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I've seen the equation to get from the momentum wavefunction to the position wavefunction as

$$\Psi(x)=\frac{1}{\sqrt{2{\hbar}\pi}}\int_{-\infty}^\infty{e^{ipx/\hbar}}\Phi(p)dp$$

I was wondering if the above equation to get from the momentum wavefunction to the position wavefunction would be the same in any number of dimensions or if it only applies in one dimension. As I understand when talking about wavefunctions and the Schrodinger Equation it's sometimes assumed that we're talking about the one dimensional case unless specified otherwise. This is why when showing the Schrodinger Equation the laplacian in one dimension is often shown instead of the laplacian in three dimensions. The thing that makes me think it's likely specific to a one dimensional case is that often triple integrals are often used for three dimensions and the above equation has a single integral. Assuming it's specific to the one dimensional case I'm not sure whether in the 3d case changing a single integral to a triple integral is all that changes to the equation or if other things about the equation also change.

Does it apply in any number of dimensions and if not, what would be the generalized version for $n$ dimensions?

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Since $|\Psi|^2$ is a PDF, $\Psi$ has the same dimension as $x^{-n/2}$ in $n$-dimensional space; by the same logic, $\Phi$ has the same dimension as $p^{-n/2}$. And since $d^np$ has the dimension of $p^n$, or equivalently of $\hbar^n/x^n$, we need to replace the factor of $(2\pi\hbar)^{-1/2}$ with $(2\pi\hbar)^{-n/2}$. (Meanwhile $px$ become the dot product $p\cdot x$, and $\int_{\Bbb R}$ becomes $\int_{\Bbb R^n}$.)

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  • $\begingroup$ So does this mean that in 3d the equation becomes $$\Psi(x_1,x_2,x_3)=\frac{1}{(2\hbar\pi)^\frac{3}{2}}\int_{-\infty}^\infty\left(\int_{-\infty}^\infty\left(\int_{-\infty}^\infty\left({e^{i(p_1x_1+p_2x_2+p_3x_3)/\hbar}}\Phi(p_1,p_2,p_3)\right)dp_1\right)dp_2\right)dp_3$$? $\endgroup$ Aug 19, 2021 at 19:25
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    $\begingroup$ @AndersGustafson Yes, although we'd normally write it more concisely as$$\Psi(\vec{x})=(2\pi\hbar)^{-3/2}\int_{\Bbb R^3}e^{ip\cdot x/\hbar}\Phi(\vec{p})d^3\vec{p}.$$ $\endgroup$
    – J.G.
    Aug 19, 2021 at 19:55
  • $\begingroup$ Thank You for the information! $\endgroup$ Aug 19, 2021 at 22:37

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