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A unital quantum channel preserves the identity operator: $\Phi (I)=I$. In equal input and output dimensions, it therefore preserves the maximally mixed state. For amplitude damping channel,Hence the channel is not unital for $p>0$; it is unital only in the identity case $p=0$. Trace preservation involves the opposite products $K_i^{†}{K}_i$ and does not imply unitality.

Successive uses of the same quantum channel mean coupling to a fresh vacuum environment each time, or resetting the environment between uses. They do not mean repeatedly applying the dilation unitary to one unreinitialized two-state environment. Direct composition givesInductively, with $q_n=1-(1-p{)}^n$,For $0<p\le 1$, this tends to the pure ground state $|0⟩⟨0|$, whose Von Neumann entropy is zero. For $p=0$, the channel is the identity: the state remains $\rho$ and its Von Neumann entropy remains $S(\rho )$. Thus the repeated amplitude damping limit must include this no-decay exception.

A zero-Von Neumann entropy output means there is no uncertainty about the final atomic state. It does not mean that the receiver has gained information about the initial preparation: all initial states have the same limiting output, so their distinguishability has been erased from the atom. Under the global unitary evolution, information is carried into the environment. Without observing an emission record, this is a nonselective dissipative process, not an inference about which input was sent. Indeed, starting from the excited pure state, one intermediate output has Von Neumann entropy equal to the binary entropy of $p$, which is positive for $0<p<1$; the atom's Von Neumann entropy need not decrease monotonically during the approach to the final ground state.

Model the environment's vacuum and one-photon states by $|0_E⟩,|1_E⟩$. A unitary dilation of amplitude damping acts on the initially vacant environment asThese images are orthonormal. One unitary completion takes $U|0,1_E⟩=-\sqrt{p}|1,0_E⟩+\sqrt{1-p}|0,1_E⟩$ and fixes $|1,1_E⟩$. Taking environment matrix elements $⟨k_E|U|0_E⟩$ gives $K_0,K_1$.

Evolve $\rho \otimes |0_E⟩⟨0_E|$ by $U$ and take the partial trace over the environment. Orthogonality of the two environment records removes the cross-branch terms, giving ${\mathcal{A}}_p(\rho )=K_0\rho K_0^{†}+K_1\rho K_1^{†}$. Entrywise,Excited population is transferred to the ground state, while coherence is reduced by the square root of the survival probability.

Choose $|0⟩$ as the ground state and $|1⟩$ as the excited state. The relevant quantum channel is the amplitude damping channel. Its Kraus operators areThey satisfy $K_0^{†}{K}_0+K_1^{†}{K}_1=I$. The $K_0$ branch corresponds to no emitted photon: the ground-state amplitude is unchanged and the excited-state amplitude is attenuated. The $K_1$ branch corresponds to photon emission and transition to the ground state. These are unnormalized branch states; the squared norm or matrix trace of each branch gives its probability.

A Poisson random measure with intensity the sigma-finite measure $\mu$ is a random nonnegative integer-valued measure $M$ such that, for each outcome outside a single null set, $A\mapsto M(A)$ is countably additive, and for each measurable $A$, $M(A)$ is a measurable random variable. For every measurable $A$ with $\mu (A)<\infty$,and for every finite collection of pairwise disjoint finite-intensity measurable sets, their counts are independent random variables. These are integer-valued counts, with possible value infinity on infinite-intensity sets. In particular $\mu (A)=0$ implies $M(A)=0$ almost surely. The intensity identity is $\mathbb{E}M(A)=\mu (A)$, also in the extended sense.

For completeness, choose an increasing finite-intensity exhaustion $E_m\uparrow E$. Each $M(E_m)$ is finite almost surely, simultaneously for all $m$, so $M$ is a sigma-finite measure almost surely. If $\mu (A)=\infty$, then $\mu (A\cap E_m)\to \infty$, and for any fixed integer $K$,Hence $M(A)=\infty$ almost surely. Independence for counts on arbitrary disjoint measurable sets follows by this exhaustion; an infinite count is a constant extended value. This convention completes the definition on a general measurable space, not just on bounded subsets of Euclidean space.

