# To get the complete Radial Wavefunction, why do we multiply by Y(l,m)?

1. Feb 23, 2014

### unscientific

Hi guys, I'm doing a quantum course at the moment, and there's one thing in Binney's book which I don't really understand:

Why must you multiply the radial eigenfunction by $Y_l^m$ to get the "complete wavefunction" ?

They do this to normalize the eigenfunction, and to do things like calculate mean energies and mean distance from the nucleus:

Last edited: Feb 23, 2014
2. Feb 23, 2014

Staff Emeritus
Because you need a 3-dimensional wavefunction because objects move in 3 dimensions.

3. Feb 23, 2014

### unscientific

Why does multiplying $Y_l^m$ to eigenfunction $u_n$ make it a 3-dimensional wavefunction?

The "1-D" eigenfunction is obtained by solving:

4. Feb 23, 2014

### WannabeNewton

You need the energy eigenfunctions to form a complete set. You need to include the spherical harmonics in the radial eigenfunctions in order to make sure these eigenfunctions do form a complete set. All of this just comes from solving the 3D time-independent Schrodinger equation for the Coulomb potential. What we do is use separation of variables to break the eigenvalue equation down into a radial equation, giving us the solutions you cited, and an angular equation that gives us spherical harmonics. In the end multiply the radial solutions and the spherical harmonics together to get the complete set of solutions just like for all separation of variables problems.

5. Feb 23, 2014

### dextercioby

The complete radial function is made up of 2 completely different parts, the part for the discrete energy spectrum and the part of the continuous energy spectrum. For a certain spectral value of the Hamiltonian, you can write down the full wavefunction of the H-atom as a product of the radial eigenfunction and the spherical harmonic. Check out formulas 4.12 and 4.13 of Bethe and Salpeter.

6. Feb 23, 2014

### unscientific

Thanks I got it. But when i try to calculate $u_{n}^{l=n-2}$, something goes wrong:

Starting, we define operator A by:

$$A_{n-2} = \frac{a_0}{\sqrt 2}\left(\frac{i}{\hbar}p_r + \frac{1-n}{r} + \frac{Z}{(n-1)a_0}\right)$$

Substituting $p_r = -i\hbar (\frac{\partial}{\partial r} + \frac{1}{r})$:

$$A_{n-2} = \frac{a_0}{\sqrt 2}\left( \frac{\partial}{\partial r} + \frac{2-n}{r} + \frac{Z}{(n-1)a_0}\right)$$

Thus, we want to solve:

$$\left(\frac{\partial}{\partial r} + \frac{2-n}{r} + \frac{Z}{(n-1)a_0} \right) u_{n}^{l=n-2} = 0$$

Solving by integrating factor method, we obtain:

$$u_n^{l=n-2} = A r^{n-2} e^{\frac{Z}{(n-1)a_0}r}$$

Normalizing,

$$u_n^{l=n-2} = \frac{1}{\sqrt{[2(n-1)]!}} \left(\frac{2Z}{(n-1)a_0}\right)^{\frac{3}{2}} \left(\frac{2Z}{(n-1)a_0}\right)^{n-2} e^{-\frac{Z}{(n-1)a_0}r}$$

This is similar to $u_n^{l=n-1}$, simply replace n by n-1:

But when I substitute n = 2, so l = 0, I get $u_2^0 = \frac{1}{\sqrt 2} \left(\frac{2Z}{a_0}\right)^{\frac{3}{2}} e^{-\frac{Z}{a_0}r}$

I get a completely different result from the book:

I'm not sure what's wrong with my derivation?

Last edited: Feb 23, 2014