# Conservation of angular momentum in electronic transition

My book says the "angular momentum is always conserved in a transition". But if suppose we have a s electron then it's angular momentum is $\hbar\sqrt{l(l+1)}+\hbar\sqrt{s(s+1)}$ $\implies \hbar\frac{\sqrt{3}}{2}$ and suppose it transits to a p electron (which is in accordance with the selection rule as $\Delta l=1$) which has an angular momentum of $\hbar\sqrt2 + \hbar\frac{\sqrt{3}}{2}$. As the angular momentum of a photon is $\hbar$ then it means (mathematically) that we would have to add $\sqrt2$photons to the atom but that's not possible. Then how is the angular momentum conserved??

Also I can't see why the spin angular momentum of a photon is $\hbar$ as spin angular momentum is $\hbar\sqrt{s(s+1)}$ and as $s=1$ for a photon so it should be $\hbar\sqrt2$.

P.S. the book I am using is Elements of Physical Chemistry by Peter Atkins and Julia de Paula.

I think you have misunderstood some concept. The angular momentum of spin is just $\sqrt{s(s+1)}\hbar$. The spin which we always say is spin projection quantum number in fact not the angular momentum. And I'm not familiar with chemistry, but I know that we can not add up the orbital and spin angular momentum directly because they are not in the same direction. Sometimes we can add up the quantum number. If the spin and orbital angular momentum in opposite direction, we should subtract.

I have some idea about your example. The electron transition is "electron jumps from one energy level to another". So the energy must change. Where does the energy come from? If it is photon, this example has no paradox. But can the energy come from other way? I have no idea.

• I had that in the back of my mind because orbital and spin angular momentum are both vectors but I thought it would be better to clarify here. – user4973 Mar 25 '14 at 12:17
• @ShrayanshJyoti I doesn't understand what you say. In fact, angular momentum in quantum is not a classical angular momentum. – Ben Mar 25 '14 at 12:21
• "The classical definition of angular momentum as $\mathbf{L}=\mathbf{r}\times\mathbf{p}$ can be carried over to quantum mechanics, by reinterpreting r as the quantum position operator and p as the quantum momentum operator. L is then an operator, specifically called the orbital angular momentum operator."--Says Wikipedia. – user4973 Mar 25 '14 at 13:16
• Yes, that's right. – Ben Mar 25 '14 at 13:17