Quantum Operations

A quantum operation is a physically allowed map that takes an input density operator to an output density operator, or to an unnormalized density operator associated with a selected outcome. Quantum operations are the common language behind noise channels, unread measurements, postselected branches, state preparation, loss, and reduced open-system dynamics.

In finite-dimensional standard quantum theory, the central definition is:

$$\text{quantum operation} \quad=\quad \text{completely positive, trace-nonincreasing linear map}.$$

A trace-preserving operation is a quantum channel. A general trace-nonincreasing operation also includes selected outcomes whose probability can depend on the input; trace-preserving maps form a subset of this class.

Required background. Use Density Operators to represent input and output states by positive trace-one operators; use Partial Trace to derive subsystem dynamics from a larger evolution.

Helpful background. Use Entangled States to interpret the ancilla test that distinguishes complete positivity from positivity.

Density-Operator Maps

Let

$$\mathcal E: \mathcal B(\mathcal H_{\mathrm{in}}) \to \mathcal B(\mathcal H_{\mathrm{out}})$$

be a linear map on operators. For an input state $\rho$, the unnormalized output is

$$\sigma = \mathcal E(\rho).$$

For a deterministic operation, $\sigma$ is already normalized:

$$\operatorname{Tr}\sigma=1.$$

For a selected outcome or filter, the trace is the probability that the branch occurs:

$$p_{\mathcal E}(\rho) = \operatorname{Tr}\mathcal E(\rho), \qquad 0\le p_{\mathcal E}(\rho)\le1.$$

When $p_{\mathcal E}(\rho)\ne0$, the conditional output state is

$$\rho_{\mathcal E} = \frac{\mathcal E(\rho)} {\operatorname{Tr}\mathcal E(\rho)}.$$

The unnormalized state is often the safer object during calculations because it keeps the branch probability and the postselected state together.

Linearity from Mixing

Quantum operations are linear on density operators because classical randomization should be represented consistently. If a source prepares $\rho_1$ with probability $q$ and $\rho_2$ with probability $1-q$, then the input density operator is

$$\rho = q\rho_1+(1-q)\rho_2, \qquad 0\le q\le1.$$

A physical operation that does not know which preparation occurred should satisfy

$$\mathcal E(\rho) = q\mathcal E(\rho_1) + (1-q)\mathcal E(\rho_2).$$

Mixing consistency is affinity on normalized states; the unnormalized branch probabilities and the linear extension to all operators complete the operation description. Conditioning on success need not preserve the original mixture weights.

For example, let $K=\operatorname{diag}(1,1/2)$ and select the branch $\mathcal E(\rho)=K\rho K^\dagger$. For input $I/2$,

$$\mathcal E(I/2)=\operatorname{diag}(1/2,1/8), \qquad p=\frac58, \qquad \rho_{\mathrm{selected}}=\operatorname{diag}(4/5,1/5).$$

The original equal mixture of $|0\rangle$ and $|1\rangle$ is no longer equal among successful trials: the two preparations succeed with probabilities $1$ and $1/4$. Normalizing the two branches separately and averaging with the old weights would incorrectly give $I/2$.

Positivity

A density operator is positive, so any candidate operation must at least send positive operators to positive operators:

$$X\ge0 \quad\Longrightarrow\quad \mathcal E(X)\ge0.$$

This property is called positivity. It is necessary, but not sufficient for a local physical operation. A system may be entangled with an untouched reference system, and the operation must remain valid on the joint state. That stronger condition is complete positivity.

Complete Positivity

The map $\mathcal E$ is completely positive if, for every finite-dimensional reference system $R$,

$$X_{SR}\ge0 \quad\Longrightarrow\quad (\mathcal E_S\otimes\mathrm{id}_R)(X_{SR})\ge0.$$

The identity map $\mathrm{id}_R$ means that the reference is not acted on. Complete positivity is a compatibility condition with composite systems. It prevents a map from producing negative probabilities merely because the system was entangled with something outside the laboratory operation.

The complete-positivity page works through a transpose-map counterexample on a Bell state.

Trace Conditions

A map is trace preserving if

$$\operatorname{Tr}\mathcal E(X) = \operatorname{Tr}X$$

for all trace-class inputs $X$. A completely positive trace-preserving map is a channel, often abbreviated CPTP.

A map is trace nonincreasing if

$$\operatorname{Tr}\mathcal E(X) \le \operatorname{Tr}X$$

for all $X\ge0$. A completely positive trace-nonincreasing map is often abbreviated CPTNI. It describes a possible selected outcome.

Equivalently, the adjoint map obeys $\mathcal E^\dagger(I)=I$ for a trace-preserving map and $\mathcal E^\dagger(I)\le I$ for a trace-nonincreasing map. This is distinct from unitality, which requires $\mathcal E(I)=I$.

$$\text{trace lost} \quad=\quad \text{probability assigned to other branches}.$$

An individual selected operation should not increase trace on positive inputs. A collection of selected operations can sum to a trace-preserving channel.

Kraus Form

In finite dimensions, every completely positive map has a Kraus representation

$$\mathcal E(\rho) = \sum_\alpha K_\alpha\rho K_\alpha^\dagger.$$

The trace condition is controlled by

$$G = \sum_\alpha K_\alpha^\dagger K_\alpha.$$

For a channel,

$$G=I.$$

For a trace-nonincreasing operation,

$$0\le G\le I.$$

Then

$$p_{\mathcal E}(\rho) = \operatorname{Tr}\mathcal E(\rho) = \operatorname{Tr}(\rho G).$$

Thus the positive operator $G$ is the effect associated with this selected operation. Kraus representation derives the operator-sum theorem, representation freedom, and trace conditions.

