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

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  • $\begingroup$ The integral is really not that hard to calculate in Cartesian coordinates, what exactly is the difficulty you're facing? $\endgroup$ – Philip Nov 6 '20 at 10:49
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    $\begingroup$ The difficulty I have is that my operator is in xyz coordinates but my function is radial- dependent on (r). So how can I calculate it. Or can I just simply use r=sqrt(x^2+y^2+z^2) substitute it in radial function construct the integral and calculate it? $\endgroup$ – majyno Nov 6 '20 at 11:01
  • $\begingroup$ Does this answer your question? Using the uncertainty principle to estimate the ground state energy of hydrogen $\endgroup$ – John Rennie Nov 6 '20 at 11:26
  • $\begingroup$ Use the chain rule to take the derivative of psi(r(x)), that is easier than substituting $r$. You'll need the chain rule for functions of more than one variable. $\endgroup$ – doublefelix Nov 6 '20 at 11:55
  • $\begingroup$ Ouch my bad about the ∇ of course it is not. But could you @EmilioPisanty please help my with the problem? I am still lost in it. $\endgroup$ – majyno Nov 6 '20 at 14:20
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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.

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