Absorption, emission, and scattering

The absorption and smission discussion comes from API pages 67-93. The resonant scattering discussion comes from page 93 of API.

Absorption and emission

In emission, an atom has initial state $|\phi _{i}\rangle =|b,K,0{\rangle }$, and the final state is $|\phi _{f}\rangle =|a,K',k\epsilon {\rangle }$.

File:Absorption, emission, and scattering-c3-emission-diagram

Due to momentum conservation, we must have that $\hbar K=\hbar K'+\hbar k$. From energy conservation, we have

$E_{b}+{\frac {\hbar ^{2}K^{2}}{2M}}=E_{a}+{\frac {\hbar ^{2}{K'}^{2}}{2M}}+\hbar \omega$

What we now want to discuss is how this process changes In the center-of-mass system, the atom is initially at rest: $K=0$. Afterwards, it has momentum $K'=-k$. The photon emitted has energy $\hbar \omega =\hbar \omega _{0}-E_{rec}$, where $E_{rec}$ is the kinetic energy (of the atom) associated with the momentum of a single photon. If the atom is initially moving, with $v\neq 0$, then $K\neq 0$, and the photon emitted is doppler shifted, giving $\hbar \omega =\hbar \omega _{0}-E_{rec}+\hbar kv$. $v=\hbar K/M$ is the velocity before emission. In absorption, there is a new element, which is the radiation field, giving us several cases that must be distinguished. The diagram is as follows:

File:Absorption, emission, and scattering-c3-absorption-diagram

Before absorption, the atom is in state $a$, with momentum $K=0$. A photon of momentum $k$ is absorbed. After absorption, the atom is in state $b$, with momentum $K'=k$. If the absorption frequency is $\omega _{abs}$, then $\hbar \omega _{abs}=\hbar \omega _{0}+E_{rec}$, so that if $\omega _{em}$ is the emission frequency, then $\omega _{abs}-\omega _{em}=2E_{rec}/\hbar$; there is a energy difference. This difference is typically about $100$ kHz. It took the development of narrowband lasers to see this (as was done first by Jan Hall at Boulder in a famous experiment).

Global atom+radiation system

We have given discrete states for the initial and final states of our perturbation processes, $|\phi _{i}\rangle =|a,Nk_{0}\epsilon _{0}{\rangle }$ and $|\phi _{f}\rangle =|b,(N-1)k_{0}\epsilon _{0}{\rangle }$. The coupling between these we call absorption. Spontaneous emission, or absorption followed by spontaneous emission (which we call scattering), is coupling of these two states to a larger system. This may take us back to the initial state or to other states $|\phi _{g}\rangle =|a,(N-1)k_{0}\epsilon _{0},k\epsilon {\rangle }$, which form a continuum of states. In the end, we will not be able to obtain an exact solution to this coupling to the continuum, but we will be able to obtain useful limits and approximate solutions. When we can eliminate spontaneous emission and coupling to external states, we are left with a closed system, where the evolution is coherent, and the physics of Rabi oscillations. These Rabi oscillations are the physics of coherent two-state systems. The coupling to the external continuum is characterized by an exponential decay. It has a characteristic decay time $\tau =1/\Gamma$. When we talk about absorption, we have to consider that we are not simply coupling an initial state to a final state; we also couple to a continuum, which modifies the character of the final state. Two obvious limiting cases which are useful to consider are:

• $\Omega \gg \Gamma$ -- the Rabi frequency is much faster than the damping rate. This gives us damped Rabi oscillations. This is the strong field limit.
• $\Omega \ll \Gamma$ -- This is the weak field limit. We should thus first take care of the coupling to the continuum first; this mixes the final state with the continuum, meaning that some character of the continuum is mixed into the final discrete state. This creates, on the final side, a continuum of possible final states. By Fermi's golden rule, this gives us a rate $\Omega ^{2}/\Gamma$, where the denominator is interpreted as a density of states.

