Local Measurement Statistics

Local measurement statistics are the probabilities seen by an observer who can act only on one subsystem of a composite quantum system. The reduced density operator is physically meaningful because it reproduces exactly those statistics.

If the joint state on $\mathcal H_A\otimes\mathcal H_B$ is $\rho_{AB}$ and

$$\rho_A = \operatorname{Tr}_B\rho_{AB},$$

then every measurement performed only on $A$ can be computed from $\rho_A$ alone. The discarded subsystem $B$ can still carry correlations with $A$, but it is not needed for local outcome probabilities on $A$.

Helpful background. Reduced density operators defines the local state; partial trace supplies the contraction and cyclicity rules used below.

Local Observables

For bounded observables (in particular all finite-dimensional ones), a local measurement on $A$ has the form

$$M_A\otimes I_B.$$

The expectation value in the joint state is

$$\langle M_A\rangle = \operatorname{Tr}_{AB} \bigl[ \rho_{AB}(M_A\otimes I_B) \bigr].$$

The defining property of the partial trace gives

$$\operatorname{Tr}_{AB} \bigl[ \rho_{AB}(M_A\otimes I_B) \bigr] = \operatorname{Tr}_A(\rho_A M_A).$$

This identity is the operational meaning of $\rho_A$: it is the unique density operator on $\mathcal H_A$ that gives the same expectation values for all observables accessible on subsystem $A$.

Projective Measurements on a Subsystem

For a projective measurement on $A$ with projectors $\{P_a\}$,

$$P_aP_{a'} = \delta_{aa'}P_a, \qquad \sum_a P_a=I_A,$$

the corresponding measurement on the composite system uses projectors

$$P_a\otimes I_B.$$

The Born-rule probability for outcome $a$ is

$$\begin{aligned} p(a) &= \operatorname{Tr}_{AB} \bigl[ \rho_{AB}(P_a\otimes I_B) \bigr]\\ &= \operatorname{Tr}_A(\rho_A P_a). \end{aligned}$$

Thus the same local probabilities are obtained whether one starts from the full joint state and inserts an identity on $B$, or first reduces to $\rho_A$ and then applies the ordinary density-operator Born rule.

General Measurement Effects

The same statement holds for generalized measurement probabilities. Let $\{E_a\}$ be positive operators on $\mathcal H_A$ satisfying

$$E_a\ge0, \qquad \sum_a E_a=I_A.$$

The operators $E_a$ are the effects of a POVM. If this measurement is performed locally on $A$, the corresponding effects on the joint system are

$$E_a\otimes I_B.$$

The probability of outcome $a$ is

$$p(a) = \operatorname{Tr}_{AB} \bigl[ \rho_{AB}(E_a\otimes I_B) \bigr] = \operatorname{Tr}_A(\rho_A E_a).$$

This is why reduced density operators are not tied to projective measurements only. They summarize all statistics obtainable from any measurement whose effects act on the local subsystem.

What Local Statistics Determine

If enough local measurements are performed on identically prepared copies, their statistics can reconstruct $\rho_A$. For a qubit, for example,

$$\rho_A = \frac12 \bigl( I+\mathbf r\cdot\boldsymbol\sigma \bigr),$$

where the Bloch vector components are

$$r_i = \operatorname{Tr}(\rho_A\sigma_i), \qquad i\in\{x,y,z\}.$$

Measurements of $\sigma_x$, $\sigma_y$, and $\sigma_z$ on subsystem $A$ determine $\mathbf r$ and therefore determine $\rho_A$.

These are three convenient settings, not a universal minimum: a single informationally complete POVM can also reconstruct a qubit from many repeated preparations.

They do not determine $\rho_{AB}$. Many different joint states can have the same reduced state. Local tomography reconstructs the local state, not the global state.

A maximally mixed state need not give uniform POVM outcomes

For one qubit of a Bell pair, $\rho_A=I/2$. Consequently

$$p(a)=\frac12\operatorname{Tr}E_a.$$

Every rank-one projective spin outcome has probability $1/2$, irrespective of its axis. General POVM effects need not have equal traces. For example,

$$E_0=\frac12|0\rangle\langle0|,\quad E_1=\frac12|0\rangle\langle0|,\quad E_2=|1\rangle\langle1|$$

are positive and sum to $I$. Their probabilities on $I/2$ are $1/4,1/4,1/2$. This device can be realized by a $Z$ measurement followed by a fair random split of the $0$ record into two labels.

The Bell state and the separable mixture $(|00\rangle\langle00|+|11\rangle\langle11|)/2$ give the same probabilities for every local POVM, including this one. Their joint correlations differ, as calculated in classical correlation versus entanglement.

Joint Measurements and Marginals

If a local measurement on $A$ has effects $\{E_a\}$ and a local measurement on $B$ has effects $\{F_b\}$, then the joint probability is

$$p(a,b) = \operatorname{Tr}_{AB} \bigl[ \rho_{AB}(E_a\otimes F_b) \bigr].$$

The marginal probability for $A$ is obtained by summing over $B$:

$$\begin{aligned} p_A(a) &= \sum_b p(a,b)\\ &= \operatorname{Tr}_{AB} \left[ \rho_{AB} \left( E_a\otimes\sum_b F_b \right) \right]\\ &= \operatorname{Tr}_{AB} \bigl[ \rho_{AB}(E_a\otimes I_B) \bigr]\\ &= \operatorname{Tr}_A(\rho_A E_a). \end{aligned}$$

The local marginal does not depend on which complete measurement $\{F_b\}$ is chosen on $B$. The joint distribution can depend strongly on both measurements, but the unconditioned local distribution on $A$ cannot.

