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I was recently introduced to wavefunctions in my freshman Modern Physics class.

I understand that these waves do not- on their own- have a physical interpretation but the square of their magnitude does. But I am confused about how the wavefunction we find depends on the uncertainty in momentum. I say this because, as far as I understand, the more certain I am about momentum, the less certain I should be about position. So in my mind, if momentum is exactly known the wavefunction should show that the probability of finding the particle anywhere to be the same non-zero value. Is this somewhat how it works?

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Indeed, the eigenfunction of the momentum operator corresponds to equal probability of finding the particle anywhere in space:

The (one-dimensional) momentum operator is: $$\hat{p} = -i\hbar\frac{d}{dx}.$$ Its eigenfunction is: $$\psi_p(x) = Ce^{ipx/\hbar},$$ where $C$ is the normalization constant. The magnitude square of this wave function is: $$|\psi(x)|^2 = |A|^2,$$ i.e. it is constant everywhere.

Remark This goes beyond the question, but may be relevant. What may pose here a conceptual difficulty is that the wave function is not normalizable, as integral $$\int_{-\infty}^{+\infty} |A|^2 dx$$ diverges. One often uses periodic boundary conditions in the box of length $L$, by demanding that $\psi(x+L) = \psi(x)$, which means that the momentum can take only the values $p_n = \frac{2\pi\hbar}{L}n$, where $n$ is an integer. Then the wave function takes form: $$\psi_n(x) =\frac{1}{\sqrt{L}}e^{ip_nx/\hbar},$$ whereas its amplitude squared is $$|\psi_n(x)|^2 = \frac{1}{L},$$ i.e. the same everywhere in the box.

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Yes. Position and momentum are conjugate variables under the Fourier Transform. The position space wave function $f(\mathbf x)$ and the momentum space wave function $f(\mathbf p)$ are related by $$ f(\mathbf x) = (\frac{1}{2\pi})^{3/2} \int d\mathbf p^3 e^{i \mathbf x . \mathbf p} f(\mathbf p)$$ $$ f(\mathbf p) = (\frac{1}{2\pi})^{3/2} \int d\mathbf x^3 e^{-i \mathbf x . \mathbf p} f(\mathbf x)$$ (other conventions for the FT are possible, the function and FT are here distinguished by the choice of argument).

Strictly speaking states of exact position and momentum are not physically possible, as they would lead to meaningless formulae. This can be fixed using distribution theory, at least for non-relativistic quantum mechanics but it is a source of divergence problems in quantum electrodynamics. Generally ad hoc, non-rigorous, treatments can be used (renormalisation).

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    $\begingroup$ "This can be fixed using distribution theory, at least for non-relativistic quantum mechanics but it is a source of divergence problems in quantum electrodynamics." What do you mean? QED is a relativistic QFT that doesn't even have straightforward position operators. $\endgroup$
    – ACuriousMind
    Commented Mar 29, 2020 at 11:15
  • $\begingroup$ I refer to the use of the FT. QED was originally formulated as relativistic quantum mechanics. The only reason for reformulating it as a QFT is to sweep mathematical problems under the carpet, such as mathematically undefinable quantities which arise in the products of field operators. ejtp.com/articles/ejtpv10i28p27.pdf $\endgroup$ Commented Mar 29, 2020 at 11:20

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