Searching for a shell

Polarizability from quantum TD-LRT

DFT,  Math 

Previously we have shown how :link:quantum TD-LRT works, now let’s put that into work by calculating the polarizability of a quantum system. Here we start by considering the dipole-electric field interaction energy

V^(t)=−d^⋅E(t)=er^⋅E(t)(1)\hat{V}(t) = -\mathbf{\hat d} \cdot \mathbf{E}(t) = e\mathbf{\hat r} \cdot \mathbf{E}(t) \tag{1}

where the external electric field is still

E(t)=E0e−iωt+E0∗e+iωt.\mathbf{E}(t) = \mathbf{E}_0 e^{-i\omega t} + \mathbf{E}_0^* e^{+i\omega t}.

Substituting Eq. 1 into Eq. 3 of :link:quantum TD-LRT, we get:

c˙n(t)=−1iℏ∑mcm(t)eiωnmtλ⟨n∣d^∣m⟩⋅E(t)\dot{c}_n(t) = \frac{-1}{i\hbar}\sum_m c_m(t) e^{i\omega_{nm}t} \lambda \braket{n|\mathbf{\hat d}|m} \cdot \mathbf{E}(t)

Again, if we initialize system to $\ket{0}$ at $t_0=-\infty$ (see Eq. 6 in :link:quantum TD-LRT ). From now assuming $n\neq 0$, to first order, we have

c˙n(t)=−1iℏeiωn0t⟨n∣d^∣0⟩⋅E(t)\dot{c}_n(t) = \frac{-1}{i\hbar} e^{i\omega_{n0}t} \braket{n|\mathbf{\hat d}|0} \cdot \mathbf{E}(t)

Explicitly

c˙n(t)=−1iℏeiωn0t⟨n∣d^∣0⟩⋅E(t)=−1iℏeiωn0t⟨n∣d^∣0⟩⋅(E0e−iωt+E0∗e+iωt)=iℏeiωn0t⟨n∣d^∣0⟩⋅E0e−iωt+iℏeiωn0t⟨n∣d^∣0⟩⋅E0∗e+iωt=F−(t)+F+(t)(2)\begin{aligned} \dot{c}_n(t) &= \frac{-1}{i\hbar} e^{i\omega_{n0}t} \braket{n|\mathbf{\hat d}|0} \cdot \mathbf{E}(t)\\ &=\frac{-1}{i\hbar} e^{i\omega_{n0}t} \braket{n|\mathbf{\hat d}|0} \cdot \left( \mathbf{E}_0e^{-i\omega t} + \mathbf{E}_0^*e^{+i\omega t}\right)\\ &=\frac{i}{\hbar} e^{i\omega_{n0}t} \braket{n|\mathbf{\hat d}|0} \cdot \mathbf{E}_0e^{-i\omega t} + \frac{i}{\hbar} e^{i\omega_{n0}t} \braket{n|\mathbf{\hat d}|0} \cdot \mathbf{E}_0^*e^{+i\omega t}\\ &=F_{-}(t) + F_{+}(t) \tag{2} \end{aligned}

Note: Time ordering

$e^{\mu t'}$ is a damping factor added to the E-field so that we have perturbation adiabatically turned on in the remote past, and the integration converges as $t_0 \to -\infty$. For more, see [this post](https://chengcheng-xiao.github.io/post/2021/08/16/Greens_function_2.html).

Now we want to get $c_n(t)$, for that we need to integrate c˙n(t)\dot{c}_n(t). Here we assume the field was adiabatically turned on in the remote past $t_0=-\infty$ by multiplying the E-field by $e^{\mu t’}$ with $\mu \to 0^+$(or equivalently by adding a small positive imaginary part to the frequency). Then integrate from $t’=-\infty$ to $t’=t$, for $F_{-}(t)$ (similar to Eq. 6 in :link:quantum TD-LRT) we get

