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?


1 Answer 1


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.

  • $\begingroup$ Yup, this was clear, thanks so much! $\endgroup$
    – chris
    Jun 14 at 9:37
  • $\begingroup$ sorry there is still a mistake I'll rectify it (it's tricky to get everything right) $\endgroup$
    – lpz
    Jun 14 at 9:38
  • $\begingroup$ 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... $\endgroup$
    – chris
    Jun 14 at 10:44

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