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.
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 ...
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].