08 Jun 2026
V: Survival Curvature Beyond the Linear Covariance Response
Covariances describe widths and pairwise correlations. They do not describe an entire non-Gaussian wavefunction. A state can therefore have no linear change in any cross covariance and still begin moving away from its initial quantum ray. Survival probability supplies the cleanest test.
This chapter develops that new test from the Schrödinger equation. It uses the preparation and quench already constructed in Chapter III and the exact covariance response derived in Chapter IV; neither derivation is repeated here. The new physics is the relation between departure from the full state, energy uncertainty, and a quadratic response invisible to linear covariance.
1. Prerequisites carried forward
Use $\hbar=2M=1$ and $[q_\alpha,p_\beta]=i\delta_{\alpha\beta}$. The complete input from the preparation and quench chapters is
\[\begin{gathered} H_0=h_+(q_+,p_+)+h_-(q_-,p_-),\\ |\Psi_0\rangle =|m_+,\ell_+\rangle\otimes|m_-,\ell_-\rangle, \qquad H_0|\Psi_0\rangle=E_0|\Psi_0\rangle,\\ H_f=H_0-\omega_cL_z, \qquad L_z=q_+p_- -q_-p_+. \end{gathered}\]The one-mode factors are normalized, real, definite-parity exceptional-Hermite eigenstates. Consequently,
\[\langle q_\alpha\rangle_0 =\langle p_\alpha\rangle_0=0, \qquad \left\langle \frac{q_\alpha p_\alpha+p_\alpha q_\alpha}{2} \right\rangle_0=0.\]The confinement scales $\Omega_\pm$ remain fixed while the signed angular coupling $\omega_c$ is activated. For the magnetic realization, stability requires $\Omega_\pm^2>\omega_c^2$.
Chapter IV established, without Gaussian closure,
\[\left.\frac{d}{dt}\operatorname{Cov}(q_+,q_-)\right|_0 =\omega_c(V_{q,-}-V_{q,+}),\] \[\left.\frac{d}{dt}\operatorname{Cov}(p_+,p_-)\right|_0 =\omega_c(V_{p,-}-V_{p,+}),\]with both mixed position-momentum cross rates zero. These formulas are the motivation for the present question: if the two modes have matching local variances, all these first derivatives vanish, but is the state actually stationary?
2. Survival amplitude, probability, and curvature
The time-evolved state is
\[|\Psi(t)\rangle=e^{-iH_ft}|\Psi_0\rangle.\]The survival amplitude
\[\mathcal A(t)=\langle\Psi_0|\Psi(t)\rangle\]is a complex overlap. Its modulus squared,
\[\boxed{ \mathcal S(t)=|\mathcal A(t)|^2, }\]is the survival probability. It answers a precise experimental question: if the system is projected onto the initially prepared state at time $t$, what is the probability of obtaining that state again? It does not ask whether a selected observable happens to retain its initial value. Taking the modulus also removes an overall phase, because vectors that differ only by a global phase represent the same physical quantum ray.
Assume that $H_f$ is self-adjoint and
\[|\Psi_0\rangle\in D(H_f^2).\]This condition gives a finite fourth energy moment and controls the remainder used below. Expanding the exponential,
\[\mathcal A(t) =1-i\langle H_f\rangle_0t -\frac12\langle H_f^2\rangle_0t^2+\cdots .\]Multiplying by its complex conjugate gives
\[\begin{aligned} \mathcal S(t) &=\left(1-i\langle H_f\rangle_0t -\frac12\langle H_f^2\rangle_0t^2+\cdots\right)\\ &\quad\times \left(1+i\langle H_f\rangle_0t -\frac12\langle H_f^2\rangle_0t^2+\cdots\right)\\ &=1- \left(\langle H_f^2\rangle_0-\langle H_f\rangle_0^2\right)t^2 +O(t^4). \end{aligned}\]The odd powers vanish because $\mathcal A(-t)=\mathcal A(t)^*$ and hence $\mathcal S(-t)=\mathcal S(t)$. Define the energy variance
\[(\Delta H_f)^2 =\operatorname{Var}_0(H_f) =\langle H_f^2\rangle_0-\langle H_f\rangle_0^2.\]Then
\[\boxed{ \mathcal S(t)=1-(\Delta H_f)^2t^2+O(t^4). }\]Two curvature conventions occur in the literature. To avoid ambiguity, we will use the positive initial-loss coefficient
\[\boxed{ \kappa_{\mathcal S} =\lim_{t\to0}\frac{1-\mathcal S(t)}{t^2} =(\Delta H_f)^2. }\]The ordinary second derivative is
\[\mathcal S''(0)=-2\kappa_{\mathcal S}.\]The associated Zeno time is
\[\tau_Z=\frac{1}{\Delta H_f},\]provided $\Delta H_f\ne0$. This is a local time scale, not a predicted oscillation period. The reason energy variance appears is physical as well as algebraic: different post-quench energy components acquire different phases. An exact energy eigenstate has no relative dephasing, whereas a broader energy distribution leaves its initial ray more rapidly.
