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If $i$ and $j$ are two variables then Kronecker delta is written as

$$\delta_{i,j}~:=~\begin{cases}1 \hspace{3mm} \mbox{if} \hspace{3mm} i=j,\\ 0 \hspace{3mm}\mbox{if} \hspace{3mm}i \neq j. \end{cases}$$

same definition is used for orthogonality.

My question is when will we use Kronecker delta function and when will we use orthogonality?

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closed as not a real question by Emilio Pisanty, akhmeteli, Manishearth Mar 5 '13 at 6:21

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

I am not 100% sure what you are asking. As you wrote yourself, the delta function is notation that compactly states orthogonality. – DJBunk Mar 3 '13 at 15:20
The Kronecker Delta Function is just a special case of the Dirac Delta Function and is simply equivalent to orthogonality. – Cheeku Mar 3 '13 at 15:32
I could be completely off about what you're asking, but orthogonality refers to a property/relation between things, while the Kronecker delta is a function. Things which are orthogonal, can be expressed as a delta function when you take (e.g.) the inner product. – DilithiumMatrix Mar 3 '13 at 18:11

I'm not sure of what you want to know, but I'll try my best to explain: The Kronecker Delta can be introduced by the following relation: given the euclidean $n$-space with usual base $\left\{e_i\right\}$, then we have $\delta_{ij}= \left\langle e_i, e_j\right\rangle$, in other words, we define the Kronecker Delta as simply the inner product of the vectors from the base.

In truth, the inner product is bilinear, which means that if we have two vectors $v, w$, then we have their expansion in the usual base as being:



Hence their inner product between those vectors is

$$\left\langle v,w\right\rangle = \left\langle \sum_{i=1}^nv^ie_i,\sum_{j=1}^nw^je_j\right\rangle$$

But since it's bilinear, we can take the sums out and the components of the vector, so that:

$$\left\langle v,w\right\rangle = \sum_{i=1}^n\sum_{j=1}^nv^iw^j\left\langle e_i,e_j\right\rangle$$

So, bilinearity implies that the inner product is completely specified if we define the inner product on the basis vectors, and so we simply set the Kronecker Delta as you've said:

$$\delta_{ij}=~\begin{cases}1 \hspace{3mm} \mbox{if} \hspace{3mm} i=j,\\ 0 \hspace{3mm}\mbox{if} \hspace{3mm}i \neq j. \end{cases}$$

Because that's what we want that inner product to be: $e_i$ and $e_j$ should be orthogonal if $i\neq j$. Hence, introducing the Kronecker Delta is equivalent to introducing the usual inner product between two vectors. As I said, we build the Kronecker Delta to reflect the usual inner product completely. Also there's a relation with the Dirac Delta, but I think what you wanted to know is that. I hope it helps you.

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