Computational Physics - Department of Physics

474 14 Quantum Monte Carlo Methods

the exact result, and the variance is zero, as it should. The reason is that we have used the
exact wave function, and the action of the hamiltionan on thewave function

Hψ=constant×ψ,

yields just a constant. The integral which defines various expectation values involving mo-
ments of the Hamiltonian becomes then

〈Hn〉=

∫

dR∫ΨT∗(R)Hn(R)ΨT(R)

dRΨT∗(R)ΨT(R)

=constant×

∫

∫dRΨT∗(R)ΨT(R)

dRΨT∗(R)ΨT(R)

=constant.

This explains why the variance is zero forα= 1. However, the hydrogen atom and the har-

Table 14.4Result for ground state energy of the harmonic oscillator asfunction of the variational parameter
α. The exact result is forα= 1 with an energyE= 1. We list the energy and the varianceσ^2 as well. The
variableNis the number of Monte Carlo samples. In this calculation we setN= 100000 and a step length of 2
was used in order to obtain an acceptance of≈50%.
α 〈H〉 σ^2
5.00000E-01 2.06479E+00 5.78739E+00
6.00000E-01 1.50495E+00 2.32782E+00
7.00000E-01 1.23264E+00 9.82479E-01
8.00000E-01 1.08007E+00 3.44857E-01
9.00000E-01 1.01111E+00 7.24827E-02
1.00000E-00 1.00000E+00 0.00000E+00
1.10000E+00 1.02621E+00 5.95716E-02
1.20000E+00 1.08667E+00 2.23389E-01
1.30000E+00 1.17168E+00 4.78446E-01
1.40000E+00 1.26374E+00 8.55524E-01
1.50000E+00 1.38897E+00 1.30720E+00

-1

-0.8

-0.6

-0.4

-0.2

0

0.2 0.4 0.6 0.8 1 1.2 1.4

E 0

α

MC simulation with N=100000

Exact result

Fig. 14.2Result for ground state energy of the hydrogen atom as function of the variational parameterα.
The exact result is forα= 1 with an energyE=− 1 / 2. See text for further details.