**Theory of the half-filled Landau level\
N. Read\
*Departments of Physics and Applied Physics, P.O. Box 208284,\
Yale University, New Haven, CT 06520\
***

A recent theory of a compressible Fermi-liquid like state at Landau level filling factors $\nu=1/q$ or $1-1/q$, $q$ even, is reviewed, with emphasis on the basic physical concepts. 

    PACS Nos.  73.40 Hm, 03.70+k
    February, 1994

The physics of interacting electrons in 2-dimensions in very high magnetic fields has proved to be a rich subject since the discovery of the FQHE in 1982 [@tsui].  The initial motivation for study in this area was the expectation that a Wigner crystal should form when the (areal) density of electrons is low and the magnetic field is high, not only because, as at low magnetic field, the Coulomb energy of repulsion dominates the kinetic energy at low density, but also because if the magnetic field is high enough so that mixing of excitations to higher Landau levels can be neglected then the kinetic energy itself becomes essentially constant.  In this limit the electrons would behave classically, with their dynamics given by the ${\bf E}\times{\bf B}$ drift of their guiding centre coordinates, [*i.e.*]{} drift along the equipotentials, the potential being here given by the Coulomb potential of the other electrons.  The Wigner crystal is presumably the unique lowest energy, stationary state at a given density in this classical problem.  The surprise was that in fact, at accessible densities, the quantum fluctuations cause this crystal to melt (as parameters are varied) into some quantum fluid, and furthermore the nature of this fluid depends on the commensuration between the density and the magnetic field, in an essentially quantum mechanical way.  In terms of the Landau level filling factor $\nu\equiv n\Phi_0/B$ (where $n$ is the density, $B$ is the magnetic field strength and $\Phi_0=hc/e$ is the flux quantum), this was manifested by quantum Hall effect (QHE) plateaus at $\nu=p/q$ a rational number with an odd denominator $q$ ($p$ and $q$ have no common factors) in all except a handful of special cases observed more recently. 

A theory that explained many features of this picture, including the “odd denominator rule” was quickly forthcoming [@laugh; @haldhier; @halphier].  Laughlin’s states at $\nu=1/q$, $q$ odd, and the hierarchical extension to all $\nu=p/q$, $q$ odd, are incompressible fluids capable of exhibiting a QHE plateau at $\sigma_{xy}=(p/q)e^2/h$. They also possess novel elementary excitations called quasiparticles, which carry a fraction $\pm 1/q$ of the charge on an electron, and have fractional statistics, that is the phase of the wavefunction changes by $e^{i\theta}$, $\theta/\pi=\pm p'/q$ when two quasiparticles are exchanged adiabatically [@halphier; @arovas; @readhier; @blokwen].  A finite energy $\Delta$ is required to create one of these excitations.  The fractional QHE associated with a particular state will be destroyed if the energy scale characterizing the rms disorder exceeds about $\Delta$.

Some questions remained unanswered, however.  While the hierarchy theory predicted that states at $\nu$ with larger denominators would have smaller gaps, it would seem, at least naively, that all fractions with the same denominator would be equally strong, though it is possible that those near a stronger, smaller denominator state might be overwhelmed, for example $3/11$ near $1/3$. As samples improved, it could be seen that in fact most fractions observed were of the restricted form $\nu=p/(mp+1)$, $p\neq0$, $m\geq0$ even, $=2$, $4$. It was not clear if the hierarchy could explain this, although it could be simply a quantitative fact, without deeper explanation.  Furthermore, in the region near $\nu=1/2$, where $m=2$, $p$ is large in the above formula, no fractional plateaus were seen, but there was a shallow minimum in $\rho_{xx}$ that remained $\neq0$ as $T\rightarrow0$. A possibly more profound theoretical question was, what is the nature of the ground state at fillings not of the hierarchy form $p/q$, $q$ odd?  Indeed, the construction of states at $\nu=p/q$, $q$ odd that exhibit FQHE does not prove that FQHE cannot occur at even $q$. The whole question of what [*does*]{} occur received almost no attention in published work until the last few years (a notable exception is the FQHE at $\nu=5/2$ and the Haldane-Rezayi proposal for its explanation [@will1; @haldrez]).  One possibility is that there is no well defined state at these fillings in the thermodynamic limit, that phase separation into domains of two nearby, stable FQHE states occurs instead.  Another possibility is a well-defined pure phase constructed by taking a FQHE state at a nearby filling, and adding a low density of, say, quasiholes which at low density could form a Wigner crystal.  Such a state could exist (with varying lattice constant) over a range of densities.  These possibilities may actually occur near some particularly stable (large $\Delta$) fractions such as $1/3$, but seem less likely near $\nu=1/2$ since neighbouring states would have very small energy gaps.  If liquid states at non-FQHE filling factors exist, then they lay outside the then-current theoretical understanding. 

