Is it possible to derive the Aharonov-Bohm effect using translation operators alone?

Question

There are multiple proofs of the Aharonov-Bohm effect. Arguably the most simple proof is the original one by Aharonov and Bohm which appears to be proven by inspection. Two other methods I have seen use the machinery of the Berry phase and the path integral. I was wondering whether it is possible to derive the effect using translation operators because, after all, the motion of a charged particle around a solenoid can be broken up into successive translations.

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Canonical Formalism for a charged particle in a magnetic field

If we work with a non-relativistic particle for simplicity, the canonical momentum of a charged particle in a magnetic field with vector potential $\mathbf{A}$ is deduced from the Lagrangian as

$$ L = \frac{1}{2}m \dot{\mathbf{x }}^2+q \dot{\mathbf{x }} \cdot \mathbf{A}(\mathbf{x}) \quad \Rightarrow \quad \mathbf{p}=\frac{\partial L}{\partial \dot{\mathbf{x }}} = m\dot{\mathbf{x }}+q\mathbf{A}(\mathbf{x}).$$

The canonical momentum $\mathbf{p}$ is the generator of translations as it has the Poisson bracket $\{ x^i, p_j \}= \delta^i_j $. When we quantise the theory we would write down the translation operator by exponentiating the canonical momentum operator as

$$ T(\mathbf{a}) = e^{-i \mathbf{a} \cdot \mathbf{p} }$$

where in the position representation I would make the identification that the canonical momentum is given by $\mathbf{p} = -i \nabla$. I would expect that I could take the limit of successive applications of $T(\mathbf{a})$ along a closed path enclosing a solenoid to achieve the Aharonov-Bohm phase however the canonical momenta commute amongst themselves (as their Poisson brackets vanish) so successive translations around a closed loop would just give me the identity. It appears the problem is that when quantising the canonical momentum operator by writing $\mathbf{p} = -i\nabla$, the vector potential $\mathbf{A}$ has completely disappeared!

One would hope that the momentum operator would have some form of dependence on $\mathbf{A}$ in order for the translation operator to give us a non-trivial phase. I could explicitly substitute in the expression for the canonical momentum to get

$$ T(\mathbf{a}) = e^{-i \mathbf{a} \cdot(m\dot{\mathbf{x }} + q\mathbf{A}(\mathbf{x})) }$$

However, I do not know what the mechanical momentum operator $m \dot{\mathbf{x}}$ is represented as so I am unsure how to proceed.

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Magnetic Translation operator

In these notes at the bottom of page 57 there is a "magnetic translation operator" which is given by

$$ T(\mathbf{a}) = e^{-i \mathbf{a} \cdot ( i \nabla + e\mathbf{A} )}$$

which seems promising as it gives us phases, however I am not sure where the author got this from. Comparing to the canonical formalism above, it would appear he has made the identification that $ m\dot{\mathbf{x}} \equiv i \nabla$. This does not seem correct to me because one represents the canonical momentum $\mathbf{p}$ as $-i\nabla$ in the position representation in order to satisfy the canonical commutator $[x^i,p_j]=i\delta^i_j$, whereas $m \dot{\mathbf{x}}$ is not the canonical momentum operator and is the mechanical momentum. I am also uncomfortable with this operator being interpreted as a translation operator because the operator depends upon space through $\mathbf{A}$, so surely this would only make sense infinitesimally and I should have some integral of $\mathbf{A}$ in the exponent for a finite translation?

Magnetic translation operators appear when we talk about lattice gauge theories and are used to show the Aharonov-Bohm effect there, but I would like to do this in the continuum.

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My question

Can one prove the Aharonov-Bohm effect using the machinery of translation operators derived in the canonical formalism?

Answer 1

The question raised this objection to the "magnetic translation operator":

I am also uncomfortable with this operator being interpreted as a translation operator because the operator depends upon space through $\mathbf{A}$, so surely this would only make sense infinitesimally and I should have some integral of $\mathbf{A}$ in the exponent for a finite translation?

I'll address this objection by deriving a different expression for the magnetic translation operator, one that does have an integral of $\mathbf{A}$ in the exponent. Then I'll suggest a perspective that might help answer the bottom-line question. I'll set $e=1$ to reduce clutter.

The true nature of the magnetic translation operator

Consider the operators $T(\mathbf{c})$ and $U(\mathbf{c})$ defined by the different-looking conditions \begin{align} \newcommand{\bfA}{\mathbf{A}} \newcommand{\bfc}{\mathbf{c}} \newcommand{\bfx}{\mathbf{x}} \newcommand{\pl}{\partial} T(\bfc)\psi(\bfx) &=\exp\big(\bfc\cdot(\nabla-i\bfA)\big)\psi(\bfx) \tag{1} \\ U(\bfc)\psi(\bfx) &=\exp\left(-i\int_\bfx^{\bfx+\bfc} d\bfx'\cdot\bfA(\bfx')\right)\psi(\bfx+\bfc) \tag{2} \end{align} where the integral is along the straight path from $\bfx$ to $\bfx+\bfc$. The operator (1) is the unnatural-looking "magnetic translation operator." The operator (2) has an obvious generalization to any path from $\bfx$ to $\bfx+\bfc$, and then it is essentially the definition of the gauge field: the gauge field provides a so-called connection that tells us how to compare the phases of charged objects at different locations in space. The comparison is path-dependent.

