Local and Global Observables

This is the canonical treatment of local, joint, correlation, and genuinely global observables on composite systems. For only the embedding needed by the composition postulate, use Subsystems and Local Observables.

A statement such as “measure observable $A$ on subsystem $A$” is shorthand for an operator on the whole composite Hilbert space. If

$$\mathcal H_{AB} = \mathcal H_A\otimes\mathcal H_B,$$

then an observable $M_A$ available to the first subsystem is represented globally as

$$M_A\otimes I_B.$$

The identity factor is not decorative. It records which degrees of freedom the operation leaves untouched, makes dimensions and operator products well-defined, and distinguishes a local observable from a correlation observable such as $A\otimes B$.

Local operators

Let $X_A$ be a linear operator on $\mathcal H_A$. Its natural embedding into the composite operator algebra is

$$\iota_A(X_A) = X_A\otimes I_B.$$

On a product vector,

$$\begin{aligned} &(X_A\otimes I_B) \left( \lvert\psi\rangle_A\otimes\lvert\phi\rangle_B \right) \\ &\qquad= \left( X_A\lvert\psi\rangle_A \right) \otimes \lvert\phi\rangle_B. \end{aligned}$$

The action extends by linearity to every vector in $\mathcal H_A\otimes\mathcal H_B$, including entangled vectors. Similarly, an operator local to $B$ is embedded as

$$\iota_B(Y_B) = I_A\otimes Y_B.$$

The embedding preserves the operator algebra:

$$\begin{aligned} \iota_A(X_A Y_A) &= \iota_A(X_A)\iota_A(Y_A), \\ \iota_A(X_A^\dagger) &= \iota_A(X_A)^\dagger, \\ \iota_A(I_A) &= I_A\otimes I_B. \end{aligned}$$

Consequently, Hermitian operators remain Hermitian, projectors remain projectors, and unitary operators remain unitary after embedding.

Spectra and degeneracy

Suppose

$$M_A = \sum_a m_a P_a$$

is the spectral decomposition of a finite-dimensional observable. Then

$$M_A\otimes I_B = \sum_a m_a \left( P_a\otimes I_B \right).$$

The numerical eigenvalues are unchanged, but their multiplicities grow. If the eigenspace of $m_a$ in $\mathcal H_A$ has dimension $g_a$, then its eigenspace in the composite system has dimension

$$g_a\,\dim\mathcal H_B.$$

This extra degeneracy reflects that the local observable does not distinguish any states of subsystem $B$.

Operators on distinct factors commute

For arbitrary operators $X_A$ and $Y_B$,

$$\begin{aligned} &(X_A\otimes I_B)(I_A\otimes Y_B) \\ &\qquad= X_A\otimes Y_B \\ &\qquad= (I_A\otimes Y_B)(X_A\otimes I_B). \end{aligned}$$

Therefore

$$\left[ X_A\otimes I_B,\, I_A\otimes Y_B \right] =0.$$

This compatibility is an algebraic fact about tensor factors. It does not, by itself, establish relativistic locality: a tensor-factor label need not represent a spacetime region, and relativistic theories impose additional causal structure.

Product observables

If $A$ and $B$ are Hermitian, then

$$A\otimes B$$

is a Hermitian observable on the composite system. It acts on both factors:

$$\begin{aligned} &(A\otimes B) \left( \lvert\psi\rangle_A\otimes\lvert\phi\rangle_B \right) \\ &\qquad= \left( A\lvert\psi\rangle_A \right) \otimes \left( B\lvert\phi\rangle_B \right). \end{aligned}$$

If $A\lvert a\rangle=a\lvert a\rangle$ and $B\lvert b\rangle=b\lvert b\rangle$, then

$$(A\otimes B) \lvert a\rangle\lvert b\rangle = ab\, \lvert a\rangle\lvert b\rangle.$$

Thus the eigenvalues of a product observable are products of local eigenvalues.

A product observable is not the same thing as a local observable. It usually probes a joint correlation. When $A$ and $B$ are measured locally on the two subsystems, multiplying the two outcomes estimates $\langle A\otimes B\rangle$.