With these motivating remarks, I now abandon the historical approach in order to develop the overall theoretical picture [@HLR; @jain; @rezread] as logically as possible.  I will describe two alternative approaches.  The first begins close to Laughlin’s original ideas, and contains explicit trial wavefunctions, as well as other notions of the last few years.  The other approach is field theoretical and eventually arrives at the same conclusions, but may seem less physical at first.  However, it is much more appealing for explicit analytical calculation.  I will also explain the relation with Jain’s work on the hierarchy states. 

Let us turn, then, to the first of these approaches.  We make the usual assumption that, at $T=0$, interaction energies $\sim \nu^{1/2}e^2/
\epsilon\ell$ are weak compared with the Landau level splitting $\hbar\omega_c$, and so the electrons should all be in the lowest Landau level (LLL) with spins polarized when $\nu<1$. We work in the plane with complex coordinate $z_j=x_j+iy_j$ for the $j$th electron.  In the symmetric gauge [@laugh], single particle wavefunctions in the LLL are $z^m \exp\{-\frac{1}{4}|z|^2\}$, where we set $\hbar=1$, and the magnetic length $\ell=\sqrt{\hbar c/eB}=1$. $m$ is the angular momentum of the state.  An $N$ particle wavefunction has the form $$%% FOLLOWING LINE CANNOT BE BROKEN BEFORE 80 CHAR
\Psi(z_1,\ldots,z_N,\bar{z}_1,\ldots,\bar{z}_N)=f(z_1,\ldots,z_N)e^{-\frac{1}{4}
\sum_i|z_i|^2}
\label{LLLstates}$$ where $f$ is complex analytic and totally antisymmetric.  If $f$ is homogeneous of degree $M$ in each $z_i$, it has definite total angular momentum and describes a (not necessarily uniform) droplet of radius $\sim\sqrt{2M}$. As a function of each $z_i$, $f$ has $M$ zeroes which is also the number of flux quanta $N_\phi$ enclosed by the circle of radius $\sqrt{2M}$. (Similarly, on a closed surface such as the sphere, the number of zeroes of the LLL single particle wavefunctions equals the number of flux quanta through the surface.)  Thus the filling factor $\nu=N/N_\phi$ $=$ the number of particles per flux quantum. 

Now if $f$ contains a factor $$U(z)=\prod_i(z_i-z)
\label{U(z)}$$ then [*one*]{} zero for each particle is located at $z$. ($U(z)$ can be viewed as an operator, Laughlin’s quasihole operator, acting on the remainder of the wavefunction by multiplication to produce $f$.) This means that there are no particles in the immediate vicinity of $z$, there is a depletion of charge there.  We describe this as a [*vortex*]{} at $z$, since as in a vortex in a superfluid, the phase of the wavefunction winds by $2\pi$ as any $z_i$ makes a circuit around $z$.

Given a state, we can obtain a valid new state by multiplying by $U(z)$. (This increases $M$ by $1$, if $z=0$, but otherwise by an indefinite amount.  But it always increases $N_\phi$ by $1$, if $N_\phi$ is defined as the flux through the region occupied by the fluid.)  For reasonably uniform fluid states (such as Laughlin’s, or the Fermi liquid below) the vortex can be considered as an excitation of the fluid.  Now add in addition an electron.  Clearly it is attracted to the centre of the vortex, due to the density deficiency there.  Since there is no kinetic energy, it can certainly form a bound state.  Similarly it can bind to multiple vortices $U(z)^q$.

