I am now reading a book "Superconductivity, Superfluids, and Condensates" by Jame F. Annett. I am stuck at the part of writing dispersion relation of the quasiparticle in superfluid here. The author does not seem to give any explanation about this. He somehow mentioned the Fermi's golden rule previously, but I could not see how we can use it to prove this dispersion relation.

Here I attached the picture of that page. The equation that I could not follow is (2.83)

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1 Answer 1


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According to the dispersion relation above, which was obtained by neutron scattering, we can see firstly see that the quasiparticle has the linear dispersion relation at low energy.

Imagine now we drap an object of mass $M$ in the fluid (that object can be a defect at the tube wall), the momentum of the object is initially $P$. It can emit the phonon (quasiparticle - zero sound mode) with the energy $\hbar \omega_k$ and momentum $\hbar\vec{k}$.

Write down conservation of energy, we obtain

$$ \frac{\vec{P}^2}{2M}=\frac{(\vec{P}-\hbar\vec{k})^2}{2M} + \hbar \omega_k $$


$$ \omega_k = \vec{V}\cdot\vec{k}-\frac{\hbar k^2}{2M} $$

In the large $M$ limit, we finally arrived at

$$ \omega_k = \vec{V}\cdot\vec{k} $$


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