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Note: No, this is NOT a homework question. I am struggling to understand how two physical concepts are related and truly think this could be helpful to a broader audience. Also, I already have the solution to the question but fail to understand some parts of it, so it's not at all a "do my homework" question.


We're asked to calculate the expectation value of the potential energy $$V=-\frac{e^2}{4\pi \epsilon_0}\frac{1}{r}$$
And the wavefunction is

$$\psi_2(\vec{r}^\ ,t)= \frac{i}{4a^2 \sqrt{2\pi a}}r e^{-\frac{r}{2a}}\sin\theta \sin \phi \ e^{-\frac{i E_{211}t}{\hbar}}$$


The solution says:
$$\langle V\rangle = \int |\psi_2(\vec{r}^\ ,t)|^2 (-\frac{e^2}{4\pi \epsilon_0}\frac{1}{r}) d^3\vec{r}^\ =... = $$
$$ \frac{1}{32 a^5 \pi}(-\frac{e^2}{4\pi \epsilon_0})\int_{0}^{\infty} r^3 e^{\frac{-r}{a}}dr \int_{0}^{\pi} \sin^3{\theta}d\theta \int_{0}^{2\pi}\sin^2{\phi}d\phi= ...= $$
$$ - \frac{1}{4a}(-\frac{e^2}{4\pi \epsilon_0})=- \frac{1}{4a}(-\frac{\hbar ^2}{ma})=-6.8 \,\mathrm{eV}$$

Now, I don't see how they go from $- \frac{1}{4a}(-\frac{e^2}{4\pi \epsilon_0}) $ to $- \frac{1}{4a}(-\frac{\hbar ^2}{ma})$. Why does it seem like $(-\frac{e^2}{4\pi \epsilon_0})=(-\frac{\hbar ^2}{ma})$ ?

I looked up all terms : $\hbar$ is of course the reduced Planck constant, $e$ is the elementary charge, $a$ is the Bohr radius I guess, $\epsilon_0$ the vacuum permittivity, and $m$ the mass.

But I just don't see how they go from $(-\frac{e^2}{4\pi \epsilon_0})$ to $(-\frac{\hbar ^2}{ma})$ ? Even after searching for several hours on the internet, I couldn't find how the two terms are related.

Any help/ hint would be appreciated. Thank you.

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    $\begingroup$ It is just a matter of definitions. Being $a=\hbar^2 4\pi\epsilon_0/me^2$ you can invert this formula to obtain $e^2/4\pi\epsilon_0$ instead. Then, put it in your equation and you are done. $\endgroup$ – Jon Nov 7 '19 at 13:14
  • $\begingroup$ @John Uh, yeah, makes sense now, Thank You. Could you post your comment as an answer, so I can accept it and the question doesn't appear in the unanswered section anymore ? $\endgroup$ – holomorphicfunction Nov 7 '19 at 13:23
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Simply, it is just a matter of definitions. Being $$ a=\frac{\hbar^24\pi\epsilon_0}{me^2}, $$ a constant generally called Bohr's radius, you can invert this formula to obtain $e^2/4\pi\epsilon_0$ instead. Then, put it in your equation and you are done.

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