# Why do rotons correspond to the minimum of the dispersion curve?

The dispersion curve for superfluid helium-4 is given above. To my knowledge, the first paper that was able to argue that the curve should take this shape from first principles was Feynman's 1954 paper (although I could be wrong).

This curve gives the energy for quasi-particle excitations of given momenta. The smallest energy excitations are phonons, i.e. sound waves. This is understood.

The paper argues that the energy of an excitation is

$$E = \frac{\hbar k^2}{2 m S(k)}$$

(equation 18). $S(k)$ is defined to be the fourier transform of $p(r)$

$$S(k) = \int p(\vec r) \exp(i \vec k \cdot \vec r) d^3 r$$

where $p(\vec r_1 - \vec r_2)$ is the probability per unit volume of finding a helium atom at position $\vec r_2$ given that an atom is present at $\vec r_1$. Therefore, the shape of the dispersion curve is an artifact of the distribution of helium atoms in the superfluid. (Please correct me if anything I said was incorrect.)

Having said that, I don't understand the relation of the minimum of E(k) to "rotons."

We are now pretty confident that rotons are in fact tiny vortex rings in the superfluid. The vorticity of the ring is quantized by the fact that the phase of a wave function is equivalent up to multiples of $2\pi$.

My main question is: What exactly does the minimum of this dispersion curve have to do with tiny vortex rings? Does it have anything to do with the group velocity of the excitation being 0?

Side question: Is there any microscopic picture of what a "maxon" is?