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fix(HJ): A.04 — Derive EM Lagrangian, expand gauge/invariance, expand WKB

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Krishna Ayyalasomayajula
2026-05-02 12:24:14 -05:00
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concepts/advanced/hj-em-coupling.tex
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@@ -2,6 +2,25 @@
This subsection extends the Hamilton--Jacobi framework to a charged particle moving in electromagnetic fields by replacing the canonical momentum with the minimal-coupling substitution $p \to p - qA$.
\nt{Where does the $q\,\vec{v}\cdot\vec{A}$ term come from?}{
We can reverse-engineer the electromagnetic Lagrangian by demanding that the Euler--Lagrange equations reproduce the Lorentz force. Start from the ansatz $\mcL = \tfrac12 m v^2 - q\varphi + q\,\vec{v}\cdot\vec{A}$ and check the $x$-component. The Euler--Lagrange equation reads
\[
\frac{\dd}{\dd t}\!\left(\pdv{\mcL}{\dot{x}}\right) = \pdv{\mcL}{x}.
\]
The canonical momentum is $\pdv{\mcL}{\dot{x}} = m\dot{x} + qA_x$. Its total time derivative expands as
\[
\frac{\dd}{\dd t}(m\dot{x}+qA_x) = m\ddot{x} + q\pdv{A_x}{t} + q\dot{x}\pdv{A_x}{x} + q\dot{y}\pdv{A_x}{y} + q\dot{z}\pdv{A_x}{z}.
\]
The spatial derivative of the Lagrangian is
\[
\pdv{\mcL}{x} = -q\pdv{\varphi}{x} + q\dot{x}\pdv{A_x}{x} + q\dot{y}\pdv{A_y}{x} + q\dot{z}\pdv{A_z}{x}.
\]
Equating left and right sides and rearranging, the terms $q\dot{x}\pdv{A_x}{x}$ cancel, leaving
\[
m\ddot{x} = q\Biggl(-\pdv{\varphi}{x} - \pdv{A_x}{t}\Biggr) + q\dot{y}\Biggl(\pdv{A_y}{x} - \pdv{A_x}{y}\Biggr) + q\dot{z}\Biggl(\pdv{A_z}{x} - \pdv{A_x}{z}\Biggr).
\]
The first grouped term is $qE_x$ using $E_x = -\pdv{\varphi}{x} - \pdv{A_x}{t}$. The next two are the $x$-component of $q(\vec{v}\times\vec{B})_x = q(\dot{y}B_z - \dot{z}B_y)$ using $B_z = \pdv{A_y}{x} - \pdv{A_x}{y}$ and $B_y = \pdv{A_x}{z} - \pdv{A_z}{x}$. Thus $m\ddot{x} = q(E_x + (\vec{v}\times\vec{B})_x)$, matching the Lorentz force. The Lagrangian $\mcL = \tfrac12 mv^2 - q\varphi + q\,\vec{v}\cdot\vec{A}$ is not a guess --- it is uniquely fixed by the requirement that the variational principle yield $\vec{F} = q(\vec{E}+\vec{v}\times\vec{B})$.}
\dfn{Lagrangian for a charged particle in electromagnetic fields}{
Let a particle of mass $m$ and charge $q$ move with velocity $\vec{v}$ in an electromagnetic field described by the scalar potential $\varphi(\vec{r},t)$ and the vector potential $\vec{A}(\vec{r},t)$. The Lagrangian is
\[
@@ -15,9 +34,10 @@ so that in vector notation,
\[
\vec{p} = m\vec{v} + q\vec{A}.
