# Problem with solution Simplify 6 Expressions with Kronecker Delta

Simplify the following expressions using the rules for calculating with Kronecker delta:

\(\delta_{31}\,\delta_{33}\)

\(\delta_{ji}\,T_{ink}\)

\(\delta_{j1}\,\delta_{ji}\,\delta_{2i}\)

\(\delta_{ik}\,\delta_{i3}\,\delta_{3k}\)

\(\delta_{jj}\) with \(j ~\in~ \{ 1,2,3,4 \} \)

\(\delta_{k\mu} \, \varepsilon_{kmn} \, \delta_{ss} \) with \(s ~\in~ \{ 1,2 \} \)

## Solution tips

Use the properties of Kronecker delta that you learned in the lesson. In all the following expressions you should keep in mind that \(\delta_{ik}\,\delta_{ij}\) can be summed up to \(\delta_{kj}\) and that you have to sum over equal indices:`$$ \delta_{ii} ~=~ 1~+~1~+~ ... ~+~ 1 ~=~ n $$`

## Solution for (a)

Here we simplify the following expression:`\begin{align}
\delta_{31}\,\delta_{33} ~&=~ 0 \cdot 1 ~=~ 0
\end{align}`

We exploited that \( \delta_{31} = 0 \) is because the indices have two different values and \( \delta_{33} = 1 \) has two equal indices.

## Solution for (b)

We want to simplify the following expression:`\begin{align}
\delta_{ji} \, T_{ink} ~&=~ T_{jnk}
\end{align}`

We exploited the rule that the Kronecker delta \( \delta_{ji} \) eliminates everything except \( T_{jnk} \), where \( i = j \).

## Solution for (c)

Here we simplify the following expression:`\begin{align}
\delta_{j1} \, \delta_{ji} \, \delta_{2i}
\end{align}`

For example, first combine the index \(j\) in \(\delta_{j1} \, \delta_{ji}\) and then combine the index \(i\):`\begin{align}
\delta_{j1} \, \delta_{ji} \, \delta_{2i} &~=~ \delta_{i1} \, \delta_{2i} \\\\
&~=~ \delta_{12} \\\\
&~=~ 0
\end{align}`

Of course, you could just as well first combine the index \(i\) in \(\delta_{ji} \, \delta_{2i}\):`\begin{align}
\delta_{j1} \, \delta_{ji} \, \delta_{2i} &~=~ \delta_{j1} \, \delta_{2j} \\\\
&~=~ \delta_{21} \\\\
&~=~ 0
\end{align}`

You get the same result.

## Solution for (d)

Proceed analogously to (c), except that at the end you get not 0 but 1 according to the Kronecker delta definition:`\begin{align}
\delta_{ik} \, \delta_{i3} \, \delta_{3k} &~=~ \delta_{k3} \, \delta_{3k} \\\\
&~=~ \delta_{33} \\\\
&~=~ 1
\end{align}`

Again, the order of simplification does not matter.

## Solution for (e)

Here we simplify the following expression, summing over the index \( j \) up to 4:`\begin{align}
\delta_{jj}
\end{align}`We get a sum with four summands:`\begin{align}
\delta_{jj} &~=~ \delta_{11} ~+~ \delta_{22} ~+~ \delta_{33} ~+~ \delta_{44} \\\\
&~=~ 1~+~1~+~1~+~1 \\\\
&~=~ 4
\end{align}`

## Solution for (f)

Here we simplify the following expression, summing over the index \( s \) up to 2:`\begin{align}
\delta_{k\mu} \, \varepsilon_{kmn} \, \delta_{ss}
\end{align}`

Let us first combine \( \delta_{k\mu} \, \varepsilon_{kmn} \) and then write out the sum \(\delta_{ss}\):`\begin{align}
\delta_{k\mu} \, \varepsilon_{kmn} \, \delta_{ss} &~=~ \varepsilon_{\mu mn} \, \delta_{ss} \\\\
&~=~ \varepsilon_{\mu mn} \, \left( \delta_{11} ~+~ \delta_{22} \right) \\\\
&~=~ \varepsilon_{\mu mn} \, \left( 1~+~ 1 \right) \\\\
&~=~ 2\, \varepsilon_{\mu mn}
\end{align}`