07 Jun 2026

IV: Exact First-Order Covariance Response Without Gaussian Closure

covariance dynamics moment hierarchy exceptional Hermite states quantum synchronization non-Gaussian dynamics

The angular quench creates dynamics in a state that was stationary before $t=0$. The first experimentally accessible question is therefore local in time: which means and covariances begin to change immediately, and which do not?

The answer can be exact even though the state is non-Gaussian and its full moment hierarchy is not closed. The reason is that a derivative at one specified instant needs only the Hamiltonian and the moments of the state at that instant; it does not require a solution for all later times.

What is carried forward, and what is new

The fixed-confinement angular quench provides the complete input:

\[H_f=p_+^2+p_-^2+U_+(q_+)+U_-(q_-) -\omega_c(q_+p_--q_-p_+),\] \[|\Psi_0\rangle =|m_+,\ell_+\rangle\otimes|m_-,\ell_-\rangle.\]

The initial factors are normalized eigenstates of their separate one-coordinate Hamiltonians. They are real and have definite parity, while the regular potentials $U_\pm$ are even. Consequently,

\[\langle q_\alpha\rangle_0 =\langle p_\alpha\rangle_0 =\langle U_\alpha'(q_\alpha)\rangle_0=0,\] \[\frac12 \langle q_\alpha p_\alpha+p_\alpha q_\alpha\rangle_0=0.\]

The product structure also factorizes every cross-coordinate expectation at $t=0$. We continue to use

\[[q_\alpha,p_\beta]=i\delta_{\alpha\beta}, \qquad \hbar=2M=1.\]

Nothing about the exceptional-Hermite construction needs to be repeated. The new task is to derive the operator flow, expose the open moment hierarchy, and isolate the complete first-order covariance response.

Heisenberg evolution from the Hamiltonian

In the Schrödinger picture the state evolves. Equivalently, in the Heisenberg picture an operator evolves as

\[O(t)=e^{iH_ft}Oe^{-iH_ft},\]

so

\[\dot O(t)=i[H_f,O(t)].\]

As one sign check, consider $q_+$ at $t=0$. The local kinetic term gives

\[i[p_+^2,q_+]=2p_+,\]

and the angular term gives

\[i[-\omega_c(q_+p_--q_-p_+),q_+] =\omega_cq_-.\]

Repeating this calculation for the other three canonical operators yields

\[\boxed{ \begin{aligned} \dot q_+&=2p_++\omega_cq_-, & \dot q_-&=2p_--\omega_cq_+,\\ \dot p_+&=-U_+'(q_+)+\omega_cp_-, & \dot p_-&=-U_-'(q_-)-\omega_cp_+. \end{aligned} }\]

The structure is physically transparent. The terms $2p_\alpha$ and $-U_\alpha’$ are the local motion. The terms proportional to $\omega_c$ rotate the two coordinates and the two momenta. Reversing $\omega_c$ reverses only this angular part of the instantaneous flow.

Why the moment hierarchy does not close

A moment is the expectation of a product of canonical operators, such as $\langle q_+^2\rangle$ or $\langle q_+p_-\rangle$. Differentiating a moment means differentiating each operator in that product. For example,

\[\begin{aligned} \frac{d}{dt}\langle q_+^2\rangle &=\langle\dot q_+q_++q_+\dot q_+\rangle\\ &=2\langle p_+q_++q_+p_+\rangle +2\omega_c\langle q_+q_-\rangle. \end{aligned}\]

The next equation already contains the force:

\[\begin{aligned} \frac{d}{dt} \langle q_+p_++p_+q_+\rangle ={}&4\langle p_+^2\rangle\\ &-\langle q_+U_+'+U_+'q_+\rangle\\ &+2\omega_c \langle q_+p_-+q_-p_+\rangle. \end{aligned}\]

For a quadratic potential,

\[U_\alpha(q_\alpha)=a_\alpha q_\alpha^2+b_\alpha, \qquad U_\alpha'=2a_\alpha q_\alpha,\]

and the force-weighted expression is another second moment. All means and second moments then form a finite linear system.

For a rational extension, $U_\alpha’(q_\alpha)$ is not linear. The value of $\langle q_\alpha U_\alpha’(q_\alpha)\rangle$ is not determined by $\langle q_\alpha^2\rangle$ alone. Differentiating it generates expectations containing $U_\alpha’’$, momentum-force products, and further functions of $q_\alpha$. This is the open moment hierarchy.

Open does not mean ill-defined. It means that a finite covariance matrix does not contain enough information to predict its own exact evolution for all times.

