Constructing spatial wave function of hydrogen

In summary, the conversation is about constructing the spatial wave function for hydrogen in the state n = 3, l =2, m = 1. The book only shows that spherical harmonics are eigenfunctions of L^2 and L_z, not how to generate them with L_{\pm} operators. The person is wondering if there is a way to generate spherical harmonics using similar operators as the harmonic oscillator operator method. They also mention having trouble with problem 4.22, where they need to solve for g(\phi) and f(\theta). They have solved for g(\phi) as expected, but are having trouble with f(\theta) and are unsure if they have made a mistake.
  • #1
CrusaderSean
44
0
I'm having problem with griffith QM problem 4.43:
Construct the spatial wave function for hydrogen in the state n = 3, l =2, m = 1. Express your answer as a function of r, [tex]\theta[/tex], [tex]\phi[/tex], and a (the Bohr radius) only.

My prof. gave hints about radial wave function, but I have no idea how to construct spherical harmonics from scratch. The book only shows spherical harmonic are eigenfunctions of [tex]L^2[/tex] and [tex]L_z[/tex], not how you can generate them with [tex]L_{\pm}[/tex] operators. So is there a way to generate spherical harmonics with operators similar to harmonic oscillator operator method? I guess more specifically, is there a way to get spherical harmonics for [tex]Y_{l=2}^{m=0}[/tex]? I know I can apply [tex]L_{+}[/tex] to get [tex]Y_{l=2}^{m=1}[/tex].

edit: nevermind, looks like Griffith made generating spherical harmonics into problem 4.22 as oppose to showing how to do it. that was very tricky of my professor to sneak another problem into homework.

edit2: I'm having problems with 4.22...
denoting spherical harmonics as [tex]Y_{l}^{m}(\theta,\phi)=f(\theta)g(\phi)[/tex], solve for [tex]g(\phi)[/tex] first.
[tex]L_{z}Y_{l}^{l}=\hbar l Y_{l}^{l}[/tex]
[tex]g(\phi)=e^{i l \phi}[/tex] as expected.

now for [tex]f(\theta)[/tex]
[tex]L_{+}Y_{l}^{l}=0[/tex]
where [tex]L_{+}=\hbar e^{i \phi}(\partial_{\theta}+i \cot \theta \partial_{\phi})[/tex]
solving this I get
[tex]\ln f(\theta) = \frac{l}{\hbar}e^{i \phi} \ln \sin \theta[/tex]
which doesn't look right... it's an exponential of an exponetial... any clues as to what I did wrong?
 
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  • #2
CrusaderSean said:
I'm having problem with griffith QM problem 4.43:
Construct the spatial wave function for hydrogen in the state n = 3, l =2, m = 1. Express your answer as a function of r, [tex]\theta[/tex], [tex]\phi[/tex], and a (the Bohr radius) only.

My prof. gave hints about radial wave function, but I have no idea how to construct spherical harmonics from scratch. The book only shows spherical harmonic are eigenfunctions of [tex]L^2[/tex] and [tex]L_z[/tex], not how you can generate them with [tex]L_{\pm}[/tex] operators. So is there a way to generate spherical harmonics with operators similar to harmonic oscillator operator method? I guess more specifically, is there a way to get spherical harmonics for [tex]Y_{l=2}^{m=0}[/tex]? I know I can apply [tex]L_{+}[/tex] to get [tex]Y_{l=2}^{m=1}[/tex].
I don't know why you could not simply use Table 4.3 or Equation 4.32 (I am assuming you have the second edition)

edit: nevermind, looks like Griffith made generating spherical harmonics into problem 4.22 as oppose to showing how to do it. that was very tricky of my professor to sneak another problem into homework.

edit2: I'm having problems with 4.22...
denoting spherical harmonics as [tex]Y_{l}^{m}(\theta,\phi)=f(\theta)g(\phi)[/tex], solve for [tex]g(\phi)[/tex] first.
[tex]L_{z}Y_{l}^{l}=\hbar l Y_{l}^{l}[/tex]
[tex]g(\phi)=e^{i l \phi}[/tex] as expected.

now for [tex]f(\theta)[/tex]
[tex]L_{+}Y_{l}^{l}=0[/tex]
where [tex]L_{+}=\hbar e^{i \phi}(\partial_{\theta}+i \cot \theta \partial_{\phi})[/tex]
solving this I get
[tex]\ln f(\theta) = \frac{l}{\hbar}e^{i \phi} \ln \sin \theta[/tex]
which doesn't look right... it's an exponential of an exponetial... any clues as to what I did wrong?

You should not have any exponential of [itex] \phi[/itex] left! You get an equation of the form [itex]L_+ Y_l^l = e^{i(l+1)\phi} (\partial_\theta - l cot \theta) f(\theta) = 0 [/itex]. So you divide by the exponential in phi which goes away and then solve for f(theta).

Pat
 

FAQ: Constructing spatial wave function of hydrogen

What is the spatial wave function of hydrogen and why is it important?

The spatial wave function of hydrogen is a mathematical representation of the probability of finding an electron at a certain position around the nucleus of a hydrogen atom. It is important because it helps us understand the behavior and properties of hydrogen atoms, which is the most abundant element in the universe.

How is the spatial wave function of hydrogen constructed?

The spatial wave function of hydrogen is constructed using Schrödinger's equation, which is a mathematical equation that describes the energy and behavior of quantum particles. This equation takes into account the potential energy of the electron and the mass of the nucleus to determine the probability distribution of finding the electron at a certain position.

What are the factors that affect the shape of the spatial wave function of hydrogen?

The main factors that affect the shape of the spatial wave function of hydrogen are the principal quantum number (n), which determines the size of the orbital, and the angular momentum quantum number (l), which determines the shape of the orbital. Additionally, the magnetic quantum number (m) and spin quantum number (s) also play a role in determining the orientation and spin of the electron.

Can the spatial wave function of hydrogen be experimentally observed?

No, the spatial wave function of hydrogen cannot be directly observed since it is a mathematical construct. However, the probability distribution that it represents can be experimentally observed through spectroscopy and other techniques, which can provide valuable information about the electron's behavior.

Why does the spatial wave function of hydrogen have multiple energy levels?

The spatial wave function of hydrogen has multiple energy levels because the electron can exist in different orbitals with different energies. These energy levels are determined by the principal quantum number (n), with higher values of n corresponding to higher energy levels. The energy levels also play a role in determining the stability and reactivity of the hydrogen atom.

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