Problems

The ground state of the Helium atom

Let us find the ground state energy of the He atom using the variational method.

Let us use as a trial function

(We are guided by the ground state wavefunction of hydrogen )

The adjustable parameter in our trial function is a=a₀/Z'. y(r₁,r₂) is a product function, y(r₁,r₂)=f₁(r₁)f₂(r₂).

f₁(r₁) is the wavefunction of an electron in a hydrogenic atom with charge Z’.
<f₁|p²₁/(2m)|f₁>=kinetic energy of an electron in a hydrogenic atom with charge Z’.
<f₁|Z'e²/r|f₁>=potential energy of an electron in a hydrogenic atom with charge Z’.
Therefore <f₁|H₁|f₁>= kinetic energy of an electron in a hydrogenic atom with charge Z’, plus 2/Z' times the potential energy of an electron in a hydrogenic atom with charge Z’.

For an electron in a hydrogenic atom with charge Z’ we have

Therefore

Using our trial wavefunction we therefore have <f₁|H₁|f₁>=<f₂|H₂|f₂>=Z'²E_I- 4Z'E_I, and

We can evaluate this integral in two different ways.

We can recognize that the same integral can represent the electrostatic interaction energy of two spherically symmetric charge distributions with charge densities -er(r₁) and -er(r₂), respectively. Here

The electrostatic interaction energy of two such distributions is U=(e/2)òr(r)f(r)d³r=(e²/2)òòr(r)(r(r')/|r-r'|)d³rd³r'.
The electrostatic interaction energy of the two charge distributions can also be found by evaluating a different integral.

The electric field produced by one of the distributions is found from Gauss’ law.

(Gaussian units).

The self energy of the distribution is given by

When both distributions are superimposed we have a distribution with twice the charge density. The self energy now is 4U. The interaction energy is the difference between the self energy of the distribution with twice the charge density and the sum of the self energies of the component distributions. Therefore

interaction energy = 4U-2U=2U.

We can find the interaction energy by evaluating the integral for U.

We can also evaluate the integral directly. We write

To carry out the angular integration we then use

The integrals for the radial integration come straight out of common integral tables.

With we have

The experimental value for the ground state energy of He is

Let us now find the ground state energy of the He atom using perturbation theory.

the ground state wavefunction for a hydrogenic ion with Z=2.

with

The variational method often yields a very good estimate for the ground state energy of a system. But an arbitrarily chosen trial ket can give a good approximation to the ground state energy but still be very different from the true eigenket. So one must be very careful when using wave functions obtained by the variational method to calculate other physical quantities of the system.

Problems:

Consider the half space linear potential in one dimension
V(x)= ¥, x<0, V(x)=Fx, x>0.
Use the variational method to estimate the ground state energy. Choose your own variational wave function.

Solution:

The eigenfunctions of H must be zero at x=0 and x=¥. The derivative dy/dx must be continuous at all x except x=0. Any function that satisfies these boundary conditions can be written as a linear combination of eigenfunctions of H and is an acceptable trial function giving <H>>E₀.
We may choose as the trial function for x>0 the eigenfunction of the first excited state of the harmonic oscillator. It satisfies the boundary conditions. We choose

Note the change in the normalization constant, since we are only looking at x>0.

since <T> = <V>.

The exact solutions of

are the Airy functions y(z)=A_i(z), with boundary condition y(x=0)=0.

ai.gif (4821 bytes)

x=0 Þ z=-E/(Fx₀). y(z) has its first zero at z=-2.338.

Calculate the splitting induced among the degenerate n=2 levels of a hydrogenic atom, when this atom is placed in a uniform electric field E pointing in the z-direction. This is the linear Stark effect. You may use the following explicit hydrogenic wave functions |nlm>:

and the fact that in general Hint: Exploit symmetries!

Solution:

The energy of the proton in the field is

The energy of the electron in the field is

The potential energy of both particles is

Let us use first-order perturbation theory.

The first excited state is 4-fold degenerate.

To find E₁² we have to diagonalize the operator in the subspace spanned by |200>, |210>, |211>, and |21-1>. The matrix elements between states with the same parity are zero. This implies that all diagonal matrix elements are zero and all elements between states with the same l are zero. The matrix elements between states with different m are zero. The only matrix elements left are <210|z|200> and <200|z|210>.