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I am trying to calculate the expected momentum of an electron in the ground state of hydrogen atom. This is the wave function.

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So far I have done this:$$\iiint_V \Psi^* (-i\hbar) \frac {d\Psi} {dr} r^2 sin\theta dr d\theta d\phi$$ But the answer I am getting is $$\frac {i\hbar}{a_b}$$ which looks wrong because it is imaginary. What am I doing wrong?

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    $\begingroup$ The expectation value of the momentum $\langle \vec{p}\rangle$ is zero. So you don't have to bother about the $i$. Your calculus have to consider that $\nabla\psi$ is a vector when calculating $\langle \vec{p}\rangle=\int_V \psi^*(-i\hbar\nabla\psi)d^3x$ $\endgroup$ – Kevin De Notariis Oct 10 '18 at 20:59
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    $\begingroup$ part of the problem might be tied to the use of spherical coordinates. In particular, for the construction of the radial momentum operator, see physics.stackexchange.com/q/9349 $\endgroup$ – ZeroTheHero Oct 11 '18 at 0:31
  • $\begingroup$ Is the atom at rest? $\endgroup$ – my2cts Oct 11 '18 at 16:50
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You simply miscalculated the action of $\vec p$ on spherically symmetric functions f(r).

In actuality, your answer should transform vectorially, $$ \vec p f(r)= -i\hbar \nabla f(r)= -i\hbar~ \hat x ~\partial_r f(r), $$ so, then, as @KevinDeNotariis suggests, $$ \langle \vec p \rangle= -i\hbar \int d^3x ~\psi^* \partial_r \psi(r) ~\frac{\vec x}{r} , $$ trivially vanishing upon integrating over all directions.

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