Explore BrainMass

Ground state energy for particle using variationa method

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

A particle of mass m is in the one-dimentional potential given by V(x) = Kx^3 for x >= 0, there K is a positive constant. There is an infinite potential barrier at x = 0, so V(0) = infinity. Use the variational principle with the trial wave function
psi(x) = xe^[-alpha x] to estimate the ground-state energy.

© BrainMass Inc. brainmass.com March 21, 2019, 8:22 pm ad1c9bdddf

Solution Preview

Let's first derive the variational principle to understand why it works. If the exact energy eigenstates are |psi_n> and the corresponding energy eigenvalues are E_n, then an arbitrary normalized state |psi> can be expanded as:

|psi> = Sum from n = 0 to infinity of c_n |psi_n>

The norm of |psi> is then:

<psi|psi> = Sum from n = 0 to infinity of |c_n|^2

Therefore, we have:

Sum from n = 0 to infinity of |c_n|^2 = 1 (1)

The expectation value of the energy in the state |psi> is:

<H> = <psi|H|psi> = Sum from n = 0 to infinity of |c_n|^2 E_n

If |psi_0> is the ground state and is ...

Solution Summary

We consider a particle trapped in thepotential V(x) = Kx^3 for x >= 0 and explain how to estimate the ground state energy using the trial wavefunction psi(x) = xe^[-alpha x].