It is natural to consider the possibility that the ground state itself contains electrons bound to vortices, since this will give a low energy.  As each vortex is added, $N_\phi$ increases by $1$. If we add $1$ electron for each $q$ vortices, we can form identical bound states, and if all zeroes are introduced in this way, we will have $N_\phi=q(N-1)$, which implies $\nu=N/N_\phi
\rightarrow1/q$ as $N\rightarrow\infty$. This case is by far the simplest to understand.  In this state, there will be $q$ -fold zeroes as one electron approaches another.  It has long been realized that Laughlin’s Jastrow-like ansatz $$f_{\rm L}=\prod_{i<j}(z_i-z_j)^q$$ can be called a “binding of zeroes to particles”. 

However, the astute reader will already be aware that $f_{\rm L}$ is antisymmetric only for $q$ odd.  To solve this problem, and obtain a deeper insight into Laughlin’s state, we must examine the nature of the bound objects more closely.  Since we know the properties of an electron, we turn to the vortices.  First, the $q$-fold vortex carries a charge $-\nu q$ in a fluid state at arbitrary filling factor $\nu$ (this includes our compressible Fermi liquid state below, as well as the usual hierarchy states).  This can be established by Laughlin’s plasma analogy [@laugh] or indirectly through adiabatic motion of the vortex [@arovas]; the latter will be more useful here.  A crude version of this argument is that when a vortex is moved around adiabatically in some given fluid state, the wavefunction picks up a phase, since the definition (\[U(z)\]) of $U(z)$ shows that it changes phase by $2\pi$ for each particle about which it makes a circuit.  Then if it makes a circuit around a (nonselfintersecting) closed loop enclosing a region $D$ of area $A$, it will pick up a phase $2\pi\int_D d^2z\,\rho(z)$ which reduces to $\nu A$ if the density $\rho(z)$ is uniform $\rho=\nu/2\pi$. This is then identified as the same result as would be obtained for a particle of charge $-\nu$ in the magnetic field seen by the electrons.  Similarly, a $q$-fold vortex has a charge deficiency of $\nu q$, which for $\nu=1/q$ is equivalent to a real hole, so the electron–$q$-vortex composite at this filling has net charge zero, and behaves like a particle in [*zero*]{} magnetic field.  But note that the vortex is actually sensitive to the density of electrons, which can vary in space and time, even when the external magnetic field and the average filling factor are fixed. 

Now for the famous fractional statistics of vortices.  If a vortex makes a circuit around a loop enclosing another vortex, with the density otherwise uniform, the missing charge around the vortex core will make a difference of $2\pi \nu$ to the phase (independent of the size and shape of the loop, as long as the vortices remain far enough apart).  But a circuit is equivalent to two exchanges, up to translations.  So a similar calculation gives that adiabatic exchange of two vortices produces a phase $\theta=\pi\nu$. For two $q$-fold vortices, the result is $\pi q^2\nu$ which at $\nu=1/q$ is $\pi q$. This shows that at these fillings, $q$-fold vortices are fermions for $q$ odd, and bosons for $q$ even.  Hence the electron–$q$-vortex bound state is a boson for $q$ odd, and a fermion for $q$ even.  For $q$ odd, we can now argue that the bosons in zero magnetic field can Bose condense (at $T=0$) into the zero-momentum state and that this is the interpretation of the Laughlin state [@read].  If $\psi^\dagger$ creates an electron in the LLL, then $\psi^\dagger(z)U^q(z)$ creates a boson, and the condensate is obtained by letting $\int d^2z\, \psi^\dagger(z)U^q(z)
\exp\{-\frac{1}{4}|z|^2\}$ act repeatedly on the vacuum.  This produces exactly the Laughlin state [@read].  Note that condensation arises because particles try to minimise their kinetic energy $\sim k^2$. We need to show that the bound objects do in fact have such an [*effective*]{} “kinetic” energy.  The true kinetic energy of the electrons has been quenched, so this term in the effective Hamiltonian for the bound objects can only arise from electron-electron interactions.  Before discussing how this arises, let us pursue the consequences.  At $\nu=1/q$, $q$ even, the bound objects are fermions in zero net field, which cannot Bose condense, but can form a Fermi sea.  This is then the proposal for a compressible state at these fillings.  It will be compressible because it can be shown that incompressibility, and the quantized Hall effect, in the case of bosonic bound states is a consequence of Bose condensation (superfluidity), which does not occur in a Fermi sea unless BCS pairing occurs, in which case the state becomes a QHE state.  The Haldane-Rezayi and Pfaffian states arise in this way [@mooreread]. 