Covariance for arbitrary quantum states

Collect the canonical operators in the fixed order

\[R=(q_+,p_+,q_-,p_-)^{\mathsf T}.\]

Let

\[\mu_i=\langle R_i\rangle, \qquad \delta R_i=R_i-\mu_i.\]

The quantum covariance matrix is the real symmetric matrix

\[\sigma_{ij} =\frac12\langle \delta R_i\delta R_j+\delta R_j\delta R_i \rangle.\]

Symmetrization is essential when two operators do not commute: it makes each entry real for Hermitian $R_i$. For operators belonging to different coordinates, it changes nothing because they commute.

Define

\[V_{q,\alpha}=\langle q_\alpha^2\rangle_0, \qquad V_{p,\alpha}=\langle p_\alpha^2\rangle_0.\]

The initial means vanish. Product factorization removes all cross entries, and reality removes the local symmetrized $qp$ entries. Thus

\[\sigma(0)= \begin{pmatrix} V_{q,+}&0&0&0\\ 0&V_{p,+}&0&0\\ 0&0&V_{q,-}&0\\ 0&0&0&V_{p,-} \end{pmatrix}.\]

This diagonal matrix does not make the state Gaussian. Covariance is defined for every state, but it determines a state completely only within special families such as Gaussian states.

Initial means and local blocks

Taking expectations of the four Heisenberg equations gives

\[\dot{\langle q_+\rangle}_0 =2\langle p_+\rangle_0+\omega_c\langle q_-\rangle_0=0,\]

and similarly for the other three means. In the momentum equations, parity also gives $\langle U_\alpha’\rangle_0=0$. Therefore the derivatives of centered and uncentered second moments agree at $t=0$.

The local covariance blocks also have zero initial derivative. One way to see this is to separate the two causes of motion. Each factor is an eigenstate of its local $h_{m_\alpha}$, so its local expectations are stationary under $H_0$. Every contribution from the angular term to a local second moment contains a cross expectation, which factorizes to zero in the initial product state.

For the local $qp$ entry, this includes the one-dimensional virial identity

\[2V_{p,\alpha} =\langle q_\alpha U_\alpha'(q_\alpha)\rangle_0,\]

obtained by setting the local derivative of $\langle q_\alpha p_\alpha+p_\alpha q_\alpha\rangle$ to zero. No quadratic or Gaussian approximation is being used.

Coordinate-coordinate covariance rate

Because the initial means and their derivatives vanish,

\[\left.\dot\sigma_{q_+q_-}\right|_0 =\left.\frac{d}{dt}\langle q_+q_-\rangle\right|_0.\]

Using the Heisenberg equations,

\[\begin{aligned} \frac{d}{dt}\langle q_+q_-\rangle ={}&2\langle p_+q_-\rangle +2\langle q_+p_-\rangle\\ &+\omega_c \left(\langle q_-^2\rangle-\langle q_+^2\rangle\right). \end{aligned}\]

The two cross moments vanish initially by product factorization. Therefore

\[\boxed{ \left.\dot\sigma_{q_+q_-}\right|_0 =\omega_c(V_{q,-}-V_{q,+}). }\]

The angular flow immediately creates coordinate correlation only when the two initial coordinate widths differ.

Momentum-momentum covariance rate

The same calculation for $\langle p_+p_-\rangle$ gives

\[\begin{aligned} \frac{d}{dt}\langle p_+p_-\rangle ={}&-\langle U_+'(q_+)p_-\rangle -\langle p_+U_-'(q_-)\rangle\\ &+\omega_c \left(\langle p_-^2\rangle-\langle p_+^2\rangle\right). \end{aligned}\]

At $t=0$, the force terms factorize:

\[\langle U_+'p_-\rangle_0 =\langle U_+'\rangle_0\langle p_-\rangle_0=0,\]

and likewise for the other term. Hence

\[\boxed{ \left.\dot\sigma_{p_+p_-}\right|_0 =\omega_c(V_{p,-}-V_{p,+}). }\]

The nonlinear force does not disappear from the dynamics. Parity and product factorization remove it only from this particular derivative at this particular instant.

