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Square root of Dirac Delta function

  1. Oct 11, 2009 #1
    1. The problem statement, all variables and given/known data

    I wonder how to deal with the square root of Dirac Delta function, [tex]\sqrt{\delta(x)}[/tex]. Actually, this comes from a problem which asking readers to calculate the wave function of a free particle and of a harmonic oscillator at time [tex]t[/tex], provided that the wave function at time [tex]t=0[/tex] is given, i.e. [tex]\psi(x,t=0) = \sqrt{\delta(x-a)} [/tex].

    The way to obtain [tex]\psi(x,t)[/tex] in this problem is by the integral with the propagators.
    However, I have no idea how to deal with [tex]\sqrt{\delta(x)}[/tex].

    2. Relevant equations

    [tex]\sqrt{\delta(x)}=?[/tex]



    3. The attempt at a solution

    I tried to differentiate it,
    [tex]\frac{d}{dx}\sqrt{\delta(x)} = \frac{1}{2\sqrt{\delta(x)}}\delta'(x)
    = -\frac{1}{2\sqrt{\delta(x)}}\frac{\delta(x)}{x} = -\frac{1}{2}\frac{\sqrt{\delta(x)}}{x} [/tex]
    [tex] \Rightarrow
    \int \sqrt{\delta(x)}f(x)dx = -2\int xf(x)d(\sqrt{\delta(x)})
    [/tex]
    But this is still no good...

    I also tried the other definition of Dirac Delta function,
    [tex]\delta(x)\sim\lim_{\epsilon\rightarrow 0}e^{-x^2/\epsilon}[/tex]
    [tex]\Rightarrow \sqrt{\delta(x)} = \lim_{\epsilon\rightarrow 0}e^{-x^2/2\epsilon}
    =\lim_{\epsilon'\rightarrow 0}e^{-x^2/\epsilon'} = \delta(x) ?? [/tex]

    still no good...

    I also tried to calculate [tex]\psi(x,t)^2[/tex] to try to get rid of the square root,
    but it seems doesn't help.

    Is there anyone who has any ideas about [tex]\sqrt{\delta(x)}[/tex] ?
    Any help will be appreciated, thanks.

    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 13, 2009 #2
    I have an idea, but i'm not sure if it's correct.
    Consider
    [tex] f(0) = \int \delta(x)f(x)dx = \int \sqrt{\delta(x)}\sqrt{\delta(x)} f(x)dx
    = \int \sqrt{\delta(x)}\sqrt{\delta(x)} dg(x)[/tex]
    where
    [tex] \frac{dg}{dx} = f [/tex]
    ,then integration by part gives,
    [tex] f(0) = -2\int g(x)\frac{\delta'(x)}{2\sqrt{\delta(x)}} dx
    = \int g(x) \frac{\sqrt{\delta(x)}}{x} dx[/tex]

    Now, let [tex] g(x) = xh(x) [/tex], we have,
    [tex] f(0) = \int h(x)\sqrt{\delta(x)} dx [/tex]
    ,i.e.

    [tex]
    \int h(x)\sqrt{\delta(x)} dx = \left[h(x) + xh'(x)\right]_{x=0}
    [/tex]

    I'm not sure if I were correct.
    Any help would be appreciated!

    -----------

    Gosh, I found one mistake just now,
    I calculated the integration by part wrong,
    I omitted one factor of [tex] \sqrt{\delta(x)} [/tex]
    So the final formula is incorrect.
     
    Last edited: Oct 13, 2009
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