📂Mathematical Physics

Kronecker Delta

Definition

The $\delta_{ij}$, defined as follows, is called the Kronecker delta.

$$\delta_{ij} := \begin{cases} 1,&i=j \\ 0, & i\ne j \end{cases}$$

Description

The Kronecker delta is used in many places, and its main role is to indicate only what one wants among all components (elements, possibilities, etc.). For students of physics, it is mainly encountered in the expression for the inner product. This might not be immediately clear, so let’s understand it through the example below.

Example

First, let’s assume that two vectors $\mathbf{A}=(A_{1}, A_{2}, A_{3})$, $\mathbf{B}=(B_{1}, B_{2}, B_{3})$ are given. Then, the inner product of the two vectors is as follows.

$$\mathbf{A} \cdot \mathbf{B} = A_{1}B_{1} + A_{2}B_{2} + A_{3}B_{3}$$

Using the summation symbol $\sum$ to express it, it looks like this.

$$\mathbf{A} \cdot \mathbf{B} = A_{1}B_{1} + A_{2}B_{2} + A_{3}B_{3} = \sum \limits_{i=1}^{3}A_{i}B_{i}$$

Then, through the following, we can see that the above equation and $\sum \limits_{i=1}^{3}\sum \limits_{j=1}^{3}\delta_{ij}A_{i}B_{j}$ are the same.

$$\begin{align*} \sum _{i=1}^{3}\sum _{j=1}^{3}\delta_{ij}A_{i}B_{j} &= \delta_{11}A_{1}B_{1} + \delta_{12}A_{1}B_{2} + \delta_{13}A_{1}B_{3} \\ & \quad+ \delta_{21}A_{2}B_{1} + \delta_{22}A_{2}B_{2} + \delta_{23}A_{2}B_{3} \\ & \quad+ \delta_{31}A_{3}B_{1} + \delta_{32}A_{3}B_{2} + \delta_{33}A_{3}B_{3} \\ &= 1\cdot A_{1}B_{1} + 0 \cdot A_{1}B_{2} + 0\cdot A_{1}B_{3} \\ & \quad+ 0\cdot A_{2}B_{1} + 1\cdot A_{2}B_{2} + 0\cdot A_{2}B_{3} \\ & \quad+ 0\cdot A_{3}B_{1} + 0\cdot A_{3}B_{2} + 1\cdot A_{3}B_{3} \\ &= A_{1}B_{1} + A_{2}B_{2} + A_{3}B_{3} \\ &= \sum \limits_{i=1}^{3}A_{i}B_{i} \\ &= \mathbf{A} \cdot \mathbf{B} \end{align*}$$

Applying the Einstein notation, which omits $\sum$ when an index appears more than once on one side, results in the following.

$$\delta_{ij}A_{i}B_{j} = \mathbf{A} \cdot \mathbf{B}$$

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So $\delta_{ij}A_{i}B_{j}$ and $\mathbf{A} \cdot \mathbf{B}$ are the same, but it may not be clear why such an expression is used. The example above is a very simple formula, so its usefulness may not stand out. However, calculating the inner product of numerous vectors and operations like cross product, gradient, divergence, curl, Laplacian, etc., in electromagnetism will reveal its convenience. If you’re a sophomore, you’ll naturally come to understand its convenience, so there’s no need to force yourself to understand it right now.

Also, since there is a value only when both subscripts are the same, if two or more Kronecker deltas are multiplied together, there is a value only when all the indices are the same.

$$\delta_{ij}\delta_{jk}$$

In such a case, a non-$0$ value exists only in the case of $i=j=k$. Also, the Kronecker delta is an example of a $2$th tensor.

Formulae

(a) $\delta_{ii} = 3$

(b) $\delta_{ij}\delta_{jl} = \delta_{il}$

(c) $\delta_{ii}\delta_{jj} = 9$

(d) $\delta_{ii}\delta_{jj} = 6 \quad (i \ne j)$

Remember that $\sum$ is omitted when the same index appears more than once in a term.

Proof

(a)

By the Einstein notation, the following holds.

$$\delta_{ii} = \sum \limits_{i=1}^{3} \delta_{ii} = \delta_{11}+\delta_{22}+\delta_{33}=3$$

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(b)

By the Einstein notation, the following holds.
$$\delta_{ij}\delta_{jl}=\sum\limits_{j=1}^{3}\delta_{ij}\delta_{jl}=\delta_{i1}\delta_{1l}+\delta_{i2}\delta_{2l}+\delta_{i3}\delta_{3l}$$

Now, let’s consider the cases where the above value is not $0$.

$$i=l=1 \quad \text{and} \quad i=l=2 \quad \text{and} \quad i=l=3$$

If it’s the first case, the following holds.

$$\delta_{i1}\delta_{1l} = 1 \quad \text{and} \quad \delta_{i2}\delta_{2l}=\delta_{i3}\delta_{3l} = 0 \\ \implies \delta_{ij}\delta_{jl} = \delta_{i1}\delta_{1l}+\delta_{i2}\delta_{2l}+\delta_{i3}\delta_{3l} = 1$$

If it’s the second case, the following holds.

$$\delta_{i2}\delta_{2l} = 1 \quad \text{and} \quad \delta_{i1}\delta_{1l}=\delta_{i3}\delta_{3l} = 0 \\ \implies \delta_{ij}\delta_{jl} = \delta_{i1}\delta_{1l}+\delta_{i2}\delta_{2l}+\delta_{i3}\delta_{3l} = 1$$

If it’s the third case, the following holds.

$$\delta_{i3}\delta_{3l} = 1 \quad \text{and} \quad \delta_{i1}\delta_{1l}=\delta_{i2}\delta_{2l} = 0 \\ \implies \delta_{ij}\delta_{jl} = \delta_{i1}\delta_{1l}+\delta_{i2}\delta_{2l}+\delta_{i3}\delta_{3l} = 1$$

Therefore, since $\delta_{ij}\delta_{jl}$ has a value of $1$ only when it is $i=l$ and otherwise is $0$, the following result is obtained.

$$\delta_{ij}\delta_{jl} = \delta_{il}$$

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(c)

As $\sum$ is omitted due to the Einstein notation, it is as follows.

$$\begin{align*} \delta_{ii}\delta_{jj} &= \sum\limits_{i=1}^{3}\sum\limits_{j=1}^3{\delta_{ii}\delta_{jj}} \\ &= \sum\limits_{i=1}^{3}{\delta_{ii} \sum\limits_{j=1}^3\delta_{jj}} \\ &= 3\cdot 3 \\ &= 9 \end{align*}$$

The third equality holds because of (a).

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(d)

As $\sum$ is omitted due to the Einstein notation, it is as follows.

$$\begin{align*} \delta_{ii}\delta_{jj} &= \sum\limits_{i=1}^{3}\sum\limits_{\substack{j=1 \\ j\ne i}}^{3}{\delta_{ii}\delta_{jj}} \\ &= \delta_{11}\delta_{22} +\delta_{11}\delta_{33} +\delta_{22}\delta_{11} +\delta_{22}\delta_{33}+\delta_{33}\delta_{11}+\delta_{33}\delta_{22} \\ &= 6 \end{align*}$$

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