In calculating the quadratic Stark effect on the ground state of hydrogen, we find that given unperturbed hamiltonian $$H^0=\frac{p^2}{2m}-\frac{Zq_e^2}{r}$$ 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 $$H^1=[\Omega,H^0].$$ 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 $C=\frac{9}{4}=2.25$, which is well out of the error range of the value we calculated before. 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.