I now return to the dynamics of the bound objects, or “quasiparticles”.  We have argued that there is an attraction between an electron and a $q$-fold vortex.  In a uniform background, this will take the form of a central potential.  The minimum is at the centre, meaning the electron is exactly at the zeroes of the many-particle wavefunction.  Now at filling factor $1/q$, we need to consider quasiparticles in plane wave states with wavevector $\bf k$. These can be created by acting on a fluid with $\int d^2z\, e^{i{\bf k}\cdot{\bf r}}
\psi^\dagger(z)U^q(z)\exp\{-\frac{1}{4}|z|^2\}$. In the wavefunction, this means a factor $e^{i{\bf k.r}_j}$ in the term where the $j$th electron is the bound in the state with wavevector $\bf k$. Now it turns out that this factor, acting by multiplication on some given state, displaces the $j$th particle by $ik$, where $k=k_x+ik_y$ (recall that the magnetic length $\ell$ is 1).  This arises because in the Hilbert space of many particle states of the form (\[LLLstates\]), $\bar{z}_j$ acts on $f$ as $2\partial/\partial z_j$ which generates displacements [@girvinjach].  A quasiparticle with wavevector zero would have the electron exactly at the zeroes of the wavefunction.  So a quasiparticle with wavevector $\bf k$ has the electron displaced by $|k|$ from the centre of the vortex.  The electron and vortex experience a potential $V(|k|)$ due to the Coulomb repulsion of the electron by the other electrons, which are excluded from the vortex core.  A good understanding is now achieved semiclassically.  The electron will drift along an equipotential of $V(|k|)$. From the preceding discussion, at $\nu=1/q$ the $q$-fold vortex experiences a magnetic field of the same strength as the electron, and so it will also drift with the same speed but in the opposite sense relative to the gradient of the potential, due to the opposite sign of its effective charge.  This means that both components of the pair drift in the same direction, perpendicular to the vector connecting their centres, so that their separation ($=|k|$) remains constant.  The picture is like that of oppositely charged particles in a magnetic field, which can drift in a straight line as a pair.  The energy of our pair is $V(|k|)$ and the velocity is $\propto \partial
V/\partial |k|$ as it should be.  Near the bottom of the potential, it will be quadratic, and we can obtain an effective mass $\sim (\partial^2
V/\partial |k|^2)^{-1}$ due to the interactions (a similar calculation was performed in [@read]).  This shows clearly that the interactions favour condensation of the quasiparticles to minimize this effective kinetic energy.  In the $q$ even case, the quasiparticles are fermions and must have distinct $\bf k$’s, filling a Fermi sea.  Then in the ground state not all the zeroes of the wavefunction are precisely on the electrons but some are displaced by amounts up to $k_F$, determined in the usual way by the density, $k_F^2/4\pi=1/2\pi q$. For $q$ large these displacements $\sim q^{-1/2}$ are small compared with the interparticle spacing $\sim q^{1/2}$. Thus a trial state of this form may not be much worse energetically than the Laughlin state at nearby odd $q$. This, I believe, is then the essential reason why this idea has a good chance of being the correct many-body ground state: it is due to the good correlations that produce a low Coulomb energy.  Of course, if $q$ is very small, this picture might break down, but in fact we have good reason to believe it holds for $q$ as small as $2$. This argument also tells us that low-lying excited states are obtained by increasing the wavevector of a quasiparticle in the Fermi sea.  For $q$ large, the Fermi velocity will be determined by the same effective mass as near the bottom of the sea; otherwise we must take the derivative at $k_F$.