3. Why only the quenched term contributes
Because \(\lvert\Psi_0\rangle\) is an eigenstate of $H_0$,
\[(H_0-E_0)|\Psi_0\rangle=0.\]Moreover, the product structure and vanishing one-mode means imply
\[\langle L_z\rangle_0 =\langle q_+\rangle_0\langle p_-\rangle_0 -\langle q_-\rangle_0\langle p_+\rangle_0 =0.\]Therefore
\[\begin{aligned} \left(H_f-\langle H_f\rangle_0\right)|\Psi_0\rangle &=\left(H_0-E_0-\omega_cL_z\right)|\Psi_0\rangle\\ &=-\omega_cL_z|\Psi_0\rangle. \end{aligned}\]Taking the squared norm yields
\[\boxed{ \operatorname{Var}_0(H_f) =\omega_c^2\langle L_z^2\rangle_0. }\]This step does not require $H_0$ to commute with $L_z$. It uses only the fact that the initial state is an eigenstate of $H_0$.
4. Deriving the angular-momentum variance
Introduce the local variances
\[V_{q,\alpha}=\langle q_\alpha^2\rangle_0, \qquad V_{p,\alpha}=\langle p_\alpha^2\rangle_0.\]Since the means vanish, these second moments are the variances. Square the angular generator without treating the operators as commuting numbers:
\[\begin{aligned} L_z^2 ={}&q_+^2p_-^2+q_-^2p_+^2\\ &-q_+p_-q_-p_+ -q_-p_+q_+p_-. \end{aligned}\]The first two expectations factor across the product state:
\[\langle q_+^2p_-^2\rangle_0=V_{q,+}V_{p,-}, \qquad \langle q_-^2p_+^2\rangle_0=V_{q,-}V_{p,+}.\]For a real state with zero symmetrized $qp$ covariance,
\[\langle q_\alpha p_\alpha+p_\alpha q_\alpha\rangle_0=0.\]Combining this with $[q_\alpha,p_\alpha]=i$ gives
\[\langle q_\alpha p_\alpha\rangle_0=\frac{i}{2}, \qquad \langle p_\alpha q_\alpha\rangle_0=-\frac{i}{2}.\]Now reorder only operators belonging to different modes:
\[\begin{aligned} \langle q_+p_-q_-p_+\rangle_0 &=\langle q_+p_+\rangle_0 \langle p_-q_-\rangle_0 =\frac14,\\ \langle q_-p_+q_+p_-\rangle_0 &=\langle p_+q_+\rangle_0 \langle q_-p_-\rangle_0 =\frac14. \end{aligned}\]Thus
\[\boxed{ \langle L_z^2\rangle_0 =V_{q,+}V_{p,-}+V_{q,-}V_{p,+}-\frac12. }\]The $-1/2$ is not an optional correction. It is the sum of two canonical ordering contributions. Replacing $q$ and $p$ by commuting random variables would miss it.
Define
\[\boxed{ \Lambda_0 =V_{q,+}V_{p,-}+V_{q,-}V_{p,+}-\frac12. }\]The exact local result is therefore
\[\boxed{ \mathcal S(t) =1-\omega_c^2\Lambda_0t^2+O(t^4), \qquad \kappa_{\mathcal S}=\omega_c^2\Lambda_0. }\]Since $L_z$ generates rotations in the $q_+:q_-$ plane, $\Lambda_0$ measures how strongly an infinitesimal rotation changes the prepared wavefunction. If \(L_z\lvert\Psi_0\rangle=0\), the angular quench initially has no direction in which to move the ray; otherwise the squared norm of that direction is precisely $\Lambda_0$.
