Computational Physics - Department of Physics

14.6 Exercises 491

chosenz-akse with electron 1 placed at a distancer 1 from a chose origo, one proton at−R/ 2

and the other atR/ 2 , the distance from proton 1 to electron 1 becomes

r 1 p 1 =r 1 +R/ 2 , (14.42)

and

r 1 p 2 =r 1 −R/ 2 , (14.43)

from proton 2. Similarly, for electron 2 we obtain

r 2 p 1 =r 2 +R/ 2 , (14.44)

and

r 2 p 2 =r 2 −R/ 2. (14.45)

These four distances define the attractive contributions tothe potential energy

−

1

r 1 p 1

−

1

r 1 p 2

−

1

r 2 p 1

−

1

r 2 p 2

. (14.46)

We can then write the total Hamiltonian as

Ĥ=−∇

(^21)
2

−

∇^22

2

−

1

r 1 p 1

−

1

r 1 p 2

−

1

r 2 p 1

−

1

r 2 p 2

+

1

r 12

+

1

|R|

, (14.47)

and if we chooseR= 0 we obtain the helium atom.

In this project we will use a trial wave function of the form

ψT(r 1 ,r 2 ,R) =ψ(r 1 ,R)ψ(r 2 ,R)exp

(

r 12

2 ( 1 +βr 12 )

)

, (14.48)

with the following trial wave function

ψ(r 1 ,R) = (exp(−αr 1 p 1 )+exp(−αr 1 p 2 )), (14.49)

for electron 1 and

ψ(r 2 ,R) = (exp(−αr 2 p 1 )+exp(−αr 2 p 2 )). (14.50)

The variational parameters areαandβ.

One can show that in the limit where all distances approach zero that

α= 1 +exp(−R/α), (14.51)

resulting inβkas the only variational parameter. The last equation is a non-linear equation

which we can solve with for example Newton’s method discussed in chapter 4.

1. Find the local energy as function ofR.


2. Set up and algorithm and write a program which computes theexpectation value of〈Ĥ〉
   using the variational Monte Carlo method with a brute force Metropolis sampling. For each
   inter-proton distanceRyou must find the parameterβwhich minimizes the energy. Plot the
   corresponding energy as function of the distanceRbetween the protons.


3. Use thereafter the optimal parameter sets to compute the average distance〈r 12 〉between the
   electrons where the energy as function ofRexhibits its minimum. Comment your results.


4. We modify now the approximation for the wave functions of electrons 1 and 2 by subtracting
   the two terms instead of adding up, viz

ψ(r 1 ,R) = (exp(−αr 1 p 1 )−exp(−αr 1 p 2 )), (14.52)