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 $$E_1=-\frac{1}{2}mc^2\alpha^2\frac{1}{n^2}+e\epsilon <100|e\epsilon Z|100>+\sum_{n=2}^\infty \frac{|<n10|e\epsilon Z|100>|^2}{E_1^0-E_n^0}$$ 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 $|100^{(1)}>$ of the ground state $|100>$? $$|100^{(1)}>\,=\sum_{n=2}^\infty \frac{|n10><n10|e\epsilon Z|100>}{E_1^0-E_n^0}$$

  • $\begingroup$ 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. $\endgroup$ – Buzz Oct 11 '18 at 21:23
  • $\begingroup$ @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? $\endgroup$ – SH Lee Oct 12 '18 at 6:54
  • $\begingroup$ 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. $\endgroup$ – Buzz Oct 12 '18 at 8:40

You need to integrate R*Y where R is radial wave function and Y is spherical harmonic with perturbation in between. if you just consider the first excited state and the ground state then the sum drops out I think?


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