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I Square root of the delta function

  1. Jan 30, 2017 #1
    Is square root of delta function a delta function again?

    $$\int_{-\infty}^\infty f(x) \sqrt{\delta(x-a)} dx$$

    How is this integral evaluated?
     
  2. jcsd
  3. Jan 30, 2017 #2

    PeroK

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    You could try approximating the delta function by a sequence of finite spike functions and see what happens.
     
  4. Mar 23, 2017 #3
    This is actually related to Bell's theorem. It is not the solution to the OP : searching A s.t. $$\int A (a,\lambda)A (\lambda,b)d\lambda=-cos (a-b) $$

    If we make the assumption $$A (a,\lambda)=f (a-\lambda) $$ and let $$u=a-\lambda $$ we get a convolution :

    $$\int f (u)f (a-b-u)du=-cos (a-b) $$

    Using the Fourier convolution theorem :

    $$2F^2 (\omega)=-\delta (\omega+1)-\delta (\omega-1)$$

    Hence the searched function f is given by

    $$f (x)=\frac {i}{\sqrt {2}}\int e^{ikx}\sqrt {\delta (k+1)+\delta (k-1)} $$

    Bell's theorem shows that there is no such f.
     
    Last edited: Mar 23, 2017
  5. Mar 26, 2017 #4
    Addendum : no such f in the range -1 to 1
     
  6. Mar 26, 2017 #5
    not every sequence of mathematical symbols makes sense
     
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