To consider collective effects it is necessary to go beyond an independent quasiparticle picture.  There are important long range interactions that can be described as gauge fields.  We will examine these in the context of the field theoretic approach later.  As a test of the above ideas, we can perform numerical diagonalization of small systems at, say, $\nu=1/2$. This has been done recently in a paper by E. Rezayi and the author [@rezread].  Excellent agreement of trial states suggested by the above ideas is found with the exact wavefunctions of low-lying states.  The trial wavefunctions were generated on the sphere, but on the plane would be roughly $$\Psi={\cal P}_{\rm LLL}\det M\,\prod_{i<j}(z_i-z_j)^q
e^{-\frac{1}{4}\sum|z_i|^2}. 
\label{trialstate}$$ The matrix $M$ has elements that are essentially plane waves $M_{ij}\sim
e^{i{\bf k}_i\cdot{\bf r}_j}$ for the quasiparticles, filling the Fermi sea.  ${\cal P}_{\rm LLL}$ projects all electrons to the LLL.  As explained above, the plane wave factors then act as operators within the LLL, on the Jastrow factor $\prod(z_i-z_j)^q$ which if not modified would be the Laughin state for bosons at $\nu=1/q$. The Slater determinant makes the state totally antisymmetric.  The simple product form is similar to Jain’s states off half-filling [@jain]; it differs a little, but inessentially, from the present author’s first idea (in 1987) of building the Fermi sea by acting with Fourier components of $\psi^\dagger(z)
U^q(z)\exp\{-\frac{1}{4}|z|^2\}$, which was suggested by the analogy with the Laughlin states at $q$ odd [@read].  The numerical calculations also showed that the two-particle correlation function $g(r)$ in the ground state possesses ([*i*]{}) a large “correlation hole” at short distances $r$, consistent with the argument that zeroes are bound close to the electrons, and ([*ii*]{}) oscillations at large $r$, perhaps of the form $r^{-\alpha}\sin 2k_Fr$ asymptotically, as in a two-dimensional Fermi gas, which has $\alpha=3$.

We now turn to the field theoretic approach.  It begins with the observation that in 2 dimensions, particles of any (fractional) statistics can be represented by charged particles of other statistics attached to $\delta$-function flux tubes of a certain size [@wilczek].  A charged particle dragged adiabatically around a flux tube (or [*vice versa*]{}) picks up an Aharonov-Bohm phase of $2\pi$ times the product of charge of the particle and the number of flux quanta in the tube.  Thus if identical particles are attached to identical flux tubes, a circuit of one composite around the other produces a phase $4\pi$ times the charge times the flux, since each particle sees the other flux.  For an exchange, we get only half this.  In addition, the exchange of identical particles produces a phase $e^{i\theta}$ due to the statistics of the particles themselves.  Thus fermions or bosons attached to fractional flux tubes can be used to model anyons, where the total phase obtained in an exchange is fractional [@wilczek].  In the fractional quantum Hall effect we change our terminology slightly because we view flux tubes as operators that act only on particles already present.  Thus a composite is introduced by adding first a flux tube, then a particle is added at the same point.  The phase produced by an exchange is then only $\pi$ times the charge times the flux in each composite, plus the phase due to exchanging the particles themselves.  Thus, for example, we can say that attaching two flux quanta to each boson in a system leaves the composites still as bosons, or doing the same to fermions leaves them fermions.  Then in the high field situation of interest here, we may represent electrons as $q$ (integer) flux tubes attached to some other particles.  The latter must be bosons if $q$ is odd, and fermions if $q$ is even, in order to reproduce the fermi statistics of the electrons.  The flux tubes can be represented by a “fictitious” vector potential $\bf a$, not to be confused with the physical vector poetential $\bf A$ representing the constant external field, which obeys $$\nabla\times {\bf a}=2\pi q \rho$$ and the density $\rho({\bf r})=\sum \delta^{(2)}({\bf r}_i-{\bf r})$.

The next step, used in several similar problems [@zhang; @laughany; @jain; @HLR], is a mean field approximation.  The new fermions (for $q$ even) see the constant background magnetic field, and the $q$ flux tubes attached to the other fermions.  If the quantum mechanical state has a uniform density (in the quantum average), the latter becomes a constant field whose sign we can choose to be opposite to the external field.  In particular, if $\nu=1/q$, the fields cancel exactly.  The fermions now see no net field, so they can form a Fermi sea, which does have a uniform density, as we assumed.  We have arrived at the physical picture of the compressible Fermi liquid-like state, where we can say loosely that the fermions are electrons plus $q$ flux quanta.  But notice that this should not be taken too literally, since we have actually just made a transformation.  The real flux due to the external field remains uniform, not bunched up into flux tubes attached to the electrons.  It is really vorticity that is bound to the electrons, as discussed above.  In mean field theory the wavefunction for the electrons is simply $$\Psi_{\rm MF}=\det M\,\prod_{i<j}(z_i-z_j)^q/|z_i-z_j|^q
\label{MFstate}$$ with $M$ as before and no projection.  This function is [*not*]{} all in the LLL.  We can see that the effect of the transformation and the mean field approximation is to build the right kind of phase factors into the wavefunction, the same as possessed by $q$ vortices on each electron.  The factor $\prod |z_i-z_j|^q$ needed to recover eq.  (\[trialstate\]) can in fact be obtained from fluctuations about mean field, at least in a long-wavelength sense [@zhangrev]. 

