5
$\begingroup$

As we know, Landau level degeneracy in a finite rectangular system is $\Phi/\Phi_0$, where $\Phi=BS$ is the total magnetic flux and $\Phi_0=h/q$ is the flux quanta. This can be easily derived using Landau gauge and assuming the periodic boundary condition.

However, if you choose the symmetric gauge, $l_z=x\hat p_y-y\hat p_x$ commutes with Hamiltonian, the corresponding quantum number $m$ is thus a good one. After some calculation, at last the energy level is written as $$E=\left[n+(m+|m|)/2+1/2\right]\hbar\omega.$$ I want to derive the degeneracy of Landau level for a finite system with radius $R$.

The thought I got is that because in a finite system the angular momentum is bounded, the maximum value of $l_z$ is $qBR^2$ when you consider the particle doing the circular motion classically. Hence the maximum $m=l_z/\hbar=2\Phi/\Phi_0$, and the degeneracy is thus at least doubled.

What we have learned is that the gauge choice won't change the observable effect, the finite degeneracy can obviously be observed. So what's wrong with my derivation? Or because they are different just because the system we considered is simply not equivalent?

$\endgroup$
  • $\begingroup$ It's not clear in your question if you are dealing with the Lowest Landau Level (LLL) in particular or with any Landau Level. However, regardless the gauge choice, one should recover the same physics. I think the way you think is ok and gives the result with a good approximation. Note that even with using Landau gauge on a rectangular system, the $\Phi/\Phi_0$ expression does not provide strictly the LLL degeneracy but only an approximation. If needed, I can provide a derivation of LLL degeneracy estimation for the symmetric gauge. $\endgroup$ – dolun Nov 8 '14 at 14:56
  • $\begingroup$ @dolan Thanks for the comment! What I am considering is the most simple quantum mechanical problem, a charged particle in a uniform magnetic field. The degeneracy of every Landau level is same in the Landau gauge when considering a rectangular sample with periodic boundary conditions, this is almost an exact result as far as I consider. However, I am very interested in your claim, i.e. the LLL stuff and the "approximation" you said. $\endgroup$ – an offer can't refuse Nov 8 '14 at 15:34
  • $\begingroup$ I found the following to be a very useful resource when trying to understand Landau levels: hitoshi.berkeley.edu/221a/landau.pdf $\endgroup$ – Siva Nov 9 '14 at 9:20
  • $\begingroup$ @Siva Thank you,this is a good resource! Actually I solve this paradox by reading this notes today. It addressed the "unphysical angular momentum" on page 8. $\endgroup$ – an offer can't refuse Nov 9 '14 at 9:31
1
$\begingroup$

The problem with my thought is that I mix the "Canonical angular momentum" with the physical one. Since $\hat p_x,\hat p_y$ is canonical momentum, they are explicit gauge dependent. See What is canonical momentum? for a little reference.

The $\hat l_z$ in my question admit no physical meaning, so the bounded condition for it is just invalid. The true physical momentum should be $$ \begin{aligned} \hat L_z=&x\hat P_y-y\hat P_x\\ =&\hat l_z-\frac{qB}{2}r^2 \end{aligned} $$

Consider classically of left hand side, $L_z=-qBr^2$, so at last we have: $$ l_z^{max}= -qBR^2/2 $$

So we have $|q|BR^2/2\hbar$ different values of $\bar m$ to choose, which give the right degeneracy: $\Phi/\Phi_0$

$\endgroup$
1
$\begingroup$

Good to see you solved the paradox yourself! Another way to find the same result is to compute the number of orbits which is possible to stack in a surface equal to the area of the system.

Let's first remind basic things on Landau levels. The hamiltonian of a particle moving in 2D $\{x,y\}\equiv\{r,\theta\}$ plan through a static magnetic field reads : $$ \mathcal{H}=\frac{(\textbf{p}-q\mathbf{\mathcal{A}})^2}{2M}=\frac{\textbf{p}^2}{2M}+\frac{1}{8}M\,\omega^2_c\,r^2-\frac{\omega_c}{2}\,L_z $$ where $\mathcal{A}=(-By/2,Bx/2,0)$ is the potential vector in the symmetric gauge, $L_z=xp_y-yp_x=-\mathrm{i}\hbar\,\partial_\theta$ the canonical angular momentum, and $\omega_c=qB/M$ the cyclotron pulsation.

