# Quantum Mechanics - Ritz variational principle

1. Oct 24, 2004

### el_hijoeputa

I was asked to do an assigment for a Chemical Physics class on the Ritz variational principle (used to calculate an approximation of an observable). We are working a simple potential, the one dimensional particle in the box (v=0 for 0<x<L, V= infinite elsewhere) and only considering the ground state. I'm asked to make approximations of the wave function with polynomials, first a linear one, then a second order, and with a third order one. We need to do this to verify that the closest the shape of our approx. wave function is to the one obtained by solving that potential, wich is:
Y(x) = (2/L)^1/2 Sin [(pi/L) x], the calculated Energy get closer to the "real" one. Therefore a third order polynomial will perform better than a first order polynomial.

Well, my problem arises when making an a proximation of the wave function with only linear polynomials, because the derivatives in the Schrodinger Eq. are of second order, yielding 0 to the value of energy. The professor said that this is incorrect, that a change has to be made to the Schrodinger Eq. (probably using chain rule) for this case.

Anyway, I chose as my trial wave function with only linear polynomials the following:

Y(x) = { Ax, for 0<x<L/2
Y(x) = { B(x-L), for L/2<x<L

I have no idea on the modification needed to the Schrodinger Eq. Can someone shed some light, or give me advise on how to solve this?

2. Nov 19, 2010

### nuclei

i also have met this problem

3. Nov 19, 2010

### fzero

For what A and B is this a continuous function? Is the first derivative continuous?