If $T$ is entanglement breaking, applying it to one half of $|\varphi ⟩⟨\varphi |$ immediately shows that its Choi matrix is separable.

Conversely supposeis separable. The Choi reconstruction formula for the normalized convention isThe trace-preserving condition ${\mathrm{Tr}}_1{C}_T=I/d$ implies ${\sum }_k{M}_k=I$, so $\{M_k\}$ is a POVM. Part (i) now proves that $T$ is entanglement breaking. Thus

For a bipartite input,Each second factor is positive and has trace equal to the probability of measurement outcome $k$. Dividing nonzero factors by their traces therefore writes the output as a convex combination of product states. Hence every measure-and-prepare channel is an entanglement-breaking channel.

Partial transposition interchanges the positions occupied by $x$ and $y$. Positivity of ${\rho }^{T_B}$ therefore requiresBecause the system is $2\otimes 2$, the positive partial transpose criterion is necessary and sufficient. Combining these inequalities with validity of the original state gives

After permuting the computational basis, $\rho$ is the direct sum ofThe stated diagonal and trace conditions already give Hermiticity and trace one. Each $2\times 2$ block is positive semidefinite exactly when its determinant is nonnegative. Thus $\rho$ is a density matrix precisely when

The positive partial transpose criterion says separability implies ${\rho }_{AB}^{T_B}\ge 0$. A nonpositive partial transpose therefore proves entanglement. Positive partial transpose is also sufficient for separability in dimensions $2\otimes 2$ and $2\otimes 3$, but not in general higher dimensions.

A bipartite density operator is a separable quantum state whenfor probabilities $p_j$ and local states. If no such convex decomposition exists, it is an entangled state.

The normalized Choi matrix of the trace-to-identity map is $I\otimes I/d$, while that of transposition is the flip operator $F/d$. Hence the Werner–Holevo channel hasthe normalized projector onto the antisymmetric subspace.

If the channel were random unitary, part (i) would express $C_T$ as a mixture of maximally entangled vectors. Every vector in such a mixture must lie in the support of $C_T$, hence in the antisymmetric subspace. Under vectorization, an antisymmetric vector corresponds to a skew-symmetric matrix $A$, while maximal entanglement requires $A{A}^{†}$ to be proportional to $I$. In odd dimension, $\det A=\det (-A^T)=-\det A$, so $\det A=0$; such an $A$ cannot be proportional to a unitary. Thus for odd $d\ge 3$, and in particular $d=3$, this unital channel is not random unitary, disproving the converse. The odd-dimensional qualification matters because antisymmetric maximally entangled vectors can exist in even dimension.

Every random unitary channel satisfiesIt is therefore a unital quantum channel.

For $T(\rho )={\sum }_k{p}_k{U}_k\rho U_k^{†}$,Each $(U_k\otimes I)|\varphi ⟩$ is a maximally entangled state, so the Choi matrix is a convex combination of maximally entangled pure states.

The Kraus representation isThe second identity is precisely trace preservation. Different Kraus families related by an isometry represent the same channel.

A quantum channel $T:\mathcal{B}(\mathcal{H})\to \mathcal{B}(\mathcal{K})$ is a linear, completely positive, trace-preserving map. Complete positivity means $T\otimes {\mathrm{id}}_n$ is positive for every ancillary dimension $n$.

For $\tau \Omega =ϵ\ll 1$, expand the pressure factor:For a stable density wave defineThe dispersion relation becomesPerturbing the two isothermal roots givesThus the leading amplitude-damping rate is

At marginal stability write $s=i\omega$ with real $\omega$. The imaginary part of the pressure factor isFor $\gamma >1$, $\tau >0$, and nonzero $k$, the dispersion relation can have zero imaginary part only when $\omega =0$. Thus every marginal mode hasThe marginal equation is consequentlyDefining the isothermal Toomre parameter $Q_0=c_0\Omega /(\pi G{\Sigma }_0)$, its two roots areThey exist when $Q_0\le 1$, and the interval between them is unstable when $Q_0<1$. Finite thermal relaxation therefore leaves the onset criterionequal to the isothermal criterion, regardless of $\gamma$ and the nonzero value of $\tau$.

