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What if we use the Hankel transform instead of the Fourier transform? $$\Psi(x)=a\int_{-\infty }^{\infty }\Phi(k) e^{ikx} dk $$ $$\Phi(k)=b\int_{-\infty }^{\infty }\Psi(x) e^{-ikx} dx .$$ How would the probability density function of the Schrödinger wave be, would it give us the same results as the Schrödinger wave using the Fourier transform? $$\Psi_{n}(x)=a\int_{0 }^{\infty }\Phi(k) J_{n}(kx)k dk $$ $$\Phi(k)=b\int_{0 }^{\infty }\Psi_{n}(x) J_{n}(kx)x dx .$$ The reason for this question is that I can express any wave as a continuous sum of Henkel functions and this is very similar to the continuous Fourier transform.

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Your formulas for Hankel transform make no sense. You can compare them with those at https://en.wikipedia.org/wiki/Hankel_transform

In general, you can use the Hankel transform, but only for $x>0$, unless you redefine the Hankel transform somehow. So you can use it, e.g., in a cylindrical system of coordinates for the radius.

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