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KDPhysics
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- Why doesn't gauge transforming the vector potential for a solenoid affect the energy levels of a particle orbiting it a fixed radius?
In p.385 of Griffiths QM the vector potential ##\textbf{A} = \frac{\Phi}{2\pi r}\hat{\phi}## is chosen for the region outside a long solenoid. However, couldn't we also have chosen a vector potential that is a multiple of this, namely ##\textbf{A} = \alpha \frac{\Phi}{2\pi r} \hat{\phi}## where ##\alpha## is some constant? The two are related by a gauge transformation:
$$\alpha\frac{\Phi}{2\pi r}\hat{\phi} = \frac{\Phi}{2\pi r} \hat{\phi}+\nabla\bigg((\alpha-1)\frac{\Phi}{2\pi}\phi\bigg)$$
When I solve the TISE with this new gauge I get that the energy levels are:
$$E_n = \frac{\hbar^2}{2mb^2}\bigg(n-\alpha \frac{\Phi}{\Phi_0}\bigg)^2$$
which is different from what Griffiths even if the magnetic flux is quantized. How is it possible that the ground state depends on the gauge choice?
$$\alpha\frac{\Phi}{2\pi r}\hat{\phi} = \frac{\Phi}{2\pi r} \hat{\phi}+\nabla\bigg((\alpha-1)\frac{\Phi}{2\pi}\phi\bigg)$$
When I solve the TISE with this new gauge I get that the energy levels are:
$$E_n = \frac{\hbar^2}{2mb^2}\bigg(n-\alpha \frac{\Phi}{\Phi_0}\bigg)^2$$
which is different from what Griffiths even if the magnetic flux is quantized. How is it possible that the ground state depends on the gauge choice?