Factorization in product states

For a product density operator

$$\rho_{AB} = \rho_A\otimes\rho_B,$$

the expectation value factorizes:

$$\begin{aligned} \langle A\otimes B\rangle &= \operatorname{Tr}_{AB} \left[ (\rho_A\otimes\rho_B)(A\otimes B) \right] \\ &= \operatorname{Tr}_A(\rho_A A)\, \operatorname{Tr}_B(\rho_B B). \end{aligned}$$

For a general joint state, define the connected correlation

$$\begin{aligned} \operatorname{Cov}_{\rho}(A,B) &= \langle A\otimes B\rangle_\rho \\ &\quad- \langle A\otimes I_B\rangle_\rho \langle I_A\otimes B\rangle_\rho. \end{aligned}$$

Every product state has zero connected correlation for every pair $A,B$. The converse requires factorization for a sufficiently complete family of operators; one vanishing covariance does not prove that the state is a product state.

Beyond a single product

Most composite observables are not one tensor product. In finite dimensions, any operator can be expanded as a sum

$$O_{AB} = \sum_{\mu,\nu} c_{\mu\nu}\, F_\mu^{(A)}\otimes G_\nu^{(B)}$$

using operator bases on the two subsystems. Whether such an observable is operationally local, a correlation measurement, or a genuinely joint measurement depends on the complete operator and on how it is implemented, not merely on the fact that it can be written as a sum of tensor products.

The detailed taxonomy and operator-basis expansion live in Operators on Composite Systems.

Sums, interactions, and locality

For noninteracting subsystems, a Hamiltonian often has the form

$$H_0 = H_A\otimes I_B + I_A\otimes H_B.$$

The first term changes only $A$ and the second changes only $B$. An interaction adds a term such as

$$H_{\mathrm{couple}} = J\,Z_A\otimes Z_B.$$

Although this interaction is a single product operator, it is not local to either subsystem. It couples them and can generate correlations or entanglement.

Locality can also depend on the dynamical picture. An operator that is local at the reference time may spread under interacting Heisenberg evolution:

$$M_A(t) = U^\dagger(t) \left( M_A\otimes I_B \right) U(t).$$

When $U(t)$ is generated by an interaction, $M_A(t)$ generally cannot be written as $M_A'(t)\otimes I_B$. This is operator spreading, not a contradiction of the original tensor-factor definition.

Expectation values from reduced states

Let

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

The defining property of the partial trace is

$$\begin{aligned} &\operatorname{Tr}_{AB} \left[ \rho_{AB}(M_A\otimes I_B) \right] \\ &\qquad= \operatorname{Tr}_A(\rho_A M_A). \end{aligned}$$

Every expectation value accessible through an observable local to $A$ can therefore be computed from $\rho_A$ alone. Conversely, in finite dimensions, these expectation values for all Hermitian $M_A$ determine $\rho_A$ uniquely. This is the operational content of a reduced state.

The reduced state does not determine

$$\langle A\otimes B\rangle,$$

because that quantity depends on correlations retained in $\rho_{AB}$. Nor does $\rho_A$ generally determine which global preparation produced it. Many different pure and mixed joint states can share the same marginal.

For a product state,

$$\rho_{AB}=\rho_A\otimes\rho_B,$$

the two reduced states contain enough information to reconstruct the joint state. For correlated states they do not.

Local measurements

Projective measurements

Let $\{P_a\}$ be a projective measurement on $A$:

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

On the composite Hilbert space, the measurement projectors are

$$P_a\otimes I_B.$$

For a joint state $\rho_{AB}$, the Born probability is

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

If outcome $a$ is selected, the Lüders update of the joint state is

$$\rho_{AB|a} = \frac{ (P_a\otimes I_B) \rho_{AB} (P_a\otimes I_B) }{p(a)}.$$

The first equation concerns outcome statistics; the second concerns post-measurement state assignment. Keeping those roles separate prevents a large class of measurement mistakes. See Projective Measurement for the canonical update treatment.

General measurement effects

For a local POVM on $A$, let

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

The corresponding joint-system effects are $E_a\otimes I_B$, and

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

The effects determine probabilities but do not, by themselves, specify the conditional output state. That requires a measurement instrument or a choice of Kraus operators. The distinction is developed in Generalized Measurements Overview.

