Stark effect

Mechanism

Overview

Imagine an atom with occupied 2s and 2p electron states. In the Bohr model, these states are degenerate. However, in the presence of an external electric field, these electron orbitals will hybridize into eigenstates of the perturbed Hamiltonian (where each perturbed hybrid state can be written as a superposition of unperturbed states). Since the 2s and 2p states have opposite parity, these hybrid states will lack inversion symmetry and will possess a time-averaged electric dipole moment. If this dipole moment is aligned with the electric field, the energy of the state will shift down; if this dipole moment is anti-aligned with the electric field, the energy of the state will shift up. Thus, the Stark effect causes a splitting of the original degeneracy.

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, resulting in a larger electric dipole moment upon hybridization.

Multipole expansion

The Stark effect originates from the interaction between a charge distribution (atom or molecule) and an external electric field. The interaction energy ${\displaystyle V_{\mathrm{i}\mathrm{n}\mathrm{t}}}$ of a continuous charge distribution ${\displaystyle \rho (\mathbf{r})}$, confined within a finite volume ${\displaystyle \mathcal{V}}$, with an external electrostatic potential ${\displaystyle \varphi (\mathbf{r})}$ is $${\displaystyle V_{\mathrm{i}\mathrm{n}\mathrm{t}}=\underset{\mathcal{V}}{\int }\varphi (\mathbf{r})\rho (\mathbf{r}){\mathrm{d}}^3\mathbf{r}}$$ This expression is valid classically and quantum-mechanically alike. If the potential varies weakly over the charge distribution, the multipole expansion converges fast, so only a few first terms give an accurate approximation. Namely, keeping only the zeroth- and first-order terms, $${\displaystyle \varphi (\mathbf{r})\approx \varphi (\mathbf{0})-\sum _{i=1}^3{F}_i{r}_i}$$ where we introduced the electric field $F_i\equiv -{\left(\frac{\mathrm{\partial }\varphi }{\mathrm{\partial }{r}_i}\right)|}_{\mathbf{r}\mathbf{=}\mathbf{0}}$ and assumed the origin ${\displaystyle \mathbf{0}}$ to be somewhere within ${\displaystyle \mathcal{V}}$. Therefore, the interaction becomes $${\displaystyle V_{\mathrm{i}\mathrm{n}\mathrm{t}}\approx \varphi (\mathbf{0})\underset{\mathcal{V}}{\int }\rho (\mathbf{r}){\mathrm{d}}^3\mathbf{r}-\sum _{i=1}^3{F}_i\underset{\mathcal{V}}{\int }\rho (\mathbf{r})r_i{\mathrm{d}}^3\mathbf{r}\equiv q\varphi (\mathbf{0})-\sum _{i=1}^3{F}_i{\mu }_i=q\varphi (\mathbf{0})-\mathbf{F}\cdot \mathit{\mu },}$$ where ${\displaystyle q}$ and ${\displaystyle \mathit{\mu }}$ are, respectively, the total charge (zero moment) and the dipole moment of the charge distribution.

Classical macroscopic objects are usually neutral or quasi-neutral (${\displaystyle q=0}$), so the first, monopole, term in the expression above is identically zero. This is also the case for a neutral atom or molecule. However, for an ion this is no longer true. Nevertheless, it is often justified to omit it in this case, too. Indeed, the Stark effect is observed in spectral lines, which are emitted when an electron "jumps" between two bound states. Since such a transition only alters the internal degrees of freedom of the radiator but not its charge, the effects of the monopole interaction on the initial and final states exactly cancel each other.

Perturbation theory

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 $${\displaystyle V_{\mathrm{i}\mathrm{n}\mathrm{t}}=-\mathbf{F}\cdot \mathit{\mu }.}$$ 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 ${\displaystyle {\psi }_1^0,\dots ,{\psi }_g^0}$. (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 $${\displaystyle ({\mathbf{V}}_{\mathrm{i}\mathrm{n}\mathrm{t}}{)}_{kl}=⟨{\psi }_k^0|V_{\mathrm{i}\mathrm{n}\mathrm{t}}|{\psi }_l^0⟩=-\mathbf{F}\cdot ⟨{\psi }_k^0|\mathit{\mu }|{\psi }_l^0⟩,k,l=1,\dots ,g.}$$ 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 ${\displaystyle \mathit{\mu }}$, $${\displaystyle E^{(1)}=-\mathbf{F}\cdot ⟨{\psi }_1^0|\mathit{\mu }|{\psi }_1^0⟩=-\mathbf{F}\cdot ⟨\mathit{\mu }⟩.}$$Since the electric dipole moment is a vector (tensor of the first rank), the diagonal elements of the perturbation matrix V_int vanish between states that have a definite 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, $${\displaystyle 16=1+3+5+7⟹n=4\text{contains}s\oplus p\oplus d\oplus f.}$$ 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 $${\displaystyle |JKM⟩=(D_{MK}^J{)}^{\ast }\text{with}M,K=-J,-J+1,\dots ,J}$$ 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 $${\displaystyle H^{(0)}{\psi }_k^0=E_k^{(0)}{\psi }_k^0,k=0,1,\dots ,E_0^{(0)}<E_1^{(0)}\le E_2^{(0)},\dots }$$ is assumed to be solved. The perturbation theory gives $${\displaystyle E_k^{(2)}=\sum _{k^{\prime }\ne k}\frac{⟨{\psi }_k^0|V_{\mathrm{i}\mathrm{n}\mathrm{t}}|{\psi }_{k^{\mathrm{\prime }}}^0⟩⟨{\psi }_{k^{\prime }}^0|V_{\mathrm{i}\mathrm{n}\mathrm{t}}|{\psi }_k^0⟩}{E_k^{(0)}-E_{k^{\prime }}^{(0)}}\equiv -\frac{1}{2}\sum _{i,j=1}^3{\alpha }_{ij}{F}_i{F}_j}$$ with the components of the polarizability tensor α defined by $${\displaystyle {\alpha }_{ij}=-2\sum _{k^{\prime }\ne k}\frac{⟨{\psi }_k^0|{\mu }_i|{\psi }_{k^{\prime }}^0⟩⟨{\psi }_{k^{\prime }}^0|{\mu }_j|{\psi }_k^0⟩}{E_k^{(0)}-E_{k^{\prime }}^{(0)}}.}$$ 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, $${\displaystyle {\alpha }_{ij}\equiv {\alpha }_0{\delta }_{ij}⟹E^{(2)}=-\frac{1}{2}{\alpha }_0{F}^2.}$$ 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).

Applications

The Stark effect is at the basis of the spectral shift measured for voltage-sensitive dyes used for imaging of the firing activity of neurons.

The Stark effect is used to slow down and manipulate molecular beams.