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.

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    $\begingroup$ Because you are only interested in the order of magnitude I think. A factor of 2 doesn't really make a difference then.. $\endgroup$ Commented May 14, 2015 at 10:27
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    $\begingroup$ If anyone can provide a calculation of $\sigma_x$ and $\sigma_p$ for the ground state hydrogen wavefunction this would make an excellent answer and I would be very interested to see it. A bounty might even be forthcoming! $\endgroup$ Commented May 14, 2015 at 11:14

3 Answers 3


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


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.

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    $\begingroup$ Thanks gonenc, I was curious to see how close to $\hbar/2$ the hydrogen atom got. As soon as the question is eligable for a bounty I'll post one and award it to you. $\endgroup$ Commented May 14, 2015 at 16:03
  • $\begingroup$ @JohnRennie That is much more rep for bounty than I expected! Thanks! $\endgroup$
    – Gonenc
    Commented May 16, 2015 at 16:54
  • $\begingroup$ @gonenc What is $\Delta p$? If you mean $\Delta p_x$ you are off by a factor $\sqrt{3}$. $\endgroup$
    – Praan
    Commented Nov 13, 2015 at 21:13

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}


The answer to the OP's question is that this is an order of magnitude estimation and the person doing the estimation used values that were known to be closer to the correct values to make the order of magnitude estimation come out closer to the true answer.

The majority of my post shows that there is a simple choice for "r" and "p", for which you could say that $pr=1$ exactly. I've added this answer mostly as a complementary example to the other answer (which runs through an explicit calculation using the ground state wave function in the position basis).

If you make the identification: $$ \left\langle{\frac{1}{r}}\right\rangle\sim \frac{1}{r}\;, $$ where the $r$ on the LHS is an operator and the $r$ on the RHS is identified with the symbol $r$ in the original post.

And if you make the identification: $$ \langle p^2\rangle \sim p^2\;, $$ where, again, the operator $p^2$ is on the LHS and its expectation value is identified with the square of the symbol $p$ from the original post.

With these definitions it is straightforward to show that $pr=\hbar$ for the ground state of the hydrogen atom.

To keep things clean, I'll use Gaussian Atomic Hartree units where $e=\hbar=m_e=1$ and where the potential due to the hydrogen nucleus is $e/r$. (There's no "$4\pi\epsilon_0$" in Gaussian units).

By the Virial Theorem for the Hydrogen atom $$ E=-\frac{1}{2}\left\langle\frac{1}{r}\right\rangle $$ so that for the ground state (for which E=-13.6eV=-1/2 (in atomic units)) $$ \left\langle\frac{1}{r}\right\rangle=1 $$ I.e., $$ r=1\;, $$ where "$r$" is now identified with the symbol in the original post as described above. [Note that it is not the case that $\langle 1/r\rangle =1/\langle r\rangle $, i.e., there is no conflict with the other answer that worked with $r\sim \langle r\rangle $ rather than $r\sim 1/\langle 1/r\rangle $.]

Similarly, the Virial Theorem says $$ 2\langle T\rangle=\langle p^2\rangle=\left\langle\frac{1}{r}\right\rangle\;, $$ which means, for the ground state, $$ \langle p^2\rangle =1\;, $$ i.e., $$ p=1 $$ where "$p$" is now identified with the symbol in the original post as described above.

Thus, for the ground state of hydrogen, with the above-mentioned identifications $$ pr=1=\hbar $$


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