# Why is $(-\frac{e^2}{4\pi \epsilon_0}) = (-\frac{\hbar ^2}{ma})$?

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.

• 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. – Jon Nov 7 '19 at 13:14
• @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 ? – holomorphicfunction Nov 7 '19 at 13:23

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.