Heisenberg inequalities provide here consistant physical constants of the problem (typical length $\ell$ and velocity $v_m$ of the movement) : $$ M\,v_m\,\ell \sim \hbar \quad\text{with}\quad v_m=\ell\,\omega_c $$ Thus, we find $\ell=\sqrt{\frac{\hbar}{M\omega_c}}$ which is often refered as the magnetic length.

One can then compute the spectrum of $\mathcal{H}$ : $$ \mathcal{Sp(H)}=\left[ E_{n,m}=\left(n-m+1\right)\frac{\hbar\omega_c}{2};n \geq 0, m=-n,-n+2 \,...\,n-2,n\right] $$ where $n$ is the quantum number associated to the $\frac{\textbf{p}^2}{2M}+\frac{1}{8}M\,\omega^2_c\,r^2$ part of $\mathcal{H}$, and $m$ is the magnetic quantum number. Note that for a movement in the entire plan $\mathbb{R}^2$, Laundau Levels degeneracy $D$ is infinite.

If we now restrain our discussion to the Lowest Landau Level (LLL) given for $n=m$, it is easy to check that the eigen wave function is something like : $$ \Psi_{n=m}(r,\theta)\propto \left(r\,e^{\mathrm{i}\theta}\right)^m e^{-(r/2\ell)^2} $$ The essential features of $\Psi_{n=m}$ is that its RMS width is nothing more than the magnetic length $\ell$ and that it is maximum around a radius $r_{\mathrm{max}}=\sqrt{2m+1}\ell$.

If we now consider that the movement of the particle is confined in a $R$ radius disk, the degeneracy $D$ of the LLL can be estimated by counting how many orbits one can put in the surface $S=\pi\,R^2$ of the system. Suppose that the system is large enough $S>>\ell^2$ to receive a large number $m>>1$ of orbitals, then according to the expression of $\Psi_{n=m}$, the condition to satisfy to be able to put these orbitals in $S$ is simply : $$ R^2 > m\,2\ell^2 $$

Thus, the degeneracy is simply : $$ D\sim\frac{R^2}{2\ell^2}=\frac{S}{2\pi\ell^2}=\frac{\Phi}{\Phi_0} $$

$2\pi\ell^2$ can easily interpreted as the typical surface of an orbital.

$\endgroup$
  • $\begingroup$ I understand your point. Basically your claim is that for the LLL case, the wave function $\psi_m$ is localized in ring shaped area, that area is a constant for different $m$'s. For larger $m$, the localized place is farther away from the center, and the ring is narrower, to keep the ring area $2\pi l^2$ unchanged. The degeneracy is derived by using the total area $S$ divide the ring area; so back to your comments under the question, you claim that this is true only for LLL, because in other case the wave function will not have this behavior. Correct me if I have misunderstood your meaning $\endgroup$ – an offer can't refuse Nov 9 '14 at 13:53
  • $\begingroup$ Yeah, you are absolutly right. Actually, the expression I gave of $\Psi_{n,m}$ is valid in 2 cases : for the LLL $n=m$ and for the other extremal angular momentum states $n=-m$. For the other states, the expression for $\Psi_{n,m}$ can be, indeed, more complicated. $\endgroup$ – dolun Nov 9 '14 at 14:00
  • $\begingroup$ This argument can also applied to the Laudau gauge case, for Landau gauge, do you think the $\Phi/\Phi_0$ degeneracy is valid only for ground state?(the answer should be yes, because gauge choice cannot affect physics). However, as my comment under the question, based on the quantization of momentum, the degeneracy is same for every LL; beside, that derivation is simple but for me it is very rigorous. Since the only assumption it used is the periodic boundary condition. $\endgroup$ – an offer can't refuse Nov 9 '14 at 14:09
  • $\begingroup$ Yeah, your derivation is elegant since it can be done very quickly. But you should be carefull when classically computing the angular momentum with $L_z\sim -qBr^2$. I think it's working fine for the LLL, but I'm not sure it would be ok to do so for other Landau Levels. And yes, obviously the physics we are doing here should be gauge invariant. $\endgroup$ – dolun Nov 9 '14 at 15:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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