Erasure and Loss Channels

Erasure and loss channels describe situations in which the intended system may fail to arrive, leave the retained Hilbert space, or be replaced by a known loss flag. The central modeling distinction is whether the loss event is flagged.

In a flagged erasure channel, the receiver knows that the system was lost because the output contains an orthogonal erasure state. In unflagged loss, the system may be attenuated, leaked, or missing without a reliable classical record. Those are different physical and information-theoretic situations.

Helpful background. Trace-nonincreasing operations represent a selected outcome before normalization. Their missing trace is a probability, not a numerical error.

Photonic Qubits applies these distinctions to photon preparation, transmission, and detection; Gaussian Channels develops the continuous-variable description of attenuation.

Flagged Erasure Channel

Let $\mathcal H_{\mathrm{in}}$ be the input Hilbert space. The output Hilbert space for an erasure channel is enlarged:

$$\mathcal H_{\mathrm{out}} = \mathcal H_{\mathrm{surv}} \oplus \mathbb C\lvert e\rangle,$$

where $\lvert e\rangle$ is orthogonal to every valid surviving output state. Let

$$V: \mathcal H_{\mathrm{in}} \to \mathcal H_{\mathrm{surv}}$$

be an isometric embedding of the input into the non-erased output subspace. With erasure probability $\epsilon$, the channel is

$$\mathcal E_\epsilon(\rho) = (1-\epsilon)V\rho V^\dagger + \epsilon\,\operatorname{Tr}(\rho) \lvert e\rangle\langle e\rvert, \qquad 0\le\epsilon\le1.$$

For normalized inputs, this says:

$$\text{survive with probability }1-\epsilon, \qquad \text{erased with probability }\epsilon.$$

The output state itself contains the flag. Measuring the projector

$$\Pi_e=\lvert e\rangle\langle e\rvert$$

reveals whether erasure occurred without confusing the flag with an ordinary input state.

Kraus Representation

Choose an orthonormal input basis

$$\{\lvert j\rangle\}_{j=1}^d.$$

A Kraus representation of the flagged erasure channel is

$$K_s = \sqrt{1-\epsilon}\,V, \qquad K_j = \sqrt{\epsilon}\, \lvert e\rangle\langle j\rvert, \quad j=1,\ldots,d.$$

The completeness relation is

$$\begin{aligned} K_s^\dagger K_s + \sum_{j=1}^dK_j^\dagger K_j &= (1-\epsilon)I + \epsilon\sum_{j=1}^d \lvert j\rangle\langle j\rvert\\ &= I. \end{aligned}$$

Therefore the flagged erasure map is trace preserving. The different Kraus labels should not be overread as unique microscopic histories; what is physical here is the orthogonal output flag.

This example is a clean application of the trace conditions discussed in Trace-Preserving and Trace-Nonincreasing Maps.

Survival Branch and Postselection

If one keeps only the successful branch, the operation is

$$\mathcal S_\epsilon(\rho) = (1-\epsilon)V\rho V^\dagger.$$

This is completely positive and trace nonincreasing (strictly decreasing on nonzero positive inputs if $\epsilon>0$):

$$\operatorname{Tr}\mathcal S_\epsilon(\rho) = (1-\epsilon)\operatorname{Tr}\rho.$$

For a normalized input, the survival probability is $1-\epsilon$. For $\epsilon<1$, conditioning on observed survival gives

$$\rho_{\mathrm{surv}} = \frac{\mathcal S_\epsilon(\rho)} {\operatorname{Tr}\mathcal S_\epsilon(\rho)} = V\rho V^\dagger.$$

This conditional normalized map is not the same object as the physical operation before conditioning. It has discarded the probability that the system was lost.

Here survival is independent of the input, so it does not bias the surviving ensemble of input states. This special normalized map is linear. In general, state-dependent loss filters the ensemble and gives a nonlinear normalization step.

For example, suppose $|0\rangle$ survives with probability $1$ and $|1\rangle$ with probability $1/4$. The survival operator is $K_s=\operatorname{diag}(1,1/2)$. Input $|+\rangle$ survives with probability $5/8$ and becomes $(2|0\rangle+|1\rangle)/\sqrt5$. A reliable flag tells us whether loss occurred, but does not make this input-dependent process the constant-probability erasure channel above.

Erasure Versus Depolarizing Noise

Erasure is not depolarizing noise. In a depolarizing channel, the output remains in the system Hilbert space and the receiver does not receive a reliable marker telling where the damage occurred. In an erasure channel, the damaged system is replaced by a known orthogonal flag.

The contrast is visible in the outputs:

$$\text{erasure:}\quad \rho \longmapsto (1-\epsilon)V\rho V^\dagger + \epsilon\lvert e\rangle\langle e\rvert,$$

while for a $d$-dimensional depolarizing replacement model,

$$\text{depolarizing:}\quad \rho \longmapsto (1-q)\rho+q\frac{I}{d}.$$

The flag makes erasures located errors. A known missing qubit, photon, or atom is often easier to handle than an unknown corrupted one. The distinction is physical: the output includes a state orthogonal to the entire input code space.

