04 Jun 2026

I: Magnetic oscillator stability and rotated-coordinate reduction

magnetic oscillator minimal coupling angular momentum stability quantum quench

This chapter develops a two-dimensional charged oscillator from the basic minimal-coupling rule. No prior magnetic-oscillator calculation is assumed. The central distinction is that a rotation can remove the $xy$ term from the potential, but it does not remove the magnetic angular-momentum coupling. Keeping those two operations separate is necessary for correct stability, state-preparation, and time-evolution calculations.

1. Physical starting point

Consider a particle of mass $M$ and electric charge $q_{\mathrm e}$ moving in the $xy$ plane. Its position and canonical momentum operators are

\[\boldsymbol r=(x,y)^{\mathsf T}, \qquad \boldsymbol p=(p_x,p_y)^{\mathsf T},\]

with

\[[x,p_x]=[y,p_y]=\mathrm i\hbar, \qquad [x,y]=[p_x,p_y]=[x,p_y]=[y,p_x]=0.\]

A uniform magnetic field points perpendicular to the plane, $\boldsymbol B=B\hat{\boldsymbol z}$. A vector potential is not unique, but a convenient symmetric-gauge choice is

\[\boldsymbol A(x,y)=\frac{B}{2}(-y,x,0), \qquad \boldsymbol\nabla\times\boldsymbol A=B\hat{\boldsymbol z}.\]

Minimal coupling replaces mechanical momentum by

\[\boldsymbol p_{\mathrm{mech}} =\boldsymbol p-\frac{q_{\mathrm e}}{c}\boldsymbol A.\]

The bare trap is allowed to be anisotropic and to have an $xy$ coupling:

\[V_0(x,y) =\frac{M}{2} \left(\omega_x^2x^2+\omega_y^2y^2-2gxy\right).\]

Here $\omega_x$ and $\omega_y$ are trap-frequency parameters, while $g$ has units of frequency squared. The starting Hamiltonian is therefore

\[H_B =\frac{1}{2M} \left(\boldsymbol p-\frac{q_{\mathrm e}}{c}\boldsymbol A\right)^2 +V_0.\]

This formula is already gauge invariant when states and canonical momenta are transformed consistently. The symmetric gauge is chosen only because it makes rotational structure visible.

2. Units and the signed magnetic scale

Set

\[\hbar=2M=1.\]

Thus $\hbar=1$, $M=1/2$, $p_j=-\mathrm i\,\partial_j$, and energy has the same dimension as frequency. Define the signed half-cyclotron frequency

\[\omega_c=\frac{q_{\mathrm e}B}{2Mc}.\]

The sign matters. Reversing either the charge or the magnetic field sends $\omega_c\mapsto-\omega_c$. Terms proportional to $\omega_c^2$ are unchanged, but the direction of angular circulation reverses.

With these conventions,

\[\frac{q_{\mathrm e}}{c}\boldsymbol A =\left(-\frac{\omega_c}{2}y,\frac{\omega_c}{2}x\right),\]

\[\begin{aligned} H_B={}& \left(p_x+\frac{\omega_c}{2}y\right)^2 +\left(p_y-\frac{\omega_c}{2}x\right)^2\\ &+\frac{\omega_x^2x^2}{4} +\frac{\omega_y^2y^2}{4} -\frac{g}{2}xy. \end{aligned}\]

Dimensional check

In these units, $q$ has scale $\text{frequency}^{-1/2}$ and $p$ has scale $\text{frequency}^{1/2}$. Consequently,

\[p^2,\quad \omega^2q^2,\quad gxy,\quad \omega_c(qp)\]

all have the dimension of energy. This quick check catches many missing factors of $\omega_c$ or $1/2$.

