Talk:Landau levels

[Untitled]
A plot of energy versus wave vector (parabolas separated by $$\hbar\Omega$$) could be helpful for the understanding. Bamse 01:29, 4 August 2006 (UTC)

I don't think a plot will help since the energy is independent of the wave vector. Chuck Yee 20:25, 6 September 2007 (UTC)


 * True, unless the motion is not confined in z in which case there is a term $$\frac{\hbar k_z^2}{2m}$$ in the energy. But probably such plot doesn't tell much indeed. Bamse 01:42, 10 September 2007 (UTC)


 * I think the plot on this page (http://www.warwick.ac.uk/~phsbm/qhe.htm) is very helpful for understanding the landau level. Also, there's the famous textbook pictures of square lattices changing into circles.  Perhaps these would the kind of picture to use Grj23 (talk) 02:32, 9 July 2009 (UTC)


 * I've created a video which I included in the quantum hall effect page. Do you think that a similar thing might be useful here? If so what? Grj23 (talk) 07:27, 4 September 2009 (UTC)

The claim that there is a limit to N in a finite geometry is quite probably wrong, see A. S. Mikhailov's arguments in an apparently unpublished paper, which seem irrefutable to me; if I had found a published version of that paper, I would have removed the above claim without further ado. Nimdaz (talk) 11:34, 21 March 2008 (UTC)

It seems that there is an error in the flux quantum see : http://en.wikipedia.org/wiki/Magnetic_flux_quantum 20080612 —Preceding unsigned comment added by 88.162.232.248 (talk) 19:32, 12 June 2008 (UTC)
 * There are two differences here. One is the use of CGS units in this article (which maybe should be changed?), and the other is that in superconductivity the pairs have charge of 2e, while here the charge is just e.Grj23 (talk) 02:32, 9 July 2009 (UTC)

I think an explicit formula for the number of electrons in each landau level would be helpful.Grj23 (talk) 02:35, 9 July 2009 (UTC)

I wonder if a second quantisation derivation would be clearer.Grj23 (talk) 02:35, 9 July 2009 (UTC)

I think that it would be useful to give the order of magnitude of some quantities, e.g. the level spacing and the degeneracy of each level. Beeblebrox-001 (talk) 17:25, 1 July 2016 (UTC)

Adding gauge transfomrations of the landau levels
My plan of is to


 * I will be updating the Landau Quantization page with the effects of the gauge transformation on the landau levels, We know that the hamiltonian is invariant under gauge transformations and so the eigenvalues of the Hamiltonian do not change but the energy eigenstates will change by a phase, I will update the page with unitory operator relating the Hamiltonian and states before and after the gauge transformations.


 * I will also update the information about 'Gauge Transformation in Electromagnetism' in the appropriate Wikipedia page.

please give me your comments Adwait.theory (talk) 03:02, 17 October 2014 (UTC)

Landau Levels in Symmetric Gauge
The given Hamiltonian (second equation) in this section seems to be incorrect. Given that H ~ [p - Q*A/c]² for an arbitrary charge Q. I believe the first term (with the x-component of momentum) should have the plus sign (since the negatives cancel out), while the second term (with the y-component of momentum) should have the negative sign. I'm not an expert in the field, so I don't want to make the edit, just in case I'm wrong. — Preceding unsigned comment added by 138.51.112.124 (talk) 17:58, 19 October 2015 (UTC)

Cyclotron Quantization
Given that the article starts out with a statement about the problem in a cyclotron orbit, it may be useful to confine the Landau levels to a circle of radius R, rather than a flat plane of length Lx and Ly. Let me know if this makes sense and I will add. Landmark ni (talk) 15:33, 9 December 2018 (UTC)

Photon Excitation
Can someone explain why a+ corresponds to a photon excitation? Doesnt the photon have an angular momentum of 1? Shouldn't selection rules state that the photon changes m_l of the angular momentum of the excited electron by 1? There is talk about an angular momentum, and that every LL has infinitely many (inf>m>-n). If i interpret this L_z to be the angular momentum, then photon transitions between just any level would be possible. Why would one even call that operator an angular momentum...? — Preceding unsigned comment added by 2A02:8070:DAE:600:53F:798A:1981:61E8 (talk) 14:20, 13 March 2019 (UTC)

SI units
Would it be an idea to use SI units instead of CGS? It would simplify the Hamiltonians, and the discussion of flux quantization and gauge transformations by removing the factor 1/c. This would also make the page more consistent with other Wikipedia pages, and with the scientific literature in general. Betohaku (talk) 08:36, 11 April 2019 (UTC)

