Must my gauge transformation be defined on all space?

Consider the set-up of the Aharonov-Bohm effect with a solenoid of radius $$a$$ aligned along the $$z$$-direction and uniform magnetic field $$B_0$$ within. One possible vector potential, in cylindrical coordinates, is

$$A = \begin{cases} \frac{B_0r}{2}\hat{\phi} \quad \text{for } ra \end{cases}$$

If I apply a gauge transformation using the gauge parameter $$\Lambda = -\frac{B_0 a^2\phi}{2}$$ so that $$\nabla \Lambda = -\frac{B_0a^2}{2r}\hat{\phi}$$, then I would be able to make the vector potential vanish outside the solenoid. Since physics (such as the phase difference measurable in the Aharonov-Bohm set-up) cannot depend on the gauge, there must be something wrong with my gauge transformation.

Am I not allowed to take $$\Lambda$$ above to be my gauge parameter? It is easy to find examples in E&M where $$A$$ is ill-defined (i.e. infinity, due to $$r=0$$ in denominator) at some point in space so I am not sure if this is a problem that invalidates the proposed new $$A$$.

• The A-B phase is $\theta = \oint_C dx^\mu A_\mu$ where $C$ is any contour surrounding the solenoid. Under gauge transformations $\theta \to \theta + \oint_C dx^\mu \partial_\mu \Lambda$. The second term vanishes IF $\Lambda$ is a continuous function on the contour $C$. With your choice of $\Lambda$, we have $A_\mu = 0$ in the new gauge so we now have $\theta = \oint_C dx^\mu \partial_\mu \Lambda$. However, this last term no longer vanishes since your $\Lambda$ has a discontinuity at $\phi=0$ (or $\phi=2\pi$). This discontinuity is now responsible for the A-B effect. – Prahar Mitra Jan 29 '20 at 19:38

There is nothing wrong with your gauge transformation.

Transforming the electromagnetic potentials (vector potential $$\vec{A}$$ and scalar potential $$\Phi$$) by $$\begin{cases} \vec{A} &\to \vec{A} - \vec{\nabla}\Lambda \\ \Phi &\to \Phi + \frac{\partial}{\partial t}\Lambda \end{cases} \tag{1}$$ doesn't change the electromagnetic fields ($$\vec{E}$$ and $$\vec{B}$$). And as long we avoid quantum mechanics the transformation (1) seems to be a perfect symmetry transformation.

But the situation changes when we add quantum mechanics to describe the charged particles. For a particle with charge $$q$$ the transformation (1) makes an additional phase shift $$\Delta\phi=q\Delta\Lambda/\hbar$$ between any two points of its trajectory. Hence it does change the physics, i.e. it is not a symmetry transformation anymore.

This has been demonstrated in the Aharanov-Bohm effect. A charged particle moving through a potential $$\vec{A}$$ aquires a phase change, even when the magnetic field $$\vec{B}$$ is zero everywhere along the trajectory of the particle. It has even been verified experimentally. Quoted from Wikipedia - Aharanov-Bohm effect - Magnetic solenoid effect:

An early experiment in which an unambiguous Aharonov–Bohm effect was observed by completely excluding the magnetic field from the electron path (with the help of a superconducting film) was performed by Tonomura et al. in 1986.

This seeming paradoxon was resolved by extending the transformation (1). The complete transformation is defined as the combination of gauge-transforming the electromagnetic potentials ($$\vec{A}$$ and $$\Phi$$) and phase-transforming the wave function ($$\psi$$) of the charged particle (see also UCSD Physics 130 - Gauge Symmetry in Quantum Mechanics).

$$\begin{cases} \vec{A} &\to \vec{A} - \vec{\nabla}\Lambda \\ \Phi &\to \Phi + \frac{\partial}{\partial t}\Lambda \\ \psi &\to e^{-iq\Lambda/\hbar}\ \psi \end{cases} \tag{2}$$

Only this combined gauge/phase transformation (2) makes up a symmetry transformation, i.e. it doesn't change the physics.

• I thought about this further & have a question. When Griffiths discussed Aharonov-Bohm (10.2.3), he discussed the scenario of a charged bead on a circular ring around the solenoid. The wave function solutions are of the form $\psi = A\exp(in\phi)$ where $n$ is an integer, required by continuity of the wavefunction around the ring. If I apply the gauge transformation in my post, then the wavefunction needs to change by $\exp(-iq\Lambda/\hbar)=\exp(iqB_0a^2\phi/2\hbar)$ but $qB_0a^2/2\hbar$ need not be an integer and continuity is ruined. So am I right to say that not all $\Lambda$ above works? – suncup224 Feb 1 '20 at 11:46