Excited states again involve creation of particle-hole pairs.  In mean field approximation, the effective mass of fermion excitations near the Fermi surface is simply the bare electron mass $m$, since that is what appears in the Hamiltonian (we do not attempt to impose the LLL constraint).  Consideration of collective oscillations of the system leads to the correct cyclotron frequency $\omega_c=eB/mc$ which according to Kohn’s theorem cannot be renormalized.  However, under the conditions stated at the beginning of this article, the low-energy fermi excitations should have an effective mass $m^*$ determined by inter-electron interactions only.  Part of the resolution of this problem is that if fluctuations renormalize the effective mass, as they usually do in Fermi liquids, then there will also be a Landau interaction paramater $F_1$ that obeys a relation $m^{-1}=m^{*-1}+F_1$ which guarantees that the collective mode frequency is unrenormalized (like the plasma frequency which is unrenormalized in the usual electron gas at zero magnetic field) [@HLR].  A more serious problem is that interactions seem to play no role in the mean field theory.  The prediction would be the same, even for noninteracting electrons.  The Fermi liquid mean field state has finite compressibility, due to the mass $m$, even though in this case a partially filled Landau level should have infinite compressibility! ( The same holds when this approach is applied to the FQHE states for $q$ odd, where the compressibility vanishes.)  These problems can be resolved by understanding in what limit the approach is valid.  Mean field theory is good, and fluctuations in the effective magnetic field $\nabla\times({\bf A}+{\bf a})$ are small, when $q$ is small.  But for us $q$ is an even integer $\geq2$. However if we replace the original problem of electrons in a magnetic field by that of anyons in a magnetic field, then we can make $q$ small.  We choose to study anyons with statistics $\theta=\pi(1+q)$ (mod $2\pi$) and filling factor $1/q$, for $q>0$. As $q\rightarrow 0$, the magnetic field goes to zero, but for all $q$ we can still map the problem of anyons to one of fermions in zero net average magnetic field.  So at $q=0$ we reach fermions at zero magnetic field and we attach zero flux to convert them to fermions!  At this point there can be no fluctuations in the effective magnetic field, since there is no attached flux.  As we increase the external field, we must attach more flux to map to fermions in zero field, so the statistics must change accordingly.  When $\theta$ has made a full circle from $\pi$ back to $\pi$, we recover fermions (electrons) but now at a high field and filling factor $1/q$, $q=2$. We can continue and reach other even denominator states for electrons.  This idea is an adaptation of that in [@greiterwilzcek], except that they use it to argue that certain states exist by adiabatic continuation, while I use it only for a perturbation expansion in the fluctations, [*i.e.*]{} in powers of $q$. Notice that in the limit $q=0$, the system is a Fermi liquid, and the effective mass may be close to the bare mass if interactions are weak compared to the Fermi energy.  A similar statement applies to the anyons for $q$ small.  It is only at $q=2$ that we can argue that the compressibility must be infinite when interactions are set to zero, because only for fermions (such as the electrons) or charged bosons, [*and not for anyons*]{}, can we argue that the many-particle states of a noninteracting system are (anti-)symmetrized products of Landau level states. 

