# How can I tell whether a normal mode is excited or not?

I came across a question in which there were 3 ions of charge $q$ and $m$ bound by a harmonic well potential $\frac{1}{2}m\omega^2$. The natural frequency suddenly changes to $\omega+\delta\omega$, and I am supposed to find the normal mode which is excited.

So far, I have calculated 3 normal modes: (-1, 0, 1), (1,1,1) and (1,-2,1).

I understand that I should be substituting initial conditions, but is it actually possible to intuitively figure out whether a normal mode is excited or not? Thanks!

## 1 Answer

My first instinct as a physicist is always to ask whether symmetry impacts you at all. From my understanding, you had three ions spaced in a quadratic potential (and, I assume, interacting with each other). It seems clear to me that you must have started in a configuration like so:

| o --- o --- o |

That is, the ions would have been arranged symmetrically with respect to mirroring about the central ion. This is because nothing in the problem breaks the difference between "left" and "right" so we should expect the ions to be in an arrangement where left and right are identical.

Now, you ask what happens to these ions after a shift in the natural frequency (which means the quadratic potential becomes steeper). This perturbation is also symmetric, since it will do the same thing for the left ion and the right ion.

So, what we're looking for is a normal mode with some kind of symmetry to it. First, consider $(1,1,1)$. This is the center-of-mass mode: all ions move in the same direction. But which direction would they choose? Since there's no way for the system to tell between left or right, I would say this would not be excited.

Next, $(1,-2,1)$. Again, this normal mode distinguishes between right and left (will the center ion move towards the left or towards the right ion?). Not excited, would be my guess.

However, $(-1,0,1)$ seems plausible. In this mode, the two ions on the outside move outwards and inwards while the central ion stays constant. This also makes sense as a physical picture: when the well gets suddenly deeper, outer ions are moved inward until the central ion repels them. My instinct is that this is the mode excited.

This isn't rigorous -- in particular, spontaneous symmetry breaking should often be considered possible. But you asked for intuition, and this is how I was able to come to an answer quickly.

• No, this is exactly correct. There's no need to guess - all your intuitions can be backed up by a formal analysis of the symmetries. There's can be symmetry breaking in such situations (in particular, when you squeeze the chain so hard that the zigzag mode becomes the ground state, in which case you need to choose between two or more degenerate ground states) but that seems way overblown compared to the OP's statement of the problem. – Emilio Pisanty Aug 16 '16 at 15:36
• @EmilioPisanty Yeah, I would have been very very surprised if I was wrong here, the ground state being the only thing I was vaguely worried about. But it seems like the OP was given a pretty "standard" initial condition. – zeldredge Aug 16 '16 at 15:39
• Yeah, the ground state can be a problem, but probably not for the question as posed. That said, there's lots of cool physics on the linear-to-zigzag transition, because it allows for very controlled phase transitions in a very quantum regime; for further reading a good starting point is this paper. – Emilio Pisanty Aug 16 '16 at 16:09
• @EmilioPisanty Hello, thank you for the comment :) could you elaborate further on the formal analysis? Thank you! – user107224 Aug 17 '16 at 0:25
• Thanks so much for the answer zeldredge, it makes so much more sense to me now :) – user107224 Aug 17 '16 at 0:26