# Understanding the “superconductivity” of composite fermions

A popular explanation for the 5/2 fractional quantum hall plateau is that it can be understood as a Cooper pairing state of some neutral composite fermions. We know there are two characteristic features of superconductors, one is the perfect electrical conductivity; the other is the perfect diamagnetism. Since the composite fermions are neutral and effectively live in zero magnetic field, which is at least true at filling fraction 5/2 or 1/2, there can not be any true superconductivity. I know the perfect diamagnetism is translated to incompressibility for composite fermions so it explains why the pairing state is a qualified FQHE state. The question is: what does the perfect conductivity in a true superconductor correspond to in the composite fermion case? It is certainly not a perfect electrical conduction as composite fermions are neutral.

• Some neutral superfluidity? – pathintegral Aug 13 '16 at 4:25
• @pathintegral:I think that is true. But I want to know physically in terms of observable quantity what that superfluidity corresponds to. – Zhiqiang Wang Aug 17 '16 at 20:52

## 1 Answer

While the composite fermions are electrically neutral, they are charged under an emergent internal U(1) gauge field - the gauge field '$a$' used to implement the flux attachment. Working through the mapping, the physical electron current $j^e$ is related to the internal gauge field as

$$j^\mu_{e} = \frac{1}{4 \pi} \epsilon^{\mu \nu \rho} \partial_\nu a_\rho + \left(\frac{1}{4 \pi \ell_B^2}, 0, 0\right)$$

In particular, $\delta n_e = \frac{1}{4 \pi} b$, which is why 'flux expulsion' in the superconductor corresponds to incompressibility, as you mentioned.

This brings us to your question - what is the signature of dissipationless current in a CF-superconductor? Suppose we set up a persistent current of composite Fermions in the CF-superconductor. According to the London equations, we have

$$j_{CF} = - \frac{ n_{CF} e^2}{m_{CF}} a$$

We then use the relation $j^{i}_e = -\frac{1}{4 \pi} \epsilon^{ij} \partial_t a_j$ to write $j_e = \frac{m_{CF}}{ 4 \pi n_{CF} e^2} \hat{z} \times \partial_t j_{CF}$. On the other hand, we know that $j_e = \partial_t P_e$, where $P_e$ is the polarization density of the electrons. We conclude that

$$P_e = \Lambda^2 \, \hat{z} \times j_{CF}$$

where $\Lambda$ is the London penetration depth. Thus, the existence of persistent currents in a superconductor corresponds to the possibility of 'persistent polarization density' in the Hall fluid.

The possibility of a persistent polarization density is peculiar to a Hall phase. In a regular band insulator, if you polarize the material by displacing all the electrons slightly, the system will relax back, heating up in the process.

• Then I think we can think about this persistent polarization density as the electron density polarization build up in the transverse direction of a Hall bar that gives the transverse Hall voltage, which is dissipation-less. – Zhiqiang Wang Sep 20 '16 at 0:19