1. Jan 4, 2015

### KFC

Hi all,
Recently I am reading an introduction on using laser to create the so-called optical lattices and a periodic potential to trap the atoms or as a grating, some applications like to act on the cold atoms. I don't have much background on laser but there are few concepts I don't understand. It is said that the laser is working on single mode and have a certain wavelength (780nm) and frequency $\omega_L$

1) I wonder when it means frequency $\omega_L$, does it mean the resonant frequency in the chamber during lasing? If so, what is the typical value for that and how big is it comparing to the atomic resonant frequency (I mean magnitude of order)? I did some search, it seems that the atomic resonant frequency is usually in GHz but most material only tell the wavelength of the laser but not the frequency

2) The article said the laser propagating along one direction and then reflected by a mirror so to form the standing wave, which plays the role of optical lattice. As I learn in other text, the standing wave has the form

$f(x,y,t) = A\cos\omega t\sin(k_x x + k_y y)$

I think $k_x$ and $k_y$ are the wave vector related to the wavelength of the laser, which used to form the spatial period of the lattice, right? So is the $\omega$ here stands for the laser frequency? That means the amplitude of the lattice is changing by time? If the laser frequency is so high, what's the net effect of the lattice? I mean will the lattice amplitude stay as an average value because of high frequency or what?

2. Jan 4, 2015

### Staff: Mentor

$$\omega_L = \frac{2\pi c}{\lambda}$$

The 780 nm you cited is the D2-line of rubidium. You will usually find the exact frequency/wavelength as a detuning with respect to an atomic transition (often expressed in terms of the natural linewidth of the corresponding level, Γ).

What is often important in optical lattices is the time average of the square of the electric field.

Last edited: Jan 5, 2015
3. Jan 4, 2015

### KFC

Do you mean $\frac{2\pi}{\lambda}c$?

That's what confusing me. In first question, we related the laser frequency to the wavelength but here wavelength is then related to the atomic transition. So does it mean the laser frequency related to the atom? But in first part, $\omega_L$ is given if wavelength is known. In some online materials, they give two frequency, $\omega_L$ and #\omega_0#, the later one refer to the atom transition, so I always thinking that $\omega_L$ is different from atomic transition.

So does it mean if some atoms was placed within the optical lattices like that above, the atoms will see the standing wave with average constant amplitude instead of the time-modulated one?

4. Jan 5, 2015

### Staff: Mentor

Yes. Thanks for spotting the error, I will edit my post.

You can express the energy difference between two levels in an atom in terms of the frequency or wavelength of the photons that will induce this transition. This is what is meant when saying that 780 nm corresponds to a transition in rubidium. This is also what $\omega_0$ represents.

The frequency $\omega_L$ of the laser is what it is, and the equation I gave is the simple conversion between angular frequency and wavelength. Now, considering that a laser has a wavelength of 780 nm, it means that it will be close to a transition in rubidium (and possibly some other atom). In other words, you have $\delta = \omega_L - \omega_0$, where $\delta$ is the detuning, with a value close to 0, probably of the order of MHz.

By the way, we have be careful when using Hz, which is only appropriate for $\nu$. $\omega = 2 \pi \nu$ has to be expressed in s-1.

Yes: time-averaged, thus constant in time, but spatially modulated. Note that since it is the square of the field that enters in the calculation of the Stark shift, the periodicity of the lattice will be twice the wavelength of the laser.

5. Jan 5, 2015

### KFC

Thanks a lot. I think your words clarify most of my questions and I have some sense what's going with the laser now though I still need some times to learn more on that.