What does "incompressible" mean in the context of an incompressible quantum liquid?
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14
"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
Xiao-Gang Wen
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The electronic compressibility of a quantum fluid is inversely proportional to the variation of the chemical potential with respect to the number of particles
$$ \dfrac{1}{\kappa} ~ \propto ~ \left( \dfrac{\mathrm{d} \mu}{\mathrm{d} n} \right) \,, \tag{1} $$
i.e. the linear response of the chemical potential with respect to the electron density. For an incompressible fluid, $\kappa =0$, thus
$$ \dfrac{\mathrm{d} n}{\mathrm{d} \mu} ~=~0 \,, $$
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 $\operatorname{Eq}{\left(1\right)}$
$\operatorname{Eq}{\left(1\right)}$ 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, $E= \left(n \epsilon \right) V$, where $\epsilon$ is the energy per particle and $n$ is the particle density. Hence
$$ \dfrac{1}{\kappa} ~=~n^2\left(\dfrac{\mathrm{d}^2 (\epsilon n)}{\partial n^2}\right) \,. $$
On the other hand, the chemical potential is defined as$$ \mu ~:=~\left(\dfrac{\partial E}{\partial N}\right)_V ~=~\dfrac{\mathrm{d}(\epsilon n)}{\mathrm{d}n} \,,$$ therefore we get$$ \dfrac{1}{\kappa}~=~n^2 \left(\dfrac{\mathrm{d} \mu}{\mathrm{d} n}\right) \,. $$
2
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.
wsc
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