Quantum Computation III Atomic Spectra

Atomic spectra are defined as the spectrum of the electromagnetic radiation emitted or absorbed by an electron during transitions between different energy levels within an atom.

Two-level atom

Classical Model

• $N_1$ : Number of atoms in $\ket{g}$ per unit volume.

• $N_2$ : Number of atoms in $\ket{e}$ per unit volume.

• $\Gamma$ : Rate of spontaneous emission. The probability of transitioning of a single atom from $\ket{e}$ to $\ket{g}$ due to spontaneous emission per unit time.

• $I$ : Intensity. $I = \frac{1}{2} c n^2 \varepsilon_0 |E|^2$. Intensity has a dimension of $\mathrm{J\cdot m^{-2} \cdot s^{-1}}$.

• $\sigma(\omega)$ : Absorption cross-section, defined such that $\sigma(\omega) I$ is the energy absorbed by a single atom per unit time when irradiated by intensity $I$ with frequency $\omega$ .

• $\Gamma_\text{SR}$ : Rate of stimulated radiation. The probability of stimulated absorption equals the probability of stimulated emission, so that we can define such a universal rate.

Einstein coefficient

The change in number of the excited atoms $\frac{\mathrm{d}N_2}{\mathrm{d}t}$ (per unit volume) comes from three sources:

1. Spontaneous emission: In a unit time, there is a probability of $\Gamma$ for a single atom to transition to $\ket{g}$. For $N_2$ atoms, the total transition number is $N_2 \Gamma$ on average.

2. Stimulated absorption: According to the definition of absorption cross-section, $\sigma(\omega) I$ is the energy absorbed by a single atom per unit time. Since the incident beam has a frequency $\omega$, each time the medium absorb a photon $\hbar \omega$ , an atom will be excited into $\ket{e}$ (although the energy gap is $\hbar \omega_0$) . Therefore, in a unit time, $\sigma(\omega) I N_1 / \hbar \omega$ is the total number of atoms in the $\ket{g}$ state being excited into $\ket{e}$.

   /* Note: $\Gamma_\text{SR} = \sigma(\omega) I / \hbar \omega$ */

3. Stimulated emission: Since stimulated absorption and stimulated emission have the equal probability, the mechanism is similar. We have a negative term $-\frac{\sigma(\omega) I}{\hbar \omega} N_2$ .

We write down

$$\frac{\mathrm{d}N_2}{\mathrm{d}t} = -N_2 \Gamma -\frac{\sigma(\omega) I}{\hbar \omega} (N_2 - N_1)$$

For steady state, spontaneous emission + stimulated emission = stimulated absorption, $\mathrm{d} N_2/\mathrm{d}t = 0$. We obtain the steady state equation

$$-\frac{\sigma(\omega) I}{\hbar \omega} (N_2 - N_1) = N_2 \Gamma$$

Thus, the steady state of a two-level system always has $N_2 < N_1$, no matter how strong the incident beam is. Which means, it is impossible to use a two-level system as the gain medium of laser, since optical pumping cannot induce population inversion.

Intensity gain

When a laser beam with a frequency of $\omega$ and an intensity of $I$ is incident on the medium, we have

$$\frac{\mathrm{d}I}{\mathrm{d} z} = \alpha(\omega) I$$

where $\frac{\mathrm{d}I}{\mathrm{d} z}$ is the gain (caused by pure emission) of energy per unit volume per unit time. At the same time, $\sigma(\omega) I(N_2 - N_1)$ is also the gain of energy of incident beam per unit volume per unit time, where $N_2 - N_1$ means stimulated emission minuses stimulated absorption, leaving the pure gain of energy of incident beam. So

$$\alpha(\omega) = \sigma(\omega)(N_2 - N_1)$$

Combining $N_2 + N_1 = N$ and the steady state equation, we could finally obtain

$$\alpha(\omega) = \frac{\sigma(\omega) N}{1 + \dfrac{2\sigma(\omega)I}{\hbar \omega \Gamma}}$$

Line shape function

We have found that for a fixed number of particles $N_1,N_2$ and a constant incident light intensity $I$, the energy absorbed by the medium per unit time varies depending on the frequency of the incident light $\omega$. This is reflected in the dependence of absorption cross-section $\sigma(\omega)$ on $\omega$. Furthermore, we can extract a dimensionless line shape function $g(\omega)$ from it, describing the absorption efficiency of the medium for different frequencies of light.

