I have been dealing with total angular momentum of the single electron which is outside the closed shells in which sum of the angular momentums is zero.

My book says that total atomic angular momentum is $\mathbf{J}=\mathbf{L+S}$ and its magnitude $J=\sqrt{j(j+1)}\hbar$ where quantum number $j$ can have half-integer values on an interval $|\ell-s|\leq j \leq|\ell+s|$. There is allso a quantum number $m_j$ which can have half-integer values on an interval $-j \leq m_j \leq j $.

So lets say I have a single electron in an orbital $\scriptsize\boxed{\ell=1}$ (p orbital). In this case $j=\tfrac{3}{2},\frac{1}{2}$ while $m_j=\tfrac{3}{2},\frac{1}{2},-\tfrac{3}{2},-\frac{1}{2}$ and I can calculate two magnitudes for $J$:

\begin{align} J&=\tfrac{\sqrt{15}}{2}\hbar\\ J&=\tfrac{\sqrt{3}}{2} \hbar \end{align}

Now there is this weird vector sum image that I don't understand completely:

enter image description here

On the left image we have $J=\tfrac{\sqrt{15}}{2}\hbar$ while on the right we have $J=\tfrac{\sqrt{3}}{2}\hbar$. It seems that vector $\mathbf{L}$ together with its magnitude $L=\sqrt{\ell(\ell+1)}\hbar = \sqrt{2}\hbar$ is the same in both cases - this makes sense as quantum number $\ell$ doesn't change. But how do we explain the change in $\mathbf{S}$?


The left image describes a spin up ($s=1/2$) and the right one a spin down ($s=-1/2$) electron. These two states are directly responsible for the fact that there are two possible values for $j$. The total angular momentum takes on two possible values, one for the case where spin is negative and one for when it is positive. This is how you can understand the relation

$$|l-s|\leq j \leq|l+s|.$$

  • $\begingroup$ Spin magnitude of an electron is allways the same $S=\sqrt{s(s+1)}\hbar=\frac{\sqrt{3}}{2}\hbar$ where $s$ is the spin quantum number which for electron equals $1/2$. By $s$ you probably meant the $z$ component of the spin $S_z=m_s\hbar$ where $m_s= \pm s = \pm\frac{1}{2}$? So in one case $S_z = \frac{1}{2}\hbar$ while in the other case $S_z = -\frac{1}{2}\hbar$? $\endgroup$
    – 71GA
    Sep 8 '13 at 14:07
  • $\begingroup$ Yes, that is how I understand the two different values in the diagram. $\endgroup$ Sep 8 '13 at 14:11
  • $\begingroup$ This means that the $z$ axis must be defined in the same direction as the total angular momentum $\mathbf{J}$. $\endgroup$
    – 71GA
    Sep 8 '13 at 14:14
  • $\begingroup$ Yes, you have to define the axis in an appropriate way. $\endgroup$ Sep 8 '13 at 14:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.