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Classically, a system of noninteracting electrons restricted to move in 2D will execute circular motion in an external magnetic field $\perp$ to the plane. But in quantum analysis, the same noninteracting electrons in the same situation gives rise to equispaced harmonic oscillator levels, called Landau levels. Why should harmonic oscillator levels arise out of the blue? Is there a relation between classical circular trajectories and quantum harmonic oscillator levels?

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    $\begingroup$ If you consider only the $x$ component (or similarly just the $y$ component) of the motion of a particle undergoing circular motion in the $x$-$y$ plane then you will find it undergoes simple harmonic motion. $\endgroup$ Aug 12, 2020 at 21:32
  • $\begingroup$ @BySymmetry but the hamiltonian looks nothing like a harmonic oscillator hamiltonian. $\endgroup$ Aug 12, 2020 at 21:44
  • $\begingroup$ It doesn't? $\endgroup$ Aug 12, 2020 at 21:45
  • $\begingroup$ @BySymmetry Please convert that comment to an answer. $\endgroup$
    – rob
    Aug 12, 2020 at 21:50
  • $\begingroup$ Linked . $\endgroup$ Aug 12, 2020 at 21:51

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Why should harmonic oscillator levels arise out of the blue?

Edit after reading new comments: if you write down the Hamiltonian of the system using the standard minimal coupling of the momentum $\vec{p} \to \vec{p} + q \vec{A}$ and you write down $\vec{A}$ for a uniform magnetic field using the gauge that you prefer, you get exactly the Hamiltonian of a harmonic oscilator. You should not be surprised, cause circular motion is a two-dimensional harmonic motion!

Is there a relation between classical circular trajectories and quantum harmonic oscillator levels?

Yes! The classical motion of a particle of mass $m$ and electric charge $q$ in an external magnetic field $B$ is given by Newton's second law: $$ m \frac{v^2}{r} = q v B \;\;\; \rightarrow \;\;\; r = \frac{mv}{qB}. $$ Let me write this radius in terms of kinetic energy $E=\frac{1}{2}mv^2$, which is more convenient for comparison with quantum motion and let me write $r^2$ rather than $r$: $$ r^2 = \frac{2mE}{(qB)^2}. $$

Now in quantum mechanics you can prove that the energy eigenvalues are $$ E_n = \frac{\hbar q B}{m} \left( n + \frac{1}{2} \right), $$ where $n$ is a non negative integer labeling the levels; and the corresponding wave function is $$ \psi_n(x,y) = N_n (x + iy)^n e^{-qB(x^2+y^2)/4\hbar}, $$ $N_n$ being a normalization constant.

Now if you compute the expectation value of the operator $r^2 = x^2+y^2$ you get the following result $$ \left\langle r^2 \right\rangle = \frac{2\hbar}{qB} ( n + 1). $$ You can compare the classical result with the quantum result at high energies, so you can assume $n \gg 1$ and simplify $ \left\langle r^2 \right\rangle \approx \frac{2\hbar n}{qB}$, and similarly $E = \frac{\hbar q B n}{m}$. If you combine these two results to get rid of $n$, you immediately get $$ \left\langle r^2 \right\rangle = \frac{2 m E}{(qB)^2}, $$ which is exactly the classical result!

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  • $\begingroup$ thank you very much. Could you please tell in which gauge you calculated the wavefunctions $\psi_n$ and the result $\langle r^2\rangle=\frac{2\hbar}{qB}(n+1)$? $\endgroup$ Aug 14, 2020 at 15:01
  • $\begingroup$ Sure: the wave function is computed in the "symmetric gauge", which means that the vector potential is $\vec{A}(x,y) = \frac{B}{2} ( -y, x , 0)$. However I believe that $\left\langle r^2 \right\rangle$ should be gauge invariant. $\endgroup$
    – Matteo
    Aug 15, 2020 at 8:11

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