Kraus Operators

Kraus operators represent quantum operations on density operators. In measurement theory, they describe how the state changes for an outcome or after averaging over outcomes. They are closely related to POVM effects, but they are not the same object.

The most common confusion is:

$$\text{Kraus operators transform states;} \qquad \text{effects determine probabilities.}$$

Quantum Operations

A finite-dimensional quantum operation has the form

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

The operators $K_\alpha$ are Kraus operators. The map $\mathcal E$ is completely positive by construction.

The trace of the output is

$$\operatorname{Tr}\mathcal E(\rho) = \operatorname{Tr} \left( \rho\sum_\alpha K_\alpha^\dagger K_\alpha \right).$$

If the operation is trace preserving, then

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

If the operation is a single outcome branch of a measurement, it is usually trace nonincreasing:

$$\sum_\alpha K_\alpha^\dagger K_\alpha\le I.$$

The missing trace is not lost probability. It is the probability that some other outcome occurred.

Outcome-Resolved Operations

For a measurement with outcomes $m$, each outcome can be assigned an operation

$$\mathcal I_m(\rho) = \sum_\alpha K_{m\alpha}\rho K_{m\alpha}^\dagger.$$

The probability of outcome $m$ is

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

If $p(m)\ne0$, the conditional output state is

$$\rho_m = \frac{\mathcal I_m(\rho)} {\operatorname{Tr}\mathcal I_m(\rho)} = \frac{ \sum_\alpha K_{m\alpha}\rho K_{m\alpha}^\dagger }{ p(m) }.$$

The nonselective output is obtained by summing over outcomes:

$$\rho' = \sum_m\mathcal I_m(\rho) = \sum_{m,\alpha} K_{m\alpha}\rho K_{m\alpha}^\dagger.$$

If all outcomes are included, the total operation is trace preserving:

$$\sum_{m,\alpha} K_{m\alpha}^\dagger K_{m\alpha}=I.$$

Associated POVM Effects

The POVM effect associated with outcome $m$ is

$$F_m = \sum_\alpha K_{m\alpha}^\dagger K_{m\alpha}.$$

Then

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

This formula explains the relation:

Kraus operators -> operation and state update
Kraus adjoint-products -> POVM effects and probabilities

The arrow goes from a chosen operation to its effect. It does not go uniquely backward from an effect to an operation.

Single-Operator Outcome

If outcome $m$ has one measurement operator $M_m$, then

$$\mathcal I_m(\rho)=M_m\rho M_m^\dagger, \qquad F_m=M_m^\dagger M_m.$$

The conditional state is

$$\rho_m = \frac{M_m\rho M_m^\dagger} {\operatorname{Tr}(M_m\rho M_m^\dagger)}.$$

This common case is useful pedagogically, but realistic detector outcomes often group many microscopic alternatives into one reported outcome. Then several Kraus operators may be needed for a single $m$.

Projective Measurement

An ideal projective measurement is recovered by taking

$$K_a=P_a.$$

Then

$$\mathcal I_a(\rho)=P_a\rho P_a, \qquad F_a=P_a^\dagger P_a=P_a.$$

The nonselective channel is

$$\rho' = \sum_a P_a\rho P_a.$$

Projective measurement is therefore not separate from the Kraus formalism. It is the sharp, orthogonal special case.

Inefficient Detector Example

For a qubit detector intended to click on $|1\rangle$ with efficiency $\eta$, one outcome model is

$$K_{\mathrm{click}} = \sqrt{\eta}\,|1\rangle\langle1|.$$

The no-click outcome can have two Kraus operators:

$$K_{\mathrm{no},0} = |0\rangle\langle0|, \qquad K_{\mathrm{no},1} = \sqrt{1-\eta}\,|1\rangle\langle1|.$$

The effects are

$$F_{\mathrm{click}} = \eta|1\rangle\langle1|,$$ $$F_{\mathrm{no}} = |0\rangle\langle0| + (1-\eta)|1\rangle\langle1|.$$

They sum to $I$. The two no-click Kraus operators encode two different microscopic histories: the system was in $|0\rangle$, or it was in $|1\rangle$ but the detector missed it. If the detector does not record which history occurred, those histories are grouped into the same reported outcome.

Jump and No-Jump Example

Amplitude damping of a two-level system can be written with

$$K_0 = |0\rangle\langle0| + \sqrt{1-\gamma}\,|1\rangle\langle1|,$$ $$K_1 = \sqrt{\gamma}\,|0\rangle\langle1|.$$

The completeness relation is

$$K_0^\dagger K_0+K_1^\dagger K_1=I.$$

If a photon-emission record is monitored, $K_1$ can represent a jump outcome and $K_0$ a no-jump outcome over the time step. If the record is ignored, the channel is

$$\Phi(\rho) = K_0\rho K_0^\dagger + K_1\rho K_1^\dagger.$$

The same Kraus formula can therefore describe a conditioned measurement update or an unconditional noise channel, depending on whether the record is retained.

