Optimizing Energy of Hydrogen Atom with 3D Oscillator Wavefunction

[oomphgalore20]

Homework Statement

Take as a trial wavefunction for the hydrogen atom the 3D oscillator ground state wavefunction
ψ(r) = N exp (-br^2 / 2). Calculate the value of b that gives the best energy and calculate this energy.

Homework Equations

Radial part of ∇^2 = 1/r2 (∂/∂r) (r^2 ∂/∂r)

The Attempt at a Solution

I am not sure what best energy is supposed to imply.

 

[Goddar]

Hi.
Is this following a chapter about the variational method?

 

[oomphgalore20]

There is no fixed textbook, so I'm not sure what chapter this precedes or follows; we've covered the simple harmonic oscillator in 1D and went into the variational method while covering perturbation theory. I thought the 3D SHO is an extension of the 1D system, but this 'best energy' is throwing me off.

 

[Goddar]

In a variational method problem it would make sense to take a trial function and vary the parameter b in order to get a higher bound for the ground state energy ("best" energy?), that's why I'm asking. i can't think of another meaning in the context you're giving...

 

[paranormal]

However, based on the given trial wavefunction, we can calculate the energy by applying the Schrödinger equation:

Hψ(r) = Eψ(r)

Where H is the Hamiltonian operator and E is the energy.

Substituting the trial wavefunction into the equation, we get:

H(N exp (-br^2 / 2)) = E(N exp (-br^2 / 2))

Expanding the Hamiltonian operator and rearranging the terms, we get:

(-h^2/8π^2m)(∂^2/∂r^2)(r^2 exp (-br^2 / 2)) + (-e^2/r)(N exp (-br^2 / 2)) = E(N exp (-br^2 / 2))

Using the given equation for the radial part of the Laplacian, we can simplify the above equation to:

(-h^2/8π^2m)(N exp (-br^2 / 2))[(4b^2r^2 - 6br + 3) + (-e^2/r)] = E(N exp (-br^2 / 2))

Simplifying further and dividing both sides by N exp (-br^2 / 2), we get:

(-h^2/8π^2m)(4b^2r^2 - 6br + 3) + (-e^2/r) = E

To find the value of b that gives the best energy, we can differentiate the above equation with respect to b and set it equal to 0. This will give us the value of b that minimizes the energy.

∂E/∂b = (-h^2/8π^2m)(8br - 6) = 0

Solving for b, we get:

b = 3/4r

Substituting this value of b back into the energy equation, we get:

E = (-h^2/8π^2m)(3/2) + (-e^2/r)

This is the best energy that can be obtained using the given trial wavefunction. However, it is worth noting that this energy is not the exact energy of the hydrogen atom ground state, as the trial wavefunction does not fully capture the complexity of the hydrogen atom. It serves as an approximation and can be improved upon by using more sophisticated wavefunctions.