Kronecker Delta summation (easy)

[joshmccraney]

Hi PF!

As outlined in my book $\delta_{ij} \delta_{jk} = \delta_{ik}$ but don't we sum over repeated indices (and the $j$ is repeated)? Can someone explain why we do not sum in this situation?

Thanks!

 

[blue_leaf77]

Yes it is summed over $j$. Written explicitly, the sum looks like
$$
\delta_{i1} \delta_{1k} + \delta_{i2} \delta_{2k} + \ldots + \delta_{iN} \delta_{Nk}
$$
You see that if $i\neq k$, all terms above will vanish. If $i=k$ there will be only one term surviving, therefore that sum can be written compactly as $\delta_{ik}$.

 

[joshmccraney]

Thanks for responding blue_leaf77! Let's assume $i,j,k$ vary from $1,2,3$. Then we have $$\delta_{11} \delta_{11} + \delta_{22} \delta_{22} + \delta_{33} \delta_{33} = 3 \neq \delta_{ik}$$

Did I miss something?

 

[PeroK]

Thanks for responding blue_leaf77! Let's assume $i,j,k$ vary from $1,2,3$. Then we have $$\delta_{11} \delta_{11} + \delta_{22} \delta_{22} + \delta_{33} \delta_{33} = 3 \neq \delta_{ik}$$

Did I miss something?

Why would $i$ and $k$ be summed?

And, if you sum them on the right hand side you would get equality.

 

[joshmccraney]

Why would $i$ and $k$ be summed?

That's what blue_leaf wrote.

$$
\delta_{i1} \delta_{1k} + \delta_{i2} \delta_{2k} + \ldots + \delta_{iN} \delta_{Nk}
$$

Since $i$ and $k$ can range from 1,2,3 we would have$$\delta_{11} \delta_{11} + \delta_{21} \delta_{11} + \delta_{31} \delta_{11}+ \delta_{12} \delta_{21}+ \delta_{22} \delta_{21}+ \delta_{32} \delta_{21}+\cdots + \delta_{33} \delta_{33}\\=\delta_{11} \delta_{11}+\delta_{22} \delta_{22}+\delta_{33} \delta_{33}$$ which is why I was summing.

 

[PeroK]

That's what blue_leaf wrote.


Since $i$ and $k$ can range from 1,2,3 we would have$$\delta_{11} \delta_{11} + \delta_{21} \delta_{11} + \delta_{31} \delta_{11}+ \delta_{12} \delta_{21}+ \delta_{22} \delta_{21}+ \delta_{32} \delta_{21}+\cdots + \delta_{33} \delta_{33}\\=\delta_{11} \delta_{11}+\delta_{22} \delta_{22}+\delta_{33} \delta_{33}$$ which is why I was summing.

blue leaf summed only $j$, which is a repeated index.

 

[joshmccraney]

blue leaf summed only $j$, which is a repeated index.

Ohhhhh I see now! Shoot, yea I was doing it all wrong! Thank you both!