Quantum Model

• $\rho_{gg}$ : The probability of a single atom on the ground state.
• $\rho_{ee}$ : The probability of a single atom on the excited state.

By solving the Optical Bloch Equation, we can obtain

$$\rho_{ee} (t\to \infty) = \frac{\Omega^2 / \Gamma^2}{1+\dfrac{4\Delta^2}{\Gamma^2}+\dfrac{2\Omega^2}{\Gamma^2}}$$

Define a saturation parameter $S_0 = 2\Omega^2/\Gamma^2 = I/I_\text{sat}$, the density matrix element becomes

$$\rho_{ee} (t\to \infty) = \frac{S_0/2}{1+S_0+\dfrac{4\Delta^2}{\Gamma^2}}$$

When on-resonance and large intensity $S_0 \gg 1$, $1/2$ of the atoms occupy the excited state, reaches its maximum. It proves again that the steady state of a two-level system always has $N_2 < N_1$, no matter how strong the incident beam is.

Resonance fluorescence

We will now consider the radiation due to a single, isolated atom driven by a monochromatic field. The radiation, viewed as scattered light, is called resonance fluorescence.

The optical Wiener-Хи́нчин theorem writes

$$I_\text{sc}(\boldsymbol{r},\omega) = \frac{1}{2\pi} c n^2 \varepsilon_0 \int_{-\infty}^{+\infty} \langle E^{(-)}(\boldsymbol{r},t) E^{(+)}(\boldsymbol{r},t+\tau) \rangle \, \mathrm{e}^{\mathrm{i}\omega\tau} \, \mathrm{d} \tau$$

where $\hat{E}$ contains $\sigma, \sigma^\dagger$. This will become

$$I_\text{sc}(\boldsymbol{r},\omega_\text{sc}) = \frac{\hbar \omega_0 \Gamma}{r^2} f(\theta,\phi) S(\omega_\text{sc})$$

where $S(\omega_\text{sc})$ is the spectrum of the scattered light with a dimension of $\mathrm{Hz^{-1}}$, which describes the probability of the fluorescence having a frequency between $\omega_\text{sc}$ and $\omega_\text{sc} + \mathrm{d}\omega_\text{sc}$. However, $S(\omega_\text{sc})$ is not normalized. The total probability is

$$\int_0^\infty S(\omega_\text{sc}) \, \mathrm{d}\omega_\text{sc} = \rho_{ee}(t\to \infty)$$

This results in

$$P_\text{sc} = \int \frac{\hbar \omega_0 \Gamma}{r^2} f(\theta,\phi) \,\mathrm{d} \Omega \int S(\omega_\text{sc}) \, \mathrm{d}\omega_\text{sc} = \Gamma \hbar \omega_0 \rho_{ee}$$

It’s worth mentioning that the spectrum of the scattered light is named by Mollow Triplet Spectrum

$$S(\omega_\text{sc})=\frac{S_0}{(1+S_0)^2} \delta\left(\omega_\text{sc}-\omega\right)+\frac{S_0}{8 \pi(1+S_0)} \frac{\Gamma}{\left[(\omega_\text{sc}-\omega)^2+(\Gamma / 2)^2\right]} + \cdots$$

where $S_0=2\Omega^2/\Gamma^2$ is the saturation parameter.

Conclusion: In spite of the fact that scattering spectrum are the superposition of waves at different frequencies $I_\text{sc}(\boldsymbol{r},\omega_\text{sc})$, the total power $P_\text{sc}$ of all the scattered beams is exactly equivalent to the power of spontaneous emission, regardless of whether the incident beam is on-resonant $\omega = \omega_0$ or off-resonant $\omega \neq \omega_0$. However, The spectrum of scattered light is centered at the frequency of incident beam $\omega$ !

