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Model a thin transparent plate by a lossless reciprocal phase screen with transmission $t(x)=e^{i\psi (x)}$ at the chosen angular frequency. Both passages multiply the local field by the same $t(x)$: reversing travel does not by itself change the sign of an accumulated optical path length. Assume the plate is stationary, the two passages use corresponding transverse positions, and the plate is adjacent to the return surface or perfectly reimaged onto it. Ignore propagation between these reference surfaces, wave absorption, clipping and intensity-dependent changes. Take the incident boundary envelope to be $U_0(x)=A(x)e^{i{\varphi }_0(x)}$. Outward transmission gives $t{U}_0$. Ideal phase conjugation and the second transmission then giveThis is a complete short proof of cancellation in the specified phase screen model. For an ideal ordinary mirror with constant reflection factor $r$, the corresponding envelope is $rt(x{)}^2{U}_0(x)$: the extra wave phase is $2\psi (x)$, rather than zero. A constant reflection wave phase changes neither conclusion.

For an original toy optical aberration, let $\xi =x/a$ on $-1\le \xi \le 1$, take ${\varphi }_0=0$, and chooseThe even term adds symmetric curvature and the odd term makes the two sides unequal. The outward wave phase is $\psi$; immediately after ideal phase conjugation it is $-\psi$; after the second passage it is zero. At $\xi =-1,0,1$, the outward values are $\pi /4,0,3\pi /4$, while an ordinary mirror's completed return has values $\pi /2,0,3\pi /2$. All wave phases here are unwrapped relative to the central ray and are otherwise equivalent modulo $2\pi$. At a stipulated vacuum wavelength $\lambda =532\mathrm{nm}$, these outward errors correspond to optical path length errors $h=\lambda \psi /(2\pi )$ of $66.5,0,199.5\mathrm{nm}$. These are chosen values; they are not observations or physical plate thicknesses.

The wave phase plots make a subtle distinction visible: the field is still curved immediately after phase conjugation, and becomes flat only after crossing the same plate again. More generally, conjugation restores $U_0^{\ast }$, not a universally flat wavefront. If the plate instead has transmission $t=\alpha (x)e^{i\psi (x)}$ with real $0\le \alpha \le 1$, the return is $\gamma \alpha (x{)}^2{U}_0^{\ast }$. Its plate-induced wave phase cancels but its lost wave amplitude does not recover. If the return plate has changed to wave phase ${\psi }_{\mathrm{r}}$, the residual factor is $e^{i({\psi }_{\mathrm{r}}-\psi )}$. Those two variations distinguish correction of a stationary optical aberration from universal reversal of loss or material evolution.

At an equilibrium ${\mathbf{r}}_{\ast }$, the total force vanishes. A small transverse displacement must produce a transverse restoring force; a small axial displacement must also restore the bead. Downstream of a tight focus, the backward optical gradient force can oppose the forward optical scattering force. Equality at one point is insufficient: the total axial force must decrease as the bead moves downstream through that point. The equilibrium can therefore lie downstream of the intensity maximum.

For uncoupled overdamped coordinates, ${\zeta }_i\dot{\delta r_i}=-k_i\delta r_i$ with ${\zeta }_i>0$ decays only when each $k_i>0$. More generally, linearizing gives $\dot{\delta \mathbf{r}}=MJ\delta \mathbf{r}$, where $M$ is the hydrodynamic mobility matrix and $J_{ij}=\partial F_i/\partial r_j$ at equilibrium. All eigenvalues of $MJ$ must have negative real parts for linear asymptotic stability. For a conservative force with symmetric positive-definite mobility, a positive-definite Hessian matrix of the potential energy supplies this condition. A general optical force need not possess a global potential energy.

Increasing numerical aperture admits more strongly converging light rays and a steeper axial variation of the field. But a larger nominal numerical aperture alone does not establish trapping: illumination of the objective pupil, optical aberration, particle properties and the actual in-liquid focus also matter. A harmonic optical trap describes local confinement after this stability question has been answered; it does not predict capture from arbitrary positions or an infinite holding force.