Beyond the Quadratic Stark Effect (Perturbation Theory)

We next consider the stability of atoms in strong electric field and field ionization.

Mixture of the states in to the ground state

$$\begin{align} \sum_m |m^{(0)} \rangle \frac{\langle m^{(0)}|H'|n^{(0)} \rangle}{E_n^{(0)} - E_m^{(0)}} \end{align}$$

For perturbation theory, mixture coefficient is small, that is

$$\begin{align} \left| \frac{\langle m^{(0)}|H'|n^{(0)} \rangle}{E_n^{(0)} - E_m^{(0)}} \right|\ll 1 \end{align}$$

in our case

$$\begin{align} \left| \frac{\langle i |e r \mathcal{E}| g \rangle}{E_i - E_g} \right|\ll 1 \\ \Rightarrow\mathcal{E} \ll \left| \frac{E_i - E_g}{e\langle i | r | g \rangle} \right| \end{align}$$

where $|i\rangle$ is the nearest state of oppsite parity to $|g\rangle$ . $\langle i | r | g \rangle$ is on the order Bohr radius $a_0$ because there are no other length scales in this problem. For hydrogen, $E_n = -\frac{1}{2}\frac{e^2}{a_0}\frac{1}{n^2}$ , so $E_i - E_g \sim e^2/ a_0$ . Then we gte an order of critical electric field beyond which the perturbation will be invalid

$$\begin{align} \mathcal{E}_{\mathrm{crit}} \sim \frac{e}{a_0^2} \approx 5\times 10^9 \mathrm{V/cm}? \\ (\frac{1.602\times 10^{-19} \times 2.998\times 10^9 \mathrm{Fr}} { \left(5.292\times 10^{-11} \times 10^2 \mathrm{cm}\right)^2} \approx 1.7\times 10^7 \mathrm{V/cm}?) \end{align}$$

is $1,000$ times larger than laboratory electric fields.

For other atoms(not hydrogen)

$$\begin{align} \langle n, l + 1 | \vec{r} | n, l\rangle \sim n^2 a_{0} \\ \Delta E = E_{n, l + 1} - E_{n, l} = \frac{ - R_H}{(n - \delta_{l + 1})^2} - \frac{ - R_H}{(n - \delta_{l})^2} \\ (n\to\infty \mathrm{expand})\approx 2 R_H (\delta_{l + 1} - \delta_l) /n^3 \end{align}$$

(a minus sign?) where $\delta_l$ is the quantum defect. So $R_H = \frac{e^2}{a_0}$

$$\begin{align} \mathcal{E}_{\mathrm{crit}} \sim \frac{\Delta E}{e \langle \mathrm{r}\rangle} = \frac{e^2}{a_0} \frac{1}{e a_0} \frac{\delta_{l + 1} - \delta_l}{n^5} \end{align}$$

$\delta_l < 10^{-5}$ for $l > l_{\mathrm{core}} + 2 , n>7$ , $\mathcal{E}_{\mathrm{crit}} < 1 \mathrm{V/cm}$ . Where $l_{\mathrm{core}}$ is angular momentum of the core. $n^{5}$ , so, for large $n$ , $\mathcal{E}_{\mathrm{crit}}$ can be reached! An example, Rydberg atom. Highly excited states of atoms behave very differently from ground state atoms. Ref Phys. Rev. Lett 41, 103-107 (1978)

Field ionization


$$\begin{align} U_{\mathrm{total}} =& - \frac{e^2}{|z|} + e\mathrm{E} z \\ z_{\mathrm{max}} =& - \left| \frac{e}{\mathcal{E}} \right|^{1/2} \\ U_{\mathrm{max}} =& -2 e^{3/2} \mathcal{E}^{1/2} \end{align}$$

if we assume $U_{\mathrm{max}} = -E_{\mathrm{Binding}} = - \mathrm{Ryd}/n^2$ , then (in atom units $e/a_0$)

$$\begin{align} \mathcal{E}_{\mathrm{ION}} = \frac{1}{16 n^4} \end{align}$$

Application: $100%$ detection efficiency for atoms, Phys. Rev. Lett. 76, 1800 (1996)

Atoms in oscillating electric fields (AC stack effect)

Use time dependent perturbation theory. An oscillating electric field

$$\begin{align} \vec{\mathcal{E}}(t) =& \mathcal{E} \hat{e} \cos(\omega t) \\ H' =& - \vec{D} \cdot \vec{\mathcal{E}}(t) = - \frac{1}{2} \left( e^{\mathrm{i}\omega t} + e^{- \mathrm{i}\omega t} \right) \mathcal{E} \hat{e} \cdot \vec{D} \end{align}$$

the we use time dependent perturbation theory (Sakurai (5.7.17))

$$\begin{align} |\psi(t) \rangle = \sum_k a_k(t) e^{- \mathrm{i} \omega_k t} |k\rangle \end{align}$$ $$\begin{align} a_k^{(1)} = \frac{\mathcal{E}}{2 \hbar} \langle k | \hat{e}\cdot \vec{D} | g\rangle \left[\frac{e^{\mathrm{i} (\omega_{kg} +\omega) t} - 1}{\omega_{kg} + \omega} + (\omega \leftrightarrow -\omega) \right] \end{align}$$

the transmit contribution $1$ , due to the sudden switch on, will damps out.

