Potential energy of a particle inside a magnetic vector potential

Question

Why is the potential energy of a particle inside a magnetic vector potential equal to $-\frac{e}{c}\cdot\vec{A}(\vec{x}(t))\cdot\dot{\vec{x}}(t)$? It appears for example inside the lagrangian of the Aharonov-Bohm effect.

Answer 1

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Classically, the electromagnetic field acts on the particle via the Lorentz force law,

\begin{equation} \mb F\plr{\mb x,t} \e q\blr{\:\mb E\plr{\mb x,t}\p \mb{\dot{x}}\bl\times\mb B\plr{\mb x,t}\Vp{A^B_C}\vp} \tl{A-01} \end{equation} where $\:\mb x \e \plr{\smsup{x}{1},\smsup{x}{2},\smsup{x}{3}}\:$ and $\:\mb{\dot{x}}\e \plr{\smsup{\dot x}{1},\smsup{\dot x}{2},\smsup{\dot x}{3}}\:$ denote the position and velocity of the particle, $\:t\:$ is the time, and $\:\mb E\:$ and $\:\mb B\:$ are the electric and magnetic fields.If no other forces are present, Newton’s second law yields the equation of motion \begin{equation} m\,\mb{\ddot x} \e q\blr{\:\mb E\plr{\mb x,t}\p \mb{\dot{x}}\bl\times\mb B\plr{\mb x,t}\Vp{A^B_C}\vp} \tl{A-02} \end{equation}

Introducing the electromagnetic scalar and vector potentials $\:\Phi\plr{\mb x,t}\:$ and $\:\mb A\plr{\mb x,t}\:$: \begin{align} \mb E\plr{\mb x,t} & \e \m\bl\nabla\Phi\plr{\mb x,t}\m\dfrac{\partial\mb A\plr{\mb x,t}}{\partial t} \tl{A-03}\\ \mb B\plr{\mb x,t} & \e \bl\nabla\bl\times\mb A\plr{\mb x,t}\vp \tl{A-04} \end{align} equation \eqref{A-02} yields \begin{equation} m\,\mb{\ddot x} \e q\plr{\!\!\!\!\plr{\m\bl\nabla\Phi\plr{\mb x,t}\m\dfrac{\partial\mb A\plr{\mb x,t}}{\partial t}\p \mb{\dot{x}}\bl\times \blr{\bl\nabla\bl\times\mb A\plr{\mb x,t}\vp} \Vp{A^B_C}\vp}\!\!\!\!} \tl{A-05} \end{equation} The classical Lagrangian \begin{equation} L\plr{\mb x,\mb{\dot{x}},t} \tl{A-06} \end{equation} by which the equations of motion \eqref{A-05} would be derived from the Euler-Lagrange equations \begin{equation} \dfrac{\mr d}{\mr dt}\plr{\dfrac{\partial L}{\partial\mb{\dot{x}}}}\m \dfrac{\partial L}{\partial\mb x} \e \bl 0 \tl{A-07} \end{equation} or component-wise \begin{equation} \dfrac{\mr d}{\mr dt}\plr{\dfrac{\partial L}{\partial\smsup{\dot x}{k}}}\m \dfrac{\partial L}{\partial\smsup{x}{k}} \e 0\,,\qquad k\e 1,2,3 \tl{A-08} \end{equation} could be determined by an appropriate method to be \begin{equation} \boxed{\:\:L\plr{\mb x,\mb{\dot{x}},t} \e \tfrac12 m\Vlr{\mb{\dot x}}^2 \p q\plr{\m \Phi \p \mb{\dot x}\bl\cdot\mb A\Vp{A^2}}\Vp{\dfrac{\tfrac{c}{d}}{\tfrac{c}{d}}}\:\:} \tl{A-09} \end{equation} This follows the usual prescription for the Lagrangian as kinetic energy minus potential energy, with $\:q\,\Phi\:$ serving as the potential energy function, except for the $\:q\,\mb{\dot x}\bl\cdot\mb A\:$ term !!!.