The Brownian reflection principle, applied to the negative increments of the Brownian motion started at $a$, giveswhere $\Phi$ is the standard normal distribution function. Hence $T_0<\infty$ almost surely. The stopped Brownian motion $M_t=B_{t\wedge T_0}$ is a continuous nonnegative martingale, starts at $a$, and eventually vanishes. The maximal identity for a continuous nonnegative local martingale tending to zero therefore gives, for $H={\sup }_{0\le t\le T_0}{B}_t$,Equivalently,There is no atom at $a$. This is the Pareto distribution of shape one, or the translated version of the maximum before a lower Brownian barrier.

On a product extension of the probability space, take a uniform distribution variable $U$ on $(0,1)$ independent of the original sigma-algebra. For $x>0$,These are the same conditional tails as in the maximal identity for a continuous nonnegative local martingale tending to zero. Taking expectations identifies the unconditional probability distributions:If $M_0=0$, define $M_0/U=0$; the endpoint $U=0$ is a null event if one uses $[0,1]$ instead. The possible atom at zero has mass $\mathbb{P}(M_0=0)$, since the conditional probability of a positive maximum is zero exactly on that event. For fixed $M_0=m>0$, the conditional law is the Pareto distribution of shape one and minimum $m$, with density $m/x^2$ on $x>m$.

Fix $x>0$ and let $A=\{M_0<x\}\in {\mathcal{F}}_0$. On $A$, continuity of the paths implies $0\le M_{t\wedge T_x}\le x$, and $M_{T_x}=x$ on $\{T_x<\infty \}$. Bounded-time optional stopping gives, for every $t$,On $A\cap \{T_x=\infty \}$, the stopped process is $M_t\to 0$; on $A\cap \{T_x<\infty \}$ it eventually equals $x$. The bound by $x$ permits conditional dominated convergence, yieldingOn $A^c$, $T_x=0$, so the conditional hitting probability is one. Finally, a continuous nonnegative path tending to zero attains any positive supremum: after a sufficiently late time it is below half that supremum, and its maximum on the earlier compact interval is attained. Thus $\{M^{\ast }\ge x\}=\{T_x<\infty \}$, including equality at the level. We have proved the conditional form of the maximal identity for a continuous nonnegative local martingale tending to zero:The restriction to $A$ before taking the limit is important: $M_0$ need not be bounded, and the unstopped martingale need not have uniform integrability.

First take $k$ to be a nonnegative integer. The exponential tilting constructed from the independent increments restricts to ${\mathbb{P}}_n^{\lambda }$ on each ${\mathcal{F}}_n$. Since its Radon-Nikodym derivative is strictly positive, finite-time change of measure givesUnder the tilted law, $L_n=(M_n^{\lambda }{)}^{-1}$ is a martingale, becauseUse the stopped likelihood ratio under exponential tilting at this finite horizon. In detail, $\{{\tau }_k=j\}\in {\mathcal{F}}_j$, soHere the last expression is defined only on the displayed event, so no value of $M_{\infty }^{\lambda }$ is being assumed. The upward skip-free random walk has integer increments bounded above by one, which imply $S_{{\tau }_k}=k$ on a finite hit: the preceding value is at most $k-1$, and there is no upward overshoot. ConsequentlyAt $\lambda ={\lambda }_0$, the mean under one-sided exponential tilting is finite and positive. The strong law of large numbers gives $S_n/n\to {\varphi }^{\prime }({\lambda }_0)>0$ under ${\mathbb{P}}^{{\lambda }_0}$, so every integer level is hit almost surely. Since $\varphi ({\lambda }_0)=1$, passing to $n\to \infty$ givesLet $H={\sup }_{n\ge 0}{S}_n$. The event $\{H\ge k\}$ is exactly the finite-hit event: an integer sequence with all its values below $k$ has supremum at most $k-1$. Also $\mathbb{P}(H=\infty )={\lim }_k{e}^{-{\lambda }_{0}k}=0$. Subtraction of consecutive tail probabilities gives the geometric maximum of an upward skip-free random walk:This is the geometric distribution counting failures before the first success, with success parameter $1-e^{-{\lambda }_0}$.