3. Expanding the kinetic energy without guessing the sign

Because $[y,p_x]=[x,p_y]=0$, no ordering correction appears:

\[\left(p_x+\frac{\omega_c}{2}y\right)^2 =p_x^2+\omega_cyp_x+\frac{\omega_c^2}{4}y^2,\]

and

\[\left(p_y-\frac{\omega_c}{2}x\right)^2 =p_y^2-\omega_cxp_y+\frac{\omega_c^2}{4}x^2.\]

Introduce the canonical angular momentum

\[L_z=xp_y-yp_x.\]

The two linear terms combine as

\[\omega_cyp_x-\omega_cxp_y=-\omega_cL_z.\]

Hence

\[\boxed{ H_B=p_x^2+p_y^2 +\frac{\omega_x^2+\omega_c^2}{4}x^2 +\frac{\omega_y^2+\omega_c^2}{4}y^2 -\frac{g}{2}xy-\omega_cL_z. }\]

The magnetic field has produced two different structures:

the diamagnetic term, proportional to $\omega_c^2(x^2+y^2)$; and
the chiral term, $-\omega_cL_z$, which is sensitive to the sign of $q_{\mathrm e}B$.

Replacing $\omega_c$ by $\lvert\omega_c\rvert$ everywhere would therefore erase real dynamical information.

4. Diagonalizing the coordinate quadratic form

Separate the coordinate part from the momentum-dependent part:

\[H_B=\boldsymbol p^{\mathsf T}\boldsymbol p +\frac14\boldsymbol r^{\mathsf T}K\boldsymbol r -\omega_cL_z,\]

where

\[K= \begin{pmatrix} \omega_x^2+\omega_c^2 & -g\\ -g & \omega_y^2+\omega_c^2 \end{pmatrix}.\]

Since $K$ is real and symmetric, an orthogonal rotation diagonalizes it. Define

\[\Delta= \sqrt{(\omega_x^2-\omega_y^2)^2+4g^2}.\]

For $\Delta>0$, choose the angle unambiguously through

\[2\theta= \operatorname{atan2}\!\left(-2g,\omega_x^2-\omega_y^2\right).\]

Equivalently,

\[\cos(2\theta)=\frac{\omega_x^2-\omega_y^2}{\Delta}, \qquad \sin(2\theta)=-\frac{2g}{\Delta}.\]

Using atan2 rather than a one-argument arctangent prevents a quadrant error and keeps the $+$ and $-$ eigenvalue labels consistent. If $\Delta=0$, then $g=0$ and $\omega_x^2=\omega_y^2$; every rotation is equally valid.

Let

\[R(\theta)= \begin{pmatrix} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta \end{pmatrix},\]

and rotate coordinates and momenta by the same matrix:

\[\begin{pmatrix}x\\y\end{pmatrix} =R(\theta)\begin{pmatrix}q_+\\q_-\end{pmatrix}, \qquad \begin{pmatrix}p_x\\p_y\end{pmatrix} =R(\theta)\begin{pmatrix}p_+\\p_-\end{pmatrix}.\]

Because $R^{\mathsf T}R=I$,

\[[q_\alpha,p_\beta]=\mathrm i\delta_{\alpha\beta}.\]

The transformation is therefore canonical, not merely a change of labels. Direct multiplication gives

\[R^{\mathsf T}KR =\operatorname{diag}(\Omega_+^2,\Omega_-^2),\]

with the ordered eigenvalues

\[\boxed{ \Omega_\pm^2 =\frac{\omega_x^2+\omega_y^2+2\omega_c^2}{2} \pm\frac{\Delta}{2}. }\]

The same rotation leaves the oriented area form invariant, so

\[L_z=q_+p_- - q_-p_+.\]

The Hamiltonian becomes

\[\boxed{ H_B=p_+^2+p_-^2 +\frac{\Omega_+^2q_+^2}{4} +\frac{\Omega_-^2q_-^2}{4} -\omega_c(q_+p_- - q_-p_+). }\]

The $q_+q_-$ term is gone, but the two axes remain dynamically coupled.

