Expected momentum ground state electron in $rm H$ atom

Question

I want to calculate $langle p_xrangle$ and $langle p_x^2rangle$ for ground state electron in $rm H$ atom.
Radial function
$$
psi(r)=Ae^{-r/a}
$$
Momentum operator in 3D:
$$
hat{vec p}=frac{hbar}{i}left(frac{partial}{partial x},frac{partial}{partial y},frac{partial}{partial z}right)=frac{hbar}{i}nabla
$$
Momentum operator 1D:
begin{align}
hat{p}_{x} & =frac{hbar}{i}frac{d}{dx}

langle p_xrangle & =int_V psi^*(r)*hat{p}_{x}*psi(r) dV
end{align}
Intuitively, $langle p_xrangle=0$ but how do I calculate it? Should I change the operator for one expressed in spherical coordinates or something else?
And for $langle p_x^2rangle$ I would just square the momentum operator and use it instead.

Emilio Pisanty · Accepted Answer

For a spherically-symmetric state:

$⟨p_x⟩=0$ by parity symmetry. This can be proved rigorously, by changing variables $vec r mapsto -vec r$ and showing the expectation value must both change sign and remain unchanged.
All three squared components, $⟨p_x^2⟩$, $⟨p_y^2⟩$ and $⟨p_z^2⟩$, must be equal, so therefore $⟨p_x^2⟩=frac13⟨p^2⟩$. The latter can be evaluated directly using the spherical-coordinates expression for the laplacian.

The rest of the work is for you to do.

Expected momentum ground state electron in $rm H$ atom

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