# Quantum harmonic question

1. Mar 19, 2008

### evildarklord1985

2. Mar 20, 2008

### malawi_glenn

Show work done and the formulas /relations you know, then someone will try to help you.

3. Mar 20, 2008

### evildarklord1985

My first thought about this problem is trying to set up a boundary condition for the wavefunction. For 0<x<A, the wavefunction takes the form of simple harmonic oscillators, i.e. N exp(-1/2 alpha x^2) . With boundary condition where psi(x) = 0 at x=0, I'm stuck with the expression of N=0. Maybe, I didn't think it right...

4. Mar 21, 2008

### malawi_glenn

N is just a normalization factor, leave that now for the moment.

But you also have Hermite polynomials in the harmonoc wave functions right?

5. Mar 21, 2008

### evildarklord1985

so, the harmonic wave function can be written as : N H(y) exp(-1/2 alpha x^2) . With the mentioned boundary condition, do I get N*H(y) = 0 ...how would I go from here on to solve the problem?

6. Mar 22, 2008

### eys_physics

The energy is calculated as the expectation value of the Hamiltonian i.e.:
$$E_{n}=\int{{\phi_{n}{(x)H\phi_{n}(x)dx}}$$
where $$\phi_{n}(x)$$ is the eigenfunction for the energy-level $$E_{n}$$
In the case of an ordinary one-dimensional harmonic oscillator
$$H=\frac{-\hbar^{2}}{2m}\frac{d^{2}\phi}{dx^{2}}+\frac{1}{2}m\omega^{2}x^{2}$$
And therefore the energy is given by
$$E_{n}=\int_{-\infty}^{\infty}{{(\phi_{n}{(x)\frac{-\hbar^{2}}{2m}\frac{d^{2}\phi}{dx^{2}}+\frac{1}{2}m\omega^{2}x^{2}\phi_{n}(x))dx}}$$
In your case you instead integrate from 0 to $$\infty$$.
So the energy is:
$$E_{n}=\int_{0}^{\infty}{{(\phi_{n}{(x)\frac{-\hbar^{2}}{2m}\frac{d^{2}\phi}{dx^{2}}+\frac{1}{2}m\omega^{2}x^{2}\phi_{n}(x))dx}}$$

7. Mar 22, 2008

### evildarklord1985

but then how do you actually carry out the intergration?

8. Mar 22, 2008

### malawi_glenn

have you tried? Where do you get stuck?

9. Mar 23, 2008

### eys_physics

Hi again

I did only give you a clue so you could continue by yourself. But maybe you need some more help. As you know the integral over the even interval $$-\infty$$ to $$\infty$$ you can use the fact that the integrand is an even function (why?), i.e.
$$f(-x)=f(x)$$

Know you want to integrate over the half interval and therefore use the rule (I suppose you have seen it in some course in mathematics):
$$\int_{-a}^{a}f(x)dx=2\int_{0}^{a}f(x)dx$$

I will give you a detailed solution because I want you to try by yourself, but kmow I suppose you have pathway at least.