I am trying to calculate the probability of finding an electron in a small volume element close to the nucleus of a ground state hydrogen atom. I then need to calculate the same but replacing the electron with a negative pion. By "close to the nucleus" it means some radius R where $R \ll a_0$ and $a_0$ is the bohr radius for an electron. I have tried to calculate the probability using the wavefunction of ground state hydrogen but that doesn't seem to be working out for me. $$\Psi = \frac {1}{\sqrt {\pi}} \left (\frac {1}{a_0} \right )^{\frac {3}{2}} \exp \left [ \frac {-r}{a_0}\right ]$$

$$P = \int_0^{R}\int_0^{\pi}\int_0^{2\pi} \Psi.\Psi^* r^2 \sin(\theta)dr d\theta d\phi = \frac {4}{a_0^3}\int_0^R r^2\exp \left [ \frac {-2r}{a_0}\right ] dr$$

$$= \frac {4}{a_0^3} \left[\frac {4}{a_0^3} - \frac 1 4 (a_0^3 + 2Ra_0^2 + 2R^2a_0)\exp \left [ \frac {-2R}{a_0}\right ] \right]$$

Multiplying in the constant:

$$=1-\left(1+\frac {2R}{a_0} + \frac {2R^2}{a_0^2}\right)\exp \left [ \frac {-2R}{a_0}\right ]$$

Then when I take the approximation $R \ll a_0$ it simply leaves me with $P = 1 - 1 = 0$ and similar for the pionic hydrogen. Since this doesn't really reveal anything about the different probability distributions of electronic hydrogen and pionic hydrogen I get the feeling I'm not using the right approach. Is there a different way I should go about this?

  • $\begingroup$ Assuming your approach is correct (I'm not well-equipped in QM), why do you insist on setting $R/a_{0}=0$ in your final expression? I understand that $R\ll a_{0}$ but shouldn't $P$ have some dependence on $R$ after all? $\endgroup$
    – Newbie
    Jan 28 at 16:18
  • $\begingroup$ Note: in v2 of the question, the third line of mathematics has a term with dimension $a_0^{-3}$ added to some terms with dimension $a_0^{+3}$. You seem to fix this when you pull out the constant. $\endgroup$
    – rob
    Jan 28 at 16:36
  • 1
    $\begingroup$ @Newbie It really should still have some R dependence, I guess I was a little over-zealous with my simplifying. My end goal was to have an expression for both systems which when compared like $\frac {P_2} {P_1}$ would cancel off the R values and leave me with a constant ratio. $\endgroup$
    – Allod
    Jan 29 at 16:32

1 Answer 1


Personally, I’d do the approximation in the exponential,

$$e^\epsilon \approx 1 + \epsilon + \frac{\epsilon^2}{2!} + \cdots, $$

before doing the integral.

\begin{align} P &= \frac4{a_0^3} \int_0^R \mathrm dr\, r^2 \exp\frac{-2r}{a_0} \\ \frac{a_0^3}{4} P &\approx \int_0^R\mathrm dr\,r^2 - \frac2{a_0}\int_0^R \mathrm dr\,r^3 \\&= \frac{R^3}{3} - \frac2{a_0}\frac{R^4}{4} \\&= \frac{R^3}{3} \left( 1 - \frac32 \frac R{a_0} \right) \\ P &\approx \frac43 \left(\frac{R}{a_0}\right)^3 \end{align}

Throwing away the second term basically says “this volume is small enough we can assume the exponential is secretly a constant.”

This result suggests you are getting zeros because you’re only keeping terms that are quadratic in $R/a_0$, but apparently they all get cancelled out.

  • $\begingroup$ thank you for the suggestion, it definitely seems like a more reasonable argument and returns a similar relationship $P \propto \frac {1}{a_0^3}$ which I found after trying some alternate approaches myself. I'll edit them into the original question later $\endgroup$
    – Allod
    Jan 29 at 16:37

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