1. Nov 22, 2015

### zhaos

1. The problem statement, all variables and given/known data

I think, to normalize a wavefunction, we integrate over the solid angle $r^2 dr d\theta d\phi$. Typically we have $R(r)Y(\theta, \phi)$ as solutions. If $Y$ is properly normalized, then the normalization condition for $R(r)$ ought to be

$$\int_0^\infty dr r^2 |R(r)|^2 = 1$$

2. Relevant equations

Now I am trying to develop a general normalization condition for the spherical Bessel functions, which are solutions to the free particle in free space, but solved using spherical symmetry. For example the 0th solution is

$$R(r) = \sin (kr)/r$$

where $k = \sqrt{2mE/\hbar}$.

Now if I were to normalize with integration over $r^2 dr$, the normalization integral will not converge. Google searching has been unhelpful so far, but I've tended to see just integration over $dr$. So I'm conceptually unsure why this would be the case.

3. The attempt at a solution

The end goal is to have a general normalization factor for the spherical Bessel functions. If I normalize

$$\int_0^\infty dr | A R_0(r)|^2 = 1$$

In this case, $A = \sqrt{2/\pi k}$.

In fact I found by just doing the normalization for the first few functions, sqrt(2l+1) works as the additional normalization factor.
But the current sticking point is the integration over dr or r^2 dr.

Thanks.

Last edited: Nov 22, 2015
2. Nov 22, 2015

### king vitamin

Free particle states aren't normalizable. You can see this pretty quickly in cartesian coordinates (showing one dimension for simplicity):
$$\psi_k(x) = Ne^{i k x}$$
where $k$ is a quantum number related to energy $E = \hbar^2 k^2/2m$. Since $|\psi(x)|^2 = 1$, clearly you can't normalize this state.

Instead, it's conventional to normalize the states as
$$\int dx \psi_{k'}^*(x) \psi_{k}(x) = \delta(k - k')$$
with the dirac delta function. Then you can show that $N = 1/\sqrt{2 \pi}$. This normalization is useful because you can then write down general normalizable states as
$$\Psi(x) = \int dk A(k) \psi_k(x)$$
and now the normalization condition on $\Psi$ becomes
$$\int dx |\Psi(x)|^2 = \int dk A(k) = 1.$$
So $A(k)$ is a normalized distribution over which $k$'s contribute to the wave function. So a good choice for normalizing for your spherical problem would be to generalize the delta function normalization above.

3. Nov 23, 2015

### zhaos

I guess i have
$$|A|^2 \int_0^\infty dr r^2 \psi^*_{0,k'}(r)\psi_{0,k}(r) = \delta(k-k')$$
and $\psi_0(r) = \sin(kr)/r = \frac{i}{r}( \exp(ikr) - \exp(-ikr))$, and the r^2 will cancel, and then one does the integral...

Last edited: Nov 23, 2015
4. Nov 23, 2015

### king vitamin

Looks good. You'll need to use the integral I used above,
$$\int_{-\infty}^{\infty} dx e^{ikx} = 2 \pi \delta(k),$$
and be careful with all of the algebra. A few points: you should actually have $j_{0}(kr) = \sin(kr)/(kr)$, and you're missing some factors for the exponential form for $\sin$.