How Does the Zeeman Effect Impact Hydrogen Atom Energy Levels?

So ##\langle j,m | \vec L + 2 \vec S|j,m\rangle = \langle j,m |l,m\rangle + \langle j,m |2S,m\rangle##In summary, we discussed the origin of an expression and how to estimate the magnetic field using quantum numbers. We also explored the Zeeman effect on states 1s and 2s, and how to find the equivalent magnetic field and evaluate the magnetic flux density. We also looked at the difference between the Zeeman Spin Hamiltonian and Spin Orbit Hamiltonian and how they are used to calculate the total external field interaction Hamiltonian. Finally, we discussed the importance of considering the electron's position when evaluating the magnetic flux density.
  • #1
unscientific
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Homework Statement



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Part (a): What's the origin of that expression?
Part(b): Estimate magnetic field, give quantum numbers to specify 2p and general nl-configuration
Part (c): What is the Zeeman effect on states 1s and 2s?

Homework Equations


The Attempt at a Solution



Part (b)
[tex]H = -\frac{e}{m^2c^2} \frac{1}{r} \frac{\partial \phi}{\partial r} \vec S . \vec L[/tex]
[tex]-\mu . B = -\frac{e}{m^2c^2} \frac{1}{r} \frac{\partial \phi}{\partial r} \vec S . \vec L[/tex]
[tex]\frac{e}{2m}\vec S . \vec B = -\frac{e}{m^2c^2} \frac{1}{r} \frac{\partial}{\partial r} \vec S . \vec L[/tex]
[tex]|\vec B| = |\vec L| \frac{e}{m^2 c^2} \frac{m}{e} \frac{1}{r} \frac{\partial \phi}{\partial r}[/tex]

Now, ##|L| = l\hbar = \hbar## and ##\frac{\partial \phi}{\partial r} = E = \frac{e}{4\pi \epsilon_0 r^2}##.

[tex]|\vec B| = \frac{e\hbar}{mc^2 4\pi \epsilon_0 r^3}[/tex]

What value of ##r## must I use? When I use ##r = 4a_0## it gives the right answer.. as B = 0.2 T. Why can't I use ##B = a_0##?

For the 2p configuration, n =2, j = 3/2 or n=2, j = 1/2.
For general nl-configuration, ##0 < l < n, j = l \pm \frac{1}{2}##.

Part (c)

[tex]\Delta H = -\frac{e^2}{m^2c^24\pi \epsilon_0 r^3} (\vec S . \vec L)[/tex]

We are supposed to find ##\langle \Delta H\rangle##:

##\vec S . \vec L## can be written as ##\frac{1}{2}(J^2 - S^2 - L^2)##, with eigenvalues ##\frac{l}{2}## for j = l + 1/2, and ##-\frac{1}{2}(l+1)## for j = l - 1/2.

Thus for j = l + 1/2, the splitting becomes:
[tex]\frac{e^2}{m^2c^2 4\pi \epsilon_0} \frac{1}{(l+1)(2l+1)}\left(\frac{1}{na_0}\right)^3 [/tex]

For j = l - 1/2, the splitting becomes:
[tex]\frac{e^2}{m^2c^2 4\pi \epsilon_0} \frac{1}{l(2l+1)}\left(\frac{1}{na_0}\right)^3 [/tex]

I'm not sure how to proceed from here..
 
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  • #2
unscientific said:
What value of ##r## must I use? When I use ##r = 4a_0## it gives the right answer.. as B = 0.2 T. Why can't I use ##B = a_0##?
The electron is not at a fixed position, you have to integrate radially the wave function, which makes the ##\langle r^{-3} \rangle## appear.
 
  • #3
DrClaude said:
The electron is not at a fixed position, you have to integrate radially the wave function, which makes the ##\langle r^{-3} \rangle## appear.

Yes I get that, but for part (c), how does the magnetic field come into play?
 