∫−∞tF−(t′)dt′=iℏ⟨n∣d^∣0⟩⋅E0∫−∞te[i(ωn0−ω)+μ]t′dt′=iℏ⟨n∣d^∣0⟩⋅E0ei[(ωn0−ω)+μ]ti(ωn0−ω)+μ\begin{aligned} \int_{-\infty}^t F_{-}(t') dt' &= \frac{i}{\hbar} \braket{n|\mathbf{\hat d}|0} \cdot \mathbf{E}_0 \int_{-\infty}^t e^{[i(\omega_{n0}-\omega)+\mu] t'} dt'\\ &=\frac{i}{\hbar} \braket{n|\mathbf{\hat d}|0} \cdot \mathbf{E}_0 \frac{e^{i[(\omega_{n0}-\omega)+\mu] t}}{i(\omega_{n0}-\omega)+\mu} \end{aligned}

Similarily, for $F_+(t)$

∫−∞tF+(t′)dt′=iℏ⟨n∣d^∣0⟩⋅E0∗e[i(ωn0−ω)+μ]ti(ωn0+ω)+μ\begin{aligned} \int_{-\infty}^t F_{+}(t') dt' = \frac{i}{\hbar} \braket{n|\mathbf{\hat d}|0} \cdot \mathbf{E}^*_0 \frac{e^{[i(\omega_{n0}-\omega)+\mu] t}}{i(\omega_{n0}+\omega)+\mu} \end{aligned}

Combining

cn(t)=iℏ⟨n∣d^∣0⟩(E0e[i(ωn0−ω)+μ]ti(ωn0−ω)+μ+E0∗e[i(ωn0+ω)+μ]ti(ωn0+ω)+μ)=iℏ⟨n∣d^∣0⟩(E0ei(ωn0−ω)ti(ωn0−ω)+0++E0∗ei(ωn0+ω)ti(ωn0+ω)+0+)(3)\begin{aligned} {c}_n(t) &= \frac{i}{\hbar} \braket{n|\mathbf{\hat d}|0} \left( \mathbf{E}_0 \frac{e^{[i(\omega_{n0}-\omega)+\mu] t}}{i(\omega_{n0}-\omega)+\mu}+ \mathbf{E}^*_0 \frac{e^{[i(\omega_{n0}+\omega)+\mu] t}}{i(\omega_{n0}+\omega)+\mu} \right)\\ &=\frac{i}{\hbar} \braket{n|\mathbf{\hat d}|0} \left( \mathbf{E}_0 \frac{e^{i(\omega_{n0}-\omega) t}}{i(\omega_{n0}-\omega)+0^+}+ \mathbf{E}^*_0 \frac{e^{i(\omega_{n0}+\omega) t}}{i(\omega_{n0}+\omega)+0^+} \right)\\ \end{aligned} \tag{3}

where we have gotten rid of $e^{it0^+}=0$. Eq. 3 shows how the wavefunction evolves (to the first order) under a small perturbation of an external E-field - dipole interaction.

We can now calculate the expectation of the dipole operator. First remember that we have the system prepared in $\ket{0}$ ($c_0(t) \approx 1$) so that

∣ψ(t)⟩=c0(t)e−iω0t∣0⟩+∑m≠0cm(t)e−iωmt∣m⟩≈e−iω0t∣0⟩+∑m≠0cm(t)e−iωmt∣m⟩\begin{aligned} \ket{\psi(t)} &= c_0(t) e^{-i\omega_0t} \ket{0} + \sum_{m\neq 0} c_m(t) e^{-i\omega_mt} \ket{m} \\ &\approx e^{-i\omega_0t} \ket{0} + \sum_{m\neq 0} c_m(t) e^{-i\omega_mt} \ket{m} \end{aligned}