Why both mixed rates vanish

Since $q_+$ commutes with $p_-$, $\sigma_{q_+p_-}=\langle q_+p_-\rangle$ at the initial instant. Its derivative is

\[\begin{aligned} \frac{d}{dt}\langle q_+p_-\rangle ={}&2\langle p_+p_-\rangle +\omega_c\langle q_-p_-\rangle\\ &-\langle q_+U_-'(q_-)\rangle -\omega_c\langle q_+p_+\rangle. \end{aligned}\]

Product factorization removes the first and third terms at $t=0$. Reality and the canonical commutator give

\[\langle q_\alpha p_\alpha\rangle_0=\frac{i}{2}.\]

The remaining two terms cancel:

\[\omega_c\frac{i}{2}-\omega_c\frac{i}{2}=0.\]

Interchanging $+$ and $-$ gives the other mixed rate. Thus

\[\boxed{ \left.\dot\sigma_{q_+p_-}\right|_0 =\left.\dot\sigma_{p_+q_-}\right|_0=0. }\]

The cancellation is a genuinely quantum operator check. Replacing $q_\alpha p_\alpha$ by an unjustified real classical product would obscure it.

The complete first-order covariance theorem

In the order $R=(q_+,p_+,q_-,p_-)$, all results combine into

\[\boxed{ \dot\sigma(0)= \omega_c \begin{pmatrix} 0&0&V_{q,-}-V_{q,+}&0\\ 0&0&0&V_{p,-}-V_{p,+}\\ V_{q,-}-V_{q,+}&0&0&0\\ 0&V_{p,-}-V_{p,+}&0&0 \end{pmatrix}. }\]

This theorem assumes:

• a sudden fixed-confinement Hamiltonian $H_f$;
• a product of local eigenstates at $t=0$;
• real, definite-parity factors in even local potentials; and
• finite second moments and force expectations.

It does not assume Gaussianity or close the later-time hierarchy.

Several conclusions should be kept separate. Field reversal changes the sign of each nonzero rate. Equality of the coordinate variances removes only the coordinate-coordinate rate; equality of the momentum variances removes only the momentum-momentum rate. If both pairs agree, then $\dot\sigma(0)=0$, but second and higher derivatives may still be nonzero.

A nonzero cross covariance is evidence of correlation in the selected quadratures. By itself it is not an entanglement criterion and not evidence of persistent synchronization.

Worked control: unequal Gaussian widths

For the undeformed ground factor at scale $\Omega$,

\[V_q=\frac1{\Omega}, \qquad V_p=\frac{\Omega}{4}.\]

Use the same stable control values

\[\Omega_+=2,\qquad \Omega_-=1,\qquad \omega_c=\frac12.\]

Then

\[\left.\dot\sigma_{q_+q_-}\right|_0 =\frac12\left(1-\frac12\right)=\frac14,\] \[\left.\dot\sigma_{p_+p_-}\right|_0 =\frac12\left(\frac14-\frac12\right)=-\frac18.\]

Therefore

\[\sigma_{q_+q_-}(t)=\frac14t+O(t^2), \qquad \sigma_{p_+p_-}(t)=-\frac18t+O(t^2),\]

while both mixed cross entries start at $O(t^2)$. Coordinate and momentum correlations initially acquire opposite signs because the narrower coordinate distribution has the broader conjugate momentum distribution.

Exact derivatives do not imply a closed solution

For any observable $O$ whose required expectations exist,

\[\left. \frac{d^n}{dt^n}\langle O(t)\rangle \right|_{t=0} =i^n \left\langle \operatorname{ad}_{H_f}^{\,n}(O) \right\rangle_0, \qquad \operatorname{ad}_{H_f}(O)=[H_f,O].\]

This identity is exact. It permits a systematic calculation of second and higher derivatives even when the moment hierarchy is open. A Taylor polynomial made from finitely many such derivatives is still only a short-time approximation.

This distinction suggests a useful research comparison. Two non-Gaussian states can be chosen with the same initial covariance matrix. The theorem then gives them the same $\dot\sigma(0)$, but force-weighted moments entering higher derivatives can separate their dynamics. The earliest separating order quantifies how much information covariance misses for the chosen potential and states.

Relative-noise response and the synchronization numerator

To compare the fluctuations of the two modes on the same dimensionless scale, choose a reference frequency $\omega_0>0$ and define

\[Q_\alpha=\sqrt{\frac{\omega_0}{2}}\,q_\alpha, \qquad P_\alpha=\sqrt{\frac{2}{\omega_0}}\,p_\alpha.\]

Then $[Q_\alpha,P_\beta]=i\delta_{\alpha\beta}$. Define the relative-noise operator

\[\mathcal D_c =\frac12\left[ (\delta Q_+-\delta Q_-)^2 +(\delta P_+-\delta P_-)^2 \right],\]

and

\[S_c=\langle\mathcal D_c\rangle^{-1}.\]

The relative quadratures obey

\[[Q_+-Q_-,P_+-P_-]=2i.\]

The uncertainty relation and the arithmetic-geometric mean inequality give $\langle\mathcal D_c\rangle\geq1$, so $0<S_c\leq1$ for states with finite variances. An equal-time value of $S_c$ measures relative noise. It is not, by itself, proof of phase locking or a stationary synchronized regime.

