Landau levels are obtained by gauging the vector potential to be

$$ \vec{A}=\left(-By,0,0\right) $$

By which the Hamiltonian:

$$ H={1\over 2m}\left(\vec{p}-q\vec{A}\right)^2 $$

can be written as: $$ H={1\over 2m}\left(p_z^2+p_y^2\right)+{1\over2}m\omega^2\left(y+y_0\right)^2 $$


$$ y_0={p_x\over qB}\;\;\;,\;\;\;\omega^2={|q|B\over m} $$

With this we can conclude that we have a shifted harmonic oscillator, with energy spectrum of the form:

$$ E_{p_z,n}={p_z^2\over 2m}+\hbar\omega\left(n+{1\over2}\right) $$

and wave function:

$$ \psi\left(\vec{r}\right)={1\over\sqrt{2\pi\hbar}}e^{ip_zz/\hbar}\phi_n(y-y_0) $$

My Problem

By the same concept I could theoretically define new momentum in the original Hamiltonian by: $\widetilde{p}_x=p_x+By$, to obtain:

$$ H={1\over 2m}\left(p_z^2+p_y^2+\widetilde{p}_x^2\right) $$

Which looks like a free particle with shifted momentum. This supposed system, will have a very different spectra, and different wave function. Hence this must be wrong (as Landau was right).

My question is, what is wrong in my derivation?

As a side note, this must be wrong, as otherwise every oscillator could have been written a shifted free particle.

  • $\begingroup$ if you are adding a term of the form $By$ to the vector potential, then $B = \nabla \times A$ will not remain the same, and your Landau levels might be formed in a different plane altogether... $\endgroup$
    – Bruce Lee
    Feb 3, 2016 at 8:25
  • $\begingroup$ I fail to see the relation to my question. The question was, what is wrong with the second derivation? $\endgroup$
    – Yair M
    Feb 3, 2016 at 11:13
  • 5
    $\begingroup$ Because now $\tilde{p}_x$ and $p_y$ do not commute. $\endgroup$
    – Meng Cheng
    Feb 3, 2016 at 13:13


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.