# (Another) Naive Question About Wavefunctions

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?

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.
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).