What is the Variational Principle for Estimating Energy of First Excited State?

[bmb2009]

Homework Statement

A particle of mass m is in a potential of V(x) = Kx⁴ and the wave function is given as ψ(x)= e^-(ax²) use the variational principle to estimate the ground state energy.

Part B:
The true ground state energy wave function for this potential is a symmetric function of x i.e. ψ₀(x)=ψ₀(-x). Use the result that <ψ₀lψ(β)> = 0 along with an approximately chosen wave function, to estimate the energy of the first excited state.

Homework Equations

The Attempt at a Solution

Ok so I know how to compute variational method approximations and I have proven the identity <ψ₀lψ(β)> = 0 earlier on my assignment and understand the identity as well. What I don't understand is the part that says "Use the result that <ψ₀lψ(β)> = 0 along with an approximately chosen wave function, to estimate the energy of the first excited state."

Again I know that when <ψ₀lψ(β)> = 0 the variational principle becomes E₁≤ <ψlHlψ>/<ψlψ> but does the problem want me to chose a different wavefunction? And if so how to I go about choosing this new wave-function?

 

[qbert]

The point is to pick a wavefunction which is orthogonal to your first
and use that as a trial to read the first excited energy [again variationally].
The clue is telling you to look at the parity, an even wf. is orthogonal to an odd wf,
so your trial function should be chosen odd.