The expectation of the dipole operator becomes

⟨d^⟩(t)=⟨ψ∣d^∣Ψ⟩=∑m,m′cm∗(t)cm′(t)ei(ωm−ωm′)t⟨m∣d^∣m′⟩=c0∗(t)c0(t)⟨0∣d^∣0⟩+∑m≠0(cm∗(t)c0(t)ei(ωm−ω0)t⟨m∣d^∣0⟩+c0∗(t)cm(t)ei(ω0−ωm)t⟨0∣d^∣m⟩)+∑m≠0,m′≠0cm∗(t)cm′(t)ei(ωm−ωm′)t⟨m∣d^∣m′⟩≈⟨0∣d^∣0⟩+∑m≠0(cm∗(t)ei(ωm−ω0)t⟨m∣d^∣0⟩+cm(t)ei(ω0−ωm)t⟨0∣d^∣m⟩)+∑m≠0,m′≠0cm∗(t)cm′(t)ei(ωm−ωm′)t⟨m∣d^∣m′⟩\begin{aligned} \braket{\mathbf{\hat d}}(t) &= \braket{\psi|\mathbf{\hat d}|\Psi}\\ &=\sum_{m,m'} c^*_m(t) c_{m'}(t) e^{i(\omega_m-\omega_{m'})t} \braket{m|\mathbf{\hat d}|m' }\\ &= c^*_0(t) c_{0}(t) \braket{0|\mathbf{\hat d}|0 }\\ &+ \sum_{m\neq0} \left( c^*_m(t) c_0(t) e^{i(\omega_m-\omega_{0})t} \braket{m|\mathbf{\hat d}|0}+ c^*_0(t) c_m(t) e^{i(\omega_0-\omega_{m})t} \braket{0|\mathbf{\hat d}|m} \right)\\ &+ \sum_{m\neq0,m'\neq0} c^*_m(t) c_{m'}(t) e^{i(\omega_m-\omega_{m'})t} \braket{m|\mathbf{\hat d}|m' }\\ &\approx \braket{0|\mathbf{\hat d}|0 }\\ &+ \sum_{m\neq0} \left( c^*_m(t) e^{i(\omega_m-\omega_{0})t} \braket{m|\mathbf{\hat d}|0}+ c_m(t) e^{i(\omega_0-\omega_{m})t} \braket{0|\mathbf{\hat d}|m} \right)\\ &+ \sum_{m\neq0,m'\neq0} c^*_m(t) c_{m'}(t) e^{i(\omega_m-\omega_{m'})t} \braket{m|\mathbf{\hat d}|m' }\\ \end{aligned}

We see there are zero-th, first, and second order terms involving $c$ coeffs. The zero-th order term - c0∗(t)c0(t)⟨0∣d^∣0⟩c^*_0(t) c_{0}(t) \braket{0|\mathbf{\hat d}|0 } is a constant and does not affect the polarizability so we can safely ignore it, and if we ignore the second order term (because it is small), only focusing on the first order term, the dipole expectation value becomes

⟨d^⟩(t)≈∑m≠0(cm∗(t)ei(ωm−ω0)t⟨m∣d^∣0⟩+cm(t)ei(ω0−ωm)t⟨0∣d^∣m⟩)=∑m≠0(cm∗(t)eiωm0t⟨m∣d^∣0⟩+cm(t)eiω0mt⟨0∣d^∣m⟩)(4)\begin{aligned} \braket{\mathbf{\hat d}}(t) &\approx \sum_{m\neq0} \left( c^*_m(t) e^{i(\omega_m-\omega_{0})t} \braket{m|\mathbf{\hat d}|0}+ c_m(t) e^{i(\omega_0-\omega_{m})t} \braket{0|\mathbf{\hat d}|m} \right) \\ &= \sum_{m\neq0} \left( c^*_m(t) e^{i\omega_{m0}t} \braket{m|\mathbf{\hat d}|0}+ c_m(t) e^{i\omega_{0m}t} \braket{0|\mathbf{\hat d}|m} \right) \end{aligned}\tag{4}

