# Questions on Stark Effect on Hydrogen

Suppose that a hydrogen atom is subject to a weak uniform electric field $$\vec{E}=\epsilon \hat{z}$$. Let's neglect the effect of electron spin. The perturbation added to the original hamiltonian $$H_0$$ is $$H^{'}=e\epsilon Z$$.

Q1. The energy shift for the ground state is given by $$E_1 =-\frac{1}{2}mc^2\alpha^2\frac{1}{n^2} +e\epsilon \langle100|e\epsilon Z|100\rangle +\sum_{n=2}^\infty \frac{|\langle n10|e\epsilon Z|100\rangle|^2}{E_1^0-E_n^0}$$ The first order perturbation on energy is $$0$$ and most of quantum mechanics textbooks explain the Stark effect only to this extent. I found on the Internet that the exact value of the second perturbation (the third term in the RHS) is $$-\frac{9}{4} a_0^3 \epsilon^2$$ where $$a_0=\frac{\hbar^2}{me^2}$$ is the Bohr radius. How can I calculate this value?

Q2. What about the first order perturbation $$|100^{(1)}\rangle$$ of the ground state $$|100\rangle$$? $$|100^{(1)}\rangle\,=\sum_{n=2}^\infty \frac{|n10\rangle\langle n10|e\epsilon Z|100\rangle}{E_1^0-E_n^0}$$

• Your equations are actually wrong, although in a subtle way. The sums you have written only run over bound $p_{z}$ states; however, there are also unbound (continuum) states that contribute as virtual intermediates.
– Buzz
Oct 11 '18 at 21:23
• @Buzz Thanks for comment. I used the parity of the hydrogen and the commutation relation of it with $L_z$ wave function to eliminate other terms that will give us 0. What exactly does the unbounded states mean?
– asdf
Oct 12 '18 at 6:54
• The spectrum of the Hamiltonian for a $1/r$ potential contains both a set of discrete states states with energies $E<0$ (bound) and a continuum of states with energies $E>0$ (unbound). (These are analogous to the elliptical and hyperbolic orbits in the classical Kepler problem.) At large distances, where the potential is almost negligible, the continuum states can look almost like the plane wave eigenstates of the free electron Hamiltonian; however, they are strongly distorted at shorter distances. The unbound states can have the right angular momentum values to contribute to the sum.
– Buzz
Oct 12 '18 at 8:40

Q1. There are at least two ways to calculate 2nd order correction. Let me first of all sketch the idea and then provide some references. First, you should keep in mind Feynman-Hellman theorem, $$\frac{\partial f_n(\lambda)}{\partial \lambda}=\left\langle n\right|\frac{\partial \hat{f}}{\partial \lambda}\left| n\right\rangle,$$ where $$f$$ is an operator and $$\lambda$$ is a parameter. In case of 1s hydrogen atom, you know that interaction with electric field and this expression becomes $$\frac{\partial \delta \epsilon}{\partial E}=\langle 0|\frac{\partial H}{\partial E}|0\rangle,$$ where $$E$$ is electric field. Then, $$\frac{\partial \delta \epsilon}{\partial E}=\alpha E\rightarrow \delta\epsilon=-\frac{\alpha^2E}{2}\rightarrow\boxed{\alpha=-\frac{2\delta\epsilon}{E^2}}$$ with $$\alpha$$ is the polarization of the atom. The last expression gives the idea how to calculate 2nd order Stark effect: you should know the 2nd order correction to energy. Then, you know how to evaluate this correction, $$\delta \epsilon=\sum_{n\neq 1}\frac{|\langle n|\hat{V}|1\rangle|^2}{\epsilon_1^{(0)}-\epsilon_n^{(0)}},$$ where $$\hat{V}=-E\hat{r}\cos\theta$$ (we set the electric field in $$z$$-direction). It is possible to perfom summation, at least check that the sum converges. However, there is more efficient way to find $$\delta\epsilon$$, $$\delta \epsilon = \langle\psi^{(1)}|\hat{V}|\psi^{(0)}\rangle,$$ where $$\psi^{(1)}$$ and $$\psi^{(0)}$$ are 1st-order corrected wave-function and wave-functon without perturbation. It is quite straightforward (if you need more details, I will add them) to find that $$\boxed{\psi^{(1)}=\frac{2Er}{\sqrt{r}}\left(1\frac{r}{2}\right)e^{-r}Y_{10}(\theta,\phi).}$$ Then, the final step is to calculate the integral (if need more detail, I will add them) and the answer is $$\delta\epsilon=-\frac{9E^2}{4}\rightarrow\alpha=\frac{9}{2}.$$