Effect of gauge transformation
Here I copy a section that was removed and had unclear relevance to the article, for those that might find it interesting:--ReyHahn (talk) 10:03, 5 August 2021 (UTC)


 * $$\mathbf{A} \to \mathbf{A}' = \mathbf{A} + \boldsymbol{\nabla} \lambda(\mathbf{x}) $$

The definition for kinematical momenta is


 * $$\hat{\boldsymbol\pi} = \hat{\mathbf{p}}-q\hat{\mathbf{A}} $$

where $$\hat{\mathbf{p}}$$ are the canonical momenta. The Hamiltonian is a gauge invariant so $$\langle\hat{\boldsymbol\pi}\rangle$$ and $$\langle \hat{\mathbf{x}}\rangle$$ will remain invariant under gauge transformations but $$\langle\hat{\mathbf{p}}\rangle$$ will depend upon gauge. For observing the effect of gauge transformation on the quantum state of the particle, consider the state with A and A ' as vector Potential, with states $$|\alpha\rangle$$ and $$|\alpha'\rangle$$.

As $$\langle \hat{\mathbf{x}}\rangle$$ and $$\langle\hat{\boldsymbol\pi}\rangle$$ is invariant under the gauge transformation we get
 * $$\langle\alpha|\hat{\mathbf x}|\alpha\rangle = \langle\alpha'|\hat{\mathbf x}|\alpha'\rangle$$


 * $$\langle\alpha|\hat{\boldsymbol\pi}|\alpha\rangle = \langle\alpha'|\hat{\boldsymbol\pi}'|\alpha'\rangle$$


 * $$\langle\alpha|\alpha\rangle = \langle\alpha'|\alpha'\rangle$$

Consider an operator $$\mathcal{G}$$ such that $$|\alpha'\rangle = \mathcal{G} |\alpha\rangle $$

from above relation we deduce that
 * $$\mathcal{G}^{\dagger}\hat{\mathbf x}\mathcal{G} = \hat{\mathbf x}$$


 * $$\mathcal{G}^{\dagger}\left(\hat{\mathbf p} - e\hat{\mathbf A} - e\boldsymbol{\nabla} \lambda(\mathbf x)\right)\mathcal{G} = \hat{p} - e\hat{\mathbf A}$$


 * $$\mathcal{G}^{\dagger}\mathcal{G} = 1$$

from this we conclude


 * $$\mathcal{G} = \exp\left(\frac{ie\lambda(\mathbf{x})}{\hbar }\right)$$

Original work: Landau (1930)
Text in German:
 * Es wurde bis jetzt mehr oder weniger stillschweigend angenommen, dass die magnetischen Eigenschaften der Elektronen außer dem Spin ausschließlich yon der Bindung der Elektronen in Atomen herrühren. Für freie Elektronen übernahm man für den Bahneffekt das klassische Nullresultat mit der Begründung, dass auch das Fermische Integral von der entsprechenden Hamiltonfunktion wie das Boltzmannsche vom magnetischen Felde unabhängig ist.
 * Dabei wird aber eine Quantenerscheinung unberücksichtigt gelassen. Bei Vorhandensein eines Magnetfeldes wird nämlich die Elektronenbewegung in der zum Felde senkrechten Ebene finit. Das führt notwendigerweise zu einer Teildiskretheit (entsprechend der Bewegung in der genannten Ebene) der Eigenwerte des Systems, was, wie im folgenden gezeigt wird, zu einem von Null verschiedenen Bahndiamagnetismus Anlaß gibt.

Translation with deepl.com (free version):
 * Until now it was more or less tacitly assumed that the magnetic properties of electrons, apart from the spin, originate exclusively from the bonding of electrons in atoms. For free electrons, the classical zero result was adopted for the orbital effect on the grounds that the Fermi integral of the corresponding Hamiltonian function, like the Boltzmannian, is also independent of the magnetic field.
 * However, a quantum phenomenon is not taken into account. In the presence of a magnetic field, the electron motion is finite in the plane perpendicular to the field. This necessarily leads to a partial discreteness (corresponding to the motion in the plane mentioned) of the eigenvalues of the system, which, as will be shown in the following, gives rise to a non-zero orbital diamagnetism.

FbiSupLabAcc (talk) 14:05, 18 March 2023 (UTC)