Partial Trace Revisited

Partial trace is characterized by equality of all bounded local expectations. That characterization makes it possible to prove consistency under regrouping, local channels, and approximations without manipulating potentially ambiguous infinite index sums. The canonical Partial Trace article introduces the basic operation, and Trace and Partial Trace proves its trace-class construction.

Helpful background. Normal States supplies density-operator continuity. Completely Positive Maps and Instruments supply normal channels and outcome operations.

Repeated reductions commute when factors are declared

Let $\rho$ be a density operator on $\mathcal H_A\otimes\mathcal H_B\otimes\mathcal H_C$, with the usual canonical associativity identification. The reduction to $A$ can be performed in either order:

$$\operatorname{Tr}_{BC}\rho =\operatorname{Tr}_B(\operatorname{Tr}_C\rho) =\operatorname{Tr}_C(\operatorname{Tr}_B\rho).$$

The spaces in the last two expressions are understood after the corresponding factors have been removed.

To prove the first equality, pair both sides with arbitrary bounded $X$ on $A$. Repeatedly using the defining identity gives

$$\operatorname{Tr}\!\left[ X\,\operatorname{Tr}_B(\operatorname{Tr}_C\rho) \right] =\operatorname{Tr}\bigl[\rho(X\otimes I_B\otimes I_C)\bigr].$$

The direct reduction gives the same value. Trace duality determines the trace-class operator uniquely. The other order follows identically.

This is a quantum analogue of integrating out independent coordinate labels in either order. The argument uses trace-class bounded-operator duality, so it does not require a pointwise density kernel or an exchange of conditionally convergent sums.

Trace-preserving local processing respects reduction

Let $\Phi_A$ and $\Phi_B$ be normal quantum channels in the Schrödinger picture, mapping trace-class inputs to trace-class outputs. Then

$$\operatorname{Tr}_{B'}[(\Phi_A\otimes\Phi_B)(\rho)] =\Phi_A(\operatorname{Tr}_B\rho).$$

The prime permits a different output space for the second subsystem. To prove the identity, use the normal adjoints. For bounded $X$ on $A'$,

$$\operatorname{Tr}\bigl[(\Phi_A\otimes\Phi_B)(\rho) (X\otimes I_{B'})\bigr] =\operatorname{Tr}\bigl[\rho (\Phi_A^*(X)\otimes\Phi_B^*(I_{B'}))\bigr].$$

Trace preservation is precisely $\Phi_B^*(I_{B'})=I_B$, reducing this to the right-hand side’s expectation.

In particular, an unread local channel on $B$ leaves $\rho_A$ unchanged. This statement includes noisy channels and replacement channels, not only local unitaries. Normality is part of the assumptions: it supplies the trace-class maps and their tensor extension in this setting.

A remote conditional state depends only on the local effect

Now select one trace-nonincreasing completely positive operation $\Phi_B$, with effect $E=\Phi_B^*(I_{B'})$. The unnormalized conditional state on $A$ is

$$\tau_A=\operatorname{Tr}_{B'}[(\operatorname{id}\otimes\Phi_B)(\rho)].$$

Its defining expectation is

$$\operatorname{Tr}(X\tau_A)=\operatorname{Tr}[\rho(X\otimes E)].$$

Thus after discarding the measured subsystem, the remote conditional state depends on the local branch only through its effect. An explicitly positive expression is

$$\tau_A=\operatorname{Tr}_B\!\left[ (I\otimes\sqrt E)\rho(I\otimes\sqrt E) \right].$$

Cyclicity with bounded factors verifies the same expectation identity. Distinct instruments with the same outcome POVM can have different joint or local output states on $B'$, while giving the same conditional marginal on the untouched $A$.

The branch probability is $p=\operatorname{Tr}\tau_A=\operatorname{Tr}(\rho_BE)$. Normalize by $p$ only when it is positive. Summing over all outcomes returns $\rho_A$ because the effects sum to identity. Outcome conditioning therefore changes a remote ensemble without enabling a change of the unread remote marginal.

Number correlation gives a worked conditional example

For

$$\Psi_q=\sqrt{1-q}\sum_{n=0}^\infty q^{n/2}|n,n\rangle, \qquad 0<q<1,$$

let the effect on $B$ be the number event $E_n=|n\rangle\langle n|$. The displayed positive expression gives

$$\tau_A(n)=(1-q)q^n|n\rangle\langle n|.$$

The conditional state is $|n\rangle\langle n|$ and the outcome probability is $(1-q)q^n$. This remains true if the device prepares an entirely different $B'$ state after reading $n$, since that preparation does not change $E_n$.

If the record is discarded, the mixture on $A$ is its original geometric marginal. If only a bin of numbers is retained, the conditional state is the corresponding normalized diagonal mixture. Neither conclusion requires treating an exact continuous-variable eigenket as a Hilbert-space state.

Local subspace cutoffs control all marginal errors

Let $P_n\to I_A$ and $Q_n\to I_B$ be increasing finite-rank projections, and put $\Pi_n=P_n\otimes Q_n$. For a fixed density operator, $p_n=\operatorname{Tr}(\rho\Pi_n)\to1$. Whenever $p_n>0$, form

$$\rho_n=\frac{\Pi_n\rho\Pi_n}{p_n}.$$

A purification of $\rho$ and its normalized projection have overlap $\sqrt{p_n}$. The fidelity–distance bounds and reduction contractivity therefore give

$$D(\rho,\rho_n)\leq\sqrt{1-p_n}.$$

Every reduced state obeys the same upper bound on its trace distance. This supplies a simultaneous error certificate for all bounded local measurement predictions.

The bound is state dependent and becomes useless when the retained probability is small. It also says nothing by itself about convergence of unbounded energy or entropy; those require extra tail control.

Exercises

1. If an outcome effect is $E=qI_B$, with fixed $0<q\leq1$, can conditioning on that outcome change the remote state?

Solution

The expectation formula gives $\tau_A=q\rho_A$. The outcome probability is $q$, so the normalized state is unchanged. The measured subsystem can nevertheless undergo a nontrivial output transformation; only its effect is fixed.

2. A local cutoff retains probability $1-10^{-6}$. What follows for a binary measurement probability on either subsystem?

Solution

The reduced-state trace distance is at most $10^{-3}$. Any effect’s probability changes by at most that distance. An energy expectation need not satisfy the same bound because energy is generally unbounded.

References

• Busch, Paul, Pekka Lahti, Juha-Pekka Pellonpää, and Kari Ylinen, Quantum Measurement, Springer, 2016 — instruments, effects, and conditional states. Publisher.
• Holevo, Alexander S., Quantum Systems, Channels, Information: A Mathematical Introduction, second revised and expanded edition, De Gruyter, 2019 — normal channels, reductions, and infinite-dimensional composite systems. Publisher.
• Watrous, John, The Theory of Quantum Information, Cambridge University Press, 2018 — purification and distinguishability estimates. Author’s text.