\]
The kinetic (mechanical) momentum is $m\vec{v} = \vec{p} - q\vec{A}$, and it is this combination that enters the kinetic-energy part of the Hamiltonian.}
The kinetic (mechanical) momentum is $m\vec{v} = \vec{p} - q\vec{A}$, and it is this combination that enters the kinetic-energy part of the Hamiltonian. The passage from a free-particle Hamiltonian to the electromagnetic one simply requires the substitutions $\vec{p} \to \vec{p} - q\vec{A}$ and $E \to E - q\varphi$. This procedure is called \emph{minimal coupling} because it represents the lowest-order way to couple electromagnetic potentials to particle motion: the potentials enter only through the simple shift of the canonical momentum, with no higher-order derivative couplings, no spin-dependent terms, and no direct coupling of the field-strength tensor to the particle coordinates.}
\nt{The canonical momentum differs from the kinetic momentum. The extra term $q\vec{A}$ in the canonical momentum is what distinguishes a charged particle's Hamiltonian dynamics from those of a free particle, even in the absence of a scalar potential. Because $\vec{A}$ generally depends on position, the canonical momentum is not simply $m\vec{v}$, and its time derivative is not equal to the mechanical force. Instead, Hamilton's equations for the canonical variables reproduce the full Lorentz-force law.}
\nt{Canonical versus kinetic momentum}{
It is crucial to distinguish two notions of momentum for a charged particle. The canonical momentum $\vec{p} = m\vec{v} + q\vec{A}$ is the formal variable that appears in Hamilton's equations and the Hamilton--Jacobi equation. It is the quantity conserved when its corresponding coordinate is cyclic (absent from the Hamiltonian). The kinetic momentum $m\vec{v} = \vec{p} - q\vec{A}$, by contrast, is the physically measured momentum: it is the mass times the actual velocity of the particle, and its time derivative equals the mechanical Lorentz force $q(\vec{E}+\vec{v}\times\vec{B})$. These two momenta differ by $q\vec{A}$. Because $\vec{A}$ generally depends on position, the canonical momentum is not simply $m\vec{v}$, and its time derivative is not equal to the mechanical force. The canonical momentum is what Hamilton's equations govern; the kinetic momentum is what a detector would measure.}
\thm{Hamiltonian for a charged particle in electromagnetic fields}{
Let $m$ denote the mass, let $q$ denote the charge, let $\varphi(\vec{r},t)$ denote the scalar potential, and let $\vec{A}(\vec{r},t)$ denote the vector potential. The canonical momentum has components $p_i$. Then the Hamiltonian of the charged particle is
@@ -86,31 +106,55 @@ The passage from a free particle to a charged particle in electromagnetic fields
\qquad
E \longrightarrow E - q\varphi.
\]
Under a gauge transformation of the potentials,
To understand why these substitutions are consistent, consider gauge transformations of the potentials. An arbitrary smooth scalar function $\chi(\vec{r},t)$ defines the gauge transformation
\[
\vec{A}' = \vec{A} + \nabla\Lambda,
\vec{A}' = \vec{A} + \nabla\chi,
\qquad
\varphi' = \varphi - \pdv{\Lambda}{t},
\varphi' = \varphi - \pdv{\chi}{t}.
\]
the HJ equation retains its form provided the principal function transforms as
The physical electric and magnetic fields expressed in terms of potentials are $\vec{E} = -\nabla\varphi - \pdv{\vec{A}}{t}$ and $\vec{B} = \nabla\times\vec{A}$. Substituting the primed potentials,
\[
\mcS'(\vec{r},t) = \mcS(\vec{r},t) - q\,\Lambda(\vec{r},t).
\vec{E}' = -\nabla\varphi' - \pdv{\vec{A}'}{t} = -\nabla\varphi + \nabla\pdv{\chi}{t} - \pdv{\vec{A}}{t} - \nabla\pdv{\chi}{t} = \vec{E},
\]
To see this, substitute $\nabla\mcS' = \nabla\mcS - q\nabla\Lambda$ into the HJ equation written in the transformed potentials:
\[
\vec{B}' = \nabla\times\vec{A}' = \nabla\times(\vec{A}+\nabla\chi) = \nabla\times\vec{A} + 0 = \vec{B}.