substituting Eq. 3 into Eq. 4, we get

⟨d^⟩(t)≈∑m≠0[−iℏ⟨0∣d^∣m⟩(E0∗e−i(ωn0−ω)t−i(ωm0−ω)+0++E0e−i(ωm0+ω)t−i(ωm0+ω)+0+)eiωm0t⟨m∣d^∣0⟩]+∑m≠0[iℏ⟨m∣d^∣0⟩(E0∗ei(ωm0−ω)ti(ωm0−ω)+0++E0ei(ωm0+ω)ti(ωm0+ω)+0+)eiω0mt⟨0∣d^∣m⟩]=∑m≠0[1ℏ⟨0∣d^∣m⟩(E0∗e−i(ωn0−ω)tωm0−ω+i0++E0e−i(ωm0+ω)tωm0+ω+i0+)eiωm0t⟨m∣d^∣0⟩]+∑m≠0[1ℏ⟨m∣d^∣0⟩(E0ei(ωm0−ω)tωm0−ω−i0++E0∗ei(ωm0+ω)tωm0+ω−i0+)eiω0mt⟨0∣d^∣m⟩](5)\begin{aligned} \braket{\mathbf{\hat d}}(t) &\approx \sum_{m\neq0} \left[ \frac{-i}{\hbar} \braket{0|\mathbf{\hat d}|m} \left( \mathbf{E}^*_0 \frac{e^{-i(\omega_{n0}-\omega) t}}{-i(\omega_{m0}-\omega)+0^+}+ \mathbf{E}_0 \frac{e^{-i(\omega_{m0}+\omega) t}}{-i(\omega_{m0}+\omega)+0^+} \right) e^{i\omega_{m0}t} \braket{m|\mathbf{\hat d}|0}\right]\\ &+\sum_{m\neq0} \left[ \frac{i}{\hbar} \braket{m|\mathbf{\hat d}|0} \left( \mathbf{E}^*_0 \frac{e^{i(\omega_{m0}-\omega) t}}{i(\omega_{m0}-\omega)+0^+}+ \mathbf{E}_0 \frac{e^{i(\omega_{m0}+\omega) t}}{i(\omega_{m0}+\omega)+0^+} \right) e^{i\omega_{0m}t} \braket{0|\mathbf{\hat d}|m}\right]\\ &= \sum_{m\neq0} \left[ \frac{1}{\hbar} \braket{0|\mathbf{\hat d}|m} \left( \mathbf{E}^*_0 \frac{e^{-i(\omega_{n0}-\omega) t}}{\omega_{m0}-\omega+i0^+}+ \mathbf{E}_0 \frac{e^{-i(\omega_{m0}+\omega) t}}{\omega_{m0}+\omega+i0^+} \right) e^{i\omega_{m0}t} \braket{m|\mathbf{\hat d}|0}\right]\\ &+\sum_{m\neq0} \left[ \frac{1}{\hbar} \braket{m|\mathbf{\hat d}|0} \left( \mathbf{E}_0 \frac{e^{i(\omega_{m0}-\omega) t}}{\omega_{m0}-\omega-i0^+}+ \mathbf{E}^*_0 \frac{e^{i(\omega_{m0}+\omega) t}}{\omega_{m0}+\omega-i0^+} \right) e^{i\omega_{0m}t} \braket{0|\mathbf{\hat d}|m}\right] \end{aligned} \tag{5}

Note: electric field definition

We use $e^{i\omega t}$ and $e^{-i\omega t}$ to construct $\cos$. So here simply taking terms related to $e^{i\omega t}$ should give us the polarizability.

