- #1

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It seems that both are 1 at a certain point and 0 otherwise...

The delta function is a eigenfunction of x and the Kronecker delta is ...

i'm kind of confused..

- Thread starter churi55
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- #1

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It seems that both are 1 at a certain point and 0 otherwise...

The delta function is a eigenfunction of x and the Kronecker delta is ...

i'm kind of confused..

- #2

mathman

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=f(a) when a is in the interval, and integral =0 if a is not in the interval.

Kronecker delta G(n-k) (usually for integer argument, not real) G=1 for n=k, =0 for n not=k.

- #3

quasar987

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Then the dot product of any two of these vectors can be expressed as

[tex](\hat{e}_{n}|\hat{e}_{k}) = \left\{\begin{array}{rcl}1 \ \mbox{if} \ n=k\\ 0 \ \mbox{otherwise}\end{array}[/tex]

So we write

[tex](\hat{e}_{n}|\hat{e}_{k}) = \delta_{nk}[/tex]

to compactly express this fact.

- #4

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The delta function, delta(x), is infinite at x=0, zero everywhere else. It is what a normalized Gaussian "hump" looks like in the limit as its width goes to zero.

In contrast, Kronecker delta is not really a function at all ... more like an element of a matrix (the identity matrix). So Kronecker[ij] = 1 (if i==j), or 0 (if i!=j).

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Daniel.

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in an easy language, they are inherently the same (they have the same/analoguous meaning) but the Kronecker delta is the DISCRETE variant of the delta dirac distribution/functional. So the indices are discrete where they are continuous (they vary continuously) in case of the delta dirac distribution.churi55 said:

It seems that both are 1 at a certain point and 0 otherwise...

The delta function is a eigenfunction of x and the Kronecker delta is ...

i'm kind of confused..

regards

marlon

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Daniel.

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[tex]\delta[/tex]

([tex]\delta[/tex]

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