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

In summary, the complete wavefunction in quantum mechanics is obtained by multiplying the radial eigenfunction by the spherical harmonic function. This is necessary for normalization and to calculate mean energies and distances. The 3-dimensional wavefunction is obtained by solving the time-independent Schrodinger equation for the Coulomb potential using separation of variables. To calculate ##u_{n}^{l=n-2}##, the operator A is defined and the differential equation is solved using the integrating factor method. However, when substituting n=2, a different result is obtained from the one in the book. The reason for this is unclear and further investigation is needed.
  • #1
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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
Because you need a 3-dimensional wavefunction because objects move in 3 dimensions.
 
  • #3
Vanadium 50 said:
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
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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  • #5
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
WannabeNewton said:
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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Related to To get the complete Radial Wavefunction, why do we multiply by Y(l,m)?

What is the purpose of multiplying by Y(l,m)?

The multiplication by Y(l,m) is necessary to obtain the complete radial wavefunction, which is a part of the total wavefunction for an atom or molecule. This ensures that the resulting wavefunction is a solution to the Schrödinger equation and accurately describes the probability of finding an electron in a specific energy state.

What is the significance of Y(l,m) in the radial wavefunction?

Y(l,m) represents the spherical harmonic function, which describes the angular distribution of the electron in an atom or molecule. Multiplying the radial wavefunction by Y(l,m) allows for the incorporation of the angular dependence of the wavefunction and results in a more accurate description of the electron's behavior.

Can the radial wavefunction be obtained without multiplying by Y(l,m)?

No, the complete radial wavefunction cannot be obtained without multiplying by Y(l,m). The Schrödinger equation, which governs the behavior of the wavefunction, requires that the wavefunction is a solution for both the radial and angular components. Therefore, both components must be multiplied together to obtain the complete wavefunction.

How does Y(l,m) affect the shape of the radial wavefunction?

The shape of the radial wavefunction is influenced by the spherical harmonic function Y(l,m). This function introduces nodes, or regions of zero probability, in the wavefunction depending on the values of l and m. The number and location of these nodes affect the shape of the wavefunction, which in turn affects the probability of finding an electron at a specific distance from the nucleus.

Is Y(l,m) the only factor that influences the radial wavefunction?

No, Y(l,m) is not the only factor that affects the radial wavefunction. Other factors such as the principal quantum number, n, and the azimuthal quantum number, l, also play a role in determining the shape and behavior of the radial wavefunction. However, Y(l,m) is a crucial component that incorporates the angular dependence of the wavefunction and allows for a more accurate description of the electron's behavior in an atom or molecule.

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