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In condensed matter physics, what does the term incompressible in incompressible quantum liquid mean?

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I know nothing about incompressible quantum liquids, but I would guess that it means that somewhere you are imposing the condition $\nabla \cdot F = 0$ where $F$ is some vector field. Does something like that appear in the prescription of the system? – Stephen Mc Ateer Mar 16 '12 at 21:43
A little further reading (here) suggests that it is an electron gas with "isotropic transport" (whatever that means). – Stephen Mc Ateer Mar 16 '12 at 22:02

The electronic compressibility of a quantum fluid is inversely proportional to the variation of the chemical potential with respect to the number of particles

$\dfrac{1}{\kappa} \propto \left( \dfrac{d \mu}{d n} \right)$ [1]

i.e. the linear response of the chemical potential with respect to the electron density. For an incompressible fluid, $\kappa =0$, thus

$\dfrac{d n}{d \mu}=0$

which means that when you compresses or expands the system it is equivalent to inject/take out particles.

This concept is important in quantum Hall liquids, where the response to the system to compressive stresses is analogous to the response of type II superconductors to the application of an external magnetic field: the system first generates a non-dissipative Hall current without compressing and then, at a critical value of the stress it nucleates a quasiparticle.

EDIT: Derivation of [1]

Formula [1] can be obtained very easily: the compressibility $\kappa$ is the relative volume change against a pressure change for fixed particle number $N$, as wsc said

$\kappa=-\dfrac{1}{V}\left(\dfrac{\partial V}{\partial P}\right)_N$

Pressure is nothing but the change in energy against an unit volume hence

$\dfrac{1}{\kappa}=V\left(\dfrac{\partial^2 E}{\partial V^2}\right)_N$

In the thermodynamic limit, $E= (n \epsilon) V$ where $\epsilon$ is the energy per particle and $n$ is the particle density. Hence

$\dfrac{1}{\kappa}=n^2\left(\dfrac{d^2 (\epsilon n)}{\partial n^2}\right)$

On the other hand, the chemical potential is defined as

$\mu :=\left(\dfrac{\partial E}{\partial N}\right)_V =\dfrac{d(\epsilon n)}{dn}$

Therefore we get

$\dfrac{1}{\kappa}=n^2 \left(\dfrac{d \mu}{d n}\right)$

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"incompressible" in incompressible quantum liquid means "gapped". An incompressible quantum liquid is a gapped quantum system where all quesiparticle excitations have a finite energy gap (such as FQH liquid). See a discussion here

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Thank you. Could you please explain why gapped liquid is thermodynamically incompressible? – Blue Feb 11 '14 at 19:52

It means precisely what it sounds like: the macroscopic compressibilty is zero, or microscopically, you cannot take a particle out of (or put a particle into) the system without paying a nonzero energy cost.

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Could you elaborate $1)$ on how compressibilty is related to the change of particles and $2)$ I don't quite know how the compressibilty in thermodynamics relates to ${\bf{\nabla V}}=0$ – NikolajK Mar 16 '12 at 23:06
As DaniH mentions, the compressibilty (typically defined proportional to $\partial P/ \partial V$) can be related to the way the density changes as you tune the chemical potential. If you can increase the chemical potential without increasing the quasiparticle density, this implies that there is a gap that must be paid to inject a quasiparticle into the system. As for 2, I have no idea what V is supposed to represent, or why you would need a divergence-free condition on it. – wsc Mar 17 '12 at 15:42
Does that mean that a gapless dispersion is compressible, while one with a gap is incompressible? – leongz Mar 19 '12 at 19:15

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