# Free particle in magnetic field / Landau quantization

I have a question concerning a possible derivation of the Landau quantization.

In our lecture notes (and some other places as well), the following ansatz is used:

$$\Psi(x,y,z) = \exp{\left(-j(\beta y + k_z z)\right)}u(x),$$ which then leads to the known results ($u(x)$ is identified as a harmonical oscillator,...)

But I'm unhappy with the ansatz itself: if I calculate the mean momentum in $y$ direction, I obtain $\hbar \beta$. From classical mechanics, I know the solution can be nothing but a spiral-like movement (should be compatible with quantum mechanics, as I could just use a particle with a huge momentum, which could be directly observed...).

So, what am I missing?? I would also expect some symmetry around some path center?!

Sorry for my maybe rough english, but I'm not too familiar with using it in physical context...

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Aren't you missing that the field points in the $z$ direction and therefore the classical trajectory is spiral? I.e. only projection to the plane perpendicular to the field (which is actually the most relevant case for applications in 2D systems) is circle. But there is still conserved momentum in the $z$ direction. Is this your problem, or is it something deeper? There are certainly other confusing points about Landau quantization, like treating $x$ and $y$ assymetrically while in reality they are equivalent... –  Marek Jun 21 '11 at 21:40
You are of course right, corrected that. It is also the net y momentum that confuses me (not z like stated initially). My problem is thus related to the different treatment of x and y, yes. But well, if they were treated assymetrically, but still in a similar way, giving a net momentum of zero in each direction, this would be okay with me ;-) –  Wolfgang Noichl Jun 21 '11 at 21:43
@Wolfgang: okay. With that being settled, I am actually looking forward towards answers myself :) I know quantum magnetic systems admit multidute of what seems like contradicting descriptions and while they provide ways to calculate stuff, one can't really say they describe what's going on physically. E.g. in this problem there is no preferred direction $x$, so if you chose any other axis (e.g. $y$) and quantized along that, you'd obtain new description again splitted into harmonic oscillator (but now along $y$ axis) and the rest. –  Marek Jun 21 '11 at 21:49
I'm not so sure why you need zero momentum... $\beta$ is just the quantum number of y-axis momentum (and can certainly be zero if you want to describe such a state). What happens to a classical electron in a magnetic field when it starts with some y-axis momentum? –  wsc Jun 21 '11 at 21:55
It starts flying along a spiral (projected in x-y plane, a circle). And thus, if averaged over long times, has no net momentum along y axis. The "classical long time average -> quantum mechanics expectation value" is not my idea, see i.e. wikipedia on harmonical oszillator. Of course, the analogy is not complete (i.e. classical electron would also lose energy...), but even for quantum mechanics, the electron moving along y should be dragged along x, which should rule out plane waves along y? –  Wolfgang Noichl Jun 21 '11 at 22:06

The key insight is that whenever (electro-)magnetism and Hamiltonians are involved, there is always a gauge fixing happening somewhere. The basic issue is that the Hamiltonian is something like $$H = \left|p-eA\right|^2,$$ and for any given magnetic field $B$ there is a variety of potentials $A$ which can fit. For a uniform magnetic field, a frequently used one is the Landau gauge where $$A = \left(0, Bx, 0\right).$$ In this gauge, the Hamiltonian is $$H = p_x^2 + (p_y - eBx)^2.$$ You then stick your ansatz in and get what you already know. The point to remember is that $p_x = i \frac{\partial}{\partial x}$ is not the mechanical momentum. It is the canonical momentum, which is defined to be the one that obey the canonical commutation relation $\left[ x, p_x \right] = i$. Ditto for $y$ and $p_y$. The mechanical momentum is exactly what's in the Hamiltonian: $p - eA$. Fairly sure if you follow that through, you get exactly what you expect, i.e. no average motion.

For a lot of applications, it's nicer to use the symmetric gauge, where $A = \frac{1}{2}\left(-By, Bx, 0\right)$. In that case, you retain the explicit $x$-$y$ symmetry and things like angular momentum keeps making sense.

Notice that things like $p$ change depending on the gauge you choose. Physical variables such as actual mechanical momentum does not. Indeed, in this case of a magnetic field in the $z$ direction, the mechanical momentum in the $x$ and $y$ directions do not commute! You cannot simultaneously know both.

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It should be stressed that going to the symmetric gauge $\vec{A} = \frac{1}{2}\left(-By, Bx, 0\right)$, mentioned in the answer (v1), just rotates the asymmetric treatment $45^\circ$ to between the $x\pm y$ directions. –  Qmechanic Jun 22 '11 at 5:49