Electron-positron distance in positronium I find that the positronium atom has a reduced mass of $\mu =4.55 \times 10^{-31}$ kg (i.e. $m_e / 2$), which gives a ground state energy ($n=1$) of 
$$E = \frac{\mu e^4}{2(4\pi \epsilon_0 \hbar)^2}=6.84 eV$$
This gives me a ground state radius of $r = 1.05 \times 10^{-10}$ m. I am asked to find the electron-positron distance. I have been told the answer is precisely the value of $r$ I just found, but intuitively I would say it should be the diameter of the atom, so $2r$. (I have in mind the picture of an electron and a positron orbiting at opposite points along the same circle around a common point, and I would think that $r$ represents the radius of said circle). Why is this not the case? Thank you.
 A: Your mental picture is incorrect. When we do the coordinate separation from the individual coordinates $\mathbf r_1$, $\mathbf r_2$ to the center-of-mass / relative coordinates
\begin{align}
\mathbf R & = \frac{m_1\mathbf r_1+m_2\mathbf r_2}{m_1+m_2} \\
\mathbf r & = \mathbf r_2-\mathbf r_1,
\end{align}
the relative coordinate is the vector from mass $1$ to mass $2$, not some (sub)multiple of it. This is the fundamental dynamical variable of hydrogenic problems, i.e. the one with the decoupled dynamics with hamiltonian
$$
H = \frac{1}{2\mu} \mathbf p^2 - \frac{Ze^2}{r}
$$
for $\mu$ the reduced mass and $[r_i,p_j]=i\hbar\delta_{ij}$, and it is this dynamical variable that carries all the results.
A: Actually your picture of an electron and a positron orbiting on opposite sides around a center is not so wrong! The conceptual problem arises only from the separation of the Hamiltonian of the Schrödinger equation into two parts. One part with the center of mass  coordinate $\vec R$ and the sum of the positron and electron masses. The second part with the coordinate difference of the particles $\vec r=\vec r_1-\vec r_2$ and an equivalent fictitious particle with the reduced mass $\mu=m_e/2=m_p/2$ with a Coulomb potential energy $$V(\vec r_1-\vec r_2)=V(\vec r)= -\frac {e^2}{4\pi \epsilon_0 |\vec r|}$$ Thus you get for this fictitious system formally a radius $a=0.106nm$ for the ground state orbital which is twice the Bohr radius $a_0$. When you look at the position vector of the individual particles $\vec r_1$ and $\vec r_2$, you will see that they are given by $$\vec r_1 =\vec R+ \vec r/2$$ and $$\vec r_2 =\vec R- \vec r/2$$ Thus the distance of the positron and electron in the ground state can be considered to be the radius of the fictitious particle orbital $r=a=0.106nm$ as stated in the other answer. But both the electron and the positron are actually orbiting on "opposite sides" around the common center of gravity $\vec R$ with a radius $$|r_1-R|=|r_2-R|=\frac {a}{2}=a_0$$ which is just the Bohr radius.
