# Landau levels in rotationally invariant gauge

I try to find wave-function of electron in external constant magnetic field in gauge $$A=\frac{B}{2}(-y,x,0).$$ I substitute anzats, $$\psi=e^{-i\omega t}e^{ip_zz}F(x,y)$$. Then, I rewrite equation in polar coordinates and obtain (I write only differential operator): $$\partial_r^2+\frac{1}{r}\partial_r+\frac{1}{r^2}\partial_{\theta}^2-ieB\partial_{\theta}-\frac{e^2B^2}{4}r^2+\Omega,$$ where $$\Omega=2m\omega-p_z^2+eBs$$ and $$s=\pm 1$$. Then, I use $$F(r,\theta)=f(r)e^{i\theta n}$$, $$\partial_r^2+\frac{1}{r}\partial_r-\frac{n^2}{r^2}+eBn-\frac{e^2B^2}{4}r^2+\Omega.$$ To solve this equation, I changle variables, $$\xi=r^2$$ and find $$\partial_{\xi}^2+\frac{1}{\xi}\partial_{\xi}+\frac{eBn+\Omega}{4\xi}-\frac{n^2}{4\xi^2}-\frac{(eB)^2}{16}.$$ Using asymptotes, I know that $$f(r)=\rho(r)e^{-\xi/2}\xi^{n/2}.$$ Finally, equation for $$\rho(r)$$ is $$\xi\rho''+(n+1-\xi)\rho'+\frac{\rho}{2}\left(\frac{\Omega+eBn}{2}-\frac{(eB)^2\xi}{8}+\frac{\xi}{2}-n-1\right)=0.$$ I know that solution of this equation should be Laguerre polynomial up to factor with exp function. Using Wolfram Mathematica, I see that solution should be $$\exp\left(\frac{\xi}{2}+\frac{eB\xi}{4}\right)L_{n}^{(\Omega-eB)/(2eB)}\left(\frac{eB\xi}{2}\right).$$ Moreover, Mathematica says me that confluent hypergeometric function is also the solution.

I do not understand several facts:

1. How to rewrite equation for $$\rho$$ in the "canonical" form and explicitly see that solutions are Laguerre polynomials with exp prefactor?
2. How can I choose the correct solution? It seems that both functions, Laguerre polynomial and confluent hypergeometric function are related to Hermite polynomials. I compare with Hermtie because I know that the solution of electron in external magnetic field in gauge $$A=B(-y,0,0)$$ is Hermite polynomial.
3. What should I do to find spectrum? It seems that all the information of spectrum should be encoded in upper index of Laguerre polynomial. So, my guess is that for specific values
4. Where I can find normalization factor? To be honest, I do not want to perform calculation for it
• Did you read 1.4.3 from damtp.cam.ac.uk/user/tong/qhe.html ? You can easily find lowest Landau level and act with the raising operator to obtain all wave functions. It seems that you solve this task in irrational way. Commented Mar 1, 2020 at 22:21
• This is called symmetric gauge, not rotationally invariant gauge! @ArtemAlexandrov
– SRS
Commented Mar 2, 2020 at 5:04
• @SRS wave-function in this gauge is invariant under rotations in transverse to magnetic field plane (in $x,y$-plane, I mean), therefore it is also sometimes called rotationally invariant gauge. Commented Mar 2, 2020 at 7:20
• Note that it is incorrect to equate Landau level wave functions in Landau and symmetric gauges because of the level degeneracy. The wave function with definite E and l_z in symmetric gauge is a superposition of those found in Landau gauge with the same E and different values of p_y, see: iopscience.iop.org/article/10.1088/0031-8949/47/6/004 Commented Mar 2, 2020 at 17:22
• @AlexeySokolik I understand your point, thank you for useful reference! Commented Mar 2, 2020 at 19:25

Starting from the equation after variables separation, I have obtained $$\partial_r^2+\frac{1}{r}\partial_r-\frac{n^2}{r^2}-\frac{e^2B^2}{4}r^2+\Omega(n),$$ where I now absorb term $$eBn$$ into $$\Omega$$. Introducing magnetic length, $$l^2=1/(eB)$$ and new variable $$\xi=r^2/(2l_0^2)$$, one can obtain $$\xi\partial_{\xi}^2+(n+1-\xi)\partial_{\xi}+\left(\lambda-\frac{n+1}{2}\right),$$ where $$\lambda=\Omega l_0^2$$. This equation is nothing more than confluent hypergeometric function equation and with arguments of finiteness of w-f at infinity, it is easy to find the spectrum and normalization factor (both are given in the paper).