The coefficient must be nonnegative because \(\Lambda_0=\lVert L_z\lvert\Psi_0\rangle\rVert^2\). It can also be checked directly:
\[\begin{aligned} V_{q,+}V_{p,-}+V_{q,-}V_{p,+} &\ge 2\sqrt{V_{q,+}V_{p,+}V_{q,-}V_{p,-}}\\ &\ge\frac12, \end{aligned}\]where the second line uses $V_{q,\alpha}V_{p,\alpha}\ge1/4$. A computed negative $\Lambda_0$ is therefore an immediate sign of a normalization, derivative, or operator-ordering error.
5. What field reversal can and cannot change
The short-time loss depends on $\omega_c^2$, so
\[\mathcal S(t;\omega_c) =\mathcal S(t;-\omega_c)+O(t^4).\]By contrast, linear cross-covariance rates generated by the angular term are proportional to $\omega_c$. Field reversal can reverse a covariance flow while leaving the survival curvature unchanged. This even-versus-odd distinction is a useful diagnostic in both symbolic calculations and code.
6. A symmetric state that defeats covariance intuition
Suppose the two prepared factors are identical:
\[V_{q,+}=V_{q,-}=V_q, \qquad V_{p,+}=V_{p,-}=V_p.\]The angular contribution to every first-order cross-covariance rate then vanishes because each such rate contains a difference of like variances. Define the one-mode uncertainty product
\[U=\Delta q\,\Delta p=\sqrt{V_qV_p}.\]The survival coefficient becomes
\[\boxed{ \Lambda_0=2U^2-\frac12. }\]For two equal ordinary oscillator vacua, $U=1/2$ and $\Lambda_0=0$. Their isotropic product wavefunction is rotationally invariant, so $L_z$ annihilates it.
For two equal codimension-two added states,
\[U_2=0.5172471466\ldots,\]and therefore
\[\Lambda_0 =2(0.5172471466\ldots)^2-\frac12 =0.03508922135\ldots>0.\]The non-Gaussian product begins leaving its initial ray even though the entire linear cross-covariance channel is silent. The correct implication is
\[\text{zero linear covariance response} \quad\not\Rightarrow\quad \text{stationary state}.\]Covariance symmetry removes one diagnostic; it does not turn covariance into a complete description of a non-Gaussian state.
7. Closed-form moments for the codimension-two state
For $\phi_{2,0}$, define
\[\mathcal M =\sqrt{\frac{\pi}{2}}e^{1/2} \operatorname{erfc}\!\left(\frac{1}{\sqrt2}\right).\]Direct integration gives
\[V_q(\Omega) =\frac{2(\mathcal M-1/2)}{\Omega},\]and
\[V_p(\Omega) =\frac{\Omega}{2} \left(\frac32+\frac{\mathcal M}{3}\right).\]Set
\[a=\mathcal M-\frac12, \qquad b=\frac32+\frac{\mathcal M}{3}.\]For codimension-two factors at two different scales,
\[\boxed{ \Lambda_0 =ab\left(\frac{\Omega_-}{\Omega_+} +\frac{\Omega_+}{\Omega_-}\right)-\frac12. }\]This form shows two sources of survival loss: the non-minimum uncertainty product $ab>1/4$ and the scale mismatch $\Omega_-/\Omega_++\Omega_+/\Omega_-\ge2$.
For
\[\Omega_+=\frac54, \qquad \Omega_-=\frac{17}{20}, \qquad \omega_c=\frac34,\]the stability margins are
\[\Omega_+^2-\omega_c^2=1, \qquad \Omega_-^2-\omega_c^2=\frac4{25},\]and the exact-moment evaluation gives
\[\Lambda_0=0.07537829213164765\ldots.\]Consequently,
\[\kappa_{\mathcal S} =\omega_c^2\Lambda_0 =0.0424002893240518\ldots,\]so the local prediction is
\[\mathcal S(t) =1-0.0424002893240518\ldots\,t^2+O(t^4).\]The direct Maxima calculation below reproduces the same coefficient from the wavefunction rather than assuming these closed-form moments.
