# Expected momentum ground state electron in $\rm H$ atom

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.

• The integral is really not that hard to calculate in Cartesian coordinates, what exactly is the difficulty you're facing? – Philip Nov 6 '20 at 10:49
• 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? – majyno Nov 6 '20 at 11:01
• Does this answer your question? Using the uncertainty principle to estimate the ground state energy of hydrogen – John Rennie Nov 6 '20 at 11:26
• 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. – doublefelix Nov 6 '20 at 11:55
• 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. – majyno Nov 6 '20 at 14:20

## 1 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.