5. Coordinate scales are not dynamical normal-mode frequencies

If $\omega_c=0$, the last term vanishes and $\Omega_\pm$ are ordinary oscillator frequencies. If $\omega_c\ne0$, calling them the normal-mode frequencies is incorrect: they diagonalize only $K$, not the complete quadratic form in $(q_+,q_-,p_+,p_-)$.

Hamilton’s equations, which agree with the Heisenberg equations for this quadratic system, are

\[\begin{aligned} \dot q_+&=2p_+ +\omega_cq_-, & \dot q_-&=2p_- -\omega_cq_+,\\ \dot p_+&=-\frac{\Omega_+^2}{2}q_+ +\omega_cp_-, & \dot p_-&=-\frac{\Omega_-^2}{2}q_- -\omega_cp_+. \end{aligned}\]

With

\[J= \begin{pmatrix}0&-1\\1&0\end{pmatrix}, \qquad D=\operatorname{diag}(\Omega_+^2,\Omega_-^2),\]

eliminating the momenta gives

\[\ddot{\boldsymbol q} +2\omega_cJ\dot{\boldsymbol q} +(D-\omega_c^2I)\boldsymbol q=0.\]

Try $\boldsymbol q(t)=\boldsymbol u e^{\mathrm i\nu t}$. A nonzero $\boldsymbol u$ exists only if

\[\left[ (\Omega_+^2-\omega_c^2)-\nu^2 \right] \left[ (\Omega_-^2-\omega_c^2)-\nu^2 \right] -4\omega_c^2\nu^2=0.\]

Define

\[S=\Omega_+^2+\Omega_-^2+2\omega_c^2, \qquad P=(\Omega_+^2-\omega_c^2) (\Omega_-^2-\omega_c^2).\]

The two positive squared frequencies in the stable regime are

\[\boxed{ \nu_{\mathrm{high,low}}^2 =\frac{S\pm\sqrt{S^2-4P}}{2}. }\]

These $\nu$ values, not $\Omega_\pm$, describe the actual small-amplitude time scales. The associated eigenvectors are generally elliptically polarized combinations of both rotated coordinates.

As a useful check, an isotropic bare trap with frequency $\omega_0$ has $\Omega_+^2=\Omega_-^2=\omega_0^2+\omega_c^2$, and the formula reduces to

\[\nu_{\mathrm{high,low}} =\sqrt{\omega_0^2+\omega_c^2}\pm|\omega_c|.\]

Changing the sign of $\omega_c$ reverses the polarization, even though these two positive frequency magnitudes are unchanged.

6. Stability from the bare trap

Before expanding the squares, the Hamiltonian was

\[H_B= \left(p_x+\frac{\omega_c}{2}y\right)^2 +\left(p_y-\frac{\omega_c}{2}x\right)^2 +\frac14\boldsymbol r^{\mathsf T}K_0\boldsymbol r,\]

where

\[K_0= \begin{pmatrix} \omega_x^2&-g\\ -g&\omega_y^2 \end{pmatrix}.\]

The covariant kinetic squares are nonnegative. A strictly confining trap therefore requires $K_0$ to be positive definite:

\[\boxed{ \omega_x^2>0, \qquad \omega_y^2>0, \qquad \omega_x^2\omega_y^2-g^2>0. }\]

For a real symmetric $2\times2$ matrix, positivity of one leading diagonal entry and the determinant is sufficient; both diagonal inequalities are written to make the physical requirement transparent.

Why is positivity tested on $K_0$, not on $K$? The extra $\omega_c^2I$ in $K$ came from expanding a kinetic square. A wave packet can be translated while its canonical momentum compensates the local vector potential, so that apparent coordinate term cannot repair a negative direction of the bare potential.

