# Momentum probability density and its normalization

Let the (normalized) wave function $$\Psi(x,y)$$ represent a free particle in the XY plane. I know $$|\Psi|^2$$ gives me the probability density function of the particle's position, which I can then integrate over a given region to calculate the probability of finding the particle in said region. However, how about momentum? If I were to calculate the probability density of momentum instead of position, could I just obtain the probability density of $$p$$ by applying the momentum operator like so?: $$p = -i\hbar \left(\frac{\partial \Psi}{\partial x} + \frac{\partial \Psi}{\partial y}\right).$$ From this doubt, 2 questions arise:

1. Would $$|p|^2$$ give me the probability density of momentum, just as $$|\Psi|^2$$ gives me the probability density of position? Or would I need to perform an inverse Fourier transform on the wave function and then evaluate $$|\Psi(p_x,p_y)|^2$$ to obtain the probability density of the particle's momentum?

2. If, indeed, I can calculate the probability density of momentum just by applying the momentum operator to $$\Psi$$ and then taking the modulus of the result, would I need to normalize the new density probability I have calculated? I'm assuming yes since the $$\hbar$$ in $$p$$ would rescale the result, unless I used natural units, which I'm not interested in at the present moment.

Yes, you need to take the Fourier transform to obtain the wavefunction in momentum space, $$\Psi(\mathbf{p})$$. Then $$\left|\Psi(\mathbf{p})\right|^2$$ is the probability density of the momentum. If you simply use $$p^2$$, you will get $$2m$$ times the Hamiltonian.
If you start with a normalized wavefunction, the momentum representation will automatically be normalized as well (provided you use the conventions of $$1/\sqrt{2\pi\hbar}$$ and $$e^{\pm ipx/\hbar}$$). Note that for free particles, the position and momentum eigenstates are not normalizable in the traditional sense. Instead, they are Dirac normalizable.