Question

Obtain an expression for the variational parameter, 2, from your minimization in (8) and substitute it back into your expression for the approximate energy, Ea to get an approximation to the ground state energy for the harmonic oscillator. 10. What is the difference between the energy you obtained in (9) and the one you got in (6b)? 11. Compare your result in (9) to the true value obtained from solving the H.O.Schrödinger equation (check McQuarrie or your H.O. activity for this value). Is it the same or different? Explain. You should plan on substituting for all constants to get a numerical value (in units of hw) to compare to. 8. Using the Additional Information, minimize Es for the harmonic oscillator. Because the Hamiltonian is a Hermitian operator, the following is true: Sfîf,dt = (Afdt where the first integral can be written as the matrix element H, (this a shorthand notation for the integral). and H, H,

Answer 1

To obtain an expression for the variational parameter, 2, we start by minimizing the energy functional, Es, with respect to 2. From equation (8), we have:

Es = <Ψ|H|Ψ> / <Ψ|Ψ> = (1/2)ω(2 + λ^2) / (1 + λ^2)

Taking the derivative of Es with respect to 2 and setting it to zero, we get:

dEs/d2 = ω(1 + λ^2) / (1 + λ^2)^2 = 0

Solving for λ^2, we get:

λ^2 = 1

Substituting this back into the expression for Es, we get the approximate ground state energy:

Ea = (1/2)ω(2 + λ^2) / (1 + λ^2) = (3/2)ω

This is the same as the result we obtained in equation (6b) using the variational principle.

The true ground state energy for the harmonic oscillator is given by E0 = (1/2)ω. Comparing this to our result for Ea, we see that the approximation is off by a factor of 3/2. This is not unexpected, as the variational method provides an upper bound on the true ground state energy.

Using the Additional Information, we can minimize Es for the harmonic oscillator and obtain a numerical value for the ground state energy. The Hamiltonian for the harmonic oscillator is given by:

H = (1/2)mω^2x^2

Using the fact that the ground state wavefunction for the harmonic oscillator is given by:

Ψ0(x) = (mω/πħ)^1/4 exp(-mωx^2/2ħ)

we can write the matrix element H as:

<H> = ∫ Ψ0(x) H Ψ0(x) dx = (1/2)ħω

Substituting this into the expression for the ground state energy, we get:

E0 = <H> / <Ψ0|Ψ0> = (1/2)ω

This is the exact ground state energy of the harmonic oscillator, which is lower than the approximation we obtained using the variational method.