We set $\hat{e} = \hat{z}$ , $\vec{D} = - e z \hat{z}$ . Consider the ground state $|g\rangle$ , then

$$\begin{align} |g^(0) \rangle = \sum_k \delta_{kg} e^{- \mathrm{i} \omega_k t}| k\rangle = e^{- \mathrm{i}\omega_g t} | g\rangle \end{align}$$

the indeuced dipole(drop the sudden switch on term)

$$\begin{align} &\left( \langle g^{(0)}| + \langle g^{(1)}| \right) D \left( |g^{(0)}\rangle + | g^{(1)}\rangle \right) \\ =& 2 \mathrm{Re} \langle g^{(0)} | D | g^{(1)} \rangle \\ =& 2 \mathrm{Re} \sum_k \langle g | D | k \rangle a_k^{(1)}(t) e^{-\mathrm{i}\omega_kt} e^{\mathrm{i}\omega_gt}\\ =& \frac{e^2}{\hbar} \mathrm{Re}\sum_k |\langle k | z| g\rangle|^2 \left[ \frac{e^{\mathrm{i} \omega t} }{\omega_{kg} + \omega} + (\omega \leftrightarrow -\omega) \right] \mathcal{E} \\ =& \frac{e^2}{\hbar} \mathrm{Re}\sum_k |\langle k | z| g\rangle|^2 \cdot 2 \frac{\omega_{kg}}{\omega_{kg}^2 - \omega^2} \cdot \cos (\omega t) \mathcal{E} \\ =& \alpha(\omega) \cdot \cos (\omega t) \mathcal{E} \end{align}$$


$$\begin{align} \alpha (\omega) \equiv \frac{e^2}{\hbar} \mathrm{Re}\sum_k |\langle k | z| g\rangle|^2 \cdot 2 \frac{\omega_{kg}}{\omega_{kg}^2 - \omega^2} \end{align}$$

it is easy to show that $\alpha (0)$ is the DC result.

About AC Stark shift, we have 3 poinst to discuss:

  • relation to dressed atom
  • oscillator strength
  • index of refraction

Relation to dressed atom

the energy shift of DC Stark effect is

$$\begin{align} \Delta E = - \frac{1}{2} \alpha(\omega) \bar{\mathcal{E}^2(t)} \end{align}$$

where $\bar{\mathcal{E}^2(t)}$ means time average, then contribute a factor 2

$$\begin{align} \Delta E =& - \frac{1}{4} \alpha(\omega) \mathcal{E}^2 \\ =& - \frac{1}{4} \frac{2e^2}{\hbar} |\langle 1 | z| 2\rangle|^2 \frac{\omega_{12}}{\omega_{12}^2 - \omega^2} \mathcal{E}^2 \\ =& - \frac{\hbar}{4} \omega_R^2 \left[ \frac{1}{\delta} + \frac{1}{2\omega - \delta} \right] \end{align}$$

where Rabi frequency $\hbar^2 \omega_R^2 = e^2|\langle 1 | z| 2\rangle|^2 \mathcal{E}^2$ , detuning $\delta = \omega_{12} - \omega$ . The second term is called Bloch–Siegert shift . This looks very similar to time-independent perturbation theory, those are two term with detuning $\delta, -(2\omega -\delta)$ absorb a photon and emit a photon correspondingly. This can be derived using the quantized EM field (photons) picture.

  • photon number states $\rightarrow$ time independent perturbation theory
  • coherent state $\mathcal{E} \cos(\omega t)$ $\rightarrow$ time dependent perturbation theory

Oscillator strength

Compare our result (AC polarizability of atom) to a classical HO(harmonic oscillator), with charge $q_k$ , mass $m_k$ , frequency $\omega_k$ .

Drivien by electric field $\mathcal{E} \cos(\omega t)$ :

For a classical HO:

\begin{align} D_z(\omega, t) = \mathcal{E} q_k z_q \end{align} where $z_{q}$ can be solved by classical mechanics of a driven oscillator, so \begin{align} D_z(\omega, t) = \mathcal{E} \frac{q_k^2}{m(\omega_k^2 - \omega^2)} \cos (\omega t) \end{align}

For a atom

\begin{align} D_z(\omega, t) = \sum_k f_{kg} \frac{e^2}{m(\omega_k^2 - \omega^2)} \mathcal{E}\cos(\omega t) \end{align} where oscillation strength defined as \begin{align} f_{kg} \equiv \frac{2m}{\hbar} \omega_{kg} |\langle k| z | j\rangle|^2 \end{align}

Atom response as a set of classical oscillators with an effective chrege

$$\begin{align} q_k^2 = f_{kg} e^2 \end{align}$$

For oscillation strength

  • classical correspondance
  • Thomas–Reiche–Kuhn sum rule: $\sum_k f_{kg} = 1$
  • it's a dimensionless parameter

For hydrogen, $1s \to 2p$ , $f = 0.4126$

We can also write the matrix element in terms of oscillation strength

$$\begin{align} |\langle k| z | j\rangle|^2 = f_{kg} \frac{1}{2} \cdot\frac{\hbar}{2m}\cdot \frac{1}{\omega_{kg}} \end{align}$$

where $\frac{\hbar}{2m}$ is the compton wave length of electron, $\frac{1}{\omega_{kg}}$ is the photon wave length.

Index of refraction


ref Jackson (10.146)