[Physics] Why are the classical electron radius, the Bohr radius and the Compton wavelength of an electron related to each other

atomic-physicselectromagnetismphysical constantsquantum mechanics

Using the definition of the fine-structure constant $\alpha = \frac{e^2}{4 \pi \epsilon_0 \hbar c}$ and the Compton wavelength of an electron $\lambda_c = \frac{h}{m_e c}$ the classical electron radius $r_e$ and the Bohr radius $a_0$ can be expressed like $$r_e = \alpha \frac{\lambda_c}{2\pi}$$ $$a_0 = \frac{1}{\alpha} \frac{\lambda_c}{2\pi} $$

This means e.g. that the classical electron radius can be expressed in terms of the Bohr radius as $r_e = \alpha^2 a_0$.

Isn't that peculiar? Why should the classical radius of the electron and the distance of an electron to the nucleus in an atom be related to each other? And why are both multiples of the Compton wavelength?

Best Answer

It is not surprising that both $r_e$ and $a_0$ are multiples of the Compton wavelength: any two positive lengths are multiples of each other. While it is true that there is more to this than that simple statement, the essential fact is that since those three lengths are composed simply and out of the same basic ingredients, there is very little leeway for how they can be different.

Let's have a look at these quantities: $$ \lambda_C=\frac{2\pi\hbar}{mc},\,\, r_e=\frac{e^2}{mc^2}\text{ and }a_0=\frac{\hbar^2}{m e^2}, $$ in Gaussian units, where $m$ is the electron mass. $%These are, respectively, the characteristic length scales of the photon momentum that matches an electron, the electron's rest mass as electrostatic energy, and the basic quantum mechanical electrostatic problem for the electron.$

Notice that they are all inversely proportional to the electron rest mass, though for different reasons: heavier electrons would require beefier photons to deflect them; they have higher rest mass and would need a more compact spherical charge to match; and a higher $m$ effectively reduces $\hbar$ in the hydrogenic Schrödinger equation, making it harder to get to the quantum regime.

Given that, you have three lengths that are determined by the three constants $\hbar$, $c$ and $e$. That's enough constants to make three different lengths, but they are few enough that any quotient must be a function of the unique dimensionless combination of these constants - the fine structure constant, $$\alpha=\frac {e^2} {\hbar c}.$$ Thus, it is necessary that any two of these three lengths must be multiples of the third and of $\alpha^{\pm1}$ (modulo $2\pi$).

This constant, however, is particularly important. It is the natural measure of the strength of electromagnetic interactions: it gives, as a pure number, the electromagnetic coupling $e^2$ between two unit charges, in natural relativistic units where $\hbar=c=1$. Thus, while the relations you remark on are algebraically necessary, it doesn't mean they are devoid of physical content:

  • $r_e=\alpha \lambda_C/2\pi$ says that for a more strongly interacting QED a more loosely bound spherical electron would suffice to match the rest mass energy.
  • $a_0=\frac 1\alpha \lambda_C/2\pi$ says that for a more strongly interacting QED the proton would hold its hydrogenic electron in tighter orbits.

Both of these are indeed most naturally phrased in terms of the Compton wavelength, as it is the characteristic quantum-relativistic length scale of the electron, and does not depend on any particular physical interaction, whereas the other two do - and are therefore obtained from the first via the strength of that interaction.

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