# Why are the classical electron radius, the Bohr radius and the Compton wavelength of an electron related to each other?

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## Solution 1

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.

## Solution 2

The three lengths you are considering are all built using only the $e$, $m_e$ and the fundamental constants. If you look at the definitions, you can notice that they all have the form $\frac{something}{m_e}$.

It's clear then that you can get one length from the others just multiplying by some factor of $e$ and fundamental constants; since all the quantities are lengths, the factor must be dimensionless: it must be a power of $\alpha$, times some number.

The $2\pi$ factors come from the fact that $\lambda_C$ involves $h$ and the others $\hbar$.

## Solution 3

Why does $$r_e = α \frac{λ}{2π} = α^2 a_0 = \frac{α^3}{4πR_∞}$$? No appeal to numerology or "naturalness" is required.

## 0) Definitions

First, switch to the "reduced" or "angular" Compton wavelength:

$$ƛ=\frac{λ}{2π}=\frac{ℏ}{m c}$$

Similarly, define the "reduced" Rydberg wavelength, as $$R_∞$$ has dimensions of linear wavenumber:

$$ƛ_∞=\frac{1}{2πR_∞}$$

The equality now reads:

$$r_e = α ƛ_e = α^2 a_0 = α^3 \frac{ƛ_∞}{2}$$

Lastly, recall the fine-structure constant $$α$$ is defined as ratio of

1. the distance between any two unit charges to

2. the reduced wavelength of a photon whose energy equals the charges' potential energy

$$α=\frac r ƛ, \mathrm{when}\;V(r)=ℏc/ƛ$$

## 1) $$r_e$$ vs $$ƛ_e$$

• The classical electron radius $$r_e$$ is the separation between two unit charges, such that their potential energy equals the electron's rest energy, $$V(r_e)=m_e c^2$$.

• The electron's reduced Compton wavelength $$ƛ_e$$ is the reduced wavelength of a photon equalling the electron's rest energy, $$ℏc/ƛ_e=m_e c^2$$.

So $$V(r_e) = ℏc/ƛ_e$$. Their ratio is defined to be

$$\frac{r_e}{ƛ_e}=α$$

## 2) $$a_0$$ vs $$ƛ_∞$$

• The Rydberg energy is half the potential energy of two unit charges separated by one Bohr radius $$a_0$$. (Only half, because $$hcR_∞$$ was defined as the ionization energy, which per the virial theorem is half the potential.)

So $$V(a_0)=2ℏc/ƛ_∞$$. Their ratio is defined to be

$$\frac{a_0}{ƛ_∞/2}=α$$

## 3) $$ƛ_e$$ vs $$a_0$$

Not as clear. $$ƛ_e$$ corresponds to no well-known charge separation for which $$a_0$$ could be the photon's reduced wavelength. The Bohr model derives $$a_0$$ by serendipitously unsound methods, but offers no help here. Instead use QM. Griffith's (4.53) gives the radial Schrödinger equation for hydrogen is

$$E\,u(r) = -\frac{ℏ^2}{2m_e} + \left(-\frac{α ℏ c}{r} + \frac{ℏ^2}{2m_e} \frac{l(l+1)}{r^2}\right)u(r),\quad u(r)=r\,R(r),\quad ψ(r,θ,φ)=R(r)Y(θ,φ)$$

First, tidy up by expressing the total (negative) energy as the angular wavenumber $$κ=ƛ^{-1}$$ of a free electron(4.54),

$$E = -\frac{ℏ^2 κ^2}{2m_e} = -\frac{ℏ^2 κ^2}{2(\frac{ℏ}{ƛ_e c})} = - κ^2 \frac{ℏ c ƛ_e}{2},$$

changing to electron units ($$r = \hat{r} ƛ_e$$, $$κ = \hat{κ} / ƛ_e$$, $$u(r) = u(ƛ_e \hat{r})$$),

$$-\hat{κ}^2 \frac{ℏ c}{2 ƛ_e} u(r)= -\frac{ℏ c}{2 ƛ_e} u''(r) + \left(-\frac{α ℏ c}{ƛ_e \hat r} + \frac{ℏ c}{2 ƛ_e} \frac{l (l+1)}{\hat{r}^2}\right)u(r),$$

and gathering terms for the electron's mass-energy ($$E_e = m_e c^2 = ℏ c / ƛ_e$$),

$$-\hat{κ}^2 \frac{E_e}{2} u(r)= -\frac{E_e}{2} u''(r) + E_e \left(-\frac{α}{\hat r} + \frac{l (l+1)}{2\hat{r}^2}\right)u(r).$$

Now set $$l=0$$ and apply the virial theorem (4.190, 4.191, $$2⟨T⟩=⟨\vec r · \vec ∇ V⟩=⟨-V⟩=-2 E$$) to get

$$⟨E⟩ = ⟨T⟩ + ⟨V⟩ = -\hat{κ}^2 \frac{E_e}{2} = \frac{E_e}{2} - E_e ⟨\frac{α}{\hat r}⟩,$$

then subtract the kinetic term, divide by $$-E_e$$, change from wavenumber to wavelength (for clarity), and without loss of generalization let $$⟨\frac{1}{\hat r}⟩ = \frac{1}{\hat r}$$,

$$-\hat{κ}^2 E_e = - E_e ⟨\frac{α}{\hat r} ⟩,\quad \hat{κ}^2 = ⟨\frac{α}{\hat r}⟩,\quad \frac 1 {\hat{ƛ}^2} = \frac{α}{\hat r}.$$

In principle these could take any value, but the solution (4.68) requires that $$ƛ=\hat r$$ at the lowest eigenstate. Thus $$\hat r = 1/α$$ (generally $$ƛ=n/Zα$$, $$r=n^2/Z^2α$$).

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### asmaier

Updated on August 01, 2022

### Comments

• asmaier 5 months

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?

• alexchandel over 2 years
This "appeal to naturalness" is woefully misleading and incomplete. Beyond the fact that dimensionless ratios can be arbitrary transcendental numbers like $ζ(1/2)$, $γ_e$, $ζ(3/2)$, $ζ(4)$, $γ$, $δ$, or even $g_e = \sqrt{4π α}$ (all of which arise in physics), this numerology flat-out ignores the defining relationship between $r_e$ and $ƛ_e$ (and between $a_0$ and $\frac{1}{4π R_∞}$) which explains the $α$ without any handwaving.
• alexchandel over 2 years
This explains nothing. Being "a power of $α$ times some number" is trivially true of every ratio, including the ratio of the Earth's mass to Jupiter's.
• Emilio Pisanty over 2 years
@alexchandel You're perfectly welcome to post your own answer.