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:

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

If the map is trace preserving, it is usually called a quantum channel. If it is trace nonincreasing, it describes an operation that may occur only with some probability, such as one outcome of a measurement.

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

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

This mixing consistency is why operations are represented as linear maps on the operator space, rather than as arbitrary nonlinear functions of density matrices.

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 channel gateway’s complete-positivity section supplies the transpose-map counterexample and the equivalent Choi, Kraus, and dilation tests.

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 is the right object for a 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. The Kraus representation section states 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 an environment prepared in $\eta_E$:

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

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. The Stinespring representation gives the working construction and explains how an environment basis produces a Kraus family.

Operations and Instruments

A quantum instrument is a family of operations

$$\{\mathcal I_m\}$$

labeled by classical outcomes. Each $\mathcal I_m$ is completely positive and trace nonincreasing. The total nonselective map

$$\Phi = \sum_m\mathcal I_m$$

is trace preserving if the measurement always returns one of the outcomes.

For an input state $\rho$,

$$p(m) = \operatorname{Tr}\mathcal I_m(\rho),$$

and, if $p(m)\ne0$,

$$\rho_m = \frac{\mathcal I_m(\rho)} {\operatorname{Tr}\mathcal I_m(\rho)}.$$

The effects of the associated POVM are

$$F_m = \mathcal I_m^\dagger(I),$$

where $\mathcal I_m^\dagger$ is the adjoint map with respect to the trace pairing. Then

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

The important distinction is:

$$\text{POVM effects give probabilities;} \qquad \text{operations give probabilities and state changes.}$$

An instrument must therefore specify both the outcome probabilities encoded by its effects and the state-update maps encoded by its operations.

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 one projective outcome $P_m$,

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

Its trace is

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

The nonselective projective measurement is 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, $\Phi(\rho)=\tau$. This is a channel, but it is not unitary unless the input space is trivial.

Noise channels

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

Common Mistakes

Normalizing too early

For selected outcomes, $\mathcal E(\rho)$ is intentionally unnormalized. Its trace is the branch probability. Normalizing before summing branches loses probability information.

Treating every operation as a channel

Every channel is an operation, but not every operation is a channel. A selected outcome is trace nonincreasing, not trace preserving.

Thinking positivity is enough

Positivity on isolated states does not guarantee physical behavior on entangled states. Complete positivity is the standard consistency requirement.

Confusing effects with operations

An effect $F_m$ gives $p(m)=\operatorname{Tr}(\rho F_m)$. It does not determine the output state after outcome $m$.

Forgetting the domain

A finite-dimensional channel is defined on a chosen input Hilbert space. Leakage, truncation, and restricted subspaces can make a fitted operation valid only within a specified model.

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 $\tau$.

Cross-Links

• Quantum Channels and Noise
• Density Operators
• Entangled States
• Partial Trace
• No-Broadcasting Theorem

References

• K. Kraus, States, Effects, and Operations: Fundamental Notions of Quantum Theory, Springer (1983).
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 10th anniversary edition (2010).
• M. M. Wilde, Quantum Information Theory, Cambridge University Press, 2nd edition (2017).
• J. Watrous, The Theory of Quantum Information, Cambridge University Press (2018).
• A. S. Holevo, Quantum Systems, Channels, Information: A Mathematical Introduction, De Gruyter (2012).
• H.-P. Breuer and F. Petruccione, The Theory of Open Quantum Systems, Oxford University Press (2002).