  • #5
DrClaude said:
You need to add an additional perturbation corresponding to the coupling with the magnetic field.
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/zeeman.html

I've been reading binney's book and I'm extremely confused. He mentioned the 'Zeeman Spin Hamiltonian' and the 'Spin Orbit Hamiltonian' - what's the difference and what are they used for?
2mee438.png


My understanding is that:

1. In the absence of an external electric field, the electron moving around the atom experiences a magnetic torque. This torque causes its spin to precess. This precession results in an interaction (energy) between the electron's orbital motion and spin.

2. Part (b) asks us to find the electron's own 'equivalent magnetic field' that's causing this effect. The spin interaction hamiltonian is simply ##H_{ZS} = -\vec \mu_s \cdot \vec B = -\gamma \vec S \cdot \vec B = -\frac{g_sq}{2m}\vec S \cdot \vec B = -\frac{e}{m} \vec S \cdot \vec B##

3. But in the presence of an external magnetic field, two things happen. Firstly, the spin interacts with the external field, giving hamiltonian : ##H_{ZS} = -\frac{e}{m} \vec S \cdot \vec B##.
Secondly, the electron's orbital angular momentum interacts with the external field, giving hamiltonian: ##H_l = - \vec \mu_l \cdot \vec B = -\frac{g_l q}{2m} \vec L \cdot \vec B = -\frac{e}{2m} \vec L \cdot \vec B##. In this case, ##g_l = 1## because there is no degeneracy.

Together, the TOTAL external field interaction Hamiltonian is given by:
[tex]H = H_{ZS} + H_l = \frac{e}{2m} \left( \vec L + 2\vec S\right)\cdot \vec B[/tex]
 
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  • #6
DrClaude said:
The electron is not at a fixed position, you have to integrate radially the wave function, which makes the ##\langle r^{-3} \rangle## appear.

Also, for part (b), the expression for the magnetic flux density is this:

[tex]|\vec B| = \frac{e\hbar}{mc^2 4\pi \epsilon_0 r^3}[/tex]

For a 2p configuration, when I use ##r = a_0## it doesn't give the right answer. Only ##r = 4a_0## gives the right answer. Why is that so?
 
  • #7
How do I evaluate ##\frac{Be}{2m}\langle j,m | L_z + 2S_z|j,m\rangle##?
 
  • #8
I think I got it. In this question, l = 0, so ##\frac{Be}{2m}\langle j,m | L_z + 2S_z|j,m\rangle = \frac{Be}{2m}\langle \pm |2S_z|\pm\rangle = \pm \frac{Be}{2m}(\hbar)##.

If ##l \neq 0##, we have to add angular momenta.
 

FAQ: How Does the Zeeman Effect Impact Hydrogen Atom Energy Levels?

1. What is the Zeeman effect in Hydrogen?

The Zeeman effect in Hydrogen is a phenomenon in which the spectral lines of Hydrogen atoms split into multiple lines when exposed to a magnetic field. This effect was first discovered by Dutch physicist Pieter Zeeman in 1896.

2. What causes the Zeeman effect in Hydrogen?

The Zeeman effect in Hydrogen is caused by the interaction between the magnetic moment of the Hydrogen atom's electron and the external magnetic field. This interaction causes the energy levels of the electron to split, resulting in the splitting of the spectral lines.

3. How does the Zeeman effect in Hydrogen provide evidence for quantum mechanics?

The Zeeman effect in Hydrogen provides evidence for quantum mechanics because it demonstrates the quantized nature of energy levels in atoms. The splitting of spectral lines can only be explained by considering the discrete energy levels predicted by quantum mechanics.

4. What are the two types of Zeeman effect in Hydrogen?

There are two types of Zeeman effect in Hydrogen: normal Zeeman effect and anomalous Zeeman effect. In normal Zeeman effect, the spectral lines split into three components, while in anomalous Zeeman effect, the lines split into more than three components.

5. How is the Zeeman effect in Hydrogen used in scientific research?

The Zeeman effect in Hydrogen is used in scientific research to study the structure and behavior of atoms and to measure magnetic fields. It has also been used to investigate the composition and properties of stars and other celestial objects.

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