Answer to Question #277919 in Physical Chemistry for Kyle

Question

A particle is in a one-dimensional potential of: V(x)= λx4 , with λ>0. The upper bound of the ground state can be calculated using the variational method. The following trial function may be used: 1 2 2 ( ) exp{ } 2 a φ x N ax = − where Na is the normalization factor. (1) [5 pts] Does this trial function satisfy the requirements of the symmetry and the asymptotic behavior/boundary conditions of the ground-state wavefunction? Explain why. (2) [5 pts] Find Na. (You can use integral equations in McQuarrie or any other textbooks directly.) (3) [5 pts] Compute the upper bound of the ground state using this trial function. You don’t need to calculate the integrals analytically or numerically. Keep the integrals in your answer.

Answer 1

Consider a particle of mass m moving in the one dimensional potential :

$V(x)=\lambda x^4$

We can obtain an upper bound on the energy of the ground state using the variational method.