# Problem in the derivation of the Landé $g$-factor

I was watching a lecture on Youtube (https://www.youtube.com/watch?v=NSac7cMQnJw&t=971s) where the professor derives the Landé $$g$$-factor for the weak-field Zeeman effect. As part of the derivation he writes the equation (time stamp 4:10) $$\alpha^{-1}[J^2,[J^2 , \mathbf S]] = (\mathbf S\cdot \mathbf J) \mathbf J - \frac12 (J^2 \mathbf S + \mathbf S J^2) \tag{*}$$ for some constant $$\alpha$$. I tried this derivation myself, but ran into some hiccups which made me question the validity of this equation. Starting with just a single component of the inner commutator on the left, \begin{align*} [J^2 , S_x] &= [L^2+S^2 + 2\mathbf L \cdot \mathbf S, S_x] = 2[\mathbf L \cdot \mathbf S, S_x]\\&= 2 [L_xS_x + L_yS_y + L_zS_z, S_x] \\&= 2(L_y S_y S_x -L_y S_xS_y + L_z S_z S_x - L_z S_x S_z) \\ &= 2(-L_y[S_x, S_y] + L_z[S_z,S_x]) \\ &= 2i\hbar(S_yL_z - S_z L_y) = 2i\hbar (\mathbf S \times \mathbf L)_x \end{align*} (using the fact that all components of $$\mathbf L$$ commute with all components of $$\mathbf S$$ to bunch the $$S$$'s together). I generalized this to $$[J^2 , \mathbf S] = 2i\hbar (\mathbf S \times \mathbf L) = 2i\hbar (\mathbf S \times (\mathbf J - \mathbf S)) = 2i\hbar (\mathbf S \times \mathbf J).$$ Moving on to a single component of the whole commutator on the left, we have \begin{align*} [J^2,[J^2 , S_x]] &= 2i\hbar [J^2, (\mathbf S \times \mathbf J)_x] \\ &=2i\hbar[J^2, S_yJ_z - S_z J_y] \\ &= 2i\hbar \{ [J^2, S_y J_z ] - [J^2, S_z J_y]\}. \end{align*} But $$[A,BC] = ABC - BCA = ABC - BAC + BAC - BCA = [A,B]C + B[A,C],$$ so \begin{align*} [J^2,[J^2 , S_x]] &= 2i\hbar \{ [J^2, S_y]J_z + [J^2, J_z]S_y - [J^2, S_z]J_y - [J^2, J_y]S_z\} \\ &= 2i\hbar\{ [J^2, S_y]J_z - [J^2, S_z]J_y \} \\ &= 2i\hbar \{2i\hbar (\mathbf S \times \mathbf J)_y J_z - 2i\hbar (\mathbf S \times \mathbf J)_z J_y \} \\ &= -4\hbar^2 \{ (\mathbf S \times \mathbf J) \times \mathbf J\}_x. \end{align*} Once again, I generalized and used the BAC-CAB rule for triple cross products to obtain \begin{align*} [J^2,[J^2 , \mathbf S]] &= -4\hbar^2[ (\mathbf S \times \mathbf J) \times \mathbf J] \\ &=-4\hbar^2[-(\mathbf J \cdot \mathbf J)\mathbf S + (\mathbf S\cdot \mathbf J)\mathbf J]\\ &=-4\hbar^2[(\mathbf S\cdot \mathbf J)\mathbf J - J^2\mathbf S], \end{align*} which concludes the LHS. On the other hand, the RHS seems to be \begin{align*} (\mathbf S\cdot \mathbf J) \mathbf J - \frac12 (J^2 \mathbf S + \mathbf S J^2) &= (\mathbf S\cdot \mathbf J) \mathbf J - \frac12 ( J^2\mathbf S + J^2\mathbf S - [J^2, \mathbf S]) \\ &= (\mathbf S\cdot \mathbf J) \mathbf J - J^2\mathbf S + i\hbar(\mathbf S \times \mathbf J). \end{align*} This is so tantalizingly close to the required result, yet I can't seem to get rid of the extra $$i\hbar(\mathbf S \times \mathbf J)$$ term. At the same time, I am inclined to trust the MIT professor more than myself, so what went wrong with this derivation?

There are two problems with your derivation. The first one is in the calculation of $$[\vec S,\vec J^2]$$. In the final step, you are implicitly using $$\vec S\times \vec S$$ which is false. A close attention to the anticommutation rules gives: $$\vec S\times\vec S = i\hbar\vec S$$ so this leads to: $$[\vec J{}^2,\vec S] = 2i\hbar((\vec S\times \vec J)-i\hbar \vec S)$$

You also have a second problem in the triple cross product formula. If you look closely, you are implicitly assuming that the three operators commute when you apply it.

Instead, you'd get: $$((\vec S\times \vec J)\times \vec J)_k = S_lJ_kJ_l-S_kJ_lJ_l\\ = S_lJ_lJ_k+S_l i\hbar\epsilon_{klm}J_m-S_kJ_lJ_l$$ so in vectorial notation: $$(\vec S\times \vec J)\times \vec J = (\vec S\cdot \vec J)\vec J+i\hbar(\vec S\times \vec J)-\vec S\vec J{}^2$$ as you can see, you forgot the extra cross product term and the ordering of your second term was a bit cavalier. You can further symmetrize the third, final term: $$\vec S\vec J{}^2 = \frac{1}{2}(\vec S\vec J{}^2+\vec J{}^2\vec S-[\vec J{}^2,\vec S])\\ =\frac{1}{2}(\vec S\vec J{}^2+\vec J{}^2\vec S)-\frac{1}{2} [\vec J{}^2,\vec S]$$ to get: $$(\vec S\times \vec J)\times \vec J = (\vec S\cdot \vec J)\vec J+i\hbar(\vec S\times \vec J)-\frac{1}{2}(\vec S\vec J{}^2+\vec J{}^2\vec S)+\frac{1}{2} [\vec J{}^2,\vec S]$$

Incorporating these modifications, you'd rather get: $$[\vec J{}^2,[\vec J{}^2,\vec S]] = 2i\hbar[\vec J{}^2,(\vec S\times \vec J)-i\hbar \vec S)]\\ = 2i\hbar([\vec J{}^2,\vec S]\times \vec J+\vec S\times[\vec J{}^2,\vec J]-i\hbar[\vec J{}^2,\vec S])\\ = -4\hbar^2(((\vec S\times \vec J)-i\hbar \vec S)\times \vec J-\frac{1}{2}[\vec J{}^2,\vec S])\\ = -4\hbar^2((\vec S\cdot \vec J)\vec J+i\hbar(\vec S\times \vec J)-\frac{1}{2}(\vec S\vec J{}^2+\vec J{}^2\vec S)+\frac{1}{2} [\vec J{}^2,\vec S]-i\hbar(\vec S\times \vec J)-\frac{1}{2}[\vec J{}^2,\vec S])\\ = -4\hbar^2((\vec S\cdot \vec J)\vec J-\frac{1}{2}(\vec S\vec J{}^2+\vec J{}^2\vec S))$$ which is the advertised form, with $$\alpha = -4\hbar^2$$ (which has the correct dimensions).

Hope this helps and tell me if something's not clear.

• Yup, this was clear, thanks so much! Jun 14 at 9:37
• sorry there is still a mistake I'll rectify it (it's tricky to get everything right)
– lpz
Jun 14 at 9:38
• Again, thanks for the answer. Those identities I falsely used just seemed so apparent from when I studied electrodynamics, which is why I guess operators are so difficult to work with... Jun 14 at 10:44