z_1 z_2 \vert 1s 1s\rangle\vert^2 }{2E_{1s}-E_n-E_m}}$   
  = $\displaystyle{\textstyle\frac{6}{R^6}}$$\displaystyle\sum_{n,m}^{}$$\displaystyle{\frac{\vert\langle np\vert z\vert 1s\rangle \langle mp\vert z\vert 1s \rangle
\vert^2 }{2E_{1s}-E_n-E_m}}$ (27)
The numerical result is given above and agrees with Pauling and Beach to about .002 atomic units.    

For students interested in continuing, the strength of the Axilrod-Teller (see B.M. Axilrod and E. Teller, J. Chem. Phys. 11, 293 (1943)) triple-dipole three-body potential between three well separated hydrogen atoms can be calculated using these same techniques and matrix elements.

next up previous contents
Next: 5 Goals and Pedagogy Up: Visual Schrödinger: A Visualizer-Solver Previous: 3 Prepared Examples