See Depolarizing Channel for the unflagged isotropic benchmark model.

Leakage and Loss from a Subspace

In many platforms the computational Hilbert space is only a subspace of a larger physical Hilbert space:

$$\mathcal H_{\mathrm{phys}} = \mathcal H_{\mathrm{comp}} \oplus \mathcal H_{\mathrm{leak}}.$$

Leakage means population leaves $\mathcal H_{\mathrm{comp}}$. If leakage is detected reliably, it can behave like an erasure flag. If not, it is an unflagged loss process on the computational subspace.

Let $\Phi_{\mathrm{phys}}$ be a channel on the full physical space and let $P_{\mathrm{comp}}$ project onto the computational subspace. The retained-subspace operation is

$$\mathcal L_{\mathrm{ret}}(\rho) = P_{\mathrm{comp}} \Phi_{\mathrm{phys}}(\rho) P_{\mathrm{comp}}.$$

This map is generally trace nonincreasing:

$$\operatorname{Tr}\mathcal L_{\mathrm{ret}}(\rho) \le \operatorname{Tr}\rho.$$

The missing trace is the probability that population leaked out of the retained subspace. If the leaked sector is included as part of the output, the full physical map can remain trace preserving.

This is why atom loss, molecular loss, ion shelving outside a detection manifold, and leakage to noncomputational states should not be modeled as pure dephasing inside a fixed qubit unless that approximation has been justified. Effective Hamiltonians in Quantum Information separates coherent logical evolution from the contraction of a projected physical propagator. The neutral-atom platform page gives a concrete example in Neutral Atoms.

A leakage flag alone does not guarantee that the surviving state is unchanged or that the loss probability is state independent. Leakage amplitudes may also return coherently on a later step. Projecting onto the computational subspace after every step generally differs from projecting only at the end. Leakage and Crosstalk explores these effects in multilevel devices.

Bosonic Pure-Loss Channel

For an optical or oscillator mode, loss is usually modeled by coupling the signal mode $a$ to an environment mode $e$ through a beam splitter of transmissivity $\eta$. In the Heisenberg picture,

$$a_{\mathrm{out}} = \sqrt{\eta}\,a_{\mathrm{in}} + \sqrt{1-\eta}\,e_{\mathrm{in}}, \qquad 0\le\eta\le1.$$

With an initially independent vacuum environment, this defines the quantum-limited pure-loss channel. A coherent state transforms as

$$\lvert\alpha\rangle \longmapsto \lvert\sqrt{\eta}\,\alpha\rangle$$

after the environment is traced out. Photon number is attenuated:

$$\langle a^\dagger a\rangle \longmapsto \eta\langle a^\dagger a\rangle.$$

For $0<\eta\le1$, a useful Kraus representation is

$$A_\ell = \sqrt{\frac{(1-\eta)^\ell}{\ell!}}\, \eta^{\hat n/2}a^\ell, \qquad \ell=0,1,2,\ldots,$$

where

$$\hat n=a^\dagger a.$$

Acting on a Fock state,

$$A_\ell\lvert n\rangle = \begin{cases} \sqrt{\binom n\ell} (1-\eta)^{\ell/2} \eta^{(n-\ell)/2} \lvert n-\ell\rangle, & \ell\le n,\\ 0, & \ell>n. \end{cases}$$

The Fock-basis formula also defines the endpoint $\eta=0$: every $|n\rangle$ loses all $n$ photons, and the channel resets any input to vacuum. The label $\ell$ counts photons lost to the environment in this representation. If the environment is not monitored, the receiver does not usually know $\ell$.

Bosonic loss is therefore not automatically the same as a flagged erasure channel. A lost photon may produce an identifiable vacuum in a fixed time bin, or it may be confused with source failure, detector inefficiency, mode mismatch, or another unobserved event. The modeling depends on what the receiver can actually distinguish.

Detector Inefficiency

For a mode-matched detector with no dark counts or dead-time effects, inefficiency can be modeled as loss before ideal detection. For a photon-counting detector with quantum efficiency $\eta$, place a beam splitter of transmissivity $\eta$ before a perfect detector and trace out the lost port.

For an ideal on/off detector after this loss, the no-click POVM element is

$$\Pi_0 = \sum_{n=0}^\infty (1-\eta)^n \lvert n\rangle\langle n\rvert,$$

and

$$\Pi_{\mathrm{click}} = I-\Pi_0.$$

This formula says that an $n$-photon input fails to produce a click with probability $(1-\eta)^n$ in the idealized independent-loss model.

Detector inefficiency belongs to the measurement model, not only to the state channel. The same physical loss can be represented as a channel before measurement or as inefficient POVM effects for outcome probabilities. Predicting a conditional remaining state additionally requires an instrument: the effects alone do not specify it. The right representation depends on whether one is predicting transmitted states, click probabilities, or conditional post-measurement states.

Relation to Amplitude Damping

Amplitude damping and loss are related but not identical.

