Integer Quantum Hall Effect: Corbino Ring Derivation

Why are we allowed to consider adding extra flux to the system? When we measure the hall resistance, we don’t add such a flux. Why is this not considered a different system? I will go through the derivation and discuss the part I have trouble with.

We initially consider an electron confined to circular disk, with a circular hole in the middle. We apply a constant magnetic field, B, perpendicular to the plane of the disk. The Hamiltonian written in cylindrical polar coordinates is,

$$H_{Phi=0}=frac{1}{2m}[-hbar^2 frac{1}{r}frac{partial}{partial r}(rfrac{partial}{partial r})+ (-frac{ihbar}{r}frac{partial}{partial phi }+ frac{eBr}{2})^2].$$

The wavefunctions of the lowest Landau Level (LLL), are

$$psi_m =e^{im phi}r^me^{-r^2/4l_B^2 }quad, $$

where we haven’t bothered to normalise and where $l_B=sqrt frac{hbar}{eB}$. Angular momentum is a good quantum number and is given by $m$. All of the states in the lowest Landau level are degenerate, as is the case for any given Landau level.

The Hamiltonian for non zero flux, $Phi$, through the hole in the disk is

$$H_{Phi}=frac{1}{2m}[-hbar^2 frac{1}{r}frac{partial}{partial r}(rfrac{partial}{partial r})+ (-frac{ihbar}{r}frac{partial}{partial phi }+ frac{eBr}{2} +frac{ePhi}{2pi r})^2].$$

To get the new LLL wavefunctions we can simply take the previous wavefunctions and multiply them by $e^{-iePhi phi/2pi hbar}$. This is essentially a gauge transformation. However, we require that the wavefunction is single valued. So this trick is only possible when $Phi$ is an integer multiple of $2pi hbar /e=h/e$. We see that multiplying the original wavefunctions by $e^{-iphi}$ reduces the $m$ label by one but keeps the radial dependence the same. The energy of the state remains the same too. Using arguments from the adiabatic theorem, if an electron starts in a LLL at $Phi=0$ and if we slowly increase the flux from zero to $h/e$ this electron will move into the original ($Phi=0$) LLL with one higher $m$ label; the electron is pushed radially outwards. If all LLLs are filled, the net effect is that an electron is moved from the inner ring to the outer ring.

So, we know that increasing the flux from zero to $h/e$ pushes a single LLL electron from the inner disk to the outer. This is true in fact for all Landau levels; for every filled Landau level, the net effect is that an electron is moved from inner to outer.

We then say that the flux, as it is slowly and constantly increased to $Phi_0=h/e$, induces an emf around the ring of $Phi_0/T=h/eT$, where $T$ is the (long) time it takes to change it. Furthermore, we know that the radial current is simply $ne/T$, where $n$ is the number of filled Landau levels, as $n$ electrons are (net) moved from inner to outer ring. Dividing the induced emf by the current we obtain the Hall resistance, $h/ne^2$.

Why are we allowed to add flux in this way? The standard way of measuring Hall resistance is to pass a current through a sample, fixed by putting the sample in series with a large resistor, then read of the Hall Voltage. The Hall voltage divided by the current is the Hall resistance. Why, in this derivation, can one use a flux to induce an emf and then effectively read of the radial current?

How do we know that adding the flux does not alter the physics of the system? Another way of asking this, I think, is what are the limitations on how one can modify the original set up when measuring/deriving the Hall resistance, a value we claim is intrinsic to the original system?

Possible Explanation

Part of my trouble is that the vector potential, due to the flux, is non zero in the disk itself. It seems sensible to use a gauge transformation to remove this. Working in the original gauge has been useful, we were able to map between wavefunctions when $Phi =0$ and when $Phi=h/e$ and ultimately find the radial current. But now we try and remove the vector potential due to the flux, which will allow to view the system as being akin to the standard set up for measuring the hall resistance.

(Note that when I write A here I really mean A associated with the flux through the centre of the ring which we vary.)

We say $Ato A+ nabla Lambda$ and $Vto V – frac{partial Lambda}{partial t}$, where $E=-nabla V – frac{partial A}{partial t}$.

In the disk we originally had $A_{phi}=frac{Phi}{2pi r}$ and $A_r=0$. Or rather, $A=(0,frac{Phi}{2pi r})$.

We let $Lambda=frac{phiPhi}{2 pi}$, this defines our specific gauge transformation. $nabla Lambda=(0,-frac{Phi}{2pi r})$. Thus our new magnetic vector potential, associated with the additional flux, is $(0,0)$ inside of the disk.

Our new $V$ is thus $V_{orig}-frac{phi}{2pi }frac{dPhi}{dt}$. This is equal to, $V_{orig} – frac{hphi}{eT 2pi}$. Thus, ignoring $V_{orig}$, around the ring there is a potential difference of $frac{h}{eT}$. The current is still $ne/T$ as the physical system is still the same. So the resistance is as desired.

I haven’t really added anything here. I’ve just appreciated that the original derivation I gave involves implicitly doing the gauge transformation and thus takes into account a zero vector potential within the ring. Before the gauge transformation we saw that $n$ electrons move from inner to outer. We then transform to a gauge in which we just have a potential difference and a current, which makes statements about Hall resistance easy.

I would still appreciate remarks on this.

I’d also appreciate a comment on the fact that I have moved to a gauge with zero vector potential in the ring due to the additional flux, $Phi$, however,

$$Phi=int Bcdot dS=intnabla times A cdot dS=int A cdot dl.$$

Something has gone wrong somewhere as if I do the last integral in the ring I get zero as I have $A$ to be zero in the ring, but I know that $Phi$ is not zero for all time!