# First order perturbation theory, quantum physics

1. Sep 19, 2008

### Antti

I have an infinite potential well with length L. The first task was to calculate the eigenvalues and -functions for the energy of a particle in the well. The requirements were
$$\psi(0, L) = 0$$ and there is no time-dependence.

I've calculated:

$$\hat{H}\psi(x) = E\psi(x)$$

$$E = \frac{(\pi\hbar n)^{2}}{2 m L^{2}}$$

$$\psi(x) = sin(\frac{\pi\n x}{L})$$

Now the question. We add a small potential "rectangle" V(x) at the center of the potential well. It has length a and height q, a << L. What are the new eigenvalues and -functions for the perturbed case? Im supposed to use first order perturbation theory.

I've done like this:

There's a formula saying:

$$E_{n_p} = \int \overline{\psi_{n}} H' \psi_{n} dV =$$
$$\int \overline{\psi_{n}} H' \psi_{n} x^{2} dx$$

Which gives the new eigenvalues. I tried just using H' = V(x) = q at first. But the dimension of E didn't match. If you substitute V(x) (which is an energy) into the above equation you see that the dimension will be Energy*distance^4 after integration. So what to do with H' ?

Last edited: Sep 19, 2008
2. Sep 19, 2008

### yaychemistry

Hello,
So, two things:
First make sure that your eigenfunctions are properly normalized, e.g. in one dimension:
$$\int \overline{\psi_n}\psi_n\,dx = 1$$.
Also, what does this equation imply about the units of $$\psi$$? (hint: the dx infinitessimal has units of length)

Second, what is the volume element dV? If we are in one dimension then dV = dx. If we are in cartesian coordinates in 3 dimensions then dV = dxdydz and if we are in spherical coordinates then $$dV = r^2drd\theta d\phi$$ which appears to be the volume element you are using without the angular terms.

Hope this helps

3. Sep 21, 2008

### Antti

Thanks! It helped. Normalizing psi and replacing x^2 dx with just dx did the trick :)