Using the uncertainty principle to estimate the ground state energy of hydrogen

$\newcommand{\d}[1]{\,\mathrm{d}#1}\newcommand{\pdv}[2]{\frac{\partial #1}{\partial #2}}\newcommand{\p}{\psi_{100}}\newcommand{\pdvt}[2]{\frac{\partial^2 #1}{\partial #2^2}}$The hydrogen ground state is the following:

$$\psi_{100}=Y_{00}\frac{2}{a_0^{3/2}}e^{-r/a_0}$$

The expectation value of the position operator on this state is the following:

\begin{align} \left<x\right>&=\iint r^3 \left| \psi_{100}\right| ^2 \d\Omega \d r \\ &= \int (Y_{00})^2 \d \Omega \int \frac{2r^3}{a_0^{3/2}}e^{-r/a_0} \d r \tag{1} \\ &= \int\frac{2r^3}{a_0^{3/2}}e^{-r/a_0} \d r = \frac{3a_0}{2} \tag{2} \end{align}

where I used the orthonormality of spherical harmonics from going $(1)$ to $(2)$. I'll omit this step in all the calculations below.

Similarly the expectation value of the square of the position operator is,

$$\left<x^2\right>=\iint r^4 \left| \psi_{100}\right| ^2 \d\Omega \d r = \int\frac{2r^4}{a_0^{3/2}}e^{-r/a_0} \d r = 3(a_0)^{2}$$

Thus $\Delta x= \sqrt{ 3(a_0)^{2} -\left( \frac{3a_0}{2} \right)^2 } = \frac{\sqrt 3}{2} a_0 $

The expectation value of the momentum operator is zero as it can be easily seen by

$$\left<p\right>=\iiint\frac{\hbar}{i}\psi_{100} \pdv{\psi_{100}}{x} \d x \d y \d z = \iiint \frac{\hbar}{2i}\pdv{\psi_{100}^2}{x} \d x \d y \d z=0$$

Note that in this case I have used the Cartesian coordinates, with which the integral is easy to evaluate in this case because the derivative of an even function is odd and thus the integral vanishes. Notice also that except for this one all the integrals are evaluated in spherical coordinates and hence the $r^2 \d r \d \Omega$ term.

The expectation value of the square of the momentum operator is then:

$$-\frac{\left<p^2\right>}{\hbar^2}= \iint r^2 \p \nabla^2 \p \d r \d \Omega = \iint r^2 \p \left( \pdvt{}{r} + \frac{2}{r} \pdv{}{r} + \frac{1}{r^2} \nabla^2_{\theta\phi} \right) \p \d r \d \Omega $$

where $\nabla^2_{\theta\phi}$ is a term involving partial derivatives of $\theta$ and $\phi$ acting directly on $\p$.

\begin{align} \implies \left<p^2\right> &= -\hbar^2 \int r^2 \p \left( \pdvt{\p}{r} + \frac{2}{r} \pdv{\p}{r} \right) \d r\\ &= \hbar^2 \int \frac{4\,r\,\left( r−2\,a_0\right) \,{e}^{−\frac{2\,r}{a_0}}}{{a_0}^{5}} \d r \\ &= \frac{\hbar^2}{{a_0}^{2}} \end{align}

Thus $\Delta p = \sqrt{\frac{\hbar^2}{{a_0}^{2}}} = \frac{\hbar}{{a_0}}$. Calculating the $\Delta x $ and $\Delta p $ took some(!) algebra. Let's see what the uncertainty relation has to say.

$$\Delta x \Delta p = \hbar \frac{\sqrt 3}{2} \approx 0.866 \hbar \approx \hbar $$

Knowing that the uncertainty relation would give an uncertainty more than $\hbar/2$ your teacher choose to give you $\Delta x \Delta p =\hbar$ so that it would be a more reasonable limit.

Note that $\Delta p_x \Delta r$ does not satisfy the uncertainty principle in the strict sense since $r$ is not conjugate to $p_x$ (or $p_y$ and $p_z$). Instead you can consider $\Delta p_x \Delta x$. The ground state of the hydrogen atom is \begin{equation} \psi_0(r) = \frac{1}{\sqrt{\pi a^3}} e^{-r/a}, \end{equation} where $a$ is the Bohr radius. First of all, $\left< x \right> = 0$, because $\psi_0$ is spherically symmetric. The fluctuation in $x$ is thus given by \begin{equation} \Delta x = \sqrt{\left< x^2 \right>} = \sqrt{\frac{\left< r^2 \right>}{3}}, \end{equation} since the ground state has spherical symmetry. For the expectation value of $r^2$, we find \begin{align} \left< r^2 \right> & = \frac{4\pi}{\pi a^3} \int_0^\infty dr \, r^4 e^{-2r/a} \\ & = \frac{4}{a^3} \frac{1}{2^4} \frac{d^4}{d^4(1/a)} \int_0^\infty dr \, e^{-2r/a} \\ & = \frac{1}{8a^3} \frac{d^4}{d^4(1/a)} \frac{1}{1/a} = \frac{4!}{8} a^2 = 3a^2, \end{align} and we obtain $\Delta x = a$. Next, $\left< \vec p \right> = 0$ because the wave function is real and normalizable ($\vec p$ is hermitian) so we have \begin{equation} \Delta p_x = \sqrt{\left< p_x^2 \right>} = \sqrt{\frac{\left< p^2 \right>}{3}}, \end{equation} similarly as before. With $\left< p^2 \right> = \frac{\hbar^2}{a^2}$ (see the answer of gonenc), you find $\Delta p_x = \frac{\hbar}{\sqrt{3}a}$. Finally, we obtain \begin{equation} \Delta x \Delta p_x = \frac{\hbar}{\sqrt{3}} \approx 0.58 \hbar > \frac{\hbar}{2}. \end{equation} Also note that since $\Delta r = \frac{\sqrt{3}}{2} a$ (see the answer of gonenc), we have \begin{equation} \Delta r \Delta p_x = \frac{\hbar}{2}. \end{equation}