# Schrödinger equation for particle on a ring in a magnetic field

## Main Question or Discussion Point

hi

i need the schrödinger equation for a particle(electron) in a ring under the influence of a magnetic field that goes through perpendicular to the plane of the ring and i want to consider the spin too.

Well, the particle in the ring is pretty easy:

$- \frac{ \hbar^2}{2mr^2} \psi''(\phi)=E \psi (\phi)$

but what is about the magnetic field?

i thought that if i take

$E_{pot}=\mu B=\frac{e m v r}{2m}B=\frac{e l B}{2m}$

and therefore:
$- \frac{ \hbar^2}{2mr^2} \psi''(\phi)+\frac{i \hbar e}{2m} \psi '(\phi) B=E \psi (\phi)$

and then i would get an additional term if i want to consider the spin:

$E_{pot}=\mu B=\frac{-g_s \mu_B \sigma }{\hbar }B$
and therefore:
$- \frac{ \hbar^2}{2mr^2} \psi''(\phi)+\frac{i \hbar e}{2m} \psi '(\phi) B+\frac{-g_s \mu_B \sigma }{\hbar }B\psi(\phi)=E \psi (\phi)$

so, is this the right equation?

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vanhees71
Gold Member
2019 Award
That's not completely correct.

The important point of electromagnetic fields is that it is a gauge field, and thus to get a gauge invariant equation, you have to introduce the covariant derivative

$$D_{\mu}=\partial_{\mu}+\mathrm{i} q A_{\mu},$$

where $A_{\mu}$ is the four potential of the electromagnetic field. In the following non-relativistic limit I write $A_0=\Phi$ for the scalar potential and $\vec{A}$ for the vector potential.

For non-relativistic without spin you start with the Schroedinger equation for a free particle

$$\mathrm{i} \frac{\partial \psi}{\partial t}=-\frac{1}{2m} \Delta \psi.$$

To couple the electromagnetic field to it, you just substitute all derivatives by their gauge-covariant derivatives, i.e.,

$$\partial_t \rightarrow D_0=\partial_t+\mathrm{i} q \Phi.$$

$$D_i = \frac{\partial}{\partial x^i} + \mathrm{i} q A_i = \frac{\partial}{\partial x^i} - \mathrm{i} q A^i,$$

i.e. the correct substitution for the nabla operator in 3D-vector analysis notation reads

$$\vec{\nabla} \rightarrow \vec{D}=\vec{\nabla}-\mathrm{i} q A^i.$$

Plugging this into the Schrödinger equation, one gets

$$\mathrm{i} \frac{\partial \psi}{\partial t}-q \Phi \psi=\frac{1}{2m} (-\mathrm{i} \vec{\nabla}-q \vec{A})^2 \psi$$

or to extract the Hamiltonian

$$\mathrm{i} \frac{\partial \psi}{\partial t}=\hat{H} \psi=\frac{1}{2m} [-\Delta \psi - \mathrm{i} \vec{\nabla} \cdot (\vec{A} \psi) - \mathrm{i} \vec{A} \cdot \vec{\nabla} \psi - q^2 \vec{A}^2 \psi]+ q \Phi \psi.$$

For a homogeneous magnetic field you set

$$\vec{A}=-\frac{1}{2} \vec{x} \times \vec{B}.$$

To consider spin you indeed only need to add

$$\hat{H}_{\text{Spin}}=-\frac{q}{2m} g_S \hat{\vec{S}} \cdot \vec{B}.$$

This leads to the Pauli equation.

For the problem of a particle restricted to a circle, just substitute the gradient and Laplacean in spherical coordinates and set $\partial/\partial r=\partial/\partial \vartheta=0$.