The PDF writes only $k\ge 0$, without explicitly declaring integer levels. If it is read as allowing real $k$, then ${\tau }_k={\tau }_{\lceil k\rceil }$ and the displayed factor must be $e^{-\lambda \lceil k\rceil }$; the hitting probability is likewise $e^{-{\lambda }_0\lceil k\rceil }$. For example, increments $+1$ and $-1$ with probabilities $1/3$ and $2/3$ have ${\lambda }_0=\log 2$: hitting level $1/2$ has probability $1/2$, not $2^{-1/2}$. The integer interpretation supplies the intended formula.

For Borel sets $A_1,\dots ,A_n$, multiply the finite-time Radon-Nikodym derivative into the product law of the increments:The factorization, extended from rectangles to the product sigma-algebra, proves independence and identical distributions. Their common law is the exponential tiltingFor $\lambda >0$, $|x|e^{\lambda x}$ is bounded on $x\le 0$, while $x{e}^{\lambda x}\le C_{\epsilon }{e}^{(\lambda +\epsilon )x}$ on $x\ge 0$. Choosing a small two-sided neighborhood of $\lambda$ still contained in $(0,\infty )$ gives an integrable dominating function for the derivative. Differentiation under the integral sign then yieldsThe mean under one-sided exponential tilting needs an endpoint qualification. At $\lambda =0$, finiteness of positive exponential moments makes $X_1^{+}$ integrable, but gives no integrability of $X_1^{-}$. The formula holds with the right derivative and the extended mean ${\varphi }_{+}^{\prime }(0)=\mathbb{E}{X}_1\in [-\infty ,\infty )$. To see this, apply monotone convergence to $(1-e^{-h{X}_1^{-}})/h$ and dominated convergence to $(e^{h{X}_1^{+}}-1)/h$ as $h\downarrow 0$. A finite mean at zero is not guaranteed by the printed hypotheses: $\mu (\{-j\})=6/({\pi }^2{j}^2)$, $j\ge 1$, has finite $\varphi (\lambda )$ for every $\lambda \ge 0$ but $\mathbb{E}{X}_1=-\infty$.

The moment-generating function satisfies $0<\varphi (\lambda )<\infty$. The proposed process is nonnegative and adapted to the natural filtration of the random walk, and independence of the increments givesMoreover $X_{n+1}$ is independent of ${\mathcal{F}}_n$, soTherefore $M^{\lambda }$ is a mean-one nonnegative martingale, the exponential martingale of a random walk. For $\lambda =0$ it is the constant process one.

The almost sure supermartingale convergence theorem applies to the $L^1$-bounded supermartingale $X$. Hence $X_n\to X_{\infty }$ almost surely, and the Fatou lemma gives $\mathbb{E}|X_{\infty }|\le {\sup }_n\mathbb{E}|X_n|<\infty$. Take the version of $X_{\infty }$ obtained from the sequence, so it is measurable for ${\mathcal{F}}_{\infty }=\sigma ({\bigcup }_n{\mathcal{F}}_n)$. SetThe tower property of conditional expectation makes $M$ a martingale, and the uniform integrability of conditional expectations makes it uniformly integrable. Explicitly, with $Z=X_{\infty }$, $C=\mathbb{E}|Z|$, and $R,K>0$, we have $|M_n|\le R+\mathbb{E}[|Z|1_{\{|Z|>R\}}\mid {\mathcal{F}}_n]$, soFirst choose $R$ large, then $K$ large. This proves uniform integrability uniformly in $n$.

The uniformly integrable martingale convergence theorem gives a limit $M_{\infty }$ both almost surely and in $L^1$. For $A\in {\mathcal{F}}_j$ and $n\ge j$, $\mathbb{E}[M_n{1}_A]=\mathbb{E}[X_{\infty }{1}_A]$. Passing to the $L^1$ limit and then extending from the algebra ${\bigcup }_j{\mathcal{F}}_j$ to ${\mathcal{F}}_{\infty }$ identifies $M_{\infty }=X_{\infty }$ almost surely. Finally,Thus $X=M+Y$ with $M$ uniformly integrable and $Y$ a supermartingale tending almost surely to zero. The terminal decomposition of an L1-bounded supermartingale does not require $Y\ge 0$ or $Y_n\to 0$ in $L^1$. Indeed, for $X$ equal to the negative of the fair-coin doubling martingale, $X_{\infty }=0$, $M=0$, and $Y=X$ is negative and has constant absolute mean one.