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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.

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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.

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