Taking only terms that contains $e^{-i\omega t}$ from Eq. 5

⟨d^⟩(t)≈1ℏ∑m≠0[⟨0∣d^∣m⟩(E0e−i(ωm0+ω)tωm0+ω+i0+)eiωm0t⟨m∣d^∣0⟩]+1ℏ∑m≠0[⟨m∣d^∣0⟩(E0ei(ωm0−ω)tωm0−ω−i0++)eiω0mt⟨0∣d^∣m⟩]=1ℏ∑m≠0[⟨0∣d^∣m⟩⟨m∣d^∣0⟩(e−iωm0teiωm0tωm0+ω+i0+)]E0e−iωt+1ℏ∑m≠0[⟨m∣d^∣0⟩⟨0∣d^∣m⟩(eiωm0teiω0mtωm0−ω−i0+)]E0e−iωt=1ℏ∑m≠0(⟨0∣d^∣m⟩⟨m∣d^∣0⟩ωm0+ω+i0++⟨m∣d^∣0⟩⟨0∣d^∣m⟩ωm0−ω−i0+)E0e−iωt(5.1)\begin{aligned} \braket{\mathbf{\hat d}}(t) &\approx \frac{1}{\hbar} \sum_{m\neq0} \left[ \braket{0|\mathbf{\hat d}|m} \left( \mathbf{E}_0 \frac{e^{-i(\omega_{m0}+\omega) t}}{\omega_{m0}+\omega+i0^+} \right) e^{i\omega_{m0}t} \braket{m|\mathbf{\hat d}|0}\right]\\ &+\frac{1}{\hbar}\sum_{m\neq0} \left[ \braket{m|\mathbf{\hat d}|0} \left( \mathbf{E}_0 \frac{e^{i(\omega_{m0}-\omega) t}}{\omega_{m0}-\omega-i0^+}+ \right) e^{i\omega_{0m}t} \braket{0|\mathbf{\hat d}|m}\right]\\ &=\frac{1}{\hbar} \sum_{m\neq0} \left[ \braket{0|\mathbf{\hat d}|m} \braket{m|\mathbf{\hat d}|0} \left( \frac{e^{-i\omega_{m0} t}e^{i\omega_{m0}t}}{\omega_{m0}+\omega+i0^+} \right) \right]\mathbf{E}_0e^{-i\omega t} \\ &+\frac{1}{\hbar}\sum_{m\neq0} \left[ \braket{m|\mathbf{\hat d}|0} \braket{0|\mathbf{\hat d}|m} \left( \frac{e^{i\omega_{m0} t}e^{i\omega_{0m}t}}{\omega_{m0}-\omega-i0^+} \right) \right]\mathbf{E}_0e^{-i\omega t}\\ &=\frac{1}{\hbar} \sum_{m\neq0} \left( \frac{\braket{0|\mathbf{\hat d}|m} \braket{m|\mathbf{\hat d}|0}}{\omega_{m0}+\omega+i0^+} +\frac{\braket{m|\mathbf{\hat d}|0} \braket{0|\mathbf{\hat d}|m}}{\omega_{m0}-\omega-i0^+} \right) \mathbf{E}_0e^{-i\omega t} \tag{5.1} \end{aligned}

According to Maxwell’s equations, an external field we have

P(t)=d(t)V=ε0α(ω)E(t,ω)\mathbf{P} (t) = \frac{\mathbf{d}(t)}{V} = \varepsilon_0 \boldsymbol{\alpha}(\omega) \mathbf{E}(t,\omega)

so

d(t)=Vε0α(ω)E(t,ω)(5.2)\mathbf{d}(t) = V\varepsilon_0\boldsymbol{\alpha}(\omega)\mathbf{E(t,\omega)} \tag{5.2}

Comparing Eq. 5.1 to Eq. 5.2, we see that

Vε0α(ω)=1ℏ∑m≠0(⟨0∣d^∣m⟩⟨m∣d^∣0⟩ωm0+ω+i0++⟨m∣d^∣0⟩⟨0∣d^∣m⟩ωm0−ω−i0+)\begin{aligned} V\varepsilon_0\boldsymbol{\alpha}(\omega) &= \frac{1}{\hbar} \sum_{m\neq0} \left( \frac{\braket{0|\mathbf{\hat d}|m} \braket{m|\mathbf{\hat d}|0}}{\omega_{m0}+\omega+i0^+} +\frac{\braket{m|\mathbf{\hat d}|0} \braket{0|\mathbf{\hat d}|m}}{\omega_{m0}-\omega-i0^+} \right)\\ \end{aligned}

Now replacing $i0^+$ with phenomenological damping factor $i\Gamma_m/2$ of mode $m$,

