- 4.1 Hydrogen Atom Polarizability Problem
- 4.2 Hydrogen Atom van der Waals Interaction
- 4.3 Solution for Hydrogen Polarizabilty and van der Waals Strength

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.

H = - -
| (16) |

When a weak constant electric field along of magnitude *E*
is applied to this hydrogen atom, the perturbing potential is

V = zE,
| (17) |

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* = *r*_{c} will have no effect on the ground state with
or without a weak field if *r*_{c} 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 *r*_{c} . 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

np_{z}|z|1s
| (18) |

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

H = - - - -
| (19) |

V = .
| (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,

V_{vdW}(r) = - ,
| (21) |

| = |1s +
| (22) |

- . | (23) |

= - 2. | (24) |

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

z = rY_{10}(,)
| (25) |

np_{z}|z|1s = dr r^{ 3}R_{1s}(r)R_{np}(r)
| (26) |

#include <stdio.h> #include <math.h> #define NBUF 1024 main() { 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 main() { 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).

#!/bin/sh (for i in 2p.out 3p.out 4p.out 5p.out 6p.out 7p.out 8p.out 9p.out 10p.out\ 11p.out do 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 = 6.497which 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

E
| = | ||

= | |||

= | (27) |

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.