After that rather technical paragraph, we return to simpler discussion.  If the filling factor for electrons deviates from $\nu=1/q$, $q$ even, then in either approach above the quasiparticles will see a nonzero net magnetic field $B_{\rm eff}=B-B_{1/2}=\nabla\times({\bf A+a})$. If the quasiparticles fill an integer number $p$ of Landau levels due to this effective field, there will be a gap in the excitation spectrum (at least in mean field theory, but in fact it survives fluctuations) and we will obtain an integer QHE for the fermions, which is an FQHE for the electrons, with $\sigma=(e^2/h)p/(qp+1)$. This of course is Jain’s picture [@jain] of the “main sequence” of fractional QHE states as observed in experiments.  Jain’s approach was a hybrid of the approaches above.  He begins with the transformation involving $\delta$-function fluxes, but then simply modifies the mean field wavefunctions (\[MFstate\]) into the form (\[trialstate\]) with $M_{ij}=$ Landau level wavefunctions.  The argument of binding zeroes to particles to obtain a low Coulomb energy is not used.  We believe the $\nu=1/q$ case with exactly $q$ vortices per particle gives valuable insight into the reasons for the stability of the states, even away from this filling.  Also the notion of the effective mass emerges clearly in the Fermi liquid state.  We predicted [@HLR] that energy gaps in FQHE states near half-filling should scale as $1/p$ as $p\rightarrow\infty$, since the gap can be interpreted as an effective cyclotron energy $\sim B_{\rm eff}/m^*$ once we recognize the existence of the Fermi liquid, with a well-defined effective mass that controls excitation energies, as the limiting behaviour.  This prediction has received experimental support [@du]. 

The above arguments about dynamics of the quasiparticles at $\nu=1/2$ also generalizes to this case.  While the electron and the $q$-fold vortex still see the same potential $V(|k|)$, the drift velocities are different because that for the vortices is fixed by the electron density, not the true magnetic field.  The separation still remains constant, so like the back wheels of a car turning a corner, the bound pair moves (semiclassically) on two concentric circles separated by $|k|$, giving an effective cyclotron radius for the quasiparticles.  This radius is given by the usual formula $\Phi_0 k/2\pi B_{\rm
eff}$ for a particle of charge 1 in a magnetic field $B_{\rm eff}$.

The reader is cautioned that away from half-filling the quasiparticles which seem to be fermions in the mean field approach in fact have their statistics modified because of screening effects of the fluctuations.  They also acquire fractional charge, whereas at half filling there is perfect screening by the response of the fluid to the $\delta$-function of flux on the fermion.  This effect is in fact described by the extra amplitude factors present in the wavefunctions (\[trialstate\]).  The arguments given above for electron–$q$-vortex bound states at $\nu=1/q$ extend to other fillings to predict a net charge $1-q\nu$ and statistics $\theta=\pi(q^2\nu-1)$. These properties of fermions excited to higher “quasiparticle Landau levels” at Jain’s fractions are identical to those of the quasiparticle excitations in the hierarchy scheme, which is one of the main evidences for the equivalence of these two approaches [@readhier; @blokwen]. 

Space allows only a brief further discussion.  An outstanding series of experiments has been performed by Willett and coworkers [@will2].  Their technique, using the propagation of surface acoustic waves across the surface of the device, probes the longitudinal conductivity of the 2 dimensional electron gas at finite wavevectors and frequencies (this and other experiments [@kang] will be discussed at this meeting by Dr. Stormer).  Since the velocity of the waves is slow compared with the Fermi velocity of our fluid, we can consider zero frequency, finite wave vector.  The theory, treating the response of the fermionic quasiparticles in an RPA-like approximation, shows that $\sigma_{xx}(k)$ increases linearly with wavevector $k$ in the compressible state, as observed.  This results from the [*transverse*]{} conductivity $\sim 1/k$ for a conventional Fermi liquid, on including the long-range effects of the gauge field $\bf a$. (Not only can $B_{\rm eff}=2\pi q(n-(2\pi q)^{-1}))$ fluctuate in space and time, so can $E_{\rm eff}=2\pi q J$, the effective electric field due to the current $J$ in the perpendicular direction.)  It also predicted resonances in the response when the wavevector is an integer multiple of the inverse cyclotron radius for the semiclassical motion of the quasiparticles close to but just off half-filling.  This provides a measure of the Fermi wavevector.  The recent experimental observation of these resonances confirms the (nontrivial!)  existence of a Fermi surface for the charge-carrying excitations and excellent agreement of $k_F$ with the expected value ($\sqrt{2}$ times that in zero magnetic field, because of spin) is obtained. 

Open questions: a direct measure of the effective mass $m^*$ at half-filling would be most welcome.  Theoretically, a controversy remains about the possible partial breakdown of Fermi liquid theory, including behaviour of $m^*$, due to the fluctuations of $\bf a$ [@HLR]. 

I am grateful to P.A. Lee, B.I. Halperin and E. Rezayi for collaborations.  This work was supported by NSF-DMR-9157484 and by the A.P. Sloan Foundation. 

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