Joint Measurability

Two observables are jointly measurable when their probabilities are marginals of a single parent POVM. Commutation characterizes compatibility when one observable is sharp, but general unsharp POVMs can be compatible without commuting. For unbiased binary qubit observables, an explicit norm inequality gives a complete criterion and exposes the amount of noise needed for compatibility.

Helpful background. POVMs supplies the event formalism; effects supplies operator order. The physical noise interpretation is developed in Unsharp Measurements.

A parent observable and its marginals

Finite POVMs $(E_i)$ and $(F_j)$ are jointly measurable if positive operators $G_{ij}$ exist with

$$\sum_jG_{ij}=E_i,\qquad \sum_iG_{ij}=F_j.$$

Their total sum is automatically $I$. Measuring $(G_{ij})$ and discarding one label reproduces either original observable on every state. This is a single apparatus with two reported labels; it does not presuppose a sequential measurement model.

For general measurable outcome spaces, the definition uses a POVM on the product sigma-algebra with the required event marginals. Existence of scalar joint distributions separately for each state is not enough: one common operator measure must generate them all.

Commuting finite POVMs are compatible via $G_{ij}=E_iF_j$. Products are positive because the factors commute. Noncommuting effects cannot be multiplied this way to manufacture a parent.

A sharp marginal forces commutation

Suppose $E_i=P_i$ is a PVM and a parent exists. Since $0\leq G_{ij}\leq P_i$, positivity implies

$$P_iG_{ij}=G_{ij}P_i=G_{ij}.$$

To prove this support statement, apply $G_{ij}\leq P_i$ to a vector in $\ker P_i$. Its positive quadratic value vanishes, so its image under $G_{ij}$ vanishes.

For $k\ne i$, the same reasoning gives $P_iG_{kj}=0$. Summing over $k$ yields

$$P_iF_j=G_{ij}=F_jP_i.$$

Thus compatibility with a PVM forces commutation with every projection of that PVM, and the parent is uniquely $P_iF_j$. The event-wise version extends to general PVMs. For unbounded observables, this means commutation of spectral projections, not merely a formal vanishing commutator on some dense domain.

Exact criterion for two unbiased qubit observables

Let

$$E_\pm=\tfrac12(I\pm a\cdot\sigma),\qquad F_\pm=\tfrac12(I\pm b\cdot\sigma), \qquad |a|,|b|\leq1.$$

They are jointly measurable exactly when

$$|a+b|+|a-b|\leq2.$$

To prove it, write the joint plus–plus effect as $G_{++}=\tfrac12(\gamma I+g\cdot\sigma)$. The remaining three effects are fixed by the marginals. Positivity of all four is equivalent to

$$\begin{aligned} |g|,\ |a+b-g|&\leq\gamma,\\ |a-g|,\ |b-g|&\leq1-\gamma. \end{aligned}$$

Triangle inequalities give $|a+b|\leq2\gamma$ and $|a-b|\leq2(1-\gamma)$, proving necessity. For sufficiency choose $g=(a+b)/2$. Any

$$\tfrac12|a+b|\leq\gamma \leq1-\tfrac12|a-b|$$

satisfies all four inequalities. The assumed criterion is precisely what makes this interval nonempty.

For orthogonal unit directions with equal sharpness $\eta$, $a=\eta\hat x$ and $b=\eta\hat z$. The criterion reduces to $\eta\leq1/\sqrt2$. These compatible POVMs still have nonzero commutators whenever $\eta>0$. Busch and Heinosaari analyze this qubit compatibility problem in detail.

Pairwise compatibility is not global compatibility

Consider all three noisy Pauli observables $E_{i,\pm}=(I\pm\eta\sigma_i)/2$. Every pair is compatible for $\eta\leq1/\sqrt2$. A joint measurement of all three exists exactly for $\eta\leq1/\sqrt3$.

Sufficiency follows from the eight effects

$$G_s=\tfrac18(I+\eta\,s\cdot\sigma), \qquad s\in\{-1,1\}^3.$$

Their eigenvalues are $(1\pm\sqrt3\eta)/8$. For necessity, any parent obeys $\sum_s s_iG_s=\eta\sigma_i$. Multiply by $\sigma_i$, trace, and sum:

$$6\eta =\sum_s\operatorname{Tr}((s\cdot\sigma)G_s) \leq\sqrt3\sum_s\operatorname{Tr}G_s =2\sqrt3.$$

Hence values between $1/\sqrt3$ and $1/\sqrt2$ give a concrete pairwise-compatible family without a common parent. This phenomenon cannot occur for a family of pairwise commuting PVMs.

Exercises

1. For $a=\eta\hat x$ and $b=\eta\hat z$, construct a parent at $\eta=1/\sqrt2$.

Solution

Choose $G_{st}=\tfrac14[I+\eta(s\sigma_x+t\sigma_z)]$. Its eigenvalues are $0,1/2$ at the threshold, so it is positive. Summing over $s$ or $t$ gives the required marginals. This is a precise noise threshold for the unsharp-spin model.

2. Does joint measurability guarantee that a chosen first-measurement instrument leaves the second observable undisturbed?

Solution

No. Compatibility asserts existence of a parent POVM. A specified instrument adds dynamics and can disturb the second observable. Its nondisturbance condition is $\mathcal I(\Omega)^*(F_j)=F_j$, which is an extra requirement on that instrument.

References

• Busch, Paul, and Teiko Heinosaari, “Approximate joint measurements of qubit observables,” Quantum Information & Computation 8, 797–818, 2008 — qubit joint-measurability conditions. Preprint.
• Heinosaari, Teiko, Daniel Reitzner, and Peter Stano, “Notes on joint measurability of quantum observables,” Foundations of Physics 38, 1133–1147, 2008 — sharp and unsharp compatibility, including pairwise versus joint structure. Preprint.