# How to get the radius of Bohr from the uncertainty principle?

I would like to know how to get the following result: $$\begin{eqnarray} a_0=\frac{4\pi\varepsilon_0 \hbar^2}{\mu e^2} \end{eqnarray}$$ from the beginning of the uncertainty of heisenberg. If possible, I would like as much detail as possible on the assumed basic hypotheses.

## 1 Answer

That result does not really come from the uncertainty principle. It comes from quantization of angular momentum, as a postulate.

Consider the two body problem between a proton of mass $$M$$ and electron of mass $$m$$. They have charges $$\pm e$$, respectively. The relative position vector, from the proton to the electron, then follows the differential equation

$$\mu\frac{d^2\mathbf r}{dt}=-\frac{1}{4\pi\epsilon_0}\frac{e^2}{r^2}\hat r$$

Which for a circular orbit of radius $$R$$ has the condition

$$\mu\frac{v^2}{R}=\frac{1}{4\pi\epsilon_0}\frac{e^2}{R^2}$$

But we can write this in terms of the center of mass angular momentum:

$$L=m\frac{M}{m+M}Rv_e+M\frac{m}{M+m}Rv_p$$

Where $$v_e=\frac{M}{m+M}v$$ $$v_p=\frac{m}{m+M}v$$

Are the center of mass velocities. As such

$$L=\frac{mMRv}{m+M}=\mu Rv$$

$$v=\frac{L}{\mu R}$$

Substituting this into the circular orbit condition we get

$$\frac{L^2}{\mu R^3}=\frac{1}{4\pi\epsilon_0}\frac{e^2}{R^2}$$

$$R=\frac{4\pi\epsilon_0L^2}{\mu e^2}$$

If we then postulate that CM angular momentum is quantized as $$L=n\hbar$$ for integer $$n$$, we get

$$R_n=n^2\frac{4\pi\epsilon_0\hbar^2}{\mu e^2}=n^2a_0$$

Where

$$a_0=\frac{4\pi\epsilon_0\hbar^2}{\mu e^2}$$

Is (the Bohr radius) defined as the smallest nonzero orbital radius possible.

HOWEVER, if we insist on using the uncertainty principle, we can get an estimate of the smallest orbital radius. When we squeeze uncetainty to a minimum, we should get something of order

$$\Delta x\Delta p \sim h$$

The position shold vary between the orbit extremes, giving us something of order

$$\Delta x\sim R$$

For the momentum, it should vary between the extremes of momentum and therefore

$$\Delta p\sim \mu v$$

This then gives us

$$\mu vR\sim h$$

Which you can see is pretty much the same as quantizing angular momentum and choosing small $$n$$. This gives us an estimate of the smallest possible radius as

$$a_0\sim \frac{4\pi\epsilon_0h^2}{\mu e^2}$$

Which is only off by a factor of $$(h/\hbar)^2=(2\pi)^2\approx40$$ which isn't that bad.

• Or you could start with the correct version, $\Delta x\Delta p \sim \hbar$, and get it right in one go. – Emilio Pisanty Jan 4 at 22:01
• @EmilioPisanty the notion of the right form of the uncertainty principle goes quite deep. It is, however, well agreed that we should use $h$ when using characteristic dimensions (as I did) and $\hbar$ when using true uncertainties. – Gabriel Golfetti Jan 4 at 22:03
• For reference, there is this really nice paper on olympiad quantum mechanics here – Gabriel Golfetti Jan 4 at 22:05