Measurements on both subsystems

If $\{E_a\}$ is a POVM on $A$ and $\{F_b\}$ is a POVM on $B$, their product measurement has joint effects

$$E_a\otimes F_b$$

and probabilities

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

Summing over $b$ gives

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

The joint distribution can depend strongly on both measurement choices, while its local marginal depends only on the reduced state.

Two-qubit examples

A local spin observable

For two spin-$1/2$ systems, the $z$ component of the first spin is

$$S_z^{(A)} = \frac{\hbar}{2} \sigma_z\otimes I_2.$$

Its eigenvalues are $+\hbar/2$ and $-\hbar/2$. Each is twice degenerate because the second spin may independently be up or down.

For example,

$$S_z^{(A)} \lvert\uparrow\downarrow\rangle = \frac{\hbar}{2} \lvert\uparrow\downarrow\rangle,$$

whereas

$$S_z^{(A)} \lvert\downarrow\uparrow\rangle = -\frac{\hbar}{2} \lvert\downarrow\uparrow\rangle.$$

Local randomness and joint order

For

$$\lvert\Phi^+\rangle = \frac{ \lvert00\rangle+\lvert11\rangle }{\sqrt2},$$

each reduced state is $I_2/2$. Consequently,

$$\langle Z\otimes I\rangle=0, \qquad \langle I\otimes Z\rangle=0.$$

Yet the correlation observables satisfy

$$\langle Z\otimes Z\rangle=1, \qquad \langle X\otimes X\rangle=1.$$

Each local result is unbiased, but the paired results are perfectly correlated in either of these bases.

Now compare the separable mixed state

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

It has the same reduced states and the same $Z\otimes Z$ correlation:

$$\langle Z\otimes Z\rangle_{\mathrm{cc}}=1.$$

However,

$$\langle X\otimes X\rangle_{\mathrm{cc}}=0.$$

No experiment confined to one subsystem distinguishes these two global states. A suitable joint correlation measurement does.

No-signaling preview

Suppose an arbitrary trace-preserving quantum operation is applied only to $B$. Choose Kraus operators $\{K_\mu\}$ satisfying

$$\sum_\mu K_\mu^\dagger K_\mu=I_B.$$

When no outcome is selected, the joint state becomes

$$\rho'_{AB} = \sum_\mu (I_A\otimes K_\mu) \rho_{AB} (I_A\otimes K_\mu^\dagger).$$

For every local observable $M_A$,

$$\begin{aligned} &\operatorname{Tr}_{AB} \left[ \rho'_{AB}(M_A\otimes I_B) \right] \\ &= \sum_\mu \operatorname{Tr}_{AB} \left[ \rho_{AB} \left( M_A\otimes K_\mu^\dagger K_\mu \right) \right] \\ &= \operatorname{Tr}_{AB} \left[ \rho_{AB}(M_A\otimes I_B) \right]. \end{aligned}$$

Because this holds for every $M_A$, the reduced state is unchanged:

$$\rho'_A=\rho_A.$$

This is the algebraic core of no-signaling for local operations: an unconditioned, trace-preserving operation on $B$ cannot alter measurement statistics available on $A$ alone.

Conditioning is different. Let $\{K_{b\mu}\}_\mu$ be the Kraus operators associated with recorded outcome $b$. If that remote outcome is learned, the unnormalized conditional state is

$$\begin{aligned} p(b)\rho_{A|b} &= \sum_\mu \operatorname{Tr}_B\Omega_{b\mu}, \\ \Omega_{b\mu} &= (I_A\otimes K_{b\mu}) \rho_{AB} (I_A\otimes K_{b\mu}^\dagger). \end{aligned}$$

For a complete instrument,

$$\sum_{b,\mu} K_{b\mu}^\dagger K_{b\mu} = I_B.$$

The individual $\rho_{A|b}$ may differ from $\rho_A$, but their probability-weighted average obeys

$$\sum_b p(b)\rho_{A|b} = \rho_A.$$

Learning $b$ requires classical communication. Remote conditioning can reveal correlations; it does not provide controllable faster-than-light signaling. The canonical local-statistics treatment is Local Measurement Statistics, and Conditional States develops the conditioned description.

