# Free particle in spherical polar coords

1. Jun 2, 2009

### cscott

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

Consider the time-independent Schrodinger equation in spherical polar coordinates for a free particle, in the case where we have an azimuthal quantum number $l=0$.

(a) Solve the radial equation to find the (unnormalized) radial wavefunction $R(r)$.
(b) Normalize $R(r)$, using the definition of the dirac delta function $\delta(k'-k)$.

2. Relevant equations

$$u(r) = rR(r)$$

$$-\frac{\hbar^2}{2m}\frac{d^2u}{dr^2} + \left[V + \frac{\hbar^2}{2m}\frac{l(l+1)}{r^2}\right]u = Eu$$

3. The attempt at a solution

For a free particle, $V=0$, and with $l = 0$ the radial equation reduces to,

$$\frac{d^2u}{dr^2}=-k^2u,~~k=\frac{\sqrt{2mE}}{\hbar}$$,

with solution,

$$u = A\sin(kr) + B\cos(kr)$$,

but $u(r)=rR(r)$, so $B=0$ for a normalizable wavefunction (considering r->0). Therefore,

$$R(r) = \frac{A}{r}\sin(kr)$$

and to normalize,

$$\int_{0}^{\inf} r^2|R(r)|^2~dr = 1$$

$$|A|^2 \int_{0}^{\inf} \sin^2(kr)~dr=1$$

How do I use the dirac delta function?

Last edited: Jun 3, 2009
2. Jun 3, 2009

### cscott

Should I keep $u$ as,

$$u(r)=Ae^{ikr}$$

$$R(r)=\frac{A}{r}e^{ikr}$$

3. Aug 5, 2009

### esorolla

Hi

Sorry, but I don't see why do you need to use the delta of k'-k. I would integrate the sine square but the problem is that you forgot the integration for the angular variables. Thus there is a "4 times pi" factor missed which should be at RHS of the last equation as denominator of 1/(4*pi) since the 4*pi is the result of the integration of

$$\int sin\theta d\theta \int d\varphi$$

This comes from the fact that to normalize you have to integrate in a volume whose differential element is

$$dv= r^2 sin\theta dr d\theta d\varphi$$

But I don't see anything else to be added.