Trace-Preserving and Trace-Nonincreasing Maps

Trace preservation is the probability-conservation condition for deterministic quantum operations. Trace nonincrease is the corresponding condition for selected outcomes, filters, loss branches, and postselected operations.

For a normalized state $\rho$,

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

If a map is deterministic, the output should also have trace $1$. If a map describes only one possible branch, the trace of the unnormalized output is the probability of that branch.

Trace-preserving maps are included among trace-nonincreasing maps. A selected operation may happen with certainty on some inputs but with smaller probability on others. Both trace conditions must be combined with complete positivity to describe a physical operation.

Helpful background. Quantum operations explains why an unnormalized output retains a branch probability; Kraus operators provide concrete representations.

Trace Preservation

Let

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

be a linear map between finite-dimensional operator spaces. The map is trace preserving if

$$\operatorname{Tr}\Phi(X) = \operatorname{Tr}X$$

for all operators $X$ in its domain.

For normalized input states, this gives

$$\operatorname{Tr}\Phi(\rho)=1.$$

A completely positive trace-preserving map is a quantum channel. It may represent unitary evolution, noise, state preparation, discard-and-reprepare dynamics, or reduced dynamics after tracing out an environment.

Trace Nonincrease

A map $\mathcal E$ is trace nonincreasing on positive inputs if

$$X\ge0 \quad\Longrightarrow\quad \operatorname{Tr}\mathcal E(X) \le \operatorname{Tr}X.$$

For an input state $\rho$, the branch probability is

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

For a completely positive trace-nonincreasing operation and $p_{\mathcal E}(\rho)>0$, the conditional output state is

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

This normalization is generally nonlinear because it conditions on an event with input-dependent probability. If the success probability is the same nonzero constant for every input, the normalized transformation can instead be a linear channel. The underlying operation $\mathcal E$ is the linear trace-nonincreasing map.

Kraus Trace Conditions

For a completely positive map with Kraus representation

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

the trace is

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

Therefore $\mathcal E$ is trace preserving exactly when

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

It is trace nonincreasing exactly when

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

The positive operator

$$G_{\mathcal E} = \sum_\alpha K_\alpha^\dagger K_\alpha$$

is the effect associated with the selected operation:

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

The details of Kraus representations are developed in Kraus Representation.

Adjoint-Map Condition

The adjoint map $\Phi^\dagger$ is defined by the trace pairing:

$$\operatorname{Tr} \!\left[ A\,\Phi(B) \right] = \operatorname{Tr} \!\left[ \Phi^\dagger(A)\,B \right].$$

With this convention, $\Phi$ is trace preserving exactly when

$$\Phi^\dagger(I_{\mathrm{out}}) = I_{\mathrm{in}}.$$

It is trace nonincreasing exactly when

$$\Phi^\dagger(I_{\mathrm{out}}) \le I_{\mathrm{in}}.$$

This form is often the cleanest way to remember the relation between Schrödinger-picture states and Heisenberg-picture observables:

$$\text{trace preservation of states} \quad\Longleftrightarrow\quad \text{identity preservation by the adjoint}.$$

Choi Trace Condition

Using the unnormalized output-input Choi matrix, a map

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

is trace preserving exactly when

$$\operatorname{Tr}_{\mathrm{out}}J_\Phi = I_{\mathrm{in}}.$$

It is trace nonincreasing exactly when

$$\operatorname{Tr}_{\mathrm{out}}J_\Phi \le I_{\mathrm{in}}.$$

More precisely, $\operatorname{Tr}_{\mathrm{out}}J_\Phi=[\Phi^\dagger(I)]^{\mathsf T}$ in the defining input basis. The transpose does not alter positivity or the identity, so the trace inequalities have the form above. A normalized Choi state $J_\Phi/d_{\mathrm{in}}$ instead has marginal $I_{\mathrm{in}}/d_{\mathrm{in}}$. Together with $J_\Phi\ge0$, these are semidefinite constraints. They are useful in channel tomography, optimization over operations, and physicality checks for fitted models.

Instruments and Missing Branches

A quantum instrument is a family of operations $\{\mathcal I_m\}$ whose sum is trace preserving:

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

Each individual operation is trace nonincreasing:

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

The total map is trace preserving because the probabilities of all outcomes sum to one:

$$\sum_m p(m|\rho) = \operatorname{Tr} \sum_m\mathcal I_m(\rho) = 1.$$

If an analysis keeps only one outcome and renormalizes, it is doing postselection. The missing trace is not a mathematical defect; it is the probability assigned to the outcomes not retained.