Even though the operators defined in (1) and (2) look different, they are actually identical: $$ T(\bfc)=U(\bfc). \tag{3} $$ This identity holds for arbitrary finite $\bfc$, not just for infinitesimal displacements, but it is limited to straight paths. Any path may be assembled from infinitesimal straight segments, which agrees with the feeling that the operator (1) becomes more "natural" when $\bfc$ is infinitesimal.

Proof of (3)

Let $V$ denote the canonical translation operator $$ V(\bfc) = \exp(\bfc\cdot\nabla). \tag{4} $$ I'll prove the identity $$ T(\bfc)V(-\bfc) =\exp\left(-i\int_\bfx^{\bfx+\bfc} d\bfx'\cdot\bfA(\bfx')\right), \tag{5} $$ which implies (3), as we can verify by applying both sides of (5) to $\psi(\bfx+\bfc)$. To prove (5), I'll show that both sides satisfy the same first-order differential equation with the same initial condition. Explicitly, the conditions $f(\theta) = 1$ and $$ \frac{d}{d\theta}f(\theta) =f(\theta)\times \big(-i\bfc\cdot\bfA(\bfx+\theta\bfc)\big) \tag{6} $$ are both satisfied by $$ f(\theta) =\exp\left(-i\int_\bfx^{\bfx+\theta\bfc} d\bfx'\cdot\bfA(\bfx')\right), \tag{7} $$ and also by $$ f(\theta) = T(\theta\bfc)V(-\theta\bfc) \tag{8} $$ so (7) and (8) must be the same function. To prove that (7) satisfies (6), just rewrite the integral as an integral over $\theta$, which we can do because the path is straight. To prove that (8) also satisfies (6), use \begin{align} \frac{d}{d\theta} T(\theta\bfc) &=T(\theta\bfc)\big(\bfc\cdot(\nabla-i\bfA(\bfx))\big) \tag{9} \\ \frac{d}{d\theta} V(\theta\bfc) &=-\bfc\cdot\nabla V(-\theta\bfc) \tag{10} \end{align} to get \begin{align} \frac{d}{d\theta} T(\theta\bfc)V(-\theta\bfc) &= T(\theta\bfc)\big(-i\bfc\cdot\bfA(\bfx)\big)V(-\theta\bfc) \\ &= T(\theta\bfc)V(-\theta\bfc)\big(-i\bfc\cdot\bfA(\bfx+\theta\bfc)\big) \tag{11} \end{align} where the familiar property of the canonical translation operator was used in the last step.

Perspective

Which is the more natural way to deduce the Aharohov-Bohm effect: using the magnetic translation operator (1), or using the canonical translation operator (4)?

The magnetic translation operator (1) is unnatural in the sense that it is gauge covariant only for straight paths. By that measure, the canonical translation operator (4) is even more unnatural, because it isn't gauge covariant at all.

Maybe the best way to think about the Aharohov-Bohm effect is to think carefully the definition of a gauge field. Gauge fields are often introduced using an infinitesimal perspective, where $\bfA$ is used to make the derivative gauge covariant. However, thinking in finite terms can be more satisfying. We can think of a gauge field as a map $C\mapsto g(C)$ that assigns an element $g(C)$ of the group to each curve $C$ in space (or spacetime), subject to some natural rules. For example, joining two paths corresponds to multiplying their group elements in the same order. The group is $U(1)$ in the case of electromagnetism, but can be any group, such as a non-abelian group (where the order of multiplication matters) or even a discrete group (where the infinitesimal perspective is useless). This is how we think about a gauge field in lattice gauge theory, and we can — I dare say should — think about it this way in continuous space, too.

Intuitively, the map $C\mapsto g(C)$ serves the same purpose in gauge theory that "parallel transport" serves in general relativity: it defines what happens when an object moves from one point to another along a given path, if nothing else influences its behavior. We don't need to work hard to understand how this leads to the Aharonov-Bohm effect, because we implicitly already did that hard work when we constructed the lagrangian. In hindsight, the lagrangian was constructed precisely to ensure that an object moving freely along the curve $C$ will be multiplied by the group element $g(C)$. So, in a way, the derivation of the Aharonov-Bohm effect is implicit in the construction of the lagrangian. Any other way of deriving the Aharonov-Bohm effect is just a double-check to make sure we constructed the lagrangian correctly.

In contrast, Berry phases — though interesting and important — do not enjoy this same fundamental status. Berry phases are relevant in adiabatic processes, whereas the Aharonov-Bohm effect is a direct manifestation of the very concept of a gauge field.

Answer 2

The Aharonov Bohm phase can be seen as the result of magnetic translation on the two dimensional cylinder,but not on the two dimensional plane. See "Line of magnetic monopoles and an extension of the Aharonov Bohm effect" in Annals of Physics, 2016.