# How does a photon know to pass through without interacting?

1. Jun 20, 2013

### tim2030

How does a photon "know" to pass through without interacting?

The usual explanation given for transparency is that when the energy of a photon is smaller than the band gap energy of an atom, the photons don't interact with the electrons and pass through, so the material is transparent.

But what stops the photon from interacting with the electron? How does it "know" that the band gap energy is too large?

Is there a model of this 'non interaction', or 'passing through' somehow - perhaps as some sort of a dynamical process, using a quantum mechanical model of the electron and the changing potential caused by the passing photon's EM wave? Or is QED needed to understand this? Whatever the theory, what is a qualitative way to describe the absence of the interaction?

2. Jun 21, 2013

### Staff: Mentor

I can't answer with a complicated, in depth example, but lets say you have a switch that requires 10 lbs of force to flip it. If you only have 5 pounds of force the switch will never flip. How does the switch, or the object applying the force, know that the 5 pounds of force isn't enough? Well, it doesn't. It simply isn't enough.

Let me flip the question around on you. If a photon passes through an atom and doesn't have enough energy to excite any electrons into another state, why would it interact at all? How could it possibly interact if there is nowhere for the energy to go?

3. Jun 21, 2013

### A.T.

The "interacting" part is tricky. When something is not altered, it might have not interacted at all, or interacted in a non-dissipative way where all the effects of the interaction cancel. But in a coarse model, which doesn't go into the details of the interaction, the two cases are equivalent.

Last edited: Jun 21, 2013
4. Jun 21, 2013

### SW VandeCarr

Afaik photons don't "know" anything. The quantum energies of the incident photons must match the available energy level gaps to be absorbed. Let $E_1<E_2$ be the available energy levels for an incident photon with energy $h\nu$ (in electron volts).

Then $\Delta E = h\nu = E_2 - E_1$ (where nu is frequency) or the photon will not be absorbed. Moreover, for transparency, the overall structure of the material must be such that photon scattering is minimized. The details of that are not something I can describe right now.

Last edited: Jun 21, 2013
5. Jun 21, 2013

### Claude Bile

It is a QM thing.

When a photon of some frequency interacts with an atom (say), it polarises the atom; however the polarisation is not static as in a capacitor, it too oscillates with a certain frequency. Now, if you dig through the theory, you find that the probability of knocking the atom into a higher energy state depends on the frequency of this oscillation in a way that is very much analogous with a damped harmonic oscillator.

The transition from the valence to the conduction band can be thought of as a very broad transition (in frequency).

Does that explanation offer any insight?

Claude.

6. Jun 21, 2013

### tim2030

I guess the answer would depend on how you model a photon. If you think of it as an EM (plane) wave passing through space, then upon reaching the electron, it should cause it to experience an (oscillatory) force and thus cause it to accelerate. That's an interaction, but I'm not sure how to get from that to the final outcome where the electron is back to the same state it was before it encountered this EM wave, in the case where the wave's energy is less than the band gap.

But still, how do this explain transparency (i.e. transmission)? If the photon polarizes the atom in some oscillatory fashion, how does it come about that if the energy of the photon (hv) is less than the band gap of any of the electron orbits, the atom returns to its original state pre-interaction, and the photon continues on its merry way as if nothing had happened? For even in this case of not enough energy, it seems like there should some interaction between the photon and the atom/electron, so it's not like the photon can be said to just "pass through".

7. Jun 21, 2013

### dipole

What you've described is the qualitative answer - I don't think anyone here will be able to provide you with a better explanation than what you've already described. There may in fact be no answer to your question, because words like "interact" have a fuzzy meaning in these kinds of situations.

Quantum Mechanically, what you would calculate is the transition probability for an electron to be excited to some final state from the initial state, and you find eventually that this transitional probability depends on the difference in energy of the states as described.

If you treat the photon as a perturbation, and apply fermi's golden rule, you'll see that the transition rate goes like a delta-function, $\delta(\Delta E - \hbar\omega)$ which is clearly zero unless the energy of the photon exactly matches the different in energy between the initial and final states.

If you want a more intuitive picture than this, then you'll probably need a deeper understanding of quantum mechanics than, I suspect, anyone on this forum has. I'd be delighted if I were wrong though.