Why a remote channel cannot change the marginal

In finite dimensions, let a trace-preserving operation on $B$ have Kraus operators $K_\mu$ with $\sum_\mu K_\mu^\dagger K_\mu=I_B$. This includes nonselective measurements and noise channels. Then

$$\begin{aligned} \rho_A' &=\sum_\mu\operatorname{Tr}_B[(I\otimes K_\mu)\rho_{AB}(I\otimes K_\mu^\dagger)]\\ &=\operatorname{Tr}_B[\rho_{AB}(I\otimes\sum_\mu K_\mu^\dagger K_\mu)]\\ &=\rho_A. \end{aligned}$$

The middle step moves only operators on the traced factor cyclically. This proof allows $B$‘s output space to differ from its input space when the partial trace is taken over the corresponding output.

A selected measurement outcome is trace decreasing before normalization and does not obey the completeness step. It can change the conditional state of $A$. Averaging all outcomes recovers $\rho_A$, and the result labels must be communicated to identify the selected subensembles. Conditional states works through this distinction; no-cloning and no-signalling discusses its communication consequences.

The statement assumes a genuinely local operation and no later interaction or transmitted system from $B$ to $A$. It concerns an unchanged unconditioned marginal, not unchanged joint correlations.

Common pitfalls

Local tomography does not reconstruct correlations. Measurements on $A$ alone determine $\rho_A$. Recording paired local outcomes on both sides can instead determine joint correlations.

Uniform spin outcomes do not imply uniform outcomes for every device. The probability on $I/2$ is proportional to the trace of the POVM effect, which can differ between labels.

A selected branch is not a trace-preserving channel. Include all success and failure outcomes when applying the no-signalling proof.

References

• Michael A. Nielsen and Isaac L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press, 2010, chapters 2 and 8.
• John Watrous, The Theory of Quantum Information, Cambridge University Press, 2018, chapters 2–3. Author-hosted text.

Exercises

1. A binary unsharp spin measurement has effects $E_\pm=(I\pm\eta\,\mathbf n\cdot\boldsymbol\sigma)/2$, with $|\mathbf n|=1$ and $0\le\eta\le1$. Find its outcome probabilities on $\rho_A=(I+\mathbf r\cdot\boldsymbol\sigma)/2$.

Solution

Use $\operatorname{Tr}\sigma_i=0$ and $\operatorname{Tr}(\sigma_i\sigma_j)=2\delta_{ij}$:

$$p(\pm)=\frac{1\pm\eta\,\mathbf r\cdot\mathbf n}{2}.$$

At $\eta=1$ this is the projective spin rule. At $\eta=0$ the device returns a fair random label independent of the state. For $\mathbf r=(0,0,1/2)$, $\mathbf n=\hat z$, and $\eta=0.6$, the probabilities are $0.65$ and $0.35$.

2. Show that every spin-direction measurement on one qubit of $\lvert\Phi^+\rangle$ gives equal probabilities.

Solution

For either qubit of $\lvert\Phi^+\rangle$,

$$\rho_A = \frac12 I.$$

A projective measurement along a unit direction $\mathbf n$ has projectors

$$P_\pm = \frac12 \bigl( I\pm\mathbf n\cdot\boldsymbol\sigma \bigr).$$

Therefore

$$p(\pm) = \operatorname{Tr}(\rho_A P_\pm) = \frac12\operatorname{Tr}(P_\pm) = \frac12,$$

because each rank-one qubit projector has trace $1$.

3. Compare $x$-basis joint measurements for $\lvert\Phi^+\rangle$ and $\rho_{\mathrm{cc}}$.

Solution

The Bell state can be written in the $x$ basis as

$$\lvert\Phi^+\rangle = \frac{1}{\sqrt2} \bigl( \lvert++\rangle+\lvert--\rangle \bigr).$$

Thus the outcomes agree: $++$ and $--$ each occur with probability $1/2$.

For

$$\rho_{\mathrm{cc}} = \frac12 \lvert00\rangle\langle00\rvert + \frac12 \lvert11\rangle\langle11\rvert,$$

each product state gives independent $x$-basis outcomes with probability $1/4$ for each pair. The mixture therefore gives

$$p(++) = p(+-) = p(-+) = p(--) = \frac14.$$

The two states have the same one-qubit local statistics but different joint statistics.

4. Explain why changing the measurement on $B$ cannot change the unconditioned probability for outcome $a$ on $A$.

Solution

Let $\{F_b\}$ be any complete POVM on $B$, so $\sum_b F_b=I_B$. Then

$$\begin{aligned} \sum_b \operatorname{Tr}_{AB} \bigl[ \rho_{AB}(E_a\otimes F_b) \bigr] &= \operatorname{Tr}_{AB} \left[ \rho_{AB} \left( E_a\otimes\sum_b F_b \right) \right]\\ &= \operatorname{Tr}_{AB} \bigl[ \rho_{AB}(E_a\otimes I_B) \bigr]\\ &= \operatorname{Tr}_A(\rho_A E_a). \end{aligned}$$

The final expression contains no reference to the particular measurement chosen on $B$.