Stark effect

An electric field pointing from left to right, for example, tends to pull nuclei to the right and electrons to the left. In another way of viewing it, if an electronic state has its electron disproportionately to the left, its energy is lowered, while if it has the electron disproportionately to the right, its energy is raised.

Other things being equal, the effect of the electric field is greater for outer electron shells, because the electron is more distant from the nucleus, so it travels farther left and farther right.

The Stark effect can lead to splitting of degenerate energy levels. For example, in the Bohr model, an electron has the same energy whether it is in the 2s state or any of the 2p states. However, in an electric field, there will be hybrid orbitals (also called quantum superpositions) of the 2s and 2p states where the electron tends to be to the left, which will acquire a lower energy, and other hybrid orbitals where the electron tends to be to the right, which will acquire a higher energy. Therefore, the formerly degenerate energy levels will split into slightly lower and slightly higher energy levels.

The Stark effect originates from the interaction between a charge distribution (atom or molecule) and an external electric field. Before turning to quantum mechanics we describe the interaction classically and consider a continuous charge distribution ρ(r). If this charge distribution is non-polarizable its interaction energy with an external electrostatic potential V(r) is

${\displaystyle E_{\mathrm {int} }=\int \rho (\mathbf {r} )V(\mathbf {r} )d\mathbf {r} ^{3}}$.

If the electric field is of macroscopic origin and the charge distribution is microscopic, it is reasonable to assume that the electric field is uniform over the charge distribution. That is, V is given by a two-term Taylor expansion,

${\displaystyle V(\mathbf {r} )=V(\mathbf {0} )-\sum _{i=1}^{3}r_{i}F_{i}}$, with the electric field: ${\displaystyle F_{i}\equiv -\left.\left({\frac {\partial V}{\partial r_{i}}}\right)\right|_{\mathbf {0} }}$,

where we took the origin 0 somewhere within ρ. Setting V(0) as the zero energy, the interaction becomes

${\displaystyle E_{\mathrm {int} }=-\sum _{i=1}^{3}F_{i}\int \rho (\mathbf {r} )r_{i}d\mathbf {r} \equiv -\sum _{i=1}^{3}F_{i}\mu _{i}=-\mathbf {F} \cdot {\boldsymbol {\mu }}}$.

Here we have introduced the dipole moment μ of ρ as an integral over the charge distribution. In case ρ consists of N point charges q_j this definition becomes a sum

${\displaystyle {\boldsymbol {\mu }}\equiv \sum _{j=1}^{N}q_{j}\mathbf {r} _{j}}$.

Electric-field perturbation applied to a classical hydrogen atom produces a distortion of the electron orbit in a direction perpendicular to the applied field.[10] This effect can be shown without perturbation theory using the relation between the angular momentum and the Laplace–Runge–Lenz vector.[11] Using the Laplace-Runge-Lenz approach, one can see both the transverse distortion and the usual Stark effect.[12] The transverse distortion is not mentioned in most textbooks. This approach can also lead to an exactly solvable approximate model Hamiltonian for an atom in a strong oscillatory field.[13] “There are few exactly-solvable problems in quantum mechanics, and even fewer with a time-dependent Hamiltonian.”[14]

Turning now to quantum mechanics an atom or a molecule can be thought of as a collection of point charges (electrons and nuclei), so that the second definition of the dipole applies. The interaction of atom or molecule with a uniform external field is described by the operator

${\displaystyle V_{\mathrm {int} }=-\mathbf {F} \cdot {\boldsymbol {\mu }}.}$

This operator is used as a perturbation in first- and second-order perturbation theory to account for the first- and second-order Stark effect.

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 x 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 a dipole moment is a polar vector, the diagonal elements of the perturbation matrix V_int vanish for systems with an inversion center (such as atoms). Molecules with an inversion center in a non-degenerate electronic state do not have a (permanent) dipole 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,

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

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

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

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.

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

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

with the components of the polarizability tensor α defined by

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

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.

It is important to note that for the ground state ${\displaystyle \alpha _{0}}$ is always positive, i.e., the quadratic Stark shift is always negative.

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).