\]
Both $\vec{E}$ and $\vec{B}$ are unchanged, as required since potentials are a redundant mathematical description and only the fields are physically measurable. How does the Lagrangian behave? Under the gauge transformation, the new Lagrangian is
\[
\mcL' = \tfrac12 m v^2 - q\varphi' + q\,\vec{v}\cdot\vec{A}' = \mcL + q\,\vec{v}\cdot\nabla\chi + q\,\pdv{\chi}{t}.
\]
The extra terms combine into a total time derivative: $q\,\vec{v}\cdot\nabla\chi + q\,\pdv{\chi}{t} = \frac{\dd(q\chi)}{\dd t}$. The action changes by $\int_{t_1}^{t_2} \frac{\dd(q\chi)}{\dd t}\,\dd t = q\chi(t_2) - q\chi(t_1)$, a pure boundary term. Since the principle of least action fixes the endpoints, the variation of this boundary term vanishes: $\delta[q\chi(t_2)-q\chi(t_1)] = 0$. The Euler--Lagrange equations, which depend only on the variation of the action, remain unchanged. This is why adding a total time derivative to any Lagrangian never alters the equations of motion. Under the gauge transformation the HJ equation retains its form provided the principal function transforms as
\[
\mcS'(\vec{r},t) = \mcS(\vec{r},t) - q\,\chi(\vec{r},t).
\]
To see this, substitute $\nabla\mcS' = \nabla\mcS - q\nabla\chi$ into the HJ equation written in the transformed potentials:
\[
\frac{1}{2m}\bigl|\nabla\mcS' - q\vec{A}'\bigr|^2 + q\varphi' + \pdv{\mcS'}{t} = \frac{1}{2m}\bigl|\nabla\mcS - q\vec{A}\bigr|^2 + q\varphi + \pdv{\mcS}{t},
\]
so the unprimed and primed equations are identical. Thus the Hamilton--Jacobi formulation respects electromagnetic gauge invariance.}
\nt{A deep connection to quantum mechanics emerges from the WKB ansatz. Write the wavefunction as $\psi(\vec{r},t) = \exp\bigl(i\mcS(\vec{r},t)/\hbar\bigr)$. Substituting this form into the Schrodinger equation,
\nt{Connection to quantum mechanics through the WKB approximation}{
The Hamilton--Jacobi equation is the classical $\hbar\to 0$ limit of the Schrödinger equation, and the bridge between them is the WKB (Wentzel--Kramers--Brillouin) approximation, a semiclassical method valid when the action is large compared to $\hbar$. The WKB ansatz writes the wavefunction as $\psi(\vec{r},t) = A(\vec{r},t)\,\exp(\mathrm{i}\mcS(\vec{r},t)/\hbar)$, where $\mcS$ plays the role of a phase function and $A$ is a slowly varying amplitude. For simplicity set $A=1$ so that $\psi = \exp(\mathrm{i}\mcS/\hbar)$ and substitute this directly into the Schrödinger equation for a charged particle,
\[
i\hbar\,\pdv{\psi}{t} = \frac{1}{2m}\bigl(-i\hbar\nabla - q\vec{A}\bigr)^2\psi + q\varphi\psi,
\mathrm{i}\hbar\,\pdv{\psi}{t} = \frac{1}{2m}\bigl(-\mathrm{i}\hbar\nabla - q\vec{A}\bigr)^2\psi + q\varphi\psi.
\]
and collecting terms order by order in $\hbar$, the leading-order equation (proportional to $\hbar^0$) is exactly the classical Hamilton--Jacobi equation for a charged particle:
Compute the derivatives: $\pdv{\psi}{t} = (\mathrm{i}/\hbar)(\pdv{\mcS}{t})\psi$, and
\[
\frac{1}{2m}\bigl|\nabla\mcS - q\vec{A}\bigr|^2 + q\varphi + \pdv{\mcS}{t} = 0.
(-\mathrm{i}\hbar\nabla - q\vec{A})\psi = -\mathrm{i}\hbar\Bigl(\frac{\mathrm{i}}{\hbar}\nabla\mcS\,\psi\Bigr) - q\vec{A}\psi = (\nabla\mcS - q\vec{A})\psi.