Because

\[K=K_0+\omega_c^2I,\]

the eigenvalues of $K_0$ in the rotated basis are

\[\Omega_+^2-\omega_c^2, \qquad \Omega_-^2-\omega_c^2.\]

Thus the equivalent strict-confinement test is

\[\boxed{ \Omega_+^2>\omega_c^2, \qquad \Omega_-^2>\omega_c^2. }\]

At equality, the Hamiltonian remains bounded below because it is a sum of nonnegative covariant kinetic and potential terms, but it is not a strictly confining two-dimensional trap. Here $P=0$, so the dynamical equation has $\nu_{\mathrm{low}}=0$ and a delocalized or infinitely degenerate direction. For localized state preparation and a discrete two-mode basis, use strict inequalities with a numerical safety margin.

7. A complete worked parameter set

Take

\[\omega_c=\frac34,\qquad \omega_x^2=\frac{104}{125},\qquad \omega_y^2=\frac{41}{125},\qquad g=\frac{42}{125}.\]

First,

\[\Delta=\frac{21}{25}, \qquad \cos(2\theta)=\frac35, \qquad \sin(2\theta)=-\frac45,\]

\[\theta=-0.463647609\ldots\ \mathrm{rad} =-26.565051^\circ.\]

The dressed coordinate scales are

\[\Omega_+^2=\frac{25}{16}, \qquad \Omega_-^2=\frac{289}{400},\]

\[\Omega_+=\frac54, \qquad \Omega_-=\frac{17}{20}.\]

The bare stability margins are

\[\Omega_+^2-\omega_c^2=1, \qquad \Omega_-^2-\omega_c^2=\frac4{25}.\]

The smaller margin is positive but close enough to the boundary to generate a slow dynamical mode. Indeed,

\[S=\frac{341}{100}, \qquad P=\frac4{25},\]

and

\[\nu_{\mathrm{high,low}}^2 =\frac{341\pm\sqrt{109881}}{200}.\]

Numerically,

\[\nu_{\mathrm{high}}=1.8336889475\ldots, \qquad \nu_{\mathrm{low}}=0.2181395054\ldots.\]

Neither frequency equals $\Omega_+$ or $\Omega_-$. This is the simplest numerical demonstration of why coordinate diagonalization and phase-space diagonalization must be kept separate.

Compact Maxima check

kill(all)$
wc:3/4$  wx2:104/125$  wy2:41/125$  g:42/125$

Delta:radcan(sqrt((wx2-wy2)^2+4*g^2))$
theta:atan2(-2*g,wx2-wy2)/2$
Op2:radcan((wx2+wy2+2*wc^2+Delta)/2)$
Om2:radcan((wx2+wy2+2*wc^2-Delta)/2)$
margins:[radcan(Op2-wc^2),radcan(Om2-wc^2)]$

S:radcan(Op2+Om2+2*wc^2)$
P:radcan(margins[1]*margins[2])$
nu:[sqrt((S+sqrt(S^2-4*P))/2),
    sqrt((S-sqrt(S^2-4*P))/2)]$

[Delta,float(theta),margins,ev(nu,numer)];

Maxima returns $\Delta=21/25$, the angle $-0.4636476090\ldots$, margins $[1,4/25]$, and the two numerical frequencies displayed above.

For numerical work, the most informative diagnostics are the smaller bare margin and $\nu_{\mathrm{low}}$. As $\Omega_-^2-\omega_c^2\rightarrow0^+$ with the other margin fixed, $P\rightarrow0$ and

\[\nu_{\mathrm{low}}^2 =\frac{P}{S}+O(P^2).\]

The slow period and spatial extent therefore grow near the stability boundary, making finite-basis calculations increasingly demanding.

Result carried forward

The physical reduction supplies the next stage with canonical preparation coordinates $(q_\pm,p_\pm)$, fixed positive scales $\Omega_\pm$, and the strict stability condition $\Omega_\pm^2>\omega_c^2$. These scales diagonalize the coordinate confinement only; the signed angular generator $L_z=q_+p_- -q_-p_+$ remains a separate dynamical operator.