Nonuniqueness

A Kraus representation is not unique. If

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

and $u_{\beta\alpha}$ is a unitary matrix mixing the Kraus labels, then

$$L_\beta = \sum_\alpha u_{\beta\alpha}K_\alpha$$

gives another representation of the same operation:

$$\mathcal E(\rho) = \sum_\beta L_\beta\rho L_\beta^\dagger.$$

Therefore individual Kraus operators are often representation-dependent. Physical meaning attaches to them only when a measurement record, detector history, environment basis, or unraveling has been specified.

This warning is essential in open-system work. The same channel may have many Kraus decompositions, but not every decomposition corresponds to the same observed measurement record.

Trace Conditions

The trace condition tells what kind of operation is being represented:

Condition	Meaning
$\sum_\alpha K_\alpha^\dagger K_\alpha=I$	trace-preserving channel or complete nonselective operation
$\sum_\alpha K_\alpha^\dagger K_\alpha\le I$	trace-nonincreasing operation, usually a selected branch
$\sum_{m,\alpha}K_{m\alpha}^\dagger K_{m\alpha}=I$	complete measurement instrument with all outcomes included

If the trace condition fails, either the model is incomplete or the listed operators are not a physical operation.

Relation to Instruments

Kraus operators are a representation of completely positive maps. A quantum instrument is the outcome-indexed collection of those maps:

$$\{\mathcal I_m\}_m.$$

The instrument is the operational object. A Kraus representation is one way to write each $\mathcal I_m$. This distinction matters because Kraus representations can change while the instrument remains the same.

For calculations, Kraus operators are indispensable. For specifying the measurement at an invariant level, the instrument is the cleaner object.

Common Mistakes

• Confusing $K_m$ with the effect $F_m=K_m^\dagger K_m$.
• Assuming there is only one Kraus operator per outcome.
• Forgetting that trace-nonincreasing operations must be normalized after conditioning.
• Treating a single selected outcome operation as a trace-preserving channel.
• Reading physical detector histories from a Kraus representation without checking how the record is defined.
• Forgetting that different Kraus representations can give the same operation.
• Assuming every decomposition of a channel corresponds to a actually monitored trajectory.

Cross-Links

• Why Generalized Measurements Are Needed
• POVMs
• State Update Rules
• Selective and Nonselective Measurements
• Generalized Measurements Overview
• POVMs: First Encounter
• Common Misconceptions

References

• K. Kraus, States, Effects, and Operations: Fundamental Notions of Quantum Theory, Springer, 1983.
• A. S. Holevo, Probabilistic and Statistical Aspects of Quantum Theory, Edizioni della Normale, 2011.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2010.
• H. M. Wiseman and G. J. Milburn, Quantum Measurement and Control, Cambridge University Press, 2010.
• H.-P. Breuer and F. Petruccione, The Theory of Open Quantum Systems, Oxford University Press, 2002.

Exercises

1. Let $K_0=\sqrt{p}\,I$ and $K_1=\sqrt{1-p}\,I$, with $0\le p\le1$. Show that these define a trace-preserving operation and describe the effect of the nonselective channel.

Solution

The completeness relation is

$$K_0^\dagger K_0+K_1^\dagger K_1 = pI+(1-p)I = I.$$

The nonselective channel is

$$\Phi(\rho) = p\rho+(1-p)\rho = \rho.$$

The outcome label is a classical random label independent of the state; ignoring it leaves the state unchanged.

2. For the amplitude-damping Kraus operators

$$K_0=|0\rangle\langle0|+\sqrt{1-\gamma}|1\rangle\langle1|, \qquad K_1=\sqrt{\gamma}|0\rangle\langle1|,$$

verify the completeness relation.

Solution

Compute

$$K_0^\dagger K_0 = |0\rangle\langle0| + (1-\gamma)|1\rangle\langle1|,$$

and

$$K_1^\dagger K_1 = \gamma |1\rangle\langle1|.$$

Adding gives

$$K_0^\dagger K_0+K_1^\dagger K_1 = |0\rangle\langle0|+|1\rangle\langle1| =I.$$

3. If one outcome has Kraus operators $K_{m1}$ and $K_{m2}$, what is the associated POVM effect?

Solution

The effect is

$$F_m = K_{m1}^\dagger K_{m1} + K_{m2}^\dagger K_{m2}.$$

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

4. Suppose $L_0=(K_0+K_1)/\sqrt2$ and $L_1=(K_0-K_1)/\sqrt2$. Show that $\{L_0,L_1\}$ gives the same operation as $\{K_0,K_1\}$.

Solution

Expand:

$$L_0\rho L_0^\dagger+L_1\rho L_1^\dagger = \frac{1}{2} (K_0+K_1)\rho(K_0^\dagger+K_1^\dagger) + \frac{1}{2} (K_0-K_1)\rho(K_0^\dagger-K_1^\dagger).$$

The cross terms cancel, leaving

$$K_0\rho K_0^\dagger+K_1\rho K_1^\dagger.$$

This is a unitary mixing of Kraus labels, so it does not change the operation.