Difference between J and l quantum numbers in rotation

[Ayame17]

[SOLVED] Quantum Numbers

What is the difference between using J and l for quantum numbers? I have some lecture notes that aren't fully explained. It was talking about rotational transitions for diatomic molecules, and said the energy of a photon going from level J to level J-1 is $$\frac{Jh^2}{4\pi^2\mu(r^2)}$$. Now, I remember from my quantum module last year, that $$E_{rot}=\frac{L^2}{2I}=\frac{l(l+1)h^2}{8\pi^2\mu(r^2)}$$. I can see the resemblance between the two equations, but I just can't figure out the link between J and l!

 

[waht]

"l" quantum is number is specifically used to describe angular momentum.

"j" can either describe angular momentum or spin.

 

[Ayame17]

Yes, I found those descriptions online, but it doesn't help me with the link between them - we were told that J was the rotational quantum number of the upper level. If these two equations ARE the same, that means that 2J=l(l+1) - is this true? And if so, what are the steps to prove it?

 

[malawi_glenn]

Many different notations and conventions are used. So you must look up the definition given in your difference.



But you have not been careful here, $$\frac{Jh^2}{4\pi^2\mu(r^2)}$$ is the energy difference of states with QM# J and J-1, and the level energies are given according to: $$E_{rot}=\frac{{\vec{J}}^2}{2I}=\frac{J(J+1)h^2}{8\pi^2\mu( r^2)}$$


So you must evaluate $$E_J - E_{J-1}$$

 

[Ayame17]

I hadn't realized it, but it's so simple when you put it like that. Thanks, that's helped a lot!

 

[malawi_glenn]

I hadn't realized it, but it's so simple when you put it like that. Thanks, that's helped a lot!

hehe