I've been working on some quantum mechanics problems and arrived to this one where I have to deal with subscripts. I got stuck doing this: I have $\epsilon_{imk}\epsilon_{ikn}=\delta_{mk}\delta_{kn}-\delta_{mn}\delta_{kk}$. But then I went to check and $\delta_{mk}\delta_{kn}-\delta_{mn}\delta_{kk}$ is equal to $\delta_{mn}-3\delta_{mn}$. Why is that so? Thank you in advance.

  • $\begingroup$ Presumably, they're using Einstein convention i.e. indices appearing twice are summed over. Hence one really has $\sum_{k=1}^3\delta_{kk}=\delta_{11}+\delta_{22}+\delta_{33}=3.$ $\endgroup$ – Semiclassical Dec 10 '16 at 19:44

Levi Civita symbol $\epsilon_{ikl}$ is defined as it follows $$\epsilon_{ikl} = \left\{ \begin{array}{cl} 1 & if\quad i\neq k\neq l\quad and \quad even \quad permutation\\ -1& if\quad i\neq k\neq l\quad and \quad odd\quad permutation\\ 0 & otherwise \end{array}\right.$$

From this definition, let's start with the contraction of $\epsilon_{ijk}$ in its first index: $$\epsilon_{ikl}\epsilon_{imn}=\delta_{km}\delta_{ln}-\delta_{kn}\delta_{lm}\tag1$$

Where $\delta_{ik}$ is the Kronecker delta (identity matrix), a symmetric isotropic tensor and it is defined as it follows $$\delta_{ik} = \left\{ \begin{array}{cl} 1 & if\quad i= k\\ 0 & otherwise \end{array}\right.$$

Contracting $(1)$ once more, $\textit{i.e.}$ multiplying it by $\delta_{km}$ we have $$\delta_{km}\epsilon_{ikl}\epsilon_{imn}=\epsilon_{ikl}\epsilon_{ikn}=\delta_{km}(\delta_{km}\delta_{ln}-\delta_{kn}\delta_{lm})=\delta_{kk}\delta_{ln}-\delta_{kn}\delta_{kl}=\delta_{kk}\delta_{ln}-\delta_{ln}\tag2$$ Recall that $\delta_{lk} = \delta_{kl}$ due to symmetry properties and $\delta_{km}\delta_{km}=\delta_{kk}=\delta_{mm}$ since they are dummy indices (repeated indices indicated summation over this index).

Now comes the term $\delta_{ii}$, this quantity is a scalar, and represents the trace of the identity matrix in a n-dimensional space, therefore $\delta_{ii}=n$ and finally $(2)$ is written as it follows $$\epsilon_{ikl}\epsilon_{ikn}=(\delta_{kk}-1)\delta_{ln}=(n-1)\delta_{ln}$$ If one contracts again, the following identity $$\epsilon_{ikl}\epsilon_{ikl}=n(n-1)$$

In your case $n=3$.

Hope this helps you


Another approach is to write $$\epsilon_{ijk}=\hat x_i\cdot(\hat x_j\times \hat x_k)$$Then, we have

$$\begin{align} \epsilon_{imk}\epsilon_{ikn}&=\hat x_i\cdot(\hat x_m\times \hat x_k)x_i\cdot(\hat x_k\times \hat x_n)\\\\ &=(\hat x_m\times \hat x_k)\cdot(\hat x_k\times \hat x_n)\\\\ &=\hat x_m \cdot (\hat x_k\times(\hat x_k\times \hat x_n))\\\\ &=\hat x_m \cdot(\delta_{nk}\hat x_k-\delta_{kk}\hat x_n)\\\\ &=\delta_{nk}\delta_{mk}-3\delta_{mn} \end{align}$$

  • $\begingroup$ It is really nice! $\endgroup$ – HBR Dec 10 '16 at 20:06
  • $\begingroup$ @hbr Thank you! Much appreciative. -Mark $\endgroup$ – Mark Viola Dec 10 '16 at 20:10
  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy Holidays! -Mark $\endgroup$ – Mark Viola Dec 17 '16 at 16:30

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.