= Linear Stark effect in the first excited hydrogen eigenspace
{c}
{title2=$\Delta E=0,0,\pm3ea_0\mathcal E$}
Consider a fixed <proton> and one <Electron>, omitting <spin>, <fine structure> and the <Lamb shift>. Let $e>0$ be the <elementary charge>, $a_0$ the <Bohr radius> in this fixed-nucleus model, and $\mathbf E=\mathcal E\widehat{\mathbf z}$. Because the <Electron>'s <electric dipole moment> is $\mathbf d=-e\mathbf r$, its <electric-dipole interaction> is
$$
H'= -\mathbf d\cdot\mathbf E=e\mathcal E z.
$$
The <first excited hydrogen eigenspace> has a common unperturbed <energy eigenvalue> $E_2$. Write $|s\rangle=|2,0,0\rangle$ and $|p_m\rangle=|2,1,m\rangle$ in the <hydrogen energy eigenbasis>. The <electric-dipole selection rules for hydrogen> leave only the $s,p_0$ coupling within this <eigenspace>. Each diagonal <matrix element> vanishes by <parity>, but that does not make the entire perturbation vanish. <Degenerate perturbation theory> requires its <eigenvalues> in the whole <eigenspace>. With the usual real <wavefunction> phases the off-diagonal position <matrix element> is $\langle s|z|p_0\rangle=-3a_0$.<Quantum Mechanics by Richard Fitzpatrick/Linear Stark Effect>[Quantum Mechanics by Richard Fitzpatrick; Fitzpatrick, Linear Stark Effect, equations (7.82)–(7.86)]{ref}
The factor three can be checked directly. Set $\rho=r/a_0$ and use normalized radial <wavefunctions> and <spherical harmonics>:
$$
R_{20}=\frac{(2-\rho)e^{-\rho/2}}{2\sqrt2\,a_0^{3/2}},
\qquad
R_{21}=\frac{\rho e^{-\rho/2}}{2\sqrt6\,a_0^{3/2}},
\qquad
Y_{00}=\frac1{\sqrt{4\pi}},
\qquad
Y_{10}=\sqrt{\frac3{4\pi}}\cos\theta.
$$
The angular <integral> is $\int Y_{00}^*\cos\theta Y_{10}\,d\Omega=1/\sqrt3$. Including $z=r\cos\theta$ and the radial volume factor $r^2$ then gives
$$
\langle s|z|p_0\rangle
=\frac{a_0}{24}\int_0^\infty\rho^4(2-\rho)e^{-\rho}\,d\rho
=\frac{a_0}{24}(2\cdot4!-5!)=-3a_0.
$$
Reversing the <quantum phase> of $|p_0\rangle$ reverses this sign, with no change to the predicted <energy eigenvalues>.
Thus the nontrivial two-dimensional block and its <determinant> equation are
$$
H'_{s,p_0}=e a_0\mathcal E
\begin{pmatrix}0&-3\\-3&0\end{pmatrix},
\qquad
\det(H'_{s,p_0}-\Delta E I)
=(\Delta E)^2-9e^2a_0^2\mathcal E^2=0.
$$
Consequently, for the displayed sign convention,
$$
|\psi_-\rangle=\frac{|s\rangle+|p_0\rangle}{\sqrt2},
\quad \Delta E_-=-3ea_0\mathcal E,
\qquad
|\psi_+\rangle=\frac{|s\rangle-|p_0\rangle}{\sqrt2},
\quad \Delta E_+=+3ea_0\mathcal E.
$$
The two remaining <energy eigenstates> $|p_{+1}\rangle,|p_{-1}\rangle$ have zero first-order shifts. The slopes correspond to opposite <electric dipole moments>, $\langle d_z\rangle=+3ea_0$ and $-3ea_0$. The <electric field> selects combinations of opposite <parity> that already have these dipole expectations at leading order. This is why applying <first-order nondegenerate perturbation theory> separately to $2s$ and $2p_0$, and obtaining two zero diagonal entries, misses the splitting.
The approximation requires $ea_0|\mathcal E|$ small compared with the separation to other <principal quantum numbers>. For real <hydrogen atoms>, it also requires the relevant <electric-dipole interaction> to dominate the small splittings omitted here, including the <Lamb shift>. At sufficiently small <electric fields> a pre-existing nonzero $2s$–$2p$ separation must be retained before taking the limit. A simple two-state illustration with unperturbed separation $\delta>0$ has shifts relative to its mean
$$
E_\pm-\overline E=\pm\sqrt{(\delta/2)^2+(3ea_0\mathcal E)^2}.
$$
For $3ea_0|\mathcal E|\ll\delta$, the lower branch moves by $-(3ea_0\mathcal E)^2/\delta+O(\mathcal E^4)$ relative to its zero-field value. At $\delta=0$ the splitting is linear. This interpolation explains the importance of the <hydrogen energy-level degeneracy>; it is not a complete model of the <fine structure> multiplet.
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