Description of quantum scattering

Question

I am looking for an explanation of quantum scattering to help me understand why the sky is blue. I have checked all previous threads on the sky being blue on this website but do not feel they give the mechanism by which a photon scatters off an electron or they answer it with Rayleigh scattering. Currently I know resonant photon frequencies are absorbed and re-emitted, and i know Raman scattering is possible.

My 3 questions are:

1. What happens to the photon and electron during scattering? absorbed and re-emitted? just 'bounces' off etc?
2. Why can scattering occur at all frequencies if the energy levels in an atom are discrete?
3. what is the quantum explanation for higher frequencies (blue light) being scattered much more (freq^4) than other frequencies?

If anyone kind enough to respond could avoid heavy maths and provide a more physical description / interpretation of the maths that would be much appreciated.

Answer 1

I'll try to answer all your questions at once. Remember that in the classical description the EM field couples to the electric dipole moment of the atom, and forces it to oscillate with the field. If the field has frequency $\omega$, the oscillation amplitude (and hence the scattered waves) will depend on the factor $(\omega - \omega_0)^{-1}$ raised to some power, where $\omega_0$ is a resonant frequency of the atom.

Now in the quantum description we analyze everything in terms of creation/annihilation of photons and jumps between energy states. In the case of Rayleigh scattering (which is a type of elastic scattering) the initial and final photon energies are equal to $\hbar \omega$, while initial and final atomic energies are also equal to, say, the ground state energy $E_0$. There's only a change in the photon's propagation direction. The atom still has a dipole moment, but it is now associated with a transition between atomic states of different energies, so whenever the atom interacts with the EM field it must 'jump' between states. So let me now describe what you do in the calculation:

We only know 2 things about the scattering, the initial state and the final state of the atom and radiation field. At the start, we have an incoming photon with energy $\hbar\omega$ and direction $\mathbf{k}$ and the atom in its ground state. Now, three things may happen:

1. the photon gets absorbed by the atom (atom annihilates photon), causing it to jump to a higher energy state called a 'virtual state' (it is 'virtual' because it is not required to conserve energy). Then the atom emits a photon (atom creates photon) with energy $\hbar \omega$ and direction $\mathbf{k}'$ and jumps back to its ground state;
2. the atom emits a photon (atom creates photon) with energy $\hbar\omega$ and direction $\mathbf{k}'$ while the initial photon is still incoming, and goes to a virtual state. The atom then absorbs the incoming photon (atom annihilates photon) and goes back to its ground state;
3. the atom creates the outgoing photon at the same time it absorbs the incoming photon (simultaneous creation and annihilation), remaining in its ground state.

Notice that in all cases the energy is always conserved at the start and at the end, but not in the middle necessarily. This is not a problem because we only have access to the initial and final states, we can't measure what happened inbetween. Now, the calculation treats the three processes on the same footing, so all of them contribute to the probability of finding a final photon with direction $\mathbf{k}'$ and energy $\hbar\omega$. The calculation to get the $\omega^4$ is a bit contrived and I couldn't find a way to explain it in simple terms, but again the interaction of the field with the atom will depend on the factor $(\omega - \omega_0)^{-1}$, and at the end of the calculation you will end up with a cross section that is a function of $\omega^4$, at least for $\omega \ll \omega_0$.

So, notice that since the atom only has to conserve energy at the start and at the end it can 'violate' energy conservation inbetween, and thus interact with photons of any frequency (but the interaction is enhanced whenever $\omega$ is close to a transition frequency $\omega_0$).

Answer 2

You are asking for a description of quantum scattering in the case of Rayleigh scattering, and why the sky is blue.

When a photon interacts with an atom, three things can happen:

1. elastic scattering (Rayleigh scattering, or for example mirror reflection), in this case the photon keeps its energy level and phase and changes angle

2. inelastic scattering (heat up material), in this case the photon gives part of its energy to the atom and changes angle

3. absorption, the photon gives all its energy to the electron/atom system, and the photon ceases to exist

Now Rayleigh scattering, the main cause of the blueness of the sky, is elastic scattering. The photon does not get absorbed, and does not cease to exist. The photon keeps its energy level and changes angle.

You are asking "What happens to the photon and electron during scattering? absorbed and re-emitted? just 'bounces' off etc?"

In the case of Rayleigh scattering, which is elastic scattering, it does not get absorbed, and does not cease to exist. You cannot imagine the photon as a little billiard ball, but in some sense, it just "bounces off" like you say.

You are asking "Why can scattering occur at all frequencies if the energy levels in an atom are discrete?"

Now in the case of absorption, you are correct, the descreteness energy levels (in the atoms/molecules) cause the quantum mechanical phenomenon where we see from experiments, that for a photon to be absorbed, the photon's energy needs to match (or exceed) the energy gap between certain energy levels of the atom/electron system. Now this is not the case for elastic scattering. Now in the case of elastic scattering, the probability of the scattering event is much more dependent upon a relation between the wavelength of the photon and the size of the atom/molecule.

The Rayleigh scattering depends on the wavelength and blue light is scattered most. That means the light we see coming from directions away from the Sun has a spectrum weighted towards the blue.

Why is the sky blue and the sun yellow?

You are asking "what is the quantum explanation for higher frequencies (blue light) being scattered much more (freq^4) than other frequencies?"

And we got to the beautiful QM phenomenon called scattering, where the relationship between the wavelength of the photon and the size of the atom/molecule matters most. In the case of Rayleigh scattering, the atoms/molecules size is much smaller then the wavelength of the photons.

is the predominantly elastic scattering of light or other electromagnetic radiation by particles much smaller than the wavelength of the radiation. The strong wavelength dependence of the scattering (~λ−4) means that shorter (blue) wavelengths are scattered more strongly than longer (red) wavelengths.

https://en.wikipedia.org/wiki/Rayleigh_scattering

Now since atoms/molecules' sizes are much smaller then the wavelength of the photons we are talking about (visible in your case), the smaller (shorter) wavelength photons scatter more then the longer, causing the blue color of the sky.

Answer 3

Why is the sky blue? The light from the sun is approximately black body radiation. It is in the form of packets of electromagnetic energy. The individual “wave packets” are produced by abrupt accelerations. These accelerations are caused by collisions of high velocity ions. The Fourier transform of these accelerations describe the shape of the wave packet. Fourier transform of the wave packet describes the frequency spectrum of the light. Qualitatively the spectrum will be broad and continuous because of the short time impulsive nature of the acceleration. A broad spectrum could be considered white in this context. Sunlight passing overhead through the atmosphere, composed of mainly nitrogen. The differential scattering of white light will occur. The effective scattering cross section of a nitrogen molecule N2, is much smaller than the mean wavelength of white light, resulting in significant scattering. The wavelength of the blue components of the white light scatters more than the red components (the scattering goes as the fourth power of the frequency). As the frequency goes up the wavelength goes down. The red sunset is simply white light that has lost more light due to scattering at the blue end of the spectrum (shorter wavelength) leaving the red end stronger or less scattered (longer wavelength). The wave packets of the blue light will have a different shape to the original white light. The red wave packet will have complementary shape. If we sum the red and blue packets we will have the original white packets. The wave packets are similar to Quantum wave functions except they can be split and divided. Photon wave functions represent probability functions and can only collapse to a point and are not divisible. A white photon wave function cannot be separated into a blue and a red wave function with different scattering angles.