I've asked this question before on physics forums to no avail. Since the heart of this problem is probably in the mathematics, I think this fits here.

The following notation is used:

In calculating the quadratic Stark effect on the ground state of hydrogen, we find that given unperturbed hamiltonian $$H^0=\frac{p^2}{2m}-\frac{q_e^2}{r}$$ with pertubation $$H^1=q_e\mathcal{E}z$$ then the second order correction to the ground state energy is $$E_{100}^{(2)}=-Ca_0^3\mathcal{E}^2$$ where $$C=\frac{q_e^2}{a_0^3}\sum_{n=2}^\infty\frac{|\langle n10|z|100\rangle|^2}{E_{1}-E_{n}}=\frac{2}{3a_0^2}\sum_{n=2}^\infty\frac{n^2}{n^2-1}\left(\int_0^\infty r^3R_{n1}(r)R_{10}(r)dr\right)^2=\sum_{n=2}^\infty\frac{2^9n^9(n-1)^{2n-6}}{3(n+1)^{2n+6}}$$ This sum is nasty. Wolfram Alpha gives $$C\approx C_{1000}=\sum_{n=2}^{1000}\frac{2^9n^9(n-1)^{2n-6}}{3(n+1)^{2n+6}}=1.8314\ldots$$ To calculate the error on this approximation, we note $$0<\frac{2^9n^9(n-1)^{2n-6}}{3(n+1)^{2n+6}}<\frac{2^9}{3e^4(n-1)^3}$$ so $$C-C_{1000}< \frac{2^9}{3e^4}\int_{999}^\infty\frac{1}{(x-1)^3}dx=\frac{2^8}{3(998)^2e^4}<1.57\times10^{-6}$$ so we can be certain about the digits shown. However, in Principles of Quantum Mechanics, 2nd ed. pg. 462, Shankar avoids the trouble of calculating the radial integrals by introducing the operator $\Omega$ given by $$\Omega = -\frac{ma_0q_e\mathcal{E}}{\hslash^2}\left(\frac{r^2\cos\theta}{2}+a_0r\cos\theta\right),$$ which has the property that $$H^1|100\rangle=[\Omega,H^0]|100\rangle.$$ Hence, $$E_{100}^{(2)}=\sum_{n=2}^\infty\frac{\langle 100|H^1|n10\rangle\langle n10|H^1|100\rangle}{E_n-E_1}\\ =\sum_{n=2}^\infty\frac{\langle 100|H^1|n10\rangle\langle n10|\Omega H^0-H^0\Omega|100\rangle}{E_n-E_1}\\ = \sum_{n=2}^\infty\langle 100|H^1|n10\rangle\langle n10|\Omega|100\rangle\\ = \langle H^1\Omega\rangle_{100}-\langle H^1\rangle_{100}\langle \Omega \rangle_{100}\\ = -\frac{9}{4}a_0^3\mathcal{E}^2$$ Now surprisingly, this gives the well quoted $C=\frac{9}{4}=2.25$, which is well out of the error range of the value we calculated before. I also find similar discrepancies in other problems like this one. For example, this paper uses very similar methods to determine $$\sum_{n=2}^\infty\frac{2^8n^5(n-1)^{2n-4}}{3(n+1)^{2n+4}}=1,$$ but Wolfram Alpha calculates $$\sum_{n=2}^{1000}\frac{2^8n^5(n-1)^{2n-4}}{3(n+1)^{2n+4}}=0.564889...$$ So my question is:

What went wrong? Why do we get two different values for $C$, when we use two equally valid methods? As far as I know, neither method makes assumptions that the other doesn't, so we should get the same answer both times.

  • $\begingroup$ Maple agrees $$\sum_{n=2}^\infty\frac{2^8n^5(n-1)^{2n-4}}{3(n+1)^{2n+4}}\approx 0.565$$and $$\sum_{n=2}^{1000}\frac{2^9n^9(n-1)^{2n-6}}{3(n+1)^{2n+6}}\approx1.83$$ $\endgroup$
    – GEdgar
    Commented Nov 7, 2021 at 13:57

1 Answer 1


I have determined that the discrepancy is due to, ironically enough, and incomplete use of the completeness relation. I've verified numerically that $$\langle H^1\Omega\rangle-\langle H^1\rangle\langle\Omega\rangle\neq\sum_{n=2}^\infty\langle 100|H^1|n10\rangle\langle n10|\Omega|100\rangle=\sum_{n=2}^\infty\frac{2^9n^9(n-1)^{2n-6}}{3(n+1)^{2n+6}}=1.8314\ldots$$ The problem is that the full completeness relation for a system is given by $$I=\sum_{n} |e_n\rangle\langle e_n|+\int|\varepsilon\rangle\langle\varepsilon|d\varepsilon$$ where the sum is over the discrete (bound) states and the integral is over the continuum (scattering) states. For the Hydrogen atom, the discrete states correspond to negative energy solutions to the Hamiltonian, while the continuum states correspond to the positive energy solutions. You can find the continuum states in atomic units here. Now it is often the case that continuum solutions are simply ignored in systems with bound states because their contribution is small, but the quadratic Stark effect on the Hydrogen atom is an example where continuum states do have a substantial effect. This is explained in detail in this article. We ignored the continuum state contribution in the completeness relation, so we ended up with a larger value for $C$ than expected. Whether the first approach is necessarily right can be difficult to say, but it is the most consistent since it only uses bound states, while the second approach makes an invalid step. The most complete approach would be to use scattering state perturbation theory to incorporate scattering state contribution to the energy corrections.

  • $\begingroup$ Are you sure you have referenced the right intermediate sum for the numerical value quoted i.e. $1.8314...$? $\endgroup$ Commented Nov 14, 2021 at 16:18
  • $\begingroup$ @JamesArathoon Thank you, that was an error. It is fixed now. $\endgroup$
    – Jacob
    Commented Nov 14, 2021 at 17:34

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