Write the perturbations as $\delta \Sigma =\sigma$, radial and azimuthal velocities $u,v$, and pressure $\delta P=p$, all proportional to $e^{ikx+st}$. Linearization in the shearing sheet givesandThe continuity and pressure equations implyEliminating $\sigma ,u,v,p,\varphi$ then yields

When $\tau |s|\ll 1$, temperature relaxes during a disturbance and the pressure response is isothermal, $p/\sigma \simeq c_0^2$. When $\tau |s|\gg 1$, relaxation is negligible and the response is adiabatic, $p/\sigma \simeq \gamma c_0^2$, so the effective sound speed is $\sqrt{\gamma }{c}_0$. At finite relaxation time the phase lag between compression and pressure also damps stable waves.

After a radial Fourier transform, Poisson's equation away from the sheet becomesThe decaying, reflection-symmetric solution is $A{e}^{-|k||z|}$. Integrating Poisson's equation across $z=0$ gives the derivative jumpso $-2|k|A=4\pi G\tilde{\Sigma }$. Thus the midplane potential is

For $r_{\mathrm{in}}\ll r\ll r_{\mathrm{out}}$, part (i) gives $\bar{\nu }\Sigma \propto r^{-1/2}$. Since ${\Omega }^2\propto r^{-3}$, the dissipative flux obeysIn the spectrum, introduce the dimensionless variableThen $rdr$ is proportional to ${\omega }^{-16/7}{x}^{9/7}dx$, soAt intermediate frequencies the inner limit is much smaller than one and the outer limit much larger than one, allowing them to be replaced by zero and infinity. Thus

The luminosity from both disk faces is the radial integral of the given viscous heating rate:Substituting the steady profile and ${\Omega }^2=GM/r^3$ givesUsing $\Theta =\dot{M}\sqrt{GM}(\sqrt{r_{\mathrm{out}}}-\sqrt{r_{\mathrm{in}}})$ turns this intoThe first term is mechanical power injected by the stellar magnetic torque. The second is $\dot{M}(E_{\mathrm{in}}-E_{\mathrm{out}})$, where $E=-GM/(2r)$ is the specific energy of a circular orbit; it is negative because outward motion makes the material less tightly bound and consumes some torque power. The remainder is radiated by viscous dissipation.

Matter reaching the outer edge still has binding energy $GM/(2r_{\mathrm{out}})$ per unit mass. A wind that reaches infinity with negligible terminal energy must therefore receivefrom the stellar radiation field. In a signed disk-energy balance, the material removed at the edge carries orbital energy flux $\dot{M}{E}_{\mathrm{out}}=-GM\dot{M}/(2r_{\mathrm{out}})$.

At the inner edge, the imposed torque isThereforeThe angular-momentum conservation law says that $\mathcal{F}h+\mathcal{G}$ is constant. The magnetic process supplies angular momentum $\Theta$ at the inner boundary; viscous stress passes it outward, and the mass leaving at $r_{\mathrm{out}}$ carries the injected angular momentum together with the angular momentum $\dot{M}{h}_{\mathrm{in}}$ that entered with the mass. A larger torque must therefore move the removal radius outward so that each unit mass can carry more specific angular momentum.

In steady state, mass conservation makes the outward flux constant, $\mathcal{F}=\dot{M}$. Angular-momentum conservation then givesThe zero-torque condition at $r_{\mathrm{out}}$ fixes the constant to $\dot{M}{h}_{\mathrm{out}}$, soFor a Kepler orbit, $h=\sqrt{GMr}$ and $d\Omega /dr=-3\Omega /(2r)$. The viscous torque is consequently $\mathcal{G}=3\pi \bar{\nu }\Sigma h$. Equating the two expressions yields the steady decretion disk structure

Assume a feature spans $N_H=5$ atmospheric scale heights.

For an Earth twin around the Sun, $R_p=R_{\oplus }$, $R_{\ast }=R_{\odot }$, $H\simeq 8\mathrm{km}$, givingFor a $1.7R_{\oplus }$ super-Earth with a heavy atmosphere around a $0.2R_{\odot }$ M dwarf, take $H\simeq 6\mathrm{km}$, givingFor a $2.6R_{\oplus }$ hydrogen-rich sub-Neptune around a $0.4R_{\odot }$ star, take $T\simeq 300\mathrm{K}$, $g\simeq 12\mathrm{m}{\mathrm{s}}^{-2}$, $\bar{m}=2.3m_H$, and hence $H\simeq 90\mathrm{km}$. ThenThe hydrogen-rich sub-Neptune is the most observable with JWST because its low molecular mass and large radius produce the largest transmission annulus. Clouds can reverse this ranking in a particular system.