No-Signaling

No-signaling means that a party cannot change another party’s observable outcome distribution merely by choosing a local measurement or operation. Standard quantum theory has this property for local trace-preserving operations on a tensor-product system. It nevertheless violates Bell inequalities, because Bell locality requires more than independence of observable marginals.

Helpful background. Partial Trace defines the reduced state seen by a subsystem.

Sum over the uncommunicated outcome

Let Alice and Bob share $\rho_{AB}$. Alice chooses $x$ and applies Kraus operators $K_{a\mid x}$ satisfying $\sum_aK_{a\mid x}^\dagger K_{a\mid x}=I_A$. Bob measures an effect $F_b$. His probability when Alice’s result is not received is

$$\begin{aligned} p(b\mid x) &=\sum_a\operatorname{Tr}\!\left[ (K_{a\mid x}^\dagger K_{a\mid x}\otimes F_b)\rho_{AB}\right]\\ &=\operatorname{Tr}[(I_A\otimes F_b)\rho_{AB}]. \end{aligned}$$

The right side is independent of $x$. The proof uses locality of the operation and trace preservation after all its outcomes are included. An interaction between the laboratories, a transmitted system or an already-signaling global channel does not satisfy these premises.

For several Kraus operators per outcome, include the extra index in the sum. Continuous outcomes require the corresponding normalized integral. The physical point is unchanged: the remote observer sees the nonselective operation unless a result is communicated.

Conditional states can change without a message

For $|\Phi^+\rangle$, Alice’s computational-basis measurement conditionally prepares Bob in $|0\rangle$ or $|1\rangle$. Her $X$-basis measurement instead conditionally prepares $|+\rangle$ or $|-\rangle$. Both unconditioned ensembles have density matrix $I/2$.

Bob can see which conditional ensemble applies only after learning Alice’s setting and outcome through an ordinary record comparison. This is the basis of quantum steering, not a signaling channel. Writing a conditional state on paper does not deliver its conditioning label to Bob.

A local filter that succeeds only for selected outcomes can change a remote state conditional on success. Including failed trials restores the no-signaling marginal. Discarding failures without declaring the selection can create a misleading apparent dependence.

Bell locality is stronger

A Bell-local model represents the entire joint distribution using a common hidden variable and local response functions. No-signaling imposes only constraints such as

$$\sum_a p(a,b\mid x,y)=p(b\mid y).$$

For binary outputs, the hypothetical PR box sets $a\oplus b=xy$ with equal weight on the two allowed pairs. Each marginal is uniform for every remote setting, yet its CHSH value is four. It is no-signaling and nonlocal, and it exceeds the quantum CHSH bound. See CHSH for the nested model sets and their derivation.

No-signaling is therefore neither sufficient for quantum theory nor equivalent to Bell-local causal explanation.

Exercise: a misleading selected sample

Alice and Bob share independent fair bits. They keep a trial only when their two bits agree. Compute the retained correlation, and explain why it is not a demonstration of a physical influence.

Solution

The retained pairs are $00$ and $11$, each with conditional probability $1/2$, so their sign correlation is one. Agreement was imposed by joint data selection, which requires comparing both records. The original unselected bits remain independent. Selection can change conditional statistics without changing the mechanism that generated the data.

References

• Ghirardi, Gian Carlo, Alberto Rimini, and Tullio Weber, “A general argument against superluminal transmission through the quantum mechanical measurement process,” Lettere al Nuovo Cimento 27, 293–298 (1980), doi:10.1007/BF02817189.
• Popescu, Sandu, and Daniel Rohrlich, “Quantum nonlocality as an axiom,” Foundations of Physics 24, 379–385 (1994), doi:10.1007/BF02058098.