Practical workflow

When a problem refers to one part of a composite system:

1. Write the full Hilbert space and fix the subsystem labels.
2. Embed a local operator with identities on every untouched factor.
3. Check whether the requested quantity is local, joint, or conditional.
4. For local expectations or probabilities, reduce first and use $\rho_A$.
5. For correlations such as $\langle A\otimes B\rangle$, retain $\rho_{AB}$.
6. For post-measurement states, specify the instrument or update rule, not only the POVM effects.
7. Distinguish an ignored remote outcome from a selected and communicated outcome.
8. Check normalization, Hermiticity, positivity, and the relevant identity resolution.

Global Observables That Are Not Local

A global observable is any Hermitian operator on $\mathcal H_{AB}$. Some global observables are sums or products of local terms. Others are tied to entangled bases and cannot be reproduced by separately measuring $A$ and $B$ in fixed local bases.

The Bell-basis projectors are the standard two-qubit example:

$$\Pi_{\Phi^+} = \lvert\Phi^+\rangle\langle\Phi^+\rvert, \qquad \Pi_{\Phi^-} = \lvert\Phi^-\rangle\langle\Phi^-\rvert,$$

with analogous projectors for $\lvert\Psi^+\rangle$ and $\lvert\Psi^-\rangle$. A Bell-basis measurement distinguishes coherent superpositions of product-basis states. Measuring both qubits separately in the computational basis does not reveal the relative phase that distinguishes $\lvert\Phi^+\rangle$ from $\lvert\Phi^-\rangle$.

Global observables also appear in angular momentum. For two spins, the total spin operator

$$\mathbf S^2 = (\mathbf S_1+\mathbf S_2)^2$$

distinguishes singlet and triplet sectors. It is not the same information as separately measuring $S_{1z}$ and $S_{2z}$.

Spin-Singlet Correlations

The spin singlet is

$$\lvert\Psi^-\rangle = \frac{ \lvert\uparrow\rangle_A\lvert\downarrow\rangle_B - \lvert\downarrow\rangle_A\lvert\uparrow\rangle_B }{\sqrt2}.$$

For any common axis $\hat n$,

$$\langle (\boldsymbol\sigma\cdot\hat n) \otimes (\boldsymbol\sigma\cdot\hat n) \rangle = -1.$$

Each local spin component has zero expectation:

$$\langle (\boldsymbol\sigma\cdot\hat n) \otimes I \rangle = 0, \qquad \langle I\otimes (\boldsymbol\sigma\cdot\hat n) \rangle = 0.$$

The singlet therefore separates three ideas: local observables, joint correlation observables, and rotationally invariant global structure.

Common mistakes

• Writing $M_A$ as though it already acts on $\mathcal H_A\otimes\mathcal H_B$ and thereby hiding a dimension mismatch.
• Confusing a product observable $A\otimes B$ with a product state $\rho_A\otimes\rho_B$.
• Calling $A\otimes B$ local to one subsystem.
• Assuming one vanishing correlation proves that a state is uncorrelated.
• Expecting $\rho_A$ to determine joint quantities such as $\langle A\otimes B\rangle$.
• Confusing POVM effects, which determine probabilities, with an instrument, which also determines state updates.
• Replacing a marginal probability by a conditional probability after a remote outcome is known.
• Treating no-signaling as the absence of entanglement correlations.
• Inferring relativistic spacelike locality solely from commuting tensor-factor operators.

References

• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, 10th anniversary ed., Cambridge University Press (2010), Ch. 2.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press (2018), Chs. 1–2.
• J. Preskill, Lecture Notes for Physics 229: Quantum Information and Computation, Chapter 2, Secs. 2.3–2.4.
• P. Busch, P. Lahti, J.-P. Pellonpää, and K. Ylinen, Quantum Measurement, Springer (2016), Chs. 3 and 5.
• A. S. Holevo, Probabilistic and Statistical Aspects of Quantum Theory, 2nd ed., Edizioni della Normale (2011), Ch. 2.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press (2020), Chs. 1 and 3.
• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press (1958).
• A. Peres, Quantum Theory: Concepts and Methods, Kluwer (1995).
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer (1994).

