Problem is probably trivial, but I can't seem to find a fault in the argument.

If we write the Dirac equation as $$(c\vec{\alpha}\cdot \mathbf{p} + \beta mc^2) \psi = i\hbar \frac{\partial \psi}{\partial t}$$ then the components of $\vec{\alpha}$ and $\beta$ anticommute and their squares are identity elements.

The $i^{\text{th}}$ component of the spin angular momentum is written as $$S_i = - \frac{i\hbar}{4} \epsilon_{ijk} \alpha_j \alpha_k.$$ If I write out $S_z$ component-wise it is trivial to show that $$S_z^2 = \frac{\hbar^2}{4}I_4.$$ I tried to prove the same result using Levi-Civita symbol as follows: $$\begin{align*} S^2 &= S_iS_i = -\frac{\hbar^2}{16} \epsilon_{ijk} \alpha_j \alpha_k\epsilon_{ilm} \alpha_l \alpha_m\\ &= -\frac{\hbar^2}{16} (\delta_{jl}\delta_{km} - \delta_{jm}\delta_{kl}) \alpha_j \alpha_k \alpha_l \alpha_m\\ &= -\frac{\hbar^2}{16} (\alpha_j \alpha_k \alpha_j \alpha_k - \alpha_j \alpha_k \alpha_k \alpha_j)\\ &= \frac{\hbar^2}{8} \alpha_j \alpha_k \alpha_k \alpha_j = \frac{9}{8}\hbar^2I_4, \end{align*}$$ which is obviously wrong. The end result is expected to be $$S^2 = \frac{3}{4}\hbar^2 I_4.$$ I think I am missing something very obvious here.

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    $\begingroup$ I think that for the multiplication by 9 you counted also the case where $j=k$ which is actually zero. So you will have only the correct $6/8=3/4$ at the end $\endgroup$ – yu-v Oct 21 '19 at 15:46
  • $\begingroup$ May be that is the reason. Expressions till line 3 had that built-in. Why we have to put that restriction artificially from thereon? Or is it just the way it is? $\endgroup$ – Abhinav Pratap Singh Oct 21 '19 at 17:46
  • $\begingroup$ @yu-v I got it. $\alpha_j \alpha_k = - \alpha_k \alpha_j$ implies $j \neq k$. Thank you, your reply helped me to realize my mistake. $\endgroup$ – Abhinav Pratap Singh Oct 21 '19 at 17:56

As pointed out by yu-v, the mistake lies in counting the cases with $j=k$. The problem lies in the negligence that as soon as you put $\alpha_j \alpha_k = - \alpha_k\alpha_j$, this implies $j \neq k$. Thus, $\alpha_j\alpha_k\alpha_k\alpha_j=6$.

Of course all this trouble could've been avoided using following anticommutation relation which holds for both (equal and unequal indices) cases: $$\{ \alpha_j, \alpha_k\} = 2\delta_{jk}.$$


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