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## 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:

[itex]- \frac{ \hbar^2}{2mr^2} \psi''(\phi)=E \psi (\phi) [/itex]

but what is about the magnetic field?

i thought that if i take

[itex]E_{pot}=\mu B=\frac{e m v r}{2m}B=\frac{e l B}{2m} [/itex]

and therefore:

[itex]- \frac{ \hbar^2}{2mr^2} \psi''(\phi)+\frac{i \hbar e}{2m} \psi '(\phi) B=E \psi (\phi) [/itex]

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

[itex]E_{pot}=\mu B=\frac{-g_s \mu_B \sigma }{\hbar }B [/itex]

and therefore:

[itex]- \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) [/itex]

so, is this the right equation?

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:

[itex]- \frac{ \hbar^2}{2mr^2} \psi''(\phi)=E \psi (\phi) [/itex]

but what is about the magnetic field?

i thought that if i take

[itex]E_{pot}=\mu B=\frac{e m v r}{2m}B=\frac{e l B}{2m} [/itex]

and therefore:

[itex]- \frac{ \hbar^2}{2mr^2} \psi''(\phi)+\frac{i \hbar e}{2m} \psi '(\phi) B=E \psi (\phi) [/itex]

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

[itex]E_{pot}=\mu B=\frac{-g_s \mu_B \sigma }{\hbar }B [/itex]

and therefore:

[itex]- \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) [/itex]

so, is this the right equation?