Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

There are a lot papers explaining why Laughlin's wavefunction are energetically favorable, but seldom explain why a lower energy state could explain the plateau at $\nu=1/3$. I met at several places claims like: a lower energy state at $\nu=1/3$ will pin the electron density at $\nu=1/3$. But why is that? And what actually it means? when we move $\nu$ from $1/3$ what happens? electrons adjust there distance or new particle been added? And is this a phase transition? Hope someone familiar this field could give me some help, thanks!

share|cite|improve this question
up vote 1 down vote accepted

The Laughlin state alone doesn't explain the plateau. There is a lot more to the story.

Firstly at filling factor=1/3 the many-body ground state of the interacting electron gas is "approximately" the Laughlin wavefunction. By this I mean that the overlap between the Laughlin state and the numerically found ground state (for any realistic interaction like coulombic) is very large, i.e. their inner product is quite close to 1. Using the plasma analogy one can show that this state corresponds to uniform electron density. (See Girvin's Les Houches notes for details on Plasma analogy.)

Secondly the transport phenomena are decided by charged excitations in the system. For the filling factors 1/3,1/5,1/7,etc. the charged excitations are quasiholes and quasielectrons. While the former has a dip in the density profile at some point Z (say) in the 2D plane, the latter has the opposite thing in its density profile (as opposed to the earlier uniform case). The plasma analogy can again be used to show that these quasiparticles will have fractional e/3 charge in our case. (Atleast for now let us avoid justifying why they are excited states.)

Now lets say we are sitting exactly at 1/3 filling factor and then we add an electron to the system. It will break into 3 quasielectrons which can be separated at no extra energy cost (the idea of fractionalization). Similarly if some more electrons are added they will produce more quasiparticles. Now start thinking in terms of the 'semiclassical percolation picture' that is applied to electrons to explain Integer QHE (Again see Girvin's notes). Instead of electrons we give the same arguments using quasiparticles to explain the plateaus around 1/3 filling factor. The conductivities stop changing when the added quasiparticles are either going into the valleys of the disorder potential or are ending up on shorelines at the 2 well separated edges.

Let me clarify things a bit more. Think of starting with the 1/3 filling factor ground state. Now let us add adiabatically 1 flux quanta through a thin solenoid at the origin of space (See Laughlin's Nobel lecture). He shows that in this process e/3 charge flows towards the origin and gets collected there. Thus we have ended up with an exact ground eigenstate of the original hamiltonian + e/3 charge. So quasiholes are 'charged excitations', not the excited state when sitting at 1/3 filling. In fact the low energy gapped excitations at 1/3 filling are 'neutral collective excitations' (Again see Girvin's notes) and the existence of this gap is necessary for adiabaticity to work fine in the above thought experiment. (In the words of Laughlin the usage of the word quasiparticles here was "unfortunate".) Now if I just move the filling factor a bit in an experiment the new ground state is made up of new "quasiparticles".

share|cite|improve this answer
Hi, Akshay,thanks for your detailed answering. The Girvin's note you mention is great,I will read it carefully. Here are two more questions, hope you could answer 1. The system is at The Laughlin's ground state only at $v=1/3$ or at the region where $v=1/3$ plateau exist? 2. What is the role of energy gap to excite the quasi-particles? does it block the scattering thus lead to the zero longitude resistence? – Knightq Mar 15 '13 at 2:19
The Laughlin's state is the (approximate) ground state only at 1/3 filling factor. Also in a loose sense the gap is connected to the fact that the particles can't hop from the shoreline at one edge to another, which in turn implies 0 longitudinal resistance. – Akshay Kumar Mar 15 '13 at 15:09
So if Laughlin's ground state is only true for 1/3, why when you add particles(filling factor moves) you still use the properties (quasi particles) of Laughlin state? And what is the ground state for v not equal to 1/3? – Knightq Mar 15 '13 at 21:54
I edited my answer above. I hope it helps. The lectures and notes are definitely a must read. – Akshay Kumar Mar 16 '13 at 4:11
I see. Thanks a lot! – Knightq Mar 17 '13 at 18:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.