1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Zeeman effect in Hydrogen

  1. May 7, 2014 #1
    1. The problem statement, all variables and given/known data

    2i8k7pw.png

    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?

    2. Relevant equations



    3. 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..
     
  2. jcsd
  3. May 8, 2014 #2

    DrClaude

    User Avatar

    Staff: Mentor

    The electron is not at a fixed position, you have to integrate radially the wave function, which makes the ##\langle r^{-3} \rangle## appear.
     
  4. May 9, 2014 #3
    Yes I get that, but for part (c), how does the magnetic field come into play?
     
  5. May 10, 2014 #4

    DrClaude

    User Avatar

    Staff: Mentor

  6. May 10, 2014 #5
    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]
     
    Last edited: May 10, 2014
  7. May 10, 2014 #6
    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?
     
  8. May 10, 2014 #7
    How do I evaluate ##\frac{Be}{2m}\langle j,m | L_z + 2S_z|j,m\rangle##?
     
  9. May 15, 2014 #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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Zeeman effect in Hydrogen
  1. Zeeman effect (Replies: 1)

  2. Zeeman Effect Question (Replies: 1)

Loading...