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Gap between different Landau level let adding an electron to Integer quantum hall fluid cost a lot. So "incompressible".

But it seems that taking one electron out QHF costs nothing. So it seems that QHF's density can be lowered without costing a lot.

So I am quite confused about "incompressible property" since classically incompressible means density is almost fixed, neither increasing or decreasing.

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  • $\begingroup$ The removed electron should be placed in a higher Landau level which requires energy at least equal to the difference between the lowest and first excited Landau levels $\Delta E = \frac{\hbar e B}{mc}$. In a strong magnetic fields, this energy is very large. $\endgroup$ – David Bar Moshe Nov 21 '17 at 11:07
  • $\begingroup$ But if removed electron is put into higher Landau level, electron number is not changed and density does not change at all. If we remove electron out of system, density decrease but it does not cost a lot. $\endgroup$ – Ke Wang Nov 21 '17 at 17:39
  • $\begingroup$ Indeed, when you excite a single electron to a higher Landau level, the change of the density is very small, while the required energy is very large. This is the indication to incompressibility. As for, removing an electron from the system: The theory is valid for a closed system, thus this case is not covered in the mathematical model. $\endgroup$ – David Bar Moshe Nov 22 '17 at 10:01
  • $\begingroup$ Cont. , However, even if we consider a hypothetical situation where we can scatter one electron out of the system, while the others remain in a quantum Hall state, the density remains almost constant (as there are a lot of electrons in the system), while the energy required to kick the electron out is large. $\endgroup$ – David Bar Moshe Nov 22 '17 at 10:02
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An explanation comes from the fact that when you are "compressing" something, the force(s) is (are) directed towards the center of the "thing", therefore the force is working against the (very strong) nucleus forces. In "removing" a part from the thing, the force is working against the (very weak) "surrounding environment" forces. This shows that "incompressiblility" is directional!

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The inverse of the compressibility is defined by $\kappa^{-1} = (\frac{\partial P}{\partial V})_N = (\frac{\partial^2 U}{\partial V^2})_N$. The behavior you noticed amounts to saying that the first derivative of $U$ with respect to $V$ changes discontinuously. Which means that the second derivative is infinite, which is precisely the statement that $\kappa=0$, i.e. incompressible.

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