Power broadening

Therefore, by measuring the power of resonance fluorescence, which is actually measuring $\rho_{ee}$ since $P_\text{sc} \propto \rho_{ee}$ , we can draw a graph

To explain the line shape, we recall the expression of $\rho_{ee}$ :

$$\rho_{ee} (t\to \infty) = \frac{S_0/2}{1+S_0+\dfrac{4\Delta^2}{\Gamma^2}}$$

Define a parameter $\Gamma^\prime = \Gamma \sqrt{1+S_0}$ where $S_0 = 2\Omega^2/\Gamma^2 = I/I_\text{sat}$ evaluates the intensity of the incident beam. The $\rho_{ee}$ is written

$$\rho_{ee} = \frac{S_0}{2(1+S_0)} \frac{1}{1+\dfrac{4\Delta^2}{\Gamma^{\prime 2}}}$$

The full width at half maximum (FWHM) is exactly $\Gamma^\prime$ (When $\Delta = \pm \Gamma^\prime/2$ we have $\frac{1}{1+\frac{4\Delta^2}{\Gamma^{\prime 2}}} = \frac{1}{2}$). Two mechanism of broadening could be found

• In the weak-field limit, $S_0 \simeq 0$, $\Gamma^\prime \simeq \Gamma$. This is called natural broadening.
• In the strong-field limit, $\Gamma^\prime = \Gamma \sqrt{1+S_0}$ . The FWHM of the transition is effectively larger due to the strong coupling to the field. This phenomenon is called power broadening of the transition.

Connection with classical model

Warning: Only self-consistent when on-resonance $\omega=\omega_0$.

Optical Bloch Equation writes

$$\dot{\rho}_{ee} = -\Gamma \rho_{ee} - \frac{\Omega^2/\Gamma}{1 + \dfrac{4\Delta^2}{\Gamma^2}} (\rho_{ee}-\rho_{gg})$$

Naturally $\rho_{ee} = N_2/N,\,\rho_{gg} = N_1 / N$. Recall that in classical model

$$\frac{\mathrm{d}N_2}{\mathrm{d}t} = -N_2 \Gamma -\frac{\sigma(\omega) I}{\hbar \omega} (N_2 - N_1)$$

thus

$$\sigma(\omega) = \frac{\hbar \omega}{I} \frac{\Omega^2/\Gamma}{1 + \dfrac{4\Delta^2}{\Gamma^2}}$$

Since

$$\Omega^2 = \frac{|\langle g |\hat{d}|e\rangle|^2}{\hbar^2} |E|^2 \qquad \Gamma=\frac{\omega^3_0}{\pi \varepsilon_0 \hbar c^3} |\langle g |\hat{d}|e\rangle|^2 \qquad I = \frac{1}{2}c \varepsilon_0 |E|^2$$

The absorption cross-section now becomes

$$\sigma(\omega) = \frac{2 \pi c^2 \omega}{\omega_0^3} \frac{1}{1 + \dfrac{4\Delta^2}{\Gamma^2}} \approx \frac{\lambda^2_0}{4} \frac{\Gamma^2}{2\pi (\Gamma^2/4 +\Delta^2)}$$

where $\lambda_0 = 2\pi c/\omega_0$. If we define a line shape function

$$g(\omega) = \dfrac{1}{2\pi} \dfrac{\Gamma}{\Gamma^2/4 +\Delta^2}$$

we find $\sigma(\omega) = \Gamma \frac{\lambda^2_0}{4} g(\omega)$. We can obtain the same result taking advantage of energy conservation:

$$-\sigma(\omega) I (N_2 - N_1) = P_\text{sc} = N\Gamma \hbar \omega_0 \rho_{ee} \implies \sigma(\omega) = \Gamma \frac{\lambda^2_0}{4} g(\omega)$$

which means, given the same incident intensity and the same population, the efficiency of energy absorption by atoms varies with the changing incident light frequency $\omega$, and reaches its maximum at resonance $\Delta = 0$.

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