Intuitively, we can understand the strong coupling case using the following picture:

File:Absorption, emission, and scattering-c3-strong-coupling-intuition

It is something like a piecewise coherent evolution. For small times, the system behaves coherently, but then has to restart a number of times. The probability that a photon has been been absorbed thus goes as $(\Omega t)^{2}$. When the coherent evolution is interrupted with a probability $1/\Gamma$, then the rate is $(\Omega /\Gamma )^{2}/(1/\Gamma )=\Omega ^{2}/\Gamma$. This picture is one possible way to intepret what goes on in the quantum process, as we shall later see when visiting the quantum monte-carlo picture. Another limiting case is when we have broadband excitation, in which case the bandwidth is large compared with the coupling to the continuum, $\gamma \gg \Gamma$. We can us Fermi's golden rule to derive that the excitation rate is $\Omega ^{2}/\gamma$, simply because the electric field is now spread out over bandwidth $\gamma$. A physical picture for this:

File:Absorption, emission, and scattering-c3-strong-coupling-intuition

is that the atom has at short time no opportunity to know if the light is narrowband or broadband. The rate thus initially coherent, going as $\Omega ^{2}t^{2}$. But after a time larger than the inverse bandwidth $1/\gamma$, the atom recognizes that only the fraction $1/(\gamma t)$ is resonant and continues to drive the coherent evolution. The longer the time, the less of the spectrum drives the atom. You can then say that the time evolution is the coherent evolution, pro-rated by the fraction regarded by the atom to be on resonance,

${\frac {\Omega ^{2}t^{2}}{\gamma t}}={\frac {\Omega ^{2}}{\gamma }}t\,,$

giving what we obtained from Fermi's golden rule. Note that we've obtained an understanding of three important cases of coupling to a global radiation system without dealing with the coupling to a continuum in depth; that will come in the next chapter.

Resonant scattering

Now let us combine our approaches; if a photon is absorbed and emitted, it is not in general appropriate to describe these as two separate processes; this combined process is known as {\em scattering}. There are several possibilities:

• Atom absorbs a photon first, then emits a photon:

File:Absorption, emission, and scattering-c3-scattering1

This is described by an interaction Hamiltonian to first order

$H_{I}=-{\frac {e}{m}}{\vec {p}}\cdot {\vec {A}}$

• Atom first emits a photon, then absorbs a photon:

File:Absorption, emission, and scattering-c3-scattering2

This is also described by an interaction Hamiltonian to first order

$H_{I}=-{\frac {e}{m}}{\vec {p}}\cdot {\vec {A}}$

• A photon is absorbed and emitted at the same time, without the atom going through any intermediate state:

File:Absorption, emission, and scattering-c3-scattering3

This is described by the interaction

$H_{I2}=-{\frac {p^{2}}{2m}}A^{2}\,,$

which is of second order.

Let us consider the two first-order processes. From an energy level process, we have these two pictures:

File:Absorption, emission, and scattering-c3-scattering1-energy

The dashed line is the energy of a virtual level, giving us the energy of the total system. The difference in energy between the virtual level and the state $b$ is exactly the energy written in the energy denominator of the perturbation expansion.

Alternatively, the state $a$ first emits a photon to a virtual level, then a photon is absorbed, returning the atom back to $a'$:

File:Absorption, emission, and scattering-c3-scattering2-energy

Note that the virtual energy level is negative, reflecting the intermediate state of the process. We want to write an expression for the transition matrix,

${\langle }\phi _{f}|U(t_{f},t_{i})|\phi _{i}{\rangle }\,.$

For energy conservation, we must satisfy $\delta (E_{f}-E_{i})=0$; overall, we can write

${\langle }\phi _{f}|U(t_{f},t_{i})|\phi _{i}\rangle =\delta _{fi}-2\pi i\delta (E_{f}-E_{i}){\mathcal {T}}_{fi}\,,$

where ${\mathcal {T}}_{fi}$ is what we'll now give. For the first case,

${\mathcal {T}}_{fi}=\sum _{b}{\frac {\langle a',k'\epsilon '|H_{I}|b,a\rangle \langle b,0|H_{I}|b,k\epsilon {\rangle }}{E_{a}+\hbar \omega -E_{b}}}\,.$

The intermediate state has only one photon, of momentum $k$ and polarization $\epsilon$. For the second case, where the photon is emitted then absorbed, we obtain:

${\mathcal {T}}_{fi}=\sum _{b}{\frac {\langle a',k'\epsilon '|H_{I}|b,a\rangle \langle b,0|H_{I}|b,k\epsilon k'\epsilon '{\rangle }}{E_{a}-\hbar \omega -E_{b}}}\,.$

Note that the intermediate state has two photons, this time. The denominator also now has a minus sign, corresponding to the energy level diagram. Simply because of the energetics, this second process contributes less to the overall matrix element for the rate.

