PHGN 450 / EGGN 502
Fall 2005
Project 10: Variational Monte Carlo Calculation of the H2 Potential
Due Dec. 5
Variational Treatment of the Hydrogen Molecule
The diatomic hydrogen molecule consists of four particles, two electrons and two protons. The quantum
dynamics for the hydrogen molecule are determined by the Hamiltonian,
X 2 X
(1)
H^ = 2p^mi + Vij ;
i i6=j
i
where p^2i = h 2 r2i , i = 1; 2 refers to the electrons and i = 3; 4 refers to the protons. The potential is taken
to be a sum of two-body Coulomb potentials:
X
e2
1
1
1
1
1
1
Vij =
(2)
40 j~r1 ~r3 j j~r1 ~r4 j j~r2 ~r3 j j~r2 ~r4 j + j~r1 ~r2 j + j~r3 ~r4 j :
i6=j
This neglects the so called \ne structure" eects, such as the spin-spin interactions among the constituents.
Since the electron mass is much smaller than the proton's, it is fruitful to separate their dynamics resulting in
what is called the Born-Oppenheimer approximation. This approximation assumes that the proton motion is
so much slower than that of the electrons that one may assume the protons are essentially at rest (zero kinetic
energy) at some xed separation, ~s, while the electrons nd their lowest energy conguration. Placing the
origin at the midpoint between the protons and letting the z-axis lie along ~s, yields the Born-Oppenheimer
Hamiltonian,
2
e2
HB O = 2hm (r21 + r22 ) + 4
e
0
1
j~r1 + 2s k^j
1
!
1
+ 1
+1 :
j~r1 2s k^j
j~r2 2s k^j j~r2 + 2s k^j j~r1 ~r2 j s
(3)
By evaluating the ground state electron energy for a range of separations, one maps out the interatomic
potential energy as a function of the proton separation. The ground state energy is the expectation value of
the Hamiltonian in the ground state, 0 (~r1 ; ~r2 ),
0 i
hE (s)i = hh0 jHj
(4)
0 j0 i
where Dirac bra-ket notation has been used to write the inner product,
1
Z
h0 jHj0 i = d3 r1 d3 r2 0 (~r1 ; ~r2 ) H 0 (~r1 ; ~r2 ):
(5)
Of course, this is of no use without the ground state wave function. The variational strategy is to approximate
the true ground state wave function with a convenient analytic form, T (~r1 ; ~r2 ; 1 : : : N ), involving some
arbitrary parameters, i . Minimizing the energy with respect to the parameters provides an upper bound to
the true ground state energy. The remaining technical diculty is the evaluation of 6-dimensional integral
which is well-suited to solution by Monte Carlo methods. To put the energy integral into a form appropriate
to a Monte Carlo sampling approach, rewrite Eq.(4) with the trial wave function replacing the ground state
wave function and rearranged into the form:
Z
hE (s; 1 : : : N )i = d3 r1 d3 r2 w(~r1 ; ~r2 ; 1 : : : N )E (~r1 ; ~r2 ; 1 : : : N );
1
(6)
where
jT (~r1 ;~r2 ;1 :::N )j2
w(~r1 ; ~r2 ; 1 : : : N ) =
hT jT i
1
E (~r1 ; ~r2 ; 1 : : : N ) = T H T (~r1 ; ~r2 ; 1 : : : N );
(7)
(8)
and where the variable dependence of T has only been written once.
Trial Wavefunction
Since electrons are fermions (spin-1/2 particles), they must obey the Pauli exclusion principle which
requires that the total wave function (space part times spin part) must be antisymmetric with respect to
exchange of labels, that is, 12 = 21 . For the problem of nding the ground state energy, the spin part
of the total wave function will be antisymmetric (the so-called \singlet" state); so the space part of the wave
function must be symmetric. Another challenge is presented by the possible innities encountered by the
Coulomb potential terms when the denominators become very small. If the form of the trial wave function
is not carefully chosen such that the kinetic energy terms track with the potential energy, the Monte Carlo
sampling process will fail as the electrons are driven to sit on top of the protons. The following form satises
both of these conditions:
(9)
T ((~r1 ; ~r2 ; ; s) = (~r1 ; s)(~r2 ; s)C (r12 ; s; )
where
(~r; s) = e j~r 2shk^j=a(s) + e j~r+ 2s k^j=ai(s)
(10)
r12
1
C (r12 ; s; ) = exp 2a0 (1+
(11)
r12 ) 2a0 where r12 = j~r1 ~r2 j, is a variational paramter, a0 = 40 h2 =(e2 me ) = 0:0529 nm is the Bohr radius, and
the remaining length parameter, a(s), is determined by the proton separation according to the transcendental
equation:
a0
a(s) =
(12)
1 + e s=a(s) :
Thus, there is only one variational parameter, . The -terms are simply symmetrized hydrogen atom-like
wave functions (i.e. exponentials), and C (r12 ; s; ) is the 2-particle correlation term which acts to keep
the two electrons apart. The correlation term depends on the variational parameter, , which determines
the distance over which the wave function \heals" from some small value for r12 small to unity for large
separation.
1. (5 points) Write a subroutine to calculate a(s). (As a test case, you should nd a(0:1) = 0:04723 nm.)
2. (15 points) For the trial wave function, Eq.(9), analytically calculate the energy function,
E (~r1 ; ~r2 ; s; ) = (~r 1; ~r ; ) H T (~r1 ; ~r2 ; ):
T 1 2
(This is a tedious, but necessary rst step in creating the Monte Carlo algorithm.) Write a subroutine to
evaluate this function. (As a test case, for f~r1 = (:06; :02; :03) nm, ~r2 = (:02; :04; :05) nm, s = :1 nm, and
= 11:4g, you should nd E = 32:89 eV.)
Monte Carlo Implementation
3. (50 points) Write a Monte Carlo algorithm (30 points) using a Metropolis, et al. accept/reject sampling
scheme to evaluate the energy integral, Eq. (4) using Eq.(9) for the trial wavefunction to 1%. To test your
coding, evaluate the integral for = 11:4 and s = 0:074 nm. You should nd hE i ' 31:3 eV (10 points).
Quantitatively justify your error estimate (10 points).
2
4m
e 2 = 13:59 eV, you should
4. (20 points) Since each hydrogen atom by itself is bound by E0 = 2(4e
)
0h
subtract 2E0 from your results to give the H2 interatomic potential energy due only to the proximity of the
two hydrogen atoms. Use your code to calculate and then plot the eective interatomic potential for proton
separations between 0.03 to 0.15 nm keeping = 11:4.
5. (10 points) Explore the dependence of your result by evaluating the interatomic potential for = 5:0
and = 15:0. Can you nd a value for that gives a statistically signicant lower energy at s = :074 nm?
6. (Extra Credit) (10 points) Once you have determined your \best" interatomic potential, t the potential
to a Morse potential form:
V (s) = V0 (1 e (s smin )= )2 1
(13)
where the three t parameters are fV0 ; smin ; g. Use this potential in the Bohr-Sommerfeld quantization
program that you created in the Project 1 to calculate a \rst principles" evaluation of the hydrogen molecule
binding energy.
3