8. Reproducing the curvature
The coefficient can be recovered without propagating the two-mode state. Write a real one-mode factor as
\[\phi(q) =\left(\frac{\Omega}{2}\right)^{1/4} \mathcal N R(z)e^{-z^2/2}, \qquad z=\sqrt{\frac{\Omega}{2}}q.\]For an exceptional state, Chapter II’s exact construction gives $R(z)=\mathcal P_{m,\ell}(z)/\mathcal H_m(z)$. Let $z_i$ and $w_i$ denote the nodes and weights of a Gauss-Hermite rule for $\int e^{-z^2}f(z)\,dz$. Then
\[\begin{aligned} \langle1\rangle &\approx\mathcal N^2\sum_iw_iR(z_i)^2,\\ V_q &\approx\frac{2\mathcal N^2}{\Omega} \sum_iw_i z_i^2R(z_i)^2,\\ V_p &\approx\frac{\Omega\mathcal N^2}{2} \sum_iw_i\left[R'(z_i)-z_iR(z_i)\right]^2. \end{aligned}\]The last line follows from $p=-i\,d/dq$. It is crucial to differentiate the rational factor $R$; reweighting the coordinate probability density does not produce a momentum variance.
For $m=2$, $\mathcal H_2=4z^2+2$ and $\mathcal N_2=(8/\sqrt\pi)^{1/2}$. The following Maxima 5.49.0 calculation integrates the normalized wavefunction directly rather than inserting the closed-form moments:
kill(all)$
display2d : false$
H2 : 4*z^2 + 2$
N2 : sqrt(8/sqrt(%pi))$
R : 1/H2$
norm : quad_qagi(N2^2*R^2*exp(-z^2), z, minf, inf)[1]$
vq(O) := (2/O)*quad_qagi(
N2^2*z^2*R^2*exp(-z^2), z, minf, inf)[1]$
vp(O) := (O/2)*quad_qagi(
N2^2*(diff(R,z)-z*R)^2*exp(-z^2), z, minf, inf)[1]$
Op : 5/4$ Om : 17/20$ wc : 3/4$
Lambda : vq(Op)*vp(Om) + vq(Om)*vp(Op) - 1/2$
kappa : wc^2*Lambda$
print("norm =", norm)$
print("Vq+ =", vq(Op), " Vp+ =", vp(Op))$
print("Vq- =", vq(Om), " Vp- =", vp(Om))$
print("Lambda =", Lambda, " kappa =", kappa)$
It returns
norm = 1.0
Vq+ = 0.24908726787007765 Vp+ = 1.074099904670583
Vq- = 0.36630480569129065 Vp- = 0.7303879351759964
Lambda = 0.07537829213164804
kappa = 0.04240028932405202
The normalization, uncertainty inequalities, and $\Lambda_0\ge0$ are physical consistency checks on the same calculation. For a new exceptional state, increase the quadrature accuracy until the desired digits of all three are stable.
An independent finite-time check expands the state in successively larger product bases, propagates the projected Hermitian Hamiltonian, and computes
\[\mathcal S_N(t) =|\langle\Psi_0|\Psi_N(t)\rangle|^2.\]For each cutoff $N$, evaluate
\[K_N(t)=\frac{1-\mathcal S_N(t)}{t^2}.\]The two relevant limits are
\[K_N(t)\longrightarrow\omega_c^2\Lambda_0 \quad(t\to0),\]and the resulting curve must also stop changing when $N$ is increased. Extremely small $t$ causes subtractive cancellation in $1-\mathcal S_N(t)$; fitting it to $a_2t^2+a_4t^4$ over a shrinking window is usually more reliable. Norm conservation is not enough, because every Hermitian truncation is unitary within its own incomplete space.
What has been established
For a real, definite-parity product eigenstate and a fixed-confinement angular quench,
\[\mathcal S(t) =1-\omega_c^2 \left( V_{q,+}V_{p,-}+V_{q,-}V_{p,+}-\frac12 \right)t^2+O(t^4)\]is exact at the stated order. It tests departure from the complete initial state, whereas covariance tests selected quadratic observables. Their disagreement is not a paradox; it is evidence that a non-Gaussian state contains dynamical information beyond its covariance matrix.
Transition to finite-time dynamics
Curvature answers what happens as $t\to0$; it does not draw the later trajectory. Chapter VI now takes the exact exceptional basis from Chapter II, the quench from Chapter III, the synchronization observable from Chapter IV, and the survival coefficient just derived, and turns them into a cutoff-tested finite-time propagation.
Discussion