How Does Symmetry Simplify Quantum Harmonic Oscillator Integrals?

[evildarklord1985]

 

[malawi_glenn]

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

 

[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...

 

[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?

 

[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?

 

[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}}

 

[evildarklord1985]

but then how do you actually carry out the intergration?

 

[malawi_glenn]

have you tried? Where do you get stuck?

 

[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.