Levi Civita and Kronecker Delta

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.

• Presumably, they're using Einstein convention i.e. indices appearing twice are summed over. Hence one really has $\sum_{k=1}^3\delta_{kk}=\delta_{11}+\delta_{22}+\delta_{33}=3.$ – 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}

• It is really nice! – HBR Dec 10 '16 at 20:06
• @hbr Thank you! Much appreciative. -Mark – Mark Viola Dec 10 '16 at 20:10
• 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 – Mark Viola Dec 17 '16 at 16:30