从冰上的水 - Atomic and Optical Physics I, 10 Atoms in External Filds II, Quadratic DC Stark effect

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Summary HFS

\begin{array}{r} H = a h \vec{I} \cdot \vec{J} + (g_{J} μ_{B} m_{J} - g_{I} μ_{N} m_{I}) B_{0} \end{array}

weak field: HFS + $μ_{B} g_{F} B_{0} m_{F}$

\begin{array}{r} \frac{a h}{2} [(F (F + 1) - J (J + 1) - I (I + 1)] \end{array}

strong field

\begin{array}{r} a h m_{I} m_{J} + (g_{J} μ_{B} m_{J} - g_{I} μ_{N} m_{I}) B_{0} \end{array}

even stronger field

\begin{array}{r} A_{F S} m_{l} m_{s} + a_{S} m_{I} m_{S} + a_{l} m_{I} m_{l} + g_{S} μ_{B} m_{S} + g_{l} μ_{B} m_{l} - g_{I} μ_{N} m_{I} \end{array}

Nice example for Hamiltonian with different scaler products: $\vec{B} \cdot \vec{S}, \vec{B} \cdot \vec{L} \vec{S} \cdot \vec{L} \vec{I} \cdot \vec{J}$
vector model (rapid procession of projection) (calculation without explicit use of CG coefficents)

Atom in external electric fields: standard theory of the DC Stark effect of the atom polarizbility

Uniform electric field $E \hat{z}$ . ref Jackson Chap 4.6, Eq(4.24)

\begin{array}{r} U (\vec{r}) = q ϕ (\vec{r}) - \vec{d} \cdot \hat{z} E - α E^{2} \end{array}

this three terms corresponds monopole, permanent dipole moment, polarizbility $α$ induced dipole momentum $d_{I N D} = α E$ .

Use perturbation operator $H^{'} = - \vec{d} \hat{z} E = e z E$ ( $\vec{d} = - e \vec{r}, d = - e z$ ) . $H^{'}$ is odd parity, we have no permanent dipole until we have degenerate energy levels.

1st order perturbation energy (no degeneration)

$E_{n}^{(1)} = ⟨ n^{(0)} | H^{'} | n^{(0)} ⟩ = 0$ .

2nd order perturbation energy

\begin{array}{r} E_{n}^{(2)} = \sum_{m}^{^{'}} \frac{⟨ n^{(0)} | e z E | m^{(0)} ⟩ ⟨ m^{(0)} | e z E | n^{(0)} ⟩}{E_{n}^{(0)} - E_{m}^{(0)}} = e^{2} E^{2} \sum_{m}^{^{'}} \frac{| ⟨ n^{(0)} | z | m^{(0)} ⟩ |^{2}}{E_{n}^{(0)} - E_{m}^{(0)}} \end{array}

where $\sum_{m}^{^{'}}$ means sum over all $m \neq n$ .

dipole in the 1st order perturbed state

\begin{aligned} ⟨ \vec{d} ⟩ = & (⟨ n^{(0)} | + ⟨ n^{(1)} |) \vec{d} (| n^{(0)} ⟩ + | n^{(1)} ⟩) \\ = & ⟨ n^{(1)} | \vec{d} | n^{(0)} ⟩ + ⟨ n^{(0)} | \vec{d} | n^{(1)} ⟩ \\ = & 2 Re [⟨ n^{(1)} | \vec{d} | n^{(0)} ⟩] \\ = & 2 Re (\sum_{m}^{^{'}} \frac{⟨ n^{(0)} | \vec{d} | m^{(0)} ⟩ ⟨ m^{(0)} | e z E | n^{(0)} ⟩}{E_{n}^{(0)} - E_{m}^{(0)}}) \\ = & - 2 \hat{z} e^{2} E \sum_{m}^{^{'}} \frac{| ⟨ n^{(0)} | z | m^{(0)} ⟩ |^{2}}{E_{n}^{(0)} - E_{m}^{(0)}} \\ = & α \vec{E} \end{aligned}

where the second equality use the parity of $\vec{d}$ . The polarizbility

\begin{array}{r} α \equiv \frac{d}{E} = - 2 e^{2} \sum_{m}^{^{'}} \frac{| ⟨ n^{(0)} | z | m^{(0)} ⟩ |^{2}}{E_{n}^{(0)} - E_{m}^{(0)}} \end{array}

so, we can rewrite the 2nd order perturbed energy as

\begin{array}{r} E_{n}^{(2)} = - \frac{1}{2} α E^{2} = - \frac{1}{2} ⟨ \vec{d} ⟩ \cdot E \end{array}

