What is the relationship between H, psi, and E in Quantum Mechanics?

[xanmas]

Homework Statement

I am currently doing undergraduate research and was assigned this as sort of an introduction. I am sure this is a very rudimentary problem and appreciate any help.

Basically, its your regular old H*psi = E*psi.

Well, knowing that H is psi(-D^2/Dx^2 + cosx + sin^2x = E*psi

and I want to prove that Ne^(lambda(1-cos)) is where E is equal to zero where N is the normalization constant and lambda is an arbitrary constant.

2. The attempt at a solution

The thing that I tried to do was divide by psi, and set E=0 to completely get rid of psi. I am not sure if I am allowed to do this but I did. This left me with

-D^2/Dx^2 + cosx + sin^2x = 0.

From there, I moved them to separate sides and doubly integrated. I got

ln|x| = cos^2(x)/4 - cos(x) +x^2/4

and exponentating I got

x = e^(Cos^2(x)/4 - cos(x) +x^2/4)

The problem is, without psi, I don't think that derivitave means anything and so I think I need to somehow keep the psi in there but I don't know what to do other than to divide out psi.

I would also rather a few hints instead of an explicit solution; I am sure that its just something that I am over looking and with a hint or two, I could do this.

I am very gracious of your help,

Thomas

 

[Dick]

H operates on psi. So the equation is -psi(x)''+(cos(x)+sin^2(x))*psi(x)=0. And you don't have to solve it. Just substitute psi(x)=N*exp(lambda*(1-cos(x)). I think you'll find it works for only one choice of lambda.