For a two-level atom, amplitude damping describes relaxation

$$\lvert1\rangle \to \lvert0\rangle$$

inside the retained two-dimensional Hilbert space. If the emitted photon is not monitored, the final atom is still present in the ground state. That is not an erasure of the atom.

The encoding decides whether vacuum is a flag:

• Single-rail encoding: $|0\rangle$ and $|1\rangle$ of one mode are the logical basis. Pure loss restricted to this subspace is amplitude damping with $\gamma=1-\eta$. Vacuum is a valid logical state, so it cannot identify loss.
• Dual-rail encoding: $|10\rangle$ and $|01\rangle$ of two modes are the logical basis. With independent equal transmissivities $\eta$, an encoded state $|\psi\rangle=\alpha|10\rangle+\beta|01\rangle$ becomes

$$\eta|\psi\rangle\langle\psi|+(1-\eta)|00\rangle\langle00|.$$

Now vacuum is orthogonal to the code space, so an ideal presence measurement can distinguish the loss sector. Actual detection must provide that reliable distinction; an inefficient no-click result is ambiguous. If the two transmissivities differ, the surviving amplitudes become $\alpha\sqrt{\eta_0}$ and $\beta\sqrt{\eta_1}$: the encoded state is filtered, not merely preserved or erased.

The finite-time qubit amplitude-damping channel is treated at Amplitude-Damping Channel. The continuous-time photon-loss master equation appears in Quantum Optical Master Equation.

Thermal Loss and Gaussian Boundary

If the environment mode in the beam-splitter model is thermal instead of vacuum, attenuation is accompanied by added thermal noise. For an initially independent thermal environment, the single-mode covariance transformation is exactly

$$V \longmapsto \eta V+(1-\eta)V_E,$$

where $V_E$ is the covariance matrix of the environment mode in the same quadrature convention.

Gaussian Channels derives the associated noise bound and compares attenuation with amplification.

Exercises

1. Trace preservation of erasure. Verify that the Kraus operators $K_s=\sqrt{1-\epsilon}\,V$ and $K_j=\sqrt{\epsilon}\lvert e\rangle\langle j\rvert$ define a trace-preserving channel.

Solution

Since $V^\dagger V=I$,

$$K_s^\dagger K_s = (1-\epsilon)I.$$

Also,

$$\sum_{j=1}^dK_j^\dagger K_j = \epsilon \sum_{j=1}^d \lvert j\rangle\langle j\rvert = \epsilon I.$$

Therefore

$$K_s^\dagger K_s+\sum_{j=1}^dK_j^\dagger K_j=I.$$

2. Survival conditioning. For the survival operation $\mathcal S_\epsilon(\rho)=(1-\epsilon)V\rho V^\dagger$, compute the success probability and the normalized surviving state for a normalized input.

Solution

The success probability is

$$p_{\mathrm{surv}} = \operatorname{Tr}\mathcal S_\epsilon(\rho) = (1-\epsilon)\operatorname{Tr}\rho = 1-\epsilon.$$

For $\epsilon<1$, the normalized surviving state is

$$\rho_{\mathrm{surv}} = \frac{(1-\epsilon)V\rho V^\dagger}{1-\epsilon} = V\rho V^\dagger.$$

The probability $1-\epsilon$ is lost if one records only the normalized conditional state.

3. Fock-state loss. Use the pure-loss Kraus operator $A_\ell$ to find the probability of losing exactly $\ell$ photons from an input number state $\lvert n\rangle$.

Solution

For $\ell\le n$,

$$A_\ell\lvert n\rangle = \sqrt{\binom n\ell} (1-\eta)^{\ell/2} \eta^{(n-\ell)/2} \lvert n-\ell\rangle.$$

The probability is the squared norm:

$$p(\ell|n) = \binom n\ell (1-\eta)^\ell \eta^{n-\ell}.$$

For $\ell>n$, the probability is zero. This is the binomial distribution for independently transmitting each of the $n$ photons with probability $\eta$ in the ideal loss model.

4. Flagged or unflagged? A receiver expects one photon in a fixed time bin. The detector does not click. Name two different physical models that could give this observation and explain why they are not automatically the same channel.

Solution

One model is ideal detection after transmission loss: the source prepares exactly one photon, the time bin is known, and the detector has unit efficiency and no dark counts. Then no click identifies vacuum in that bin. Another model is detector inefficiency: the photon may have arrived but failed to trigger the detector. A third possibility is source failure.

These models can give the same observed no-click event but imply different state transformations, different conditioning, and different error correlations. A channel model must specify which degrees of freedom are part of the output and which records are trusted.

References

• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary edition, Cambridge University Press (2010).
• J. Watrous, The Theory of Quantum Information, Cambridge University Press (2018). Author’s text.
• C. Weedbrook, S. Pirandola, R. García-Patrón, N. J. Cerf, T. C. Ralph, J. H. Shapiro, and S. Lloyd, “Gaussian quantum information,” Reviews of Modern Physics 84, 621–669 (2012). DOI; author manuscript.