First order

Let the unperturbed atom or molecule be in a g-fold degenerate state with orthonormal zeroth-order state functions ${\displaystyle \psi _{1}^{0},\ldots ,\psi _{g}^{0}}$. (Non-degeneracy is the special case g = 1). According to perturbation theory the first-order energies are the eigenvalues of the g × g matrix with general element

$${\displaystyle (\mathbf {V} _{\mathrm {int} })_{kl}=\langle \psi _{k}^{0}|V_{\mathrm {int} }|\psi _{l}^{0}\rangle =-\mathbf {F} \cdot \langle \psi _{k}^{0}|{\boldsymbol {\mu }}|\psi _{l}^{0}\rangle ,\qquad k,l=1,\ldots ,g.}$$

If g = 1 (as is often the case for electronic states of molecules) the first-order energy becomes proportional to the expectation (average) value of the dipole operator ${\displaystyle {\boldsymbol {\mu }}}$,

$${\displaystyle E^{(1)}=-\mathbf {F} \cdot \langle \psi _{1}^{0}|{\boldsymbol {\mu }}|\psi _{1}^{0}\rangle =-\mathbf {F} \cdot \langle {\boldsymbol {\mu }}\rangle .}$$

Because the electric dipole moment is a vector (tensor of the first rank), the diagonal elements of the perturbation matrix V_int vanish between states with a certain parity. Atoms and molecules possessing inversion symmetry do not have a (permanent) dipole moment and hence do not show a linear Stark effect.

In order to obtain a non-zero matrix V_int for systems with an inversion center it is necessary that some of the unperturbed functions ${\displaystyle \psi _{i}^{0}}$ have opposite parity (obtain plus and minus under inversion), because only functions of opposite parity give non-vanishing matrix elements. Degenerate zeroth-order states of opposite parity occur for excited hydrogen-like (one-electron) atoms or Rydberg states. Neglecting fine-structure effects, such a state with the principal quantum number n is n²-fold degenerate and

$${\displaystyle n^{2}=\sum _{\ell =0}^{n-1}(2\ell +1),}$$

where ${\displaystyle \ell }$ is the azimuthal (angular momentum) quantum number. For instance, the excited n = 4 state contains the following ${\displaystyle \ell }$ states,

$${\displaystyle 16=1+3+5+7\;\;\Longrightarrow \;\;n=4\;{\text{contains}}\;s\oplus p\oplus d\oplus f.}$$

The one-electron states with even ${\displaystyle \ell }$ are even under parity, while those with odd ${\displaystyle \ell }$ are odd under parity. Hence hydrogen-like atoms with n>1 show first-order Stark effect.

The first-order Stark effect occurs in rotational transitions of symmetric top molecules (but not for linear and asymmetric molecules). In first approximation a molecule may be seen as a rigid rotor. A symmetric top rigid rotor has the unperturbed eigenstates

$${\displaystyle |JKM\rangle =(D_{MK}^{J})^{*}\quad {\text{with}}\quad M,K=-J,-J+1,\dots ,J}$$

with 2(2J+1)-fold degenerate energy for |K| > 0 and (2J+1)-fold degenerate energy for K=0. Here D^J_MK is an element of the Wigner D-matrix. The first-order perturbation matrix on basis of the unperturbed rigid rotor function is non-zero and can be diagonalized. This gives shifts and splittings in the rotational spectrum. Quantitative analysis of these Stark shift yields the permanent electric dipole moment of the symmetric top molecule.

Second order

As stated, the quadratic Stark effect is described by second-order perturbation theory. The zeroth-order eigenproblem

$${\displaystyle H^{(0)}\psi _{k}^{0}=E_{k}^{(0)}\psi _{k}^{0},\quad k=0,1,\ldots ,\quad E_{0}^{(0)}<E_{1}^{(0)}\leq E_{2}^{(0)},\dots }$$

is assumed to be solved. The perturbation theory gives

$${\displaystyle E_{k}^{(2)}=\sum _{k'\neq k}{\frac {\langle \psi _{k}^{0}|V_{\mathrm {int} }|\psi _{k^{\prime }}^{0}\rangle \langle \psi _{k'}^{0}|V_{\mathrm {int} }|\psi _{k}^{0}\rangle }{E_{k}^{(0)}-E_{k'}^{(0)}}}\equiv -{\frac {1}{2}}\sum _{i,j=1}^{3}\alpha _{ij}F_{i}F_{j}}$$

with the components of the polarizability tensor α defined by

$${\displaystyle \alpha _{ij}=-2\sum _{k'\neq k}{\frac {\langle \psi _{k}^{0}|\mu _{i}|\psi _{k'}^{0}\rangle \langle \psi _{k'}^{0}|\mu _{j}|\psi _{k}^{0}\rangle }{E_{k}^{(0)}-E_{k'}^{(0)}}}.}$$

The energy E⁽²⁾ gives the quadratic Stark effect.

Neglecting the hyperfine structure (which is often justified — unless extremely weak electric fields are considered), the polarizability tensor of atoms is isotropic,

$${\displaystyle \alpha _{ij}\equiv \alpha _{0}\delta _{ij}\Longrightarrow E^{(2)}=-{\frac {1}{2}}\alpha _{0}F^{2}.}$$

For some molecules this expression is a reasonable approximation, too.

For the ground state ${\displaystyle \alpha _{0}}$ is always positive, i.e., the quadratic Stark shift is always negative.

Problems

The perturbative treatment of the Stark effect has some problems. In the presence of an electric field, states of atoms and molecules that were previously bound (square-integrable), become formally (non-square-integrable) resonances of finite width. These resonances may decay in finite time via field ionization. For low lying states and not too strong fields the decay times are so long, however, that for all practical purposes the system can be regarded as bound. For highly excited states and/or very strong fields ionization may have to be accounted for. (See also the article on the Rydberg atom).