Resonant scattering

What happens when the photon has exactly the same energy as the level difference between two states in the atom? This is the interesting case of resonant scattering. Mathematically, a problem is that the energy denominator diverges on resonance. The resolution of this problem is that we add a damping term to the denominator, which limits the divergence in the denominator. We can see how such a damping term arises from diagrammatic expansion of the atom-photon process; its origin is very subtle, since the physics includes terms of infinite order terms, and involves summation over an infinite number of diagrams. The physical picture for this process is:

File:Absorption, emission, and scattering-c3-resonant-energy

showing what happens when $\hbar \omega =E_{b}-E_{a}$, causing the energy denominator diverges. Let us compute ${\mathcal {T}}$ for this, both exactly and approximately. The exact expression is

${\mathcal {T}}_{ak'\epsilon ',ak\epsilon }=\langle a,k'\epsilon '|H_{I1}{\frac {1}{E_{a}+\hbar \omega -H+i\eta }}H_{I1}|a,k\epsilon {\rangle }\,,$

where $H$ is the total Hamiltonian. $i\eta$ is included for mathematical consistency (we can let $\eta \rightarrow 0$). In perturbation theory, this middle factor becomes

$\sum _{b}{\frac {|b,0\rangle \langle b,0|}{E_{a}+\hbar \omega -E_{b}}}$

to second order; this is the Born approximation. $E_{b}$ in the denominator comes from $H_{0}$, the bare Hamiltonian. Physically, it means we couple through intermediate states to go back to the original. The state $b$ is coupled to a continuum, which has a width $\Gamma$. Therefore, we have to now accept that the discrete state coupling to the continuum smears out the state. It can be shown that this physics can be captured in the following way. We can approximate the ${\mathcal {T}}$ matrix by an expression which can deal with the resonance. Since we are resonant, we only through the intermediate state $E_{0}$, but the energy $E_{a}-E_{b}$ must be modified. The coupling between $b$ and the continuum generally involves a shift $\hbar \Delta$ in its energy level, as well as an imaginary part $i\hbar (\Gamma /2)$, due to its new width. We started out with a divergent denominator. The coupling to the continuum gives us a damping term

${\frac {1}{\hbar (\omega -\omega _{0}+i\Gamma /2)}}=\sum _{n=0}^{\infty }(-1)^{n}{\frac {i\Gamma /2)^{n}}{(\omega -\omega _{0})^{n+1}}}\,.$

This "harmless" looking inclusion of a damping term physically means, in the perturbation theory, that we have included interactions $H_{I1}$ up to infinite order. Recall that $\Gamma \propto (H_{I1})^{2}$. What terms are included here? To lowest order, we have the process

File:Absorption, emission, and scattering-c3-resonant-energy

The free propagator in the middle (the straight line) gives rise to the divergence. There are additional terms, involving emission and re-absorption of a photon, multiple times:

File:Absorption, emission, and scattering-c3-resonant-energy-sum

Thus, what we have is the physical picture that summing over all these diagrams gives an imaginary term in the denominator. And the $\Gamma$ in the denominator is given by Fermi's golden rule. Still, though, we have not solved the problem exactly, because we have not included in the diagrams some processes. Keep in mind that the vertical axes is time; we've neglected diagrams that involve having more than one virtual photon in existence at a time. Diagrams such as:

File:Absorption, emission, and scattering-c3-resonant-energy-2photon

also exist. But they can be left out when we just want the most important part of the coupling, to excellent approximation. Consider the physics of this. We've approached this by successively re-writing our equations until finding an opportunity for approximation. We had

$U=e^{-iHt/\hbar }\,.$

There is something known as a resolvant for this, where

$G(z)={\frac {1}{z-H}}\,,$

which mathematically captures all the physics of $U$ (see 50 pages in API about this). The resolvant can be re-expressed as

$G(z)={\frac {1}{z-E_{b}-R_{b}(z)}}\,.$

In this expression, $R_{b}(z)$ can be approximated. Working backwards, this gives an approximation for the physics of $U$. This is the central approach of much of many-body physics, as we shall see again later.

