(n=3, l=2, j=5/2) -> (n=2, l=1, j=1/2) hydrogen transition?

  • Thread starter Thread starter andresordonez
  • Start date Start date
  • Tags Tags
    Hydrogen Transition
andresordonez
Messages
65
Reaction score
0
Hi, I have a doubt about the fine structure of the hydrogenic atoms. In the section 3.2 of the book Physics of Atoms and Molecules by Bransden & Joachain says:

Since the electric dipole operator \mathbf{D} = -e\mathbf{r} does not depend on the spin, the selection rule derived in Chapter 4 for the quantum number l (in the dipole approximation) remains

\Delta l = \pm 1

from which it follows that the selection rule with respect to the quantum number j is

\Delta j = 0, \pm 1

My question is, why \Delta j = \pm 2 is not mentioned here? For example, why the transition
(n=3, l=2, j=5/2) \rightarrow (n=2, l=1, j=1/2) is not considered?

(where n is the energy number, l is the orbital number, and j is the total angular momentum number)
 
Physics news on Phys.org
Why would it be ? How are j and l related ?
 
Yeah, sorry I should have posted it before; l and j are related like this:

j = l \pm \frac{1}{2}
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top