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Is the Kronecker delta linear?

  1. Mar 14, 2015 #1
    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. jcsd
  3. Mar 14, 2015 #2
    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.
     
  4. Mar 14, 2015 #3
    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. :wink:
     
    Last edited: Mar 14, 2015
  5. Mar 14, 2015 #4
    New question:
    What is the significance of a basis functional with respect to physical reality? :smile:
     
    Last edited: Mar 14, 2015
  6. Mar 14, 2015 #5

    bhobba

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    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
  7. Mar 15, 2015 #6
    Nope, I ment basis functionals. But I guess I have to learn further to what eigenvectors are first. :wink:
     
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