Hyperfine structure of deuterium

AI Thread Summary
The discussion centers on calculating the hyperfine splitting of the deuterium Balmer alpha line, specifically for the transition from the 3p3/2 to 2s1/2 states. The user identifies six allowed transition lines based on quantum numbers and calculates the transition energy without hyperfine interaction, yielding a wavelength consistent with hydrogen. However, they struggle with the hyperfine correction, particularly in determining the expectation value for the first term of Griffiths' equation due to the complexity of the 3p3/2 state. The user also questions whether they can use the hydrogen wave function for deuterium in their calculations. Ultimately, they find a general expression for hydrogen-like atoms that applies to both l=0 and l≠0 states, confirming its validity for their problem.
JulienB
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Homework Statement



Hi everybody! Here is the problem:

Hyperfine splitting of Deuterium Balmer alpha line:
Consider the ##3p_{3/2}## to ##2s_{1/2}## transition in deuterium. Find the number of transition lines and their frequency shifts with respect to the transition without hyperfine interaction.

We were also given in a previous exercise ##g_d=0.8574## and the nuclear spin ##I=1##.

Homework Equations



Which equation is relevant is kind of the knot of my problem.

The Attempt at a Solution



So first I've determined the transition lines. I think that ##F=\lbrace I-j,I-j+1,...,I+j\rbrace## (if ##I>j##) from the other quantum numbers, but we were not given a definitive formula regarding ##F##. However this seems to have given coherent results with the hydrogen splitting and with what I found on the internet:

##3p_{3/2} \implies j=3/2 \implies F={1/2,3/2,5/2}##

##2s_{1/2} \implies j=1/2 \implies F={1/2,3/2}##

Therefore I have six transition lines. I think they are all allowed (correct me if I'm wrong), because ##I## is only nucleus-related, so we can rely on the allowed transitions of ##j## and ##l##. Here for every transition ##\Delta j=1## and ##\Delta l=1## so I imagine there is no problem with that.

Then I've calculated the transition energy without hyperfine structure using the fine structure equation and the reduced mass:

##E_nj=-\frac{m_r c^2 \alpha^2}{2} \frac{1}{n^2} \left[1+\frac{\alpha^2}{n^2} \left( \frac{n}{j+1/2} - \frac{3}{4} \right) \right]##
##\implies \Delta E=E_{3\frac{3}{2}} - E_{2\frac{1}{2}}=1.8893##eV

This gives me a wavelength ##\lambda=656.3##nm just like for hydrogen, so that seems correct. But now the only thing I know about hyperfine correction is the equation from Griffiths:

##E_{hf}^1=\frac{\mu_0 g e^2}{8 \pi m_p m_e} \langle \frac{3(\vec{S}_p \cdot \hat{r})(\vec{S}_e \cdot \hat{r})-\vec{S}_p\cdot\vec{S}_e}{r^3} \rangle + \frac{\mu_0 g e^2}{3 m_p m_e} \langle \vec{S}_p \cdot \vec{S}_e \rangle |\psi(0)|^2##

The first term luckily vanishes for ##l=0##, but unfortunately the ##3p_{3/2}## has ##l=1##. I am not sure how to calculate the expectation value of the first term in that case (it looks really hard).

My second problem is that even if ##l=0## for ##2s_{1/2}##, then I don't have the wave function of deuterium for ##n=2##, and google gave me only vague articles about approximations that looked suspicious in the context of my homework. Can I take the wave function ##\psi_{200}(0)## of hydrogen here? I know I can for ##n=1##.

All in all I think there must be another way that is not so hard. Maybe I just misunderstood the question? Griffiths says nothing about hyperfine structure for ##l\neq 0## and the Wikipedia article about hyperfine structure is very long when it comes to the 1st term, and what is done there seems foreign to me.

Any suggestion on how to tackle this problem?Thanks a lot in advance.Julien.
 
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Suggestion would be to google the hyperfine interaction in hydrogen. The only difference would be the different reduced mass for deuterium as well as the different mass of the nucleus. You might find a source that does more of the calculations for you.
 
@Charles Link Hi and thanks for your answer. So you say that the wave functions are the same?
Julien.
 
If it interests somebody, at the end I took the general expression for hydrogen-like atoms of Bethe & Salpeter, derived in their book "Quantum Mechanics of One- and Two-Electron Atoms". The excerpt where the equation (eq. (22.12)) is can be found here: http://hacol13.physik.uni-freiburg....0-Positronium/Anhang/H.A.Bethe,E.Salpeter.pdf. I've tested the equation on hydrogen and deuterium, and it is equally valid for ##l=0## and ##l\neq 0##.
 
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