# Four Level Laser

1. Aug 27, 2014

### chimay

I am not sure this is the right place where I should post, I hope I've chosen the right section.

I'm studying laser's fundamentals on Svelto's "Principles of Lasers". Here I find the description of a generic four level laser by means of rate equations; it can be represented like this: (first picture)

Then, I read that the CO2 laser, whose basic scheme is in the second picture, can be modeled as a four-levele laser too; I don't understand why this is true. In the CO2 laser the radiative transition is between level 3 and 2, while in the first photo it is between 2 and 1. Moreover, level n.3 in the first picture should be pratically empty, while in the CO2 laser it isn't so.

Can anyone explain me how I can model a CO2 laser like a four-level laser like the one in the first picture?

Thanks.

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• ###### Livelli energetici laser CO2.png
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2. Aug 27, 2014

### kspace

I am not a laser expert, but maybe this helps. The pictures are describing somewhat different processes. The CO2 laser is a four level system:

v0: ground state
v1: symmetric vibrational mode
v2: bending mode
v3: anti-symmetric vibrational mode

In the CO2 laser you have a lasing from the anti-symmetric (v3) vibrational mode to the symmetric vibrational (v1) mode with the release of energy as a photon. Such a transistion satisfies conservation laws. However, you can also have relaxations between the antisymmetric mode (v3) and the "bending mode" (v2), yet this is not an optically active transition. The v3-v2 transitions are due to collisions in the gas.

the first picture (left) also shows a four level system with lasing between levels (v2) and (v1). A system like this would have a molecule excited to (v3). The molecule then relaxes to (v2) and lases from (v2 to v1). Finally the molecule may relaxes from v1 to the ground state v0 via collisions.

3. Aug 28, 2014

### chimay

So your point is "there are 4 levels in both cases, they just differ from the radiative transition, 3->2 instead of 2->1", is it?

4. Aug 28, 2014

### kspace

Yes - I think that is what is going on