Channels from Larger Unitary Dynamics

One reason quantum operations are the right language is that they arise by enlarging the system, evolving unitarily, and then discarding or selecting part of the larger system.

For a deterministic reduced dynamics model, let the system couple to a fixed environment state $\eta_E$, initially uncorrelated with the input and independent of its preparation:

$$\Phi(\rho) = \operatorname{Tr}_E \!\left[ U(\rho\otimes\eta_E)U^\dagger \right].$$

This is a channel. It is completely positive and trace preserving.

For a selected measurement outcome $m$ on an apparatus or environment, the operation can have the form

$$\mathcal E_m(\rho) = \operatorname{Tr}_E \!\left[ (I\otimes\Pi_m) U(\rho\otimes\eta_E)U^\dagger (I\otimes\Pi_m) \right].$$

The trace of $\mathcal E_m(\rho)$ is the probability of outcome $m$. The normalized conditional output state is obtained only after dividing by that probability.

This is the channel-side version of the indirect-measurement picture. Stinespring representation gives the working construction and explains how an environment basis produces a Kraus family.

Operations with a classical outcome

An instrument is a complete family of outcome operations $\{\mathcal I_m\}$, with $\sum_m\mathcal I_m$ trace preserving. Its effect $F_m=\mathcal I_m^\dagger(I)$ gives $p(m)=\operatorname{Tr}(\rho F_m)$. The operation additionally determines the output state. Two instruments with the same effects can produce different noise channels when their outcomes are ignored.

Examples

Unitary evolution

For a closed system,

$$\mathcal U(\rho) = U\rho U^\dagger.$$

This is a channel with one Kraus operator $U$.

Projective measurement branch

For the Lüders implementation of one projective outcome $P_m$,

$$\mathcal I_m(\rho) = P_m\rho P_m.$$

Its trace is

$$p(m)=\operatorname{Tr}(\rho P_m).$$

Ignoring all outcomes of this implementation gives the channel

$$\Phi(\rho) = \sum_m P_m\rho P_m.$$

State preparation

A reset operation that ignores the input and prepares a fixed state $\tau$ is

$$\Phi(\rho)=\tau\operatorname{Tr}\rho.$$

On normalized inputs, $\Phi(\rho)=\tau$. For an input space of dimension greater than one, this cannot be unitary: distinct inputs give the same output. To verify complete positivity, diagonalize $\tau=\sum_a t_a|a\rangle\langle a|$ and use Kraus operators $K_{ai}=\sqrt{t_a}|a\rangle\langle i|$, where $\{|i\rangle\}$ spans the input. Their adjoint-products sum to $I_{\mathrm{in}}$.

Noise channels

Dephasing, depolarizing noise, amplitude damping, erasure, and loss are channels or families of operations used to model uncontrolled degrees of freedom. The common-noise comparison explains the symmetry, record, and approximation represented by each model.

The chosen system and preparation matter

A channel model must specify its input and output spaces. Discarding part of a system, resetting it, or adding an erasure flag can change their dimensions. A map fitted inside a truncated subspace need not predict leakage out of that subspace.

Likewise, the product-environment construction assumes the same initially independent environment for every input. If preparation changes environment correlations, the reduced dynamics need not define one channel on arbitrary system states. This limitation concerns the proposed model, not an exception to the consistency required of a local channel.

Exercises

Single-Kraus operation

Let

$$\mathcal E(\rho)=K\rho K^\dagger.$$

Show that $\mathcal E$ is trace nonincreasing exactly when $K^\dagger K\le I$.

Solution

For positive $\rho$,

$$\operatorname{Tr}\mathcal E(\rho) = \operatorname{Tr}(K\rho K^\dagger) = \operatorname{Tr}(\rho K^\dagger K).$$

Trace nonincrease means

$$\operatorname{Tr}(\rho K^\dagger K) \le \operatorname{Tr}\rho = \operatorname{Tr}(\rho I)$$

for every positive $\rho$. This is equivalent to

$$K^\dagger K\le I.$$

Projective branches

Let $\{P_m\}$ be orthogonal projectors summing to $I$. Show that each branch $\mathcal I_m(\rho)=P_m\rho P_m$ is trace nonincreasing and that $\sum_m\mathcal I_m$ is trace preserving.

Solution

For one branch,

$$\operatorname{Tr}(P_m\rho P_m) = \operatorname{Tr}(\rho P_m) \le \operatorname{Tr}\rho$$

for $\rho\ge0$ because $0\le P_m\le I$.

For the sum,

$$\operatorname{Tr} \sum_m P_m\rho P_m = \sum_m\operatorname{Tr}(\rho P_m) = \operatorname{Tr}\rho$$

because $\sum_mP_m=I$.

Reset channel

For a fixed density operator $\tau$, verify that $\Phi(\rho)=\tau\operatorname{Tr}\rho$ is trace preserving on all trace-class inputs.

Solution

Since $\operatorname{Tr}\tau=1$,

$$\operatorname{Tr}\Phi(\rho) = \operatorname{Tr}(\tau\operatorname{Tr}\rho) = (\operatorname{Tr}\tau)(\operatorname{Tr}\rho) = \operatorname{Tr}\rho.$$

The map sends every normalized input state to the same output state $\tau$.

References

• K. Kraus, States, Effects, and Operations: Fundamental Notions of Quantum Theory, Lecture Notes in Physics 190, Springer (1983). Publisher.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press (2018), chapter 2. Author’s text.