Exercises

1. Prove that local operators on distinct factors commute:

$$\left[ X_A\otimes I_B,\, I_A\otimes Y_B \right] =0.$$

Solution

Use the tensor-product multiplication rule:

$$\begin{aligned} &(X_A\otimes I_B)(I_A\otimes Y_B) \\ &\qquad= X_A\otimes Y_B, \end{aligned}$$

and

$$\begin{aligned} &(I_A\otimes Y_B)(X_A\otimes I_B) \\ &\qquad= X_A\otimes Y_B. \end{aligned}$$

The two products are equal, so their difference vanishes.

2. Let $M_A$ have an eigenvalue $m$ with degeneracy $g$, and let $\dim\mathcal H_B=d_B$. Find the degeneracy of $m$ for $M_A\otimes I_B$.

Solution

Choose a basis $\{\lvert m,\alpha\rangle_A\}_{\alpha=1}^{g}$ for the eigenspace and any basis $\{\lvert j\rangle_B\}_{j=1}^{d_B}$ for $\mathcal H_B$. Then

$$\begin{aligned} &(M_A\otimes I_B) \lvert m,\alpha\rangle_A\lvert j\rangle_B \\ &\qquad= m\, \lvert m,\alpha\rangle_A\lvert j\rangle_B. \end{aligned}$$

There are $g d_B$ independent vectors of this form, so the composite degeneracy is $g d_B$.

3. For $\rho_{AB}=\rho_A\otimes\rho_B$, prove that every product-observable expectation factorizes.

Solution

Using multiplication and trace factorization,

$$\begin{aligned} \langle A\otimes B\rangle &= \operatorname{Tr}_{AB} \left[ (\rho_A A)\otimes(\rho_B B) \right] \\ &= \operatorname{Tr}_A(\rho_A A)\, \operatorname{Tr}_B(\rho_B B). \end{aligned}$$

Thus every connected correlation vanishes in a product state.

4. For the Bell state $\lvert\Phi^+\rangle$, calculate $\langle Z\otimes I\rangle$ and $\langle Z\otimes Z\rangle$ directly.

Solution

The local operator gives

$$(Z\otimes I)\lvert\Phi^+\rangle = \frac{ \lvert00\rangle-\lvert11\rangle }{\sqrt2},$$

which is orthogonal to $\lvert\Phi^+\rangle$. Therefore

$$\langle Z\otimes I\rangle=0.$$

By contrast, both $\lvert00\rangle$ and $\lvert11\rangle$ have eigenvalue $+1$ under $Z\otimes Z$, so

$$(Z\otimes Z)\lvert\Phi^+\rangle = \lvert\Phi^+\rangle,$$

and $\langle Z\otimes Z\rangle=1$.

5. Show that the Bell state and

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

have the same local $Z$ statistics and the same $\langle Z\otimes Z\rangle$, but different $\langle X\otimes X\rangle$.

Solution

Both reduced states are $I_2/2$, so either local $Z$ outcome is equally likely. In both global states, computational-basis outcomes agree, giving

$$\langle Z\otimes Z\rangle=1.$$

For the Bell state,

$$(X\otimes X)\lvert\Phi^+\rangle = \lvert\Phi^+\rangle,$$

so its $X\otimes X$ expectation is $1$. For either projector in $\rho_{\mathrm{cc}}$, $X\otimes X$ maps the basis vector to the other orthogonal basis vector. Hence each diagonal expectation vanishes and

$$\langle X\otimes X\rangle_{\mathrm{cc}}=0.$$

6. Let

$$\lvert\Psi_p\rangle = \sqrt p\,\lvert00\rangle + \sqrt{1-p}\,\lvert11\rangle.$$

Compute $\langle Z\otimes I\rangle$, $\langle X\otimes I\rangle$, and $\langle Z\otimes Z\rangle$.

