# What is the frequency of the harmonic potential?

1. Jan 10, 2015

### KFC

Hi there,
I am reading an introduction on trapping atoms in space with magnetic potential. The article said the lab usually use a harmonic potential to trap the atoms and the potentials is in the form

$\dfrac{m}{2}(\omega_x^2x^2 + \omega_y^2y^2 + \omega_z^2z^2)$

and $\omega_{x,y,z}$ has the unit of frequency. I wonder how do you understand the frequency in the potential from physical point of view. Why there is frequency?

2. Jan 10, 2015

### ShayanJ

Let's see what force that potential gives. For the x component, we have $F_x=-\frac{\partial V}{\partial x}=-m \omega_x^2 x$. But hey, that's Hooke's law!(With the spring constant $k_x=m \omega_x^2$.) So this potential is actually a non-isotropic harmonic oscillator potential and this is the reason you have frequencies in it.

3. Jan 10, 2015

### KFC

Thanks. So if can I say under that potential, it just like 3 harmonic oscillators along x, y and z each is oscillating at the frequency $\omega_x$, $\omega_y$ and $\omega_z$ independently?

4. Jan 10, 2015

### ShayanJ

No, its just one harmonic oscillator having three independent oscillations with different frequencies in different dimensions. But yes, in terms of degrees of freedom, its no different than having three independent harmonic oscillators with different frequencies. But you should note this resemblance may not be usable in the context you're considering.

5. Jan 10, 2015

### KFC

Got it.

One more question. Usually, if you solve the harmonic oscillator with 3 oscillating frequencies along 3 different frequency, we will get a solution in 3-dimensional also. But if the frequency along x and y are way larger than the $\omega_z$. In some articles, I saw that people simply approximate the solution along the z direction only. I stuck on the explaining this approximation in physics.

The first thing come to my mind is if the oscillator oscillating along x and y much faster than z, can we consider the system may see the average motion along x and y instead because of high frequency? So we could consider the amplitude of the solution along x and y just like a constant? Only the z direction depends on time?

But before I find the explanation, I am also thing that if $\omega_z$ is way smaller than the
$\omega_{x,t}$, can we consider the profile on the z direction is changing slowly in time, so we could consider the solution in z direction is a constant, the effective solution is along x and y direction.

I know those two statements are contradictory. But I cannot tell which one (or all) is wrong. and why?

6. Jan 10, 2015

### ShayanJ

Can you point me to one of those "articles"?