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Kronecker Delta 📂Mathematical Physics

Kronecker Delta

Definition

We define δij\delta_{ij} as the Kronecker delta.

δij:={1,i=j0,ij \delta_{ij} := \begin{cases} 1,&i=j \\ 0, & i\ne j \end{cases}

Explanation

The Kronecker delta is used in many places, primarily to highlight the desired components (elements, possibilities, etc.) among all possible options. Physics students often encounter it in the context of dot products. If this concept isn’t immediately clear, consider the following example:

Example

Suppose we are given two vectors A=(A1,A2,A3)\mathbf{A}=(A_{1}, A_{2}, A_{3}) and B=(B1,B2,B3)\mathbf{B}=(B_{1}, B_{2}, B_{3}). The dot product of these vectors is given as follows:

AB=A1B1+A2B2+A3B3 \mathbf{A} \cdot \mathbf{B} = A_{1}B_{1} + A_{2}B_{2} + A_{3}B_{3}

Expressing this using the summation symbol \sum results in:

AB=A1B1+A2B2+A3B3=i=13AiBi \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}

We can then see that the above expression is equivalent to i=13j=13δijAiBj\sum \limits_{i=1}^{3}\sum \limits_{j=1}^{3}\delta_{ij}A_{i}B_{j} as illustrated below:

i=13j=13δijAiBj=δ11A1B1+δ12A1B2+δ13A1B3+δ21A2B1+δ22A2B2+δ23A2B3+δ31A3B1+δ32A3B2+δ33A3B3=1A1B1+0A1B2+0A1B3+0A2B1+1A2B2+0A2B3+0A3B1+0A3B2+1A3B3=A1B1+A2B2+A3B3=i=13AiBi=AB \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 twice in a single term, results in:

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

While it might not be immediately clear why such expressions are useful when looking at this simple example, their utility becomes apparent in fields like electromagnetism. In these areas, calculating numerous dot products, cross products, gradients, divergences, curls, Laplacians, and others demonstrates its convenience. As a second-year undergraduate, you’ll likely come to appreciate its usefulness naturally, so there’s no need to force understanding right now.

Additionally, a value exists only when both subscripts are identical, so if more than one Kronecker delta is multiplied, a non-zero value occurs only when all subscripts are identical:

δijδjk \delta_{ij}\delta_{jk}

In such cases, a non-zero value exists only when i=j=ki=j=k, not when 00. The Kronecker delta is also an example of a 22-dimensional tensor.

It is also possible to understand this by representing it as a matrix.

Formula

The following formulas hold for i,j{1,2,3}i,j \in \left\{ 1,2,3 \right\}.

(a) δii=3\delta_{ii} = 3

(b) δijδjl=δil\delta_{ij}\delta_{jl} = \delta_{il}

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

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

Remember that \sum is omitted for expressions where any index appears more than twice.

Proof

(a)

According to Einstein notation, the following holds:

δii=i=13δii=δ11+δ22+δ33=3 \delta_{ii} = \sum \limits_{i=1}^{3} \delta_{ii} = \delta_{11}+\delta_{22}+\delta_{33}=3

(b)

According to Einstein notation, the following holds: δijδjl=j=13δijδjl=δi1δ1l+δi2δ2l+δi3δ3l \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, consider the cases where the value is not 00. The following three cases exist:

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

In the first case, the following holds:

δi1δ1l=1andδi2δ2l=δi3δ3l=0    δijδjl=δi1δ1l+δi2δ2l+δi3δ3l=1 \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

In the second case, the following holds:

δi2δ2l=1andδi1δ1l=δi3δ3l=0    δijδjl=δi1δ1l+δi2δ2l+δi3δ3l=1 \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

In the third case, the following holds:

δi3δ3l=1andδi1δ1l=δi2δ2l=0    δijδjl=δi1δ1l+δi2δ2l+δi3δ3l=1 \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, δijδjl\delta_{ij}\delta_{jl} results in 11 only when i=li=l, and in all other cases, the value is 00, leading to the following outcome:

δijδjl=δil \delta_{ij}\delta_{jl} = \delta_{il}

(c)

Because \sum is omitted according to Einstein notation, we have the following:

δiiδjj=i=13j=13δiiδjj=i=13δiij=13δjj=33=9 \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 due to (a).

(d)

Because \sum is omitted according to Einstein notation, we have the following:

δiiδjj=i=13j=1ji3δiiδjj=δ11δ22+δ11δ33+δ22δ11+δ22δ33+δ33δ11+δ33δ22=6 \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*}