Stark effect

The Stark effect is the shifting and splitting of spectral lines of atoms and molecules due to the presence of an external static electric field. The amount of splitting and or shifting is called the Stark splitting or Stark shift. In general one distinguishes first- and second-order Stark effects. The first-order effect is linear in the applied electric field, while the second-order effect is quadratic in the field.

The Stark effect is responsible for the pressure broadening (Stark broadening) of spectral lines by charged particles. When the split/shifted lines appear in absorption, the effect is called the inverse Stark effect.

The Stark effect is the electric analogue of the Zeeman effect where a spectral line is split into several components due to the presence of a magnetic field.

The Stark effect can be explained with fully quantum mechanical approaches, but it has also been a fertile testing ground for semiclassical methods.


Classical electrostatics

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

 E_{\mathrm{int}} = \int \rho(\mathbf{r}) V(\mathbf{r}) d\mathbf{r}.\,

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,

 V(\mathbf{r}) = V(\mathbf{0}) - \sum_{i=1}^3 r_i F_i \quad \hbox{with the electric field:}\quad 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(\mathbf{0}) as the zero energy, the interaction becomes

  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 qj this definition becomes a sum

 \boldsymbol{\mu} \equiv \sum_{j=1}^N  q_j \mathbf{r}_j.

Perturbation theory

Turning now to quantum mechanics we see an atom or a molecule 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

 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.

First order

Let the unperturbed atom or molecule be in a g-fold degenerate state with orthonormal zeroth-order state functions  \psi^0_1, \ldots, \psi^0_g . (Non-degeneracy is the special case g = 1). According to perturbation theory the first-order energies are the eigenvalues of the gg matrix with general element

 (\mathbf{V}_{\mathrm{int}})_{kl} = \langle \psi^0_k |  V_{\mathrm{int}} | \psi^0_l \rangle = -\mathbf{F}\cdot \langle \psi^0_k | \boldsymbol{\mu} | \psi^0_l \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 \boldsymbol{\mu},

 E^{(1)} = -\mathbf{F}\cdot \langle \psi^0_1 | \boldsymbol{\mu} | \psi^0_1 \rangle = -\mathbf{F}\cdot \langle  \boldsymbol{\mu} \rangle.

Because a dipole moment is a polar vector, the diagonal elements of the perturbation matrix Vint 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 Vint for systems with an inversion center it is necessary that some of the unperturbed functions  \psi^0_i 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. Such atoms have the principal quantum number n among their quantum numbers. The excited state of hydrogen-like atoms with principal quantum number n is n2-fold degenerate and

 n^2 = \sum_{\ell=0}^{n-1} (2 \ell + 1),

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

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

The one-electron states with even \ell are even under parity, while those with odd \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

 |JKM \rangle = (D^J_{MK})^* \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 DJMK 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 problems

 H^{(0)} \psi^0_k = E^{(0)}_k \psi^0_k, \quad k=0,1, \ldots, \quad E^{(0)}_0 < E^{(0)}_1 \le E^{(0)}_2, \dots

are assumed to be solved. It is usual to assume that the zeroth-order state to be perturbed is non-degenerate. If we take the ground state as the non-degenerate state under consideration (for hydrogen-like atoms: n = 1), perturbation theory gives

 E^{(2)} = \sum_{k>0} \frac{\langle \psi^0_0 | V_\mathrm{int} | \psi^0_k \rangle \langle \psi^0_k | V_\mathrm{int} | \psi^0_0 \rangle}{E^{(0)}_0 - E^{(0)}_k} =- \frac{1}{2} \sum_{i,j=1}^3 F_i \alpha_{ij} F_j

with the components of the polarizability tensor α defined by

 \alpha_{ij}\equiv -2\sum_{k>0} \frac{\langle \psi^0_0 | \mu_i | \psi^0_k \rangle \langle \psi^0_k | \mu_j | \psi^0_0\rangle}{E^{(0)}_0 - E^{(0)}_k}.

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

Because of their spherical symmetry the polarizability tensor of atoms is isotropic,

 \alpha_{ij} = \alpha_0 \delta_{ij} \Longrightarrow E^{(2)} = -\frac{1}{2} \alpha_0 F^2,

which is the quadratic Stark shift for atoms. For many molecules this expression is not too bad an approximation, because molecular tensors are often reasonably isotropic.


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).


Modern Quantum Mechanics (2nd Edition)

The Theory of Atomic Spectra


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s