QB_PH_SEC6_021: Atomic Spectra & Multi-Electron Atoms

In the linear Stark effect in the ground state ($n = 1, 1s$) of the hydrogen atom, the first-order energy shift $\Delta E^{(1)}$ is

Correct Answer

### Step-by-Step Derivation & Verification: 1. **Perturbation Hamiltonian**: For a uniform static electric field $\vec{\mathcal{E}} = \mathcal{E}\hat{z}$: $H' = -\vec{d} \cdot \vec{\mathcal{E}} = e z \mathcal{E}$ 2. **First-Order Expectation Value**: $\Delta E^{(1)} = \langle \psi_{100} | e z \mathcal{E} | \psi_{100} \rangle = e \mathcal{E} \int |\psi_{100}(\vec{r})|^2 z \, d^3 r$ 3. **Parity Symmetry Evaluation**: - Inversion operator $\mathcal{P}$: $\vec{r} \to -\vec{r} \implies z \to -z$. - Hydrogen ground state: $\psi_{100}(r) = \frac{1}{\sqrt{\pi a_0^3}} e^{-r/a_0}$ is spherically symmetric (parity $\pi = (-1)^l = (-1)^0 = +1$, strictly even). - Probability density $|\psi_{100}|^2$ is strictly even under inversion: $|\psi_{100}(-\vec{r})|^2 = |\psi_{100}(\vec{r})|^2$. - Operator $z$ is odd: $z(-\vec{r}) = -z$. - Integrand: $(\text{even}) \times (\text{odd}) = \text{odd}$. - Integral over all space vanishes: $\Delta E^{(1)} = 0$. 4. **Conclusion**: Identically zero. Option (A) is correct.