1. Nov 24, 2013

### hokhani

Why do we often treat the electromagnetic radiation effects on Hamiltonian of a matter as a perturbation? In the other words, why the effects of radiation is so little that is treated as a perturbation?

2. Nov 24, 2013

### Staff: Mentor

You can calculate the electric field strength in atoms and in electromagnetic waves (both with classical formulas) - in many setups, the waves are a very small contributions (<<0.1%).

3. Nov 24, 2013

### Meir Achuz

Each extra photon brings in a factor e^2, which equals 1/137 so "radiation is so little that is treated as a perturbation."

4. Nov 24, 2013

### hokhani

Ok, Thanks. But we have a great number of photons that seem to affect the system significantly. How about that? Am I wrong about that?

5. Nov 24, 2013

### Staff: Mentor

Did you calculate the values I suggested?

6. Nov 25, 2013

### hokhani

In a very simple model, we can take the electric field of an atom proportional to $q/r^2$ in which q is the charge of nucleus and r is distance from its center. An electromagnetic wave is $E=E_0 exp(-i\omega t)$.I think we can consider $E_0$ as large as possible; can't we? Could you please guide me If I am wrong?

7. Nov 26, 2013

### Staff: Mentor

E0 can be very large, but then you need really powerful, focused lasers, and the perturbative approach does not work any more.

The energy density of an electric field in vacuum is $\frac{\epsilon}{2} E^2$. Take a typical, weak laser with 1mW/mm^2, calculate the energy density, and calculate the average electric field based on that. The value is somewhere at a few V/cm.