Solution

The reduced state is

$$\rho_A = p\lvert0\rangle\langle0\rvert + (1-p)\lvert1\rangle\langle1\rvert.$$

Therefore

$$\langle Z\otimes I\rangle = \operatorname{Tr}(\rho_A Z) = 2p-1,$$

and

$$\langle X\otimes I\rangle = \operatorname{Tr}(\rho_A X) =0.$$

Both terms of the joint state have equal $Z$ outcomes, so

$$\langle Z\otimes Z\rangle=1.$$

7. Let a trace-preserving operation on $B$ have Kraus operators $\{K_\mu\}$. Prove that it leaves $\rho_A$ unchanged when its outcomes are ignored.

Solution

For any $M_A$, cyclicity of the full trace gives

$$\begin{aligned} &\operatorname{Tr}_{AB} \left[ \rho'_{AB}(M_A\otimes I_B) \right] \\ &= \sum_\mu \operatorname{Tr}_{AB} \left[ \rho_{AB} \left( M_A\otimes K_\mu^\dagger K_\mu \right) \right]. \end{aligned}$$

Trace preservation implies

$$\sum_\mu K_\mu^\dagger K_\mu=I_B,$$

so the expression equals the original local expectation

$$\operatorname{Tr}_{AB} \left[ \rho_{AB}(M_A\otimes I_B) \right].$$

Since the equality holds for every $M_A$, the operators representing the two local states are equal:

$$\rho'_A=\rho_A.$$

8. Subsystem $B$ of $\lvert\Phi^+\rangle$ is measured in the computational basis. Find the conditional states of $A$ and verify that their unconditioned average is unchanged.

Solution

The two outcomes occur with probabilities

$$p(0)=p(1)=\frac12.$$

Conditioned on outcome $0$ at $B$, subsystem $A$ is assigned $\lvert0\rangle\langle0\rvert$. Conditioned on outcome $1$, it is assigned $\lvert1\rangle\langle1\rvert$. Averaging without access to the outcome gives

$$\begin{aligned} \sum_b p(b)\rho_{A|b} &= \frac12\lvert0\rangle\langle0\rvert + \frac12\lvert1\rangle\langle1\rvert \\ &= \frac{I_2}{2} = \rho_A. \end{aligned}$$

The remote result changes the conditional state, but ignoring that result leaves the local reduced state and all local statistics unchanged.

Additional exercises retained from the earlier canonical treatment

1. Local or correlation? Classify $Z\otimes I$, $I\otimes X$, and $Z\otimes Z$ as local or correlation observables.

Solution

$Z\otimes I$ is local to the first subsystem. $I\otimes X$ is local to the second subsystem. $Z\otimes Z$ is a correlation observable because its expectation depends on joint statistics of both subsystems.

2. Bell-basis versus product-basis measurement. Why does measuring both qubits in the computational basis fail to distinguish $\lvert\Phi^+\rangle$ from $\lvert\Phi^-\rangle$?

Solution

The two states are

$$\lvert\Phi^\pm\rangle = \frac{ \lvert00\rangle\pm\lvert11\rangle }{\sqrt2}.$$

Computational-basis measurement gives probability $1/2$ for $00$ and $1/2$ for $11$ in both cases. The relative sign is phase information between the two branches, and it is not recorded by local computational-basis outcome probabilities.

3. Reduced-state statistics. If $\rho_A=I/2$ for a qubit, what is $\langle Z\otimes I\rangle$? Can this answer determine $\langle Z\otimes Z\rangle$?

Solution

The local expectation is

$$\langle Z\otimes I\rangle = \operatorname{Tr}_A \left( \frac12 I\,Z \right) = 0.$$

This does not determine $\langle Z\otimes Z\rangle$. The latter is a joint correlation and depends on $\rho_{AB}$, not only on $\rho_A$.

4. Product-state factorization. Let $\rho_{AB}=\rho_A\otimes\rho_B$. Show that product-measurement probabilities factor.

Solution

For effects $E_a$ on $A$ and $F_b$ on $B$,

$$\begin{aligned} p(a,b) &= \operatorname{Tr}_{AB} \bigl[ (\rho_A\otimes\rho_B)(E_a\otimes F_b) \bigr] \\ &= \operatorname{Tr}_{AB} \bigl[ (\rho_AE_a)\otimes(\rho_BF_b) \bigr] \\ &= \operatorname{Tr}_A(\rho_AE_a)\, \operatorname{Tr}_B(\rho_BF_b). \end{aligned}$$