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

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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:

j12f6g.png
 
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  • #2
Vanadium 50
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Because you need a 3-dimensional wavefunction because objects move in 3 dimensions.
 
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Because you need a 3-dimensional wavefunction because objects move in 3 dimensions.
Why does multiplying ##Y_l^m## to eigenfunction ##u_n## make it a 3-dimensional wavefunction?

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

33c659v.png
 
  • #4
WannabeNewton
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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.
 
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dextercioby
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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.
 
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  • #6
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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.
Thanks I got it. But when i try to calculate ##u_{n}^{l=n-2}##, something goes wrong:

Starting, we define operator A by:

[tex]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)[/tex]

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

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

Thus, we want to solve:

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

Solving by integrating factor method, we obtain:

[tex]u_n^{l=n-2} = A r^{n-2} e^{\frac{Z}{(n-1)a_0}r}[/tex]

Normalizing,

[tex]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} [/tex]

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

ajx85d.png


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:

1ftl6a.png


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