I'm quite familiar with orbital shapes for a hydrogen atom (where a potential field due to a point charge, i.e nucleus exists). I've been interested in trying to figure out the orbital shapes for an electron executing cyclotron motion in a constant magnetic field.

Of course, to obtain the orbital shapes, I must first solve the Scrodinger Equation:


However, a magnetic field does not have potential energy associated with it, and hence, the $U$ term would be constant for a magnetic field. Thinking about it, it does make sense, as there may be infinitely many possible lines parallel to the field that may be considered to be the axis around which the electrons revolve (pardon my use of classical mechanical language) around. The net wave function produced due to all these fields would eventually give a constant $\psi$ value throught the entire region.

However, to obtain a meaningful and informative orbital shape, lets restrict the electron to move only around one specific axis. The electron cannot use any other parallel axis to revolve around. I believe this restriction will enable us to actually analyze the orbitals that a revolving electron forms.

I'll now phrase this scenario with defined variables:

Suppose a magnetic field $\vec{B}=B_0\vec{z}$ exists in space, with $B_0$ simply being a scalar to indicate magnitude of the field. An electron (of mass $m_e$ and charge $e$) revolves around the $z$-axis. We must find the wavefunction and consequently the orbital associated with this electron.

Given this scenario, I can infer two things from just a glance:

  1. The orbital will have cylindrical symmetry with respect to the axis (here, the $z$-axs).

  2. The component of momentum parallel to $xy$-plane has a magnitude of $|\vec{p_x} + \vec{p_y}|=erB_0$, where $r=\sqrt{x^2+y^2}$

I am still not able to find out how to find potential energy, even with this restriction. Can someone offer a helping hand, and guide me in the proper direction?

  • $\begingroup$ What do you mean by "orbital shapes"? I thought we were considering free electrons. $\endgroup$ Mar 24, 2018 at 16:41
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    $\begingroup$ This is a standard problem, known as Landau quantization, and the resulting orbitals, known as Landau levels, are fundamental building blocks for the quantum Hall effect. Those keywords should make it much easier to make effective inroads into the literature. $\endgroup$ Mar 24, 2018 at 17:12

1 Answer 1


To solve this problem, you need to consider the correct (classical) Hamiltonian for the electron in a magnetic field: $$H=\frac {(\vec p-e\vec A)^2}{2m}+q\phi$$ $\vec p$ is the (operator of) the kinetic linear momentum, $e$ and $m$ is the electron charge and mass, respectively. Here $\vec A$ is the vector potential giving $$\vec B=\nabla \times \vec A$$ and in the present case the electrical potential is zero $$\phi=0$$ Thus with $\vec p \to -i\hbar \nabla$ the time dependent Schroedinger equation for this problem is: $$i\hbar \frac {\partial \psi}{\partial t}= \frac {(-i\hbar \nabla-e\vec A)^2}{2m}\psi$$ How to obtain the quantized cyclotron levels is indicated here.


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