1
$\begingroup$

I've been asked to prove $\left[ \vec{L}^2, x_i \right] = -2i\hbar \varepsilon_{ijk}L_j x_k -2\hbar^2 x_i $ and I don't seem to get it correctly.

I propose $\left[ \vec{L}^2, x_i \right] = \left[ L_l L_l, x_i \right] = L_l\left[ L_l, x_i \right] +\left[ L_l, x_i \right]L_l$ which leads me to $\left[ \vec{L}^2, x_i \right] = -i\hbar \varepsilon_{ilk}L_lx_k - i\hbar\varepsilon_{ilk}x_kL_l$ but I think I'm using some property not properly because I arrive to the following result $\left[ \vec{L}^2, x_i \right] = -i\hbar \varepsilon_{ilk}L_lx_k -i\hbar\varepsilon_{ilk}\left[L_lx_k -i\hbar\varepsilon_{lkm}x_m\right] =-2i\hbar \varepsilon_{ilk}L_lx_k +\hbar^2 \varepsilon_{ilk}\varepsilon_{lmk}x_m = -2i\hbar \varepsilon_{ilk}L_lx_k$

which is false. I used $\varepsilon_{ilk}\varepsilon_{lkm}=-\varepsilon_{lik}\varepsilon_{lkm} = -\left(\delta_{ik}\delta_{km}-\delta_{im}\delta_{kk}\right) = 0$ and $\left[L_l, x_k\right] = i\hbar \varepsilon_{lkm}x_m$.

Could you give me some hints? I'm starting to learn how to use Levi-Civita tensor

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ First problem: after the "which leads me to" replace $j\rightarrow l$. $\endgroup$
    – Sean E. Lake
    Commented Mar 17, 2020 at 14:49
  • 2
    $\begingroup$ Next: What is $\epsilon_{ijk}\epsilon_{klm}$ in terms of Kronecker deltas? $\endgroup$
    – Sean E. Lake
    Commented Mar 17, 2020 at 14:50
  • $\begingroup$ @SeanE.Lake I edited the question. I have corrected the fact with the $j$ and the $l$. I continue having the same problem :( $\endgroup$
    – Pablo
    Commented Mar 17, 2020 at 15:19

1 Answer 1

Reset to default
1
$\begingroup$

Found my error. $\delta_{ik}\delta_{km}-\delta_{im}\delta_{kk} = (1-3)\delta_{im} \neq 0$.

Thank you all!

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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