Frequency of scattered light

Back to something more mundane. We've talked about resonant excitation. What is the frequency of the scattered light? Consider a single atom. We've discussed doppler and recoil shifts; these can all be disregarded by letting the mass be infinity, $M\rightarrow \infty$. This freezes out the translational degree of freedom, and in the laboratory can be realized by trapped ions. If the ion is excited by frequency $\omega$, what is the spectrum of the light emitted by the ion? This spectrum is the distribution of $\omega '$, the frequency of the scattered photon:

File:Absorption, emission, and scattering-c3-scattered-freq

Does the process lead to broadening? Or shift the frequency? Consider three choices:

• It is a delta function at $\omega _{0}$
• Lorentzian at $\omega _{0}$
• Delta function at $\omega$
• Lorentzian at $\omega$

By energy conservation, the only possibility is that $\hbar \omega '=\hbar \omega$. The spectrum must this be centered not at $\omega _{0}$ -- it must be centered at $\omega$. This is a driven harmonic oscillator; driven at $\omega$, it responds at $\omega$. The atom is nothing but an electron on a spring. In practical experience, though, the line is broadened. When does that come about? First, when $\hbar \omega$ is input, $\hbar \omega$ comes out. When we scan the input frequency, the intensity of the scattered light varies. Then, of course, we are changing terms in the denominator for ${\mathcal {T}}$, so we obtain the excitation spectrum, which is a Lorentzian centered at $\omega _{0}$. Second, if we replace the fixed frequency laser by broadband excitation, which as all frequencies at the same time, then the atomic response will be Lorentzian, giving an excitation spectrum that is also a Lorentzian centered at $\omega _{0}$. Two important variants of this scenario are weak and strong pulsed excitation. With pulsed excitation, having a time that is much narrower than the inverse linewidth, we get this excitation and emission spectrum:

File:Absorption, emission, and scattering-c3-pulsed1

The emission is Lorentzian, corresponding to an exponentially damped sinusoid in the time-domain. The time constant of the decay is the inverse linewidth. This is simple to understand from the classical harmonic oscillator. If you hit the oscillator with a hammer, then it will be excited during the impact, then it will ring down with the natural decay. With such short pulse excitation, we can approximate photon scattering as a two-step process, with photons coming in first, then as photons coming out. This is not appropriate for normal scattering processes. If you do this same experiment at higher intensity (see p106 of API), the perturbation diagrams for the process change:

File:Absorption, emission, and scattering-c3-pulsed2

The only way to have energy conservation and get different frequency in the scattered light is to have two or more photons scattered in the interaction. This requires higher atom density. The diagram is

File:Absorption, emission, and scattering-c3-pulsed3

We absorb a photon, go to a virtual state, emit, absorb again, and emit once again. The sum of the photon energies emitted and absorbed must match, giving a degree of freedom for the output light frequency to be different from that absorbed: $\omega '+\omega ''=2\omega$. If $b$ has width $\Gamma$, then the resonance condition $2\omega -\omega '=\omega _{0}$ can be fulfilled with width $\Gamma$. That means that by the higher order diagram process, we obtain this emission spectrum (for both photons at $\omega '$ and $\omega ''$:

File:Absorption, emission, and scattering-c3-pulsed4

The first photon gives a delta function; the others give sidebands.

This result is the famous Mollow triplet. One sideband is exactly on resonance; the other sideband is a mirror image needed for energy conservation. Conceptually, this can also be understood in the following way: when the system is driven strongly, it Rabi oscillates, creating sidebands at plus and minus the oscillation frequency, leading to sidebands.