Vε0α(ω)=1ℏ∑m≠0(⟨0∣d^∣m⟩⟨m∣d^∣0⟩ωm0+ω+Γm/2+⟨m∣d^∣0⟩⟨0∣d^∣m⟩ωm0−ω−iΓm/2)=1ℏ∑m≠0(⟨0∣d^∣m⟩⟨m∣d^∣0⟩(ωm0−ω−iΓm/2)(ωm0+ω+Γm/2)(ωm0−ω−iΓm/2)+⟨m∣d^∣0⟩⟨0∣d^∣m⟩(ωm0+ω+Γm/2)(ωm0+ω+Γm/2)(ωm0−ω−iΓm/2))=1ℏ∑m≠0(⟨0∣d^∣m⟩⟨m∣d^∣0⟩ωm0ωm02−ω2−iΓmω−Γm2/4+⟨m∣d^∣0⟩⟨0∣d^∣m⟩ωm0ωm02−ω2−iΓmω−Γm2/4)\begin{aligned} V\varepsilon_0\boldsymbol{\alpha}(\omega) &= \frac{1}{\hbar} \sum_{m\neq0} \left( \frac{\braket{0|\mathbf{\hat d}|m} \braket{m|\mathbf{\hat d}|0}}{\omega_{m0}+\omega+\Gamma_m/2} +\frac{\braket{m|\mathbf{\hat d}|0} \braket{0|\mathbf{\hat d}|m}}{\omega_{m0}-\omega-i\Gamma_m/2} \right)\\ &= \frac{1}{\hbar} \sum_{m\neq0} \left( \frac{\braket{0|\mathbf{\hat d}|m} \braket{m|\mathbf{\hat d}|0} (\omega_{m0}-\omega-i\Gamma_m/2)}{(\omega_{m0}+\omega+\Gamma_m/2)(\omega_{m0}-\omega-i\Gamma_m/2)} +\frac{\braket{m|\mathbf{\hat d}|0} \braket{0|\mathbf{\hat d}|m}(\omega_{m0}+\omega+\Gamma_m/2)} {(\omega_{m0}+\omega+\Gamma_m/2)(\omega_{m0}-\omega-i\Gamma_m/2)} \right)\\ &= \frac{1}{\hbar} \sum_{m\neq0} \left( \frac{\braket{0|\mathbf{\hat d}|m} \braket{m|\mathbf{\hat d}|0} \omega_{m0}}{\omega_{m0}^2-\omega^2-i\Gamma_m\omega - \Gamma_m^2/4} + \frac{\braket{m|\mathbf{\hat d}|0} \braket{0|\mathbf{\hat d}|m} \omega_{m0}}{\omega_{m0}^2-\omega^2-i\Gamma_m\omega - \Gamma_m^2/4} \right)\\ \end{aligned}

and we can now calculate the polarizability tensor $\boldsymbol{\alpha}$

α(ω)=1V1ε01ℏ∑m≠0(⟨0∣d^∣m⟩⟨m∣d^∣0⟩ωm0ωm02−ω2−iΓmω−Γm2/4+⟨m∣d^∣0⟩⟨0∣d^∣m⟩ωm0ωm02−ω2−iΓmω−Γm2/4)\begin{aligned} \boldsymbol{\alpha}(\omega) &= \frac{1}{V}\frac{1}{\varepsilon_0}\frac{1}{\hbar} \sum_{m\neq0} \left( \frac{\braket{0|\mathbf{\hat d}|m} \braket{m|\mathbf{\hat d}|0} \omega_{m0}}{\omega_{m0}^2-\omega^2-i\Gamma_m\omega - \Gamma_m^2/4} + \frac{\braket{m|\mathbf{\hat d}|0} \braket{0|\mathbf{\hat d}|m} \omega_{m0}}{\omega_{m0}^2-\omega^2-i\Gamma_m\omega - \Gamma_m^2/4} \right)\\ \end{aligned}

α\boldsymbol{\alpha} is a 3 by 3 rank-2 tensor, because ⟨m∣d^∣0⟩\braket{m\vert \mathbf{\hat{d}}\vert 0} comes from ⟨m∣d^∣0⟩E0\braket{m\vert \mathbf{\hat d}\vert 0}\mathbf{E}_0, ⟨0∣d^∣m⟩\braket{0\vert \mathbf{\hat d}\vert m} can lie in a different direction, and the components of α\boldsymbol{\alpha} become:

αij(ω)=1V1ε01ℏ∑m≠0(⟨0∣d^i∣m⟩⟨m∣d^j∣0⟩ωm0ωm02−ω2−iΓmω−Γm2/4+⟨m∣d^j∣0⟩⟨0∣d^i∣m⟩ωm0ωm02−ω2−iΓmω−Γm2/4)\begin{aligned} {\alpha}_{ij}(\omega) &= \frac{1}{V}\frac{1}{\varepsilon_0}\frac{1}{\hbar} \sum_{m\neq0} \left( \frac{\braket{0|\mathbf{\hat d}_i|m} \braket{m|\mathbf{\hat d}_j|0} \omega_{m0}}{\omega_{m0}^2-\omega^2-i\Gamma_m\omega - \Gamma_m^2/4} + \frac{\braket{m|\mathbf{\hat d}_j|0} \braket{0|\mathbf{\hat d}_i|m} \omega_{m0}}{\omega_{m0}^2-\omega^2-i\Gamma_m\omega - \Gamma_m^2/4} \right)\\ \end{aligned}