Let

\[A_\alpha=V_{Q,\alpha}+V_{P,\alpha}.\]

At the quench, the local variance rates vanish and the initial cross entries are zero. The covariance theorem therefore gives

\[\langle\mathcal D_c\rangle_0 =\frac12(A_++A_-),\] \[\left. \frac{d}{dt}\langle\mathcal D_c\rangle \right|_0 =\omega_c(A_+-A_-).\]

The difference

\[\mathcal N_{\mathrm{sync}}=A_+-A_-\]

is the synchronization numerator: it fixes the signed linear response of the relative noise. Expanding the reciprocal gives

\[\boxed{ S_c(t) =\frac{2}{A_++A_-} -\frac{4\omega_c(A_+-A_-)} {(A_++A_-)^2}\,t +O(t^2). }\]

This is a local response formula, not a finite-time synchronization law.

Non-Gaussian codimension-two example

For a codimension-two ground factor, direct integration of the normalized state gives

\[\mathcal M =\sqrt{\frac{\pi}{2}}e^{1/2} \operatorname{erfc}\!\left(\frac1{\sqrt2}\right),\] \[V_q=\frac{2(\mathcal M-1/2)}{\Omega}, \qquad V_p=\frac{\Omega}{2} \left(\frac32+\frac{\mathcal M}{3}\right).\]

The dimensionless sum at an arbitrary reference scale is therefore

\[A^{(2)}(\Omega;\omega_0) =\frac{\omega_0(\mathcal M-1/2)}{\Omega} +\frac{\Omega}{\omega_0} \left(\frac32+\frac{\mathcal M}{3}\right).\]

Set $\omega_0=1$ and choose the stable values

\[\Omega_+=\frac54, \qquad \Omega_-=\frac{17}{20}, \qquad \omega_c=\frac34.\]

They give

\[A_+^{(2)}=2.2727434432762048\ldots, \qquad A_-^{(2)}=1.6439282731976381\ldots,\]

and hence

\[\boxed{ S_c(0)=0.5106376395008639\ldots, \qquad \dot S_c(0)=-0.1229730434570357\ldots. }\]

The following compact Maxima calculation evaluates the closed forms directly:

kill(all)$
fpprec : 25$
M : sqrt(%pi/2)*exp(1/2)*erfc(1/sqrt(2))$
A(O) := (M-1/2)/O + O*(3/2+M/3)$
Op : 5/4$  Om : 17/20$  wc : 3/4$
S0  : ratsimp(2/(A(Op)+A(Om)))$
dS0 : ratsimp(-4*wc*(A(Op)-A(Om))/(A(Op)+A(Om))^2)$
bfloat([A(Op),A(Om),S0,dS0]);

The expected output is

[2.272743443276204807045043b0,
 1.643928273197638160910612b0,
 5.106376395008638874009324b-1,
 -1.229730434570356467813996b-1]

The two coefficients are exact consequences of the stated moments and parameters. A later finite-basis propagation may be compared with them, but agreement must improve as the basis and time-step errors are reduced.

Scope and useful generalizations

The derivation identifies precisely where each assumption enters. Product factorization removes initial cross expectations. Parity removes mean forces. Reality fixes the local symmetrized $qp$ expectation. Local stationarity removes the initial derivatives of the local covariance blocks.

Relaxing any of these assumptions is possible, but the omitted terms must then be restored. A displaced state has nonzero means; a current-carrying state can have nonzero symmetrized $qp$ covariance; an indefinite-parity state can have a nonzero mean force. The four Heisenberg equations remain valid, while the boxed first-order matrix must be recomputed from the general covariance definition.

The main result is deliberately local and exact: it provides a benchmark for symbolic algebra, numerical propagation, and experimental finite-difference estimates without pretending that second moments close the rational dynamics.

Series navigation

1. Magnetic oscillator stability and reduction
2. Exceptional-Hermite preparation
3. Fixed-confinement angular quench
4. Exact covariance response — current chapter
5. Survival curvature
6. Spectral propagation and synchronization
7. Non-Gaussian mutual information