Trace Preserving Is Not Unital

A map is unital if

$$\Phi(I_{\mathrm{in}}) = I_{\mathrm{out}}.$$

Trace preserving and unital are different conditions. Trace preservation says normalized states keep total probability. Unitality says the identity operator is fixed by the Schrödinger-picture map.

For a channel $\Phi$, trace preservation is equivalent to unitality of the adjoint map:

$$\Phi^\dagger(I_{\mathrm{out}}) = I_{\mathrm{in}}.$$

But $\Phi(I_{\mathrm{in}})=I_{\mathrm{out}}$ need not hold. Amplitude damping is trace preserving but not unital: it preserves total probability while relaxing excited-state population toward the ground state. If the input and output dimensions differ, a map cannot satisfy both conditions: taking the trace of $\Phi(I_{\mathrm{in}})=I_{\mathrm{out}}$ would require $d_{\mathrm{in}}=d_{\mathrm{out}}$.

Loss: Two Equivalent Descriptions

For state-independent loss that leaves a surviving encoded state unchanged, there are two useful descriptions. General loss need not have this form; unequal survival probabilities can filter the surviving state.

First, if one keeps only the surviving subspace, the survival operation may be trace decreasing:

$$\mathcal E_{\mathrm{surv}}(\rho) = \eta\rho, \qquad 0\le\eta\le1.$$

The trace is

$$\operatorname{Tr}\mathcal E_{\mathrm{surv}}(\rho) = \eta.$$

Second, if one includes an explicit erasure flag $\lvert e\rangle$, the full map can be trace preserving:

$$\Phi(\rho) = \eta\rho \oplus (1-\eta) \operatorname{Tr}(\rho) \lvert e\rangle\langle e\rvert.$$

The two descriptions answer different modeling questions. The trace-decreasing map describes the selected survival branch. The flagged channel includes both survival and loss as outcomes. See Erasure and Loss Channels for physical examples.

Checking every input without testing every state

A trace condition is an operator identity or inequality. For $K=\operatorname{diag}(1,1/2)$, the input $|0\rangle$ survives with probability one, but $|1\rangle$ survives with probability $1/4$. Testing only $|0\rangle$ would incorrectly suggest trace preservation.

Diagonalizing $G=K^\dagger K=\operatorname{diag}(1,1/4)$ immediately shows that the map is trace nonincreasing but not trace preserving. The extremal eigenvalues of $G$ are the smallest and largest success probabilities over all normalized inputs. Checking all normalized states would be sufficient; checking one favored state is not.

Exercises

Kraus proof

Prove that a Kraus map

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

is trace preserving if and only if $\sum_\alpha K_\alpha^\dagger K_\alpha=I$.

Solution

Using cyclicity of trace,

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

This equals $\operatorname{Tr}\rho=\operatorname{Tr}(\rho I)$ for every $\rho$ exactly when

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

Projective measurement branches

Let $\{P_m\}$ be a projective measurement. Show that each branch $\mathcal I_m(\rho)=P_m\rho P_m$ is trace nonincreasing, while $\sum_m\mathcal I_m$ is trace preserving.

Solution

For one branch,

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

because $0\le P_m\le I$.

For the sum,

$$\operatorname{Tr} \sum_mP_m\rho P_m = \sum_m\operatorname{Tr}(\rho P_m) = \operatorname{Tr} \!\left( \rho\sum_mP_m \right) = \operatorname{Tr}\rho.$$

Amplitude damping is not unital

For the qubit amplitude-damping Kraus operators

$$K_0 = \lvert0\rangle\langle0\rvert + \sqrt{1-\gamma}\, \lvert1\rangle\langle1\rvert, \qquad K_1 = \sqrt{\gamma}\, \lvert0\rangle\langle1\rvert,$$

with $0\le\gamma\le1$, show that the channel is trace preserving but not unital when $\gamma\ne0$.

Solution

First,

$$K_0^\dagger K_0 = \lvert0\rangle\langle0\rvert + (1-\gamma) \lvert1\rangle\langle1\rvert,$$

and

$$K_1^\dagger K_1 = \gamma \lvert1\rangle\langle1\rvert.$$

Therefore

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

so the channel is trace preserving.

For unitality, compute

$$\Phi(I) = K_0K_0^\dagger+K_1K_1^\dagger = (1+\gamma)\lvert0\rangle\langle0\rvert + (1-\gamma)\lvert1\rangle\langle1\rvert.$$

This equals $I$ only when $\gamma=0$.

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.