Another veiw, 2nd perturbed total energy is (there is a problem about the nomalization of perturbed wave function, a little funny)

\begin{aligned} E_{n}^{(0)} + Δ E = & E_{n}^{(0)} + E_{n}^{(1)} + E_{n}^{(2)} = (⟨ n^{(0)} | + ⟨ n^{(1) |}) (H_{0} + H^{'}) (| n^{(0)} ⟩ + | n^{(1)} ⟩) \\ = & ⟨ n^{(0)} | H_{0} | n^{(0)} ⟩ + ⟨ n^{(1)} | H_{0} | n^{(1)} ⟩ + 2 Re [⟨ n^{(0)} | H^{'} | n^{(1)} ⟩] \end{aligned}

the second term

\begin{aligned} ⟨ n^{(1)} | H_{0} | n^{(1)} ⟩ = & \sum_{l m}^{^{'}} (\frac{⟨ n^{(0)} | H^{'} | l^{(0)} ⟩ ⟨ l^{(0)} |}{E_{n}^{(0)} - E_{l}^{(0)}}) (H_{0}) (\frac{| m^{(0)} ⟩ ⟨ m^{(0)} | H^{'} | n^{(0)} ⟩}{E_{n}^{(0)} - E_{m}^{(0)}}) \\ = & \sum_{m}^{^{'}} \frac{| ⟨ n^{(0)} | H^{'} | m^{(0)} ⟩ |^{2}}{(E_{n}^{(0)} - E_{m}^{(0)})^{2}} E_{m}^{(0)} \end{aligned}

if we set $E_{n}^{(0)} = 0$ , then

\begin{array}{r} ⟨ n^{(1)} | H_{0} | n^{(1)} ⟩ = \frac{1}{2} α E^{2} \end{array}

then

\begin{array}{r} E_{n}^{(0)} + Δ E = E_{n}^{(0)} ⟨ n^{(0)} | n^{(0)} ⟩ + \frac{1}{2} α E - α E^{2} \end{array}

and $Δ E = - \frac{1}{2} α E$ , the same as before.

Unit of $α$

dimension of $[α] = [\frac{q^{2} l^{2}}{E}] = [\frac{q^{2}}{l} \frac{1}{E} l^{3}] = L^{3}$ , is volum.

For hydrogen, only consider the matrix element between 1s and 2p, then we get

\begin{array}{r} α \approx 2.96 a_{0}^{3} \end{array}

where $α = 4.5 a_{0}^{3}$ if we use all matrix elements (include the positive energy continuum states of hydrogen).

Unsold's approximation

something like Sakurai page 315.

Compare with classical EM for conducting sphere

dipole moment of a conducting sphere in a uniform electric filed is (Jackson 4.56) $E R^{3}$

\begin{aligned} (Jackson (4.56)) & p = & 4 π ϵ_{0} (\frac{ϵ / ϵ_{0} - 1}{ϵ / ϵ_{0} + 2}) R^{3} E \\ p_{conducting sphere} = & lim_{ϵ \to \infty} p = 4 π ϵ_{0} R^{3} E_{0} (SI) \\ p_{conducting sphere} \sqrt{4 π ϵ_{0}} = & 4 π ϵ_{0} R^{3} E_{0} \frac{1}{\sqrt{4 π ϵ_{0}}} (Gauss) \\ p_{conducting sphere} = & R^{3} E_{0} (Gauss) \end{aligned}

so, atoms $\Leftrightarrow$ conducting sphere.

When it comes to dipole moments and to polarizbility, atoms pretty much behave like metallic conducting sphere of the same volum.

Reference

Jackson, J. D. Classical electrodynamics. (Wiley, 1999)
Jun John Sakurai, Jim Napolitano, Modern Quantum Mechanics. (Cambridge University Press, 2017)

Atomic and Optical Physics I, 10 Atoms in External Filds II, Quadratic DC Stark effect