Quantum Steering

In 1935 Einstein, Podolsky and Rosen wrote down a maximally entangled pair and noticed something that unsettled them: depending on which observable Alice chooses to measure on her half, Bob's half is left in one of two incompatible families of eigenstates — as if Alice could reach across space and decide, after the fact, what kind of state Bob holds. Schrödinger, replying that same year, gave the effect its name — Steuerung, steering — and called it "the characteristic trait of quantum mechanics." For seven decades it stayed a piece of foundational folklore, vivid but informal, overshadowed by Bell's sharper theorem about nonlocality. Then in 2007 Wiseman, Jones and Doherty turned Schrödinger's intuition into a precise operational task, and steering became a resource — a third member of a strict hierarchy that runs entanglement ⊃ steering ⊃ Bell nonlocality. The object that carries the whole theory is the assemblage, the collection of unnormalized conditional states $\sigma_{a|x}$ that Bob is left holding after Alice announces outcome $a$ for measurement setting $x$ . A state is unsteerable if every assemblage Alice can produce is explainable by a mundane local-hidden-state (LHS) model: Bob was secretly holding one of a fixed catalogue of states all along, and Alice's black box merely tells him which. Steering is the demonstrable impossibility of that story. What makes steering distinctive is an asymmetry of trust. Bob owns a fully calibrated quantum lab and writes down genuine density operators; Alice is an uncharacterized black box that only emits classical labels. Steering is exactly the correlation we can certify when we trust one side — it is the native resource of one-sided device-independent protocols, and it is the only member of the trio that is directional: a state can be steerable from Alice to Bob yet provably not the other way. Best of all, deciding steerability is a semidefinite program, which makes it computationally sharper than either of its neighbours.

Key takeaways

Quantum steering, named by Schrödinger in 1935, is Alice's ability to remotely shape the conditional ensemble on Bob's half of an entangled state by her choice of measurement — with no signal ever sent.
It is formalized through the assemblage $\{\sigma_{a|x}\}$ , Bob's subnormalized conditional states; a state is steerable exactly when no local-hidden-state (LHS) model reproduces that assemblage.
Steering sits strictly between entanglement and Bell nonlocality: every steerable state is entangled and every Bell-nonlocal state is steerable, but both inclusions are strict.
Steering is one-sided device-independent — only Bob's apparatus is trusted — and, uniquely, directional: a state can steer A→B but not B→A (one-way steering).
Deciding unsteerability is a semidefinite feasibility program; steering inequalities are its dual hyperplanes, and violation certifies one-sided device-independent quantum key distribution.
No-signalling is automatic: $\rho_B=\sum_a\sigma_{a|x}$ is independent of $x$ , so Alice controls only the decomposition of a fixed $\rho_B$ , never $\rho_B$ itself.
Key idea: steering is the failure of Bob's assemblage $\{\sigma_{a|x}\}$ to be reproduced by any local hidden state model $\sigma_{a|x}=\sum_\lambda p(\lambda)\,p(a|x,\lambda)\,\rho_\lambda$ — a directional, one-sided device-independent resource lying strictly between entanglement and Bell nonlocality.

The core insight

Steering is the failure of Bob's conditional-state assemblage $\{\sigