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.

Helpful background. Density operators provide the input states, and POVMs describe the outcome probabilities.

The distinction to keep in mind is:

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

Kraus representations of quantum operations

For finite-dimensional input and output spaces, every completely positive map has a Kraus representation. A quantum operation is such a map with nonincreasing trace:

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

Each $K_\alpha$ maps an input vector to the output space, so it need not be square. The map is completely positive because applying it to one part of an entangled state still produces a positive operator. Kraus (1983) and Watrous, chapter 2, develop this representation theorem.

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

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

A selected outcome branch must be trace nonincreasing for every input state:

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

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

The Lüders implementation of a 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

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

The same sharp outcome effects can also accompany a different update. Specifying the projectors alone does not require these particular Kraus operators.

No click: two distinct physical models

Write $P_j=|j\rangle\langle j|$ and let $0\le\eta\le1$ be the probability of a reported click for input $|1\rangle$. Suppose the apparatus first distinguishes $0$ from $1$ sharply, but sometimes fails to report the $1$ result. Its operations may be represented by

$$K_{\mathrm{click}}=\sqrt\eta P_1, \qquad K_{\mathrm{no},0}=P_0, \qquad K_{\mathrm{no},1}=\sqrt{1-\eta}P_1.$$

The no-click effect is $F_{\mathrm{no}}=P_0+(1-\eta)P_1$. The two no-click Kraus terms are added as density operators, not coherently as amplitudes: the apparatus has already distinguished their alternatives, even though the final report does not.

A different apparatus can have this same effect with the single operator

$$\widetilde K_{\mathrm{no}}=P_0+\sqrt{1-\eta}P_1.$$

Together with the same click operator this is also complete, but it preserves conditional coherence. For input $|+\rangle=(|0\rangle+|1\rangle)/\sqrt2$, both apparatuses give $p(\mathrm{no})=1-\eta/2$. Their conditional states are respectively

$$\rho_{\mathrm{no}}= \frac{1}{2-\eta} \begin{pmatrix}1&0\\0&1-\eta\end{pmatrix}, \qquad \widetilde\rho_{\mathrm{no}}= \frac{1}{2-\eta} \begin{pmatrix}1&\sqrt{1-\eta}\\\sqrt{1-\eta}&1-\eta\end{pmatrix}.$$

For $0\le\eta<1$, the first state is mixed and the second is pure. Detector efficiency and outcome probabilities alone do not decide which model applies. Nor does a mathematical Kraus label, by itself, establish that the input possessed a definite value before measurement.

Jump and No-Jump Example

For $0\le\gamma\le1$, 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:

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

Representations with different lengths can first be padded with zero operators and then related by a unitary mixing. Mixing Kraus operators within one reported outcome leaves that operation unchanged; mixing operators across different outcomes generally changes the instrument even if it preserves the nonselective channel.

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

These are operator identities or inequalities, not tests on one chosen state. A sum smaller than $I$ can be a valid selected operation; a sum with an eigenvalue larger than one gives an impossible probability for some input. A claimed complete measurement must include every omitted branch.

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.

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:

$$L_0\rho L_0^\dagger+L_1\rho L_1^\dagger =\tfrac12(K_0+K_1)\rho(K_0^\dagger+K_1^\dagger) +\tfrac12(K_0-K_1)\rho(K_0^\dagger-K_1^\dagger).$$

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.

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.
• H. M. Wiseman and G. J. Milburn, Quantum Measurement and Control, Cambridge University Press (2010). Publisher.