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Carlo Beenakker
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Allow me to replace $k$ by $x$. The kernel $$G(x,t)=(4\pi it)^{-1/2}e^{ix^2/4t}$$ is the Green function of the Schrödinger equation, which can be written in the integral form $$G(x,t)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{-ikx}e^{-ik^2 t}dk.$$ In the limit $t\rightarrow 0$ we then have an integral representation of the delta function, $$2\pi\delta(x)=\int_{-\infty}^\infty e^{-ikx}dk.$$ Both these integral representations need to be understood in the distributional sense (meaning thatthey are formally divergent for real $x$ and $t$; they need to be multiplied by a test function and integrated for a finite answer).

Allow me to replace $k$ by $x$. The kernel $$G(x,t)=(4\pi it)^{-1/2}e^{ix^2/4t}$$ is the Green function of the Schrödinger equation, which can be written in the integral form $$G(x,t)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{-ikx}e^{-ik^2 t}dk.$$ In the limit $t\rightarrow 0$ we then have an integral representation of the delta function, $$2\pi\delta(x)=\int_{-\infty}^\infty e^{-ikx}dk.$$ Both these integral representations need to be understood in the distributional sense (meaning that they need to be multiplied by a test function and integrated).

Allow me to replace $k$ by $x$. The kernel $$G(x,t)=(4\pi it)^{-1/2}e^{ix^2/4t}$$ is the Green function of the Schrödinger equation, which can be written in the integral form $$G(x,t)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{-ikx}e^{-ik^2 t}dk.$$ In the limit $t\rightarrow 0$ we then have an integral representation of the delta function, $$2\pi\delta(x)=\int_{-\infty}^\infty e^{-ikx}dk.$$ Both these integral representations need to be understood in the distributional sense (they are formally divergent for real $x$ and $t$; they need to be multiplied by a test function and integrated for a finite answer).

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Carlo Beenakker
  • 188.1k
  • 18
  • 448
  • 651

Allow me to replace $k$ by $x$. The kernel $$G(x,t)=(4\pi it)^{-1/2}e^{ix^2/4t}$$ is the Green function of the Schrödinger equation, which can be written in the integral form $$G(x,t)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{-ikx}e^{-ik^2 t}dk.$$ In the limit $t\rightarrow 0$ we then have an integral representation of the delta function, $$2\pi\delta(x)=\int_{-\infty}^\infty e^{-ikx}dk.$$ Both these integral representations need to be understood in the distributional sense (meaning that they need to be multiplied by a test function and integrated).