Size of positronium

Positronium consists of an electron and a positron. By what factor is a positronium atom bigger than a hydrogen atom?

The solution has been explained to me. The characteristic length when solving for the energy levels of the hydrogenic atom is the Bohr radius: $$a_0 \equiv \frac{4 \pi \epsilon_0 \hbar^2}{\mu e^2}$$

For positronium, we can calcuate the reduced mass, $\mu$:

\begin{equation} \mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{m_e m_e}{m_e + m_e} = \frac{1}{2}m_e \end{equation}

giving a reduced mass of about half that for hydrogen. Therefore the Bohr radius for positronium is almost twice that of hydrogen.

However this is the distance between the two particles, rather than the center of rotation of the electron. The size of the atom is double this since the atom is symmetric, meaning that an atom of positronium is in fact the same size as a hydrogen atom.

My question is:

Does this mean that an atom of, say, muonium will also be the same size as a hydrogen atom, or was the positronium atom a special case because the two particles have exactly the same mass?

If we calculate $\mu$ for muonium, we get a value of

\begin{equation} \mu = \frac{m_1 m_2}{m_1 + m_2} = \frac{m_\mu m_e}{m_\mu + m_e} = 0.995 m_e \end{equation}

So the Bohr radius for a muonium atom is $\frac{1}{0.995} = 1.005$ times larger than a that of a hydrogen atom.

But this, again, is the distance between the two particles rather than the size of the atom.

So we multiply by $\frac{\mu}{m_e}$ again to get the distance of the electron to the barycenter of the system (as we did for positronium). We end up cancelling our previous factor of $\frac{1}{\mu}$, giving us the result of muonium being the same size as hydrogen.

This seems wrong!

• The (quite correct) explanation you were given for the relative size of positronium focuses on the reduced mass, right? So how does the reduced mass depend on the particles involved? Surely this is something that you can compute for a muonium atom. – dmckee Feb 23 '15 at 14:19
• I have added more working to the question. I get the same result as before- muonium is the same size as hydrogen. The idea that the size of an atom does not depend on the mass of the electron or nucleus seems counterintuitive. I am considering the physical case when we make the nucleus very light. It clearly moves very far away from the electron, meaning the atom should get very big. – Tarrare Feb 23 '15 at 15:49
• Shouldn't your calculation for the reduced mass of muonium use the proton mass instead of the electron mass? – Emilio Pisanty Apr 30 '15 at 18:25
• Has positronium been created? – Jiminion Apr 30 '15 at 18:37
• @Jiminion see en.wikipedia.org/wiki/Positronium#History – anna v Jan 22 '16 at 4:49

Let us see how the notion of the "electron cloud" enters the calculations. In the first Born approximation, the elastic scattering is determined with the atomic form-factor $F$ containing the reduced mass in the wave function and the "electron coordinate" in the exponential: $$d\sigma\propto|Zf(\mathbf{q})-F(\mathbf{q})|^2,\qquad (1)$$ $$F(\mathbf{q})=\int |\psi(\mathbf{r}_a)|^2 e^{-i\mathbf{q}\mathbf{r}_a(1-m_e/M_A)}d^3r_a,\qquad (2)$$ $$f(\mathbf{q})=\int |\psi(\mathbf{r}_a)|^2 e^{i\frac{m_e}{M_A}\mathbf{q}\mathbf{r}_a}d^3r_a,\qquad (3)$$ where $\mathbf{r}_a$ is the relative distance between the electron and the nucleus. Form-factor $F$ describes the negative charge cloud, and form-factor $f$ describes the positive charge cloud. This is how a fast charged projectile interacts with the system and "sees" its size.
The same cross section can equally be expressed via an effective potential $$U(\mathbf{r})=\int|\psi(\mathbf{r}_a)|^2V(\mathbf{r},\mathbf{r}_a)d^3r_a,\qquad (4)$$ which also depends on the reduced mass: $$d\sigma\propto\left|\int U(\mathbf{r})e^{-i\mathbf{q}\mathbf{r}}d^3r\right|^2,\qquad (5)$$ where $\mathbf{r}$ is the distance between the projectile and the target center of mass. Playing with $\mu$ helps understand what we (or the charged projectiles) deal with, without artificial size definitions. For example, in case of a "light" nucleus, the atomic size is large indeed since it is determined with these formulas.