Photon absorption and momentum recoil

[Niles]

Hi

When talking about atoms absorbing a photon, a classical picture is utilized. The intuitive picture often presented is that when an atom absorbs a photon, a recoil of \hbar k is obtained by the atom. After a time \tau=1/\Gamma the atom emits the photon again. So the force in this picture is (under the condition that the atom is in resonance with the light so that the probability of absorbing a photon is maximal)
<br /> F = \frac{\Delta p}{\Delta t} = \frac{\hbar k }{\tau} = \hbar k \Gamma <br />
However utilizing another picture, the force will be given by the scattering rate multiplied by the momentum recoil, i.e.
<br /> F = \hbar k \Gamma \rho_e<br />
where ρ_e is the probability to be in the excited state, which saturates to the maximum value 1/2.

Why does these two expressions differ? What is it that my first scenario has not taken into account?Niles.

 

[haruspex]

The first argument only considers the period during which the atom is excited, so if you're using it to compute an average force it is as though the atom will be immediately re-excited. The extra factor in the second expression says otherwise.

 

[Niles]

Thanks. So a two-level atom that has emitted a photon cannot absorb a photon instantly afterwards? I guess that is what the second expression tells us. It is just counter-intuitive?Niles.

 

[haruspex]

So a two-level atom that has emitted a photon cannot absorb a photon instantly afterwards?

I've no idea - I'm just saying that appears to be the difference between the equations.

 

[Niles]

Ah, I see. Thanks for chiming in though.

 

[Darwin123]

Hi

When talking about atoms absorbing a photon, a classical picture is utilized. The intuitive picture often presented is that when an atom absorbs a photon, a recoil of \hbar k is obtained by the atom. After a time \tau=1/\Gamma the atom emits the photon again. So the force in this picture is (under the condition that the atom is in resonance with the light so that the probability of absorbing a photon is maximal)
<br /> F = \frac{\Delta p}{\Delta t} = \frac{\hbar k }{\tau} = \hbar k \Gamma <br />
However utilizing another picture, the force will be given by the scattering rate multiplied by the momentum recoil, i.e.
<br /> F = \hbar k \Gamma \rho_e<br />
where ρ_e is the probability to be in the excited state, which saturates to the maximum value 1/2.

Why does these two expressions differ? What is it that my first scenario has not taken into account?Niles.

The instantaneous "force" can't be measured. The measuring devices don't have a time resolution of τ, which is determined by the bandwidth. Actual devices measure the change in momentum using an integration time longer than the mean time between absorptions.
This is not very much different from the classical case of an ideal gas in a container. The thermodynamic pressure is the determined by the change in momentum of the molecules that hit the walls of the container averaged over a very long time period and divided by the area of the wall. Thus, the thermodynamic pressure is inversely proportional to the time between collisions. However, there is an instantaneous pressure caused by the force of the molecule on the wall during the time it is in contact with the wall and divided by the area of the molecule. This "instantaneous" pressure is far greater than the thermodynamic pressure.
The instantaneous pressure of a molecule hitting the wall of a container can't really be measured. What is measurable is the thermodynamic pressure, which by definition has to be time averaged over a very long period of time. What we call radiation pressure is really the thermodynamic pressure of photons.
One can measure instantaneous changes only if the change in quantity persists long enough to be measured. "Instantaneous" changes in momentum are measurable because total momentum is conserved. Instantaneous changes in force are not measurable, because total force is not conserved.
This is not specifically a quantum mechanics problem. It exists in classical physics as well. The time scale over which a measurement is performed determines what is measured. Measurements also take place over distance scales and area scales.

 

[M Quack]

A two level atom cannot absorb a (second) photon while in the excited state, i.e. after absorbing a first photon and before (re)emitting a photon.

What surprises me in the second formula is that the "statistics factor" rho is the probability of being in the excited state, and not in the ground state. Only in the ground state a photon can be absorbed.

 

[Darwin123]

Thanks. So a two-level atom that has emitted a photon cannot absorb a photon instantly afterwards? I guess that is what the second expression tells us. It is just counter-intuitive?


Niles.

Not if you apply classical electrodynamics to the problem. Let us look from the point of view of a physicist who doesn't believe in photons. Light is a wave.
If you illuminate the atom with a monochromatic steady-state beam, then the atom will respond as though it is a forced oscillator where the force oscillates at a fixed frequency. When the light is resonanat with the atom, the frequency of the light equals the natural frequency of the atom. The second formula applies.
A light pulse that has a duration less than τ is a superposition of light from many frequencies. In most atoms and states, τ<1 ps. Some of these frequencies will be equal to the natural frequency of the atom, and others will not. So the absorption of light is greatly decreased, because most of the components won't even be absorbed. Then the first formula applies.
In terms of quantum electrodynamics, the particle view is at least as valid as the classical view.
In the case of a steady state light beam, the light is a train of separate photons. The atoms is continuously being hit by photons. Then the absorption of energy follows the second formula, as though we were measuring the thermodynamic pressure of a photon gas.
In the case of the femtosecond pulse, a single photon hits the atoms. The duration of the hit is proportional to the width of the atom, which is cτ. The speed of light is c. The impulse of the atom is proportional to τ. So the first formula applies.
So I think both formulas are intuitive if you examine it carefully. It doesn't matter if you look at it classically or with quantum mechanics. The two formulas apply to different types of measurement.