14.6 Exercises 489whereαis the variational parameter,ψT 2 (r 1 ,r 2 ,r 12 ) =exp(−α(r 1 +r 2 ))( 1 +βr 12 ), (14.30)withβas a new variational parameter andψT 3 (r 1 ,r 2 ,r 12 ) =exp(−α(r 1 +r 2 ))exp(
r 12
2 ( 1 +βr 12 ))
. (14.31)
a) Find the closed-form expressions for the local energy forthe above trial wave function for the
helium atom. Study the behavior of the local energy with these functions in the limitsr 1 → 0 ,
r 2 → 0 andr 12 → 0.
b) Compute
〈Ĥ〉=∫
d∫RΨT∗(R)Ĥ(R)ΨT(R)
dRΨT∗(R)ΨT(R), (14.32)
for the helium atom using the variational Monte Carlo methodemploying the Metropolis
algorithm to sample the different states using the trial wave functionψT 1 (r 1 ,r 2 ,r 12 ). Compare
your results with the closed-form expression〈Ĥ〉=
h ̄^2
me
α^2 −27
32
e^2
π ε 0
α. (14.33)c) Use the optimal value ofαfrom the previous point to compute the ground state of the helium
atom using the other two trial wave functionsψT 2 (r 1 ,r 2 ,r 12 )andψT 3 (r 1 ,r 2 ,r 12 ). In this case
you have to vary bothαandβ. Explain briefly which functionψT 1 (r 1 ,r 2 ,r 12 ),ψT 2 (r 1 ,r 2 ,r 12 )
andψT 3 (r 1 ,r 2 ,r 12 )is the best.
d) Use the optimal value for all parameters and all wave functions to compute the expectation
value of the mean distance〈r 12 〉between the two electrons. Comment your results.
e) We will now repeat point 1c), but we replace the helium atomwith the ions LiIIand BeIII.
Perform first a variational calculation using the first ansatz for the trial wave function
ψT 1 (r 1 ,r 2 ,r 12 )in order to find an optimal value forα. Use then this value to start the varia-
tional calculation of the energy for the wave functionsψT 2 (r 1 ,r 2 ,r 12 )andψT 3 (r 1 ,r 2 ,r 12 ). Com-
ment your results.
14.2.The H+ 2 molecule consists of two protons and one electron, with binding energyEB=
− 2. 8 eV and an equilibrium positionr 0 = 0. 106 nm between the two protons.
We define our system through the following variables. The electron is at a distancerfrom
a chosen origo, one of the protons is at the distance−R/ 2 while the other one is placed at
R/ 2 from origo, resulting in a distance to the electron ofr−R/ 2 andr+R/ 2 , respectively.
In our solution of Schrödinger’s equation for this system weare going to neglect the kinetic
energies of the protons, since they are 2000 times heavier than the electron. We assume
thus that their velocities are negligible compared to the velocity of the electron. In addition
we omit contributions from nuclear forces, since they act atdistances of several orders of
magnitude smaller than the equilibrium position.
We can then write Schrödinger’s equation as follows
{
− ̄
h^2 ∇^2 r
2 me−
ke^2
|r−R/ 2 |−
ke^2
|r+R/ 2 |+
ke^2
R}
ψ(r,R) =Eψ(r,R), (14.34)where the first term is the kinetic energy of the electron, thesecond term is the potential
energy the electron feels from the proton at−R/ 2 while the third term arises from the po-
tential energy contribution from the proton atR/ 2. The last term arises due to the repulsion
between the two protons.