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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
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    $\begingroup$ 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. $\endgroup$
    – Nikita
    Commented Mar 1, 2020 at 22:21
  • $\begingroup$ This is called symmetric gauge, not rotationally invariant gauge! @ArtemAlexandrov $\endgroup$
    – SRS
    Commented Mar 2, 2020 at 5:04
  • $\begingroup$ @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. $\endgroup$ Commented Mar 2, 2020 at 7:20
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    $\begingroup$ 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 $\endgroup$ Commented Mar 2, 2020 at 17:22
  • $\begingroup$ @AlexeySokolik I understand your point, thank you for useful reference! $\endgroup$ Commented Mar 2, 2020 at 19:25

1 Answer 1

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The answer is given in the recent paper by Orion Ciftja, "Detailed solution of the problem of Landau states in a symmetric gauge".

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).

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