<s>[INST] Using the uncertainty principle to estimate the ground state energy of hydrogen I have been reading through this estimate of the ground state energy of hydrogen  and others like it. In this one it says it is using the uncertainty principle but then proceeded to use the following:
$$pr=\hbar$$
But why is it not using:
$$pr=\frac{\hbar}{2}$$
which is more in line with the uncertainty principle and what comes earlier in the derivation.
 [/INST] 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}
</s>