Using number density $N = 1/V$ which gives the number of this dipole moments in side a unit volume, and ignoring the second order term $- \Gamma_m^2/4$ since dampening is small, we have

αij(ω)=1V1ε01ℏ∑m≠0(⟨0∣d^i∣m⟩⟨m∣d^j∣0⟩ωm0ωm02−ω2−iΓmω+⟨m∣d^j∣0⟩⟨0∣d^i∣m⟩ωm0ωm02−ω2−iΓmω)\begin{aligned} {\alpha}_{ij}(\omega) &= \frac{1}{V}\frac{1}{\varepsilon_0}\frac{1}{\hbar} \sum_{m\neq0} \left( \frac{\braket{0|\mathbf{\hat d}_i|m} \braket{m|\mathbf{\hat d}_j|0} \omega_{m0}}{\omega_{m0}^2-\omega^2-i\Gamma_m\omega } + \frac{\braket{m|\mathbf{\hat d}_j|0} \braket{0|\mathbf{\hat d}_i|m} \omega_{m0}}{\omega_{m0}^2-\omega^2-i\Gamma_m\omega } \right)\\ \end{aligned}

For a single state transition from $\ket{0}$ to $\ket{m}$ contribution, we have

αij(ω)=Nε0ωm0ℏ⟨0∣d^i∣m⟩⟨m∣d^j∣0⟩+⟨0∣d^j∣m⟩⟨m∣d^i∣0⟩ωm02−ω2−iΓmω(6){\alpha}_{ij}(\omega) =\frac{N}{\varepsilon_0} \frac{\omega_{m0} }{\hbar} \frac{\braket{0|\mathbf{\hat d}_i|m} \braket{m|\mathbf{\hat d}_j|0} + \braket{0|\mathbf{\hat d}_j|m} \braket{m|\mathbf{\hat d}_i|0} }{\omega_{m0}^2-\omega^2-i\Gamma_m\omega} \tag{6}

Averaging over all three directions, we get

αˉ(ω)=∑i=x,y,zαii/3=Nε02ωm03ℏ∣⟨0∣d^∣m⟩∣2ωm02−ω2−iΓmω(7)\bar{\alpha} (\omega)= \sum_{i=x,y,z}\mathbf{\alpha}_{ii}/3 = \frac{N}{\varepsilon_0} \frac{2\omega_{m0} }{3\hbar} \frac{|\braket{0|\mathbf{\hat d}|m}|^2}{\omega_{m0}^2-\omega^2-i\Gamma_m\omega} \tag{7}

Oscillator strength

Comparing to the result from the Lorentz model

α(ω)=Nq2ε0m1ω02−ω2−iωΓ\alpha(\omega) = \frac{Nq^2}{\varepsilon_0 m} \frac{1}{\omega_0^2-\omega^2-i\omega\Gamma}

we see that with quantum version (Eq. 7), we need to have

2ωm0∣⟨0∣d^∣m⟩∣23ℏ→q2m\frac{2\omega_{m0}|\braket{0|\mathbf{\hat d}|m}|^2}{3\hbar} \to \frac{q^2}{m}

hence the oscillator strength $S_m$ can be defined as

Sm=mq22ωm0∣⟨0∣d^∣m⟩∣23ℏS_m= \frac{m}{q^2} \frac{2\omega_{m0}|\braket{0|\mathbf{\hat d}|m}|^2}{3\hbar}

so that

q2mSm=2ωm0∣⟨0∣d^∣m⟩∣23ℏ\frac{q^2}{m} S_m = \frac{2\omega_{m0}|\braket{0|\mathbf{\hat d}|m}|^2}{3\hbar}

and the corresponding modified Lorentz model is

αm(ω)=Nq2ε0mSmω02−ω2−iωΓ\alpha_m(\omega) = \frac{Nq^2}{\varepsilon_0 m} \frac{S_m}{\omega_0^2-\omega^2-i\omega\Gamma}

Author | Chengcheng Xiao

Currently a postdoctoral research associate at Imperial College London. Predicting electron behaviour since 2016.