\]
Higher-order terms in $\hbar$ account for quantum corrections. In this sense, the classical Hamilton--Jacobi PDE is the $\hbar\to 0$ limit of the Schrodinger equation.}
Applying the operator a second time,
\[
\bigl(-\mathrm{i}\hbar\nabla - q\vec{A}\bigr)^2\psi = (-\mathrm{i}\hbar\nabla)\bigl[(\nabla\mcS - q\vec{A})\psi\bigr] - q\vec{A}\cdot(\nabla\mcS - q\vec{A})\psi.
\]
The gradient of the product gives $(\nabla\mcS - q\vec{A})(\mathrm{i}/\hbar)(\nabla\mcS)\psi + (\mathrm{i}/\hbar)(\nabla\mcS)(\nabla\mcS - q\vec{A})\psi$ plus the term involving $\nabla\cdot(\nabla\mcS-q\vec{A})$. Multiplying by $-\mathrm{i}\hbar$, the leading-order piece (proportional to $\hbar^0$) is $|\nabla\mcS - q\vec{A}|^2\psi$, while a correction proportional to $\hbar$ involves $\nabla\cdot(\nabla\mcS - q\vec{A})$. The right-hand side of the Schrödinger equation becomes
\[
\frac{1}{2m}\Bigl[|\nabla\mcS - q\vec{A}|^2 - \mathrm{i}\hbar\,\nabla\cdot(\nabla\mcS - q\vec{A})\Bigr]\psi + q\varphi\psi.
\]
The left-hand side is $-\pdv{\mcS}{t}\,\psi$. Equating and dividing by $\psi$, the $\hbar^0$ (leading-order) terms give exactly
\[
\frac{1}{2m}|\nabla\mcS - q\vec{A}|^2 + q\varphi + \pdv{\mcS}{t} = 0,
\]
which is the classical Hamilton--Jacobi equation for a charged particle. The $\hbar^1$ term $-\mathrm{i}\hbar/(2m)\,\nabla\cdot(\nabla\mcS - q\vec{A}) = 0$ yields a continuity equation for the amplitude $A$. Higher-order terms in $\hbar$ provide successive quantum corrections. In this way, the Hamilton--Jacobi equation is the zeroth-order term in a systematic semiclassical expansion of quantum mechanics.}
\ex{Separation for time-independent fields}{
Suppose the electromagnetic potentials are time-independent. Then the Hamiltonian has no explicit time dependence and the total energy $E$ is conserved. The time variable separates from the action as $\mcS = W(\vec{r}) - Et$, and the Hamilton--Jacobi equation reduces to the time-independent form
@@ -187,15 +231,13 @@ m^2\omega^2 = (q B_0)^2,
\]
The HJ separation constant analysis thus recovers the cyclotron frequency exactly, independent of the guiding-center location and the transverse energy. The transverse energy $E_\perp = 1.60\times 10^{-17}\,\mathrm{J}$ sets the gyroradius but does not affect the frequency.
Therefore, the Hamiltonian is
Therefore,
\[
\mcH = \frac{1}{2m}\Bigl[p_x^2 + \bigl(p_y - q B_0 x\bigr)^2 + p_z^2\Bigr],
\]
the Hamilton--Jacobi equation is
\[
\frac{1}{2m}\left[\left(\pdv{\mcS}{x}\right)^2 + \left(\pdv{\mcS}{y} - q B_0 x\right)^2 + \left(\pdv{\mcS}{z}\right)^2\right] + \pdv{\mcS}{t} = 0,
\]
the cyclic coordinates are $y$ and $z$, and the cyclotron frequency is
\[
\omega_c = 1.76\times 10^{11}\,\mathrm{rad/s}.
\text{cyclic coordinates: } y, z,\qquad \omega_c = 1.76\times 10^{11}\,\mathrm{rad/s}.
\]