# Is the Kronecker delta linear?

1. Mar 14, 2015

### entropy1

I don't see how the kronecker delta function is linear, and hence I don't see how it can serve as a dual basis for a ket space.

How is the kronecker delta linear? How can it serve as a dual basis?

Last edited: Mar 14, 2015
2. Mar 14, 2015

### Jazzdude

There are different meanings of "linear" depending on the context. It seems you are mixing them up, as well as Dirac and Kronecker. The functions spanning a function space don't have to be linear, but the space is a linear space, because it consists of linear superpositions of those functions.

In case of the Dirac delta you don't even have a function however. But the associated functional that extracts a single value from a test function is linear.

Look up these concepts, maybe that will clear up your confusion.

3. Mar 14, 2015

### entropy1

I think I understand now. Thanks. I mistook the Kronecker delta for being defined 0 for any vector which was not a particular basevector, but in fact it is only defined such for certain basevectors. And the Kronecker delta is not a function. The functionals with the Kronecker delta property are.

Last edited: Mar 14, 2015
4. Mar 14, 2015

### entropy1

New question:
What is the significance of a basis functional with respect to physical reality?

Last edited: Mar 14, 2015
5. Mar 14, 2015

### bhobba

I presume you mean basis vectors.

I think 'physical reality' is far too vague and undefined a concept to be of any value here.

Its value lies in the Born rule. If you expand the state |u> in terms of an observables O = Σyi |bi><bi| eigenvectors, |u> = Σ ci |bi>, the coefficients of that expansion, ci, give the probability of outcome i by |ci|^2.

Thanks
Bill

Last edited: Mar 15, 2015
6. Mar 15, 2015

### entropy1

Nope, I ment basis functionals. But I guess I have to learn further to what eigenvectors are first.