Gauge in the Aharonov Bohm effect

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SUMMARY

The discussion centers on the gauge invariance in the Aharonov-Bohm effect, specifically addressing the vector potential choice outside a long solenoid as described in Griffiths' "Quantum Mechanics." The proposed vector potential, ##\textbf{A} = \alpha \frac{\Phi}{2\pi r} \hat{\phi}##, leads to different energy levels, ##E_n = \frac{\hbar^2}{2mb^2}\bigg(n-\alpha \frac{\Phi}{\Phi_0}\bigg)^2##, which raises questions about gauge dependence. It is concluded that this transformation does not qualify as a gauge transformation in the interior of the solenoid, as physical observables remain gauge invariant, particularly the magnetic flux. References to Cohen-Tannoudji and K.-H. Yang provide further insights into gauge invariance in quantum mechanics.

PREREQUISITES
  • Understanding of gauge transformations in quantum mechanics
  • Familiarity with the Aharonov-Bohm effect
  • Knowledge of the time-independent Schrödinger equation (TISE)
  • Basic concepts of magnetic flux quantization
NEXT STEPS
  • Study gauge invariance in quantum mechanics using Cohen-Tannoudji's textbook
  • Examine K.-H. Yang's papers on gauge-invariant interpretations of quantum mechanics
  • Learn about the implications of gauge transformations on energy eigenvalues
  • Explore the relationship between magnetic flux and observable quantities in quantum systems
USEFUL FOR

Quantum physicists, graduate students in physics, and researchers interested in gauge theories and the Aharonov-Bohm effect.

KDPhysics
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TL;DR
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?
 
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To study a gauge transformation one must consider the vector potential everywhere, i.e. not only outside of the solenoid, but also inside of it. Since the magnetic field does not vanish in the interior, it can be shown that a multiplication by ##\alpha## is not really a gauge transformation in the interior. Hence your transformation is not really a gauge transformation. But nice try!
 
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Physically observable quantities are always gauge invariant. Whatever you calculate are not the energy eigenvalues of a gauge-invariant Hamiltonian. In the AB effect what's observable is a relative phase, but this phase is not gauge dependent but is given by the magnetic flux inside the solenoid, which is a gauge-invariant quantity. The gauge transformation must also fulfill the boundary conditions at the boundary of the solenoid.

A very concise treatment of gauge invariance in quantum mechanics can be found in the textbook by Cohen-Tannoudji et al.

Very illuminating is also

K.-H. Yang, Gauge-invariant interpretations of quantum mechanics, Ann. Phys. (NY) 101, 62 (1976)
https://doi.org/10.1016/0003-4916(76)90275-X

K.-H. Yang, Physical interpretation of classical gauge transformations, Ann. Phys. (NY) 101, 97 (1976)
https://doi.org/10.1016/0003-4916(76)90276-1
 
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Thanks for the answers, very enlightening!
 
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Note also that energy spectrum is gauge invariant. See e.g. Ballentine's book, Eq. (11.23).
 
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