From this Lagrangian we derive the Hamiltonian \begin{equation} \boxed{\:\:H\plr{\mb x,\mb p} \e \dfrac{\Vlr{\mb p \m q\mb A}^2}{2m}\p q\,\Phi\Vp{\dfrac{\tfrac{c}{d}}{\tfrac{c}{d}}}\:\:} \tl{A-10} \end{equation} This looks a lot like the Hamiltonian for a non-relativistic particle in a scalar potential, \begin{equation} H \e \dfrac{\Vlr{\mb p}^2}{2m}\p V\plr{\mb x,t} \tl{A-11} \end{equation} In equation \eqref{A-10}, the term $\:q\,\Phi\:$ acts like a potential energy, which is no surprise. More interestingly, the vector potential appears via the substitution \begin{equation} \mb p \quad\bl{-\!\!\!-\!\!\!-\!\!\!\longrightarrow}\quad\mb p \m q\mb A \tl{A-12} \end{equation} What does this mean? Think about what “momentum” means for a charged particle in an electromagnetic field. Noether’s theorem states that each symmetry of a system (whether classical or quantum) is associated with a conservation law. Momentum is the quantity conserved when the system is symmetric under spatial translations. One of Hamilton’s equations states that \begin{equation} \dfrac{\mr dp_i}{\mr dt}\e \dfrac{\partial H}{\partial \smsup{x}{i}} \tl{A-13} \end{equation} which implies that if $\:H\:$ is $\:\mb x\m$independent, then $\:\mr d\mb p/\mr dt\e \bl 0$. But when the electromagnetic potentials are $\:\mb x\m$independent, the quantity $\:m\mb{\dot{x}}\:$ (which we usually call momentum) is not necessarily conserved!

On the other hand, if the electromagnetic potentials $\:\Phi, \mb A\:$ are $\:\mb x\m$independent then : \begin{equation} \dfrac{\partial L}{\partial\mb x}\e q\bl\nabla\plr{\m\Phi\p \mb{\dot{x}}\bl\cdot \mb A \vp} \e \bl 0 \tl{A-14} \end{equation} so from \eqref{A-07} \begin{equation} \dfrac{\mr d}{\mr dt}\plr{\dfrac{\partial L}{\partial\mb{\dot{x}}}}\e \dfrac{\partial L}{\partial\mb x} \e \bl 0 \tl{A-15} \end{equation} In this case the quantity \begin{equation} \boxed{\:\:\mb p \e \dfrac{\partial L}{\partial\mb{\dot{x}}} \e m\,\mb{\dot x}\p q\mb A\:\:\Vp{\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}}} \tl{A-16} \end{equation} is conserved and this is the appropriate canonical momentum for a particle in an electromagnetic field.

Conclusively, the vector potential $\:\mb A\:$ appears in the canonical momentum $\:\mb p\:$ and not in a term of the potential energy.

$\bl{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}$

Reference : ''Quantum Mechanics III'' by Y. D. Chong, LibreTexts 2023, $\bl{\S 5.1}$ Quantization of the Lorentz Force Law.

Answer 2

To call this quantity the potential energy is misleading. When a particle is subject to an external (non-dynamical) field we can define a potential energy $U$ for the particle if doing so gives a total energy $E$ that is conserved. For example this will be the case for a particle that is subject to solely a gravitational field with $U = mgz$ since the sum of that potential energy and the kinetic energy $T = \frac{1}{2}mv^2$ will be conserved.

For a particle subject to an external electromagnetic field, the potential energy is simply $U = q\varphi$; the conserved total energy is $E = \frac{1}{2}mv^2 + q\varphi$. If you insert a term corresponding with the magnetic vector potential then you no longer end up with a conserved energy. For example consider if $\varphi = 0$ and $A = kx\hat{y}$. This magnetic vector potential gives a uniform magnetic field, $B = k\hat{z}$. If a charged particle's initial velocity lies in the xy plane then it will travel in a circular path with constant speed around an axis that is parallel to the z axis. The particle's kinetic energy won't change along this path, but the quantity $-qv \cdot A$ will certainly change because at some points along the trajectory $v$ will be parallel to the y-axis and at other points it will be perpendicular to the y-axis. So adding the two, you get a quantity that's not conserved. This means $\frac{1}{2}mv^2 - qv\cdot A$ is not conserved so $-qv\cdot A$ cannot be interpreted as a potential energy.

In general if we have a Lagrangian of the form $L = \frac{1}{2}mv^2 - U$ then $U$ has an interpretation as potential energy if it doesn't depend on $v$ (it can depend on the position $x$). $U$ that depends on $v$ is sometimes called a "generalized potential".

In an answer by Hans de Vries it is mentioned that the physical interpretation of the Lagrangian really has to do with the change in a particle's quantum-mechanical phase per unit proper time along its trajectory. Since the Aharonov–Bohm effect tells us that two charged particles experience different amounts of phase shifts when one of them travels in the same direction as $A$ and one travels in the opposite direction, we expect to see a term proportional to $v \cdot A$ in the Lagrangian.

Answer 3

The vector potential represents potential momentum. As the hamiltonian contains the square of the momentum an interaction term proportional to $\vec p \cdot \vec A$ results.