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.
Best Answer
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.