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Summary HFS

$$\begin{align} H = a h \vec{I}\cdot \vec{J} + (g_J \mu_B m_J - g_I \mu_N m_I) B_0 \end{align}$$

weak field: HFS + $\mu_B g_F B_0 m_F$

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

strong field

$$\begin{align} ah m_I m_J + (g_J \mu_B m_J - g_I \mu_N m_I) B_0 \end{align}$$

even stronger field

$$\begin{align} A_{FS} m_l m_s + a_S m_I m_S + a_l m_I m_l + g_S \mu_B m_S + g_l \mu_B m_l - g_I \mu_N m_I \end{align}$$
  • Nice example for Hamiltonian with different scaler products: $\vec{B}\cdot \vec{S} , \quad \vec{B}\cdot \vec{L} \quad \vec{S}\cdot \vec{L} \quad \vec{I} \cdot \vec{J}$
  • 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 $\mathcal{E} \hat{z}$ . ref Jackson Chap 4.6, Eq(4.24)

$$\begin{align} U(\vec{r}) = q \phi(\vec{r}) - \vec{d}\cdot \hat{z} \mathcal{E} - \alpha \mathcal{E}^2 \end{align}$$

this three terms corresponds monopole, permanent dipole moment, polarizbility $\alpha$ induced dipole momentum $d_{IND} = \alpha \mathcal{E}$ .

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

1st order perturbation energy (no degeneration)

$E_n^{(1)} = \langle n^{(0)} | H' | n^{(0)}\rangle = 0$ .

2nd order perturbation energy

$$\begin{align} E_n^{(2)} = \sum_m ^{'} \frac{\langle n^{(0)} | ez\mathcal{E}| m^{(0)}\rangle \langle m^{(0)} | ez\mathcal{E} | n^{(0)}\rangle} {E_n^{(0)} - E_m^{(0)}} = e^2 \mathcal{E}^2 \sum_m^{'} \frac{|\langle n^{(0)} | z| m^{(0)}\rangle|^2}{E_n^{(0)} - E_m^{(0)}} \end{align}$$

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

dipole in the 1st order perturbed state

$$\begin{align} \langle \vec{d} \rangle =& (\langle n^{(0)}| + \langle n^{(1)}|) \vec{d} (| n^{(0)}\rangle + | n^{(1)}\rangle) \\ = & \langle n^{(1)} | \vec{d} | n^{(0)} \rangle + \langle n^{(0)} | \vec{d} | n^{(1)} \rangle \\ = & 2 \mathrm{Re} \left[\langle n^{(1)} | \vec{d} | n^{(0)} \rangle \right] \\ = & 2 \mathrm{Re} \left( \sum_m^{'} \frac{\langle n^{(0)} | \vec{d} | m^{(0)} \rangle \langle m^{(0)} | ez\mathcal{E} | n^{(0)} \rangle}{E_n^{(0)} - E_m^{(0)}} \right) \\ = & - 2 \hat{z} e^2 \mathcal{E} \sum_m^{'} \frac{|\langle n^{(0)} |z| m^{(0)}\rangle |^{2}}{E_n^{(0)} - E_m^{(0)}} \\ =& \alpha \vec{\mathcal{E}} \end{align}$$

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

$$\begin{align} \alpha \equiv \frac{d}{\mathcal{E}} = - 2 e^2 \sum_m^{'} \frac{|\langle n^{(0)} |z| m^{(0)}\rangle |^{2}}{E_n^{(0)} - E_m^{(0)}} \end{align}$$

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

$$\begin{align} E_n^{(2)} = - \frac{1}{2}\alpha \mathcal{E}^2 = - \frac{1}{2}\langle \vec{d} \rangle \cdot \mathcal{E} \end{align}$$

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

$$\begin{align} E_n^{(0)} + \Delta E =& E_n^{(0)} + E_n^{(1)} + E_n^{(2)} = \left( \langle n^{(0)} | + \langle n^{(1) |} \right) \left( H_0 + H' \right) \left( | n^{(0)} \rangle + |n^{(1)} \rangle \right) \\ =& \langle n^{(0)} | H_0 | n^{(0)}\rangle + \langle n^{(1)} | H_0 | n^{(1)}\rangle + 2 \mathrm{Re} \left[ \langle n^{(0)} | H' | n^{(1)}\rangle \right] \end{align}$$

the second term

$$\begin{align} \langle n^{(1)} | H_0 | n^{(1)}\rangle =& \sum_{lm}^{'} \left( \frac{\langle n^{(0)} | H' | l^{(0)} \rangle\langle l^{(0)} | } {E_n^{(0)} - E_l^{(0)}} \right) \left( H_0 \right) \left( \frac{| m^{(0)} \rangle\langle m^{(0)} | H' | n^{(0)} \rangle } {E_n^{(0)} - E_m^{(0)}} \right) \\ =& \sum_m^{'} \frac{|\langle n^{(0)} | H' | m^{(0)} \rangle|^2}{(E_n^{(0)} - E_m^{(0)})^2} E_m^{(0)} \end{align}$$

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

$$\begin{align} \langle n^{(1)} | H_0 | n^{(1)}\rangle = \frac{1}{2}\alpha \mathcal{E}^2 \end{align}$$

then

$$\begin{align} E_n^{(0)} + \Delta E = E_n^{(0)} \langle n^{(0)} | n^{(0)} \rangle + \frac{1}{2}\alpha \mathcal{E} - \alpha \mathcal{E}^2 \end{align}$$

and $\Delta E = - \frac{1}{2}\alpha \mathcal{E}$ , the same as before.

Unit of $\alpha$

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

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

$$\begin{align} \alpha \approx 2.96 a_0^3 \end{align}$$

where $\alpha = 4.5 a_0^3$ 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) $\mathcal{E} R^3$

$$\begin{align} \tag{Jackson (4.56)} p =& 4\pi \epsilon_0 \left( \frac{\epsilon/\epsilon_0 - 1}{\epsilon / \epsilon_0 + 2} \right)R^3\mathcal{E} \\ p_{\mathrm{conducting sphere}} =& \lim_{\epsilon\to \infty} p = 4\pi \epsilon_0 R^3 E_0 \quad(\mathrm{SI}) \\ p_{\mathrm{conducting sphere}} \sqrt{4\pi\epsilon_0} =&4\pi \epsilon_0 R^3 E_0 \frac{1}{\sqrt{4\pi\epsilon_0}} \quad(\mathrm{Gauss}) \\ p_{\mathrm{conducting sphere}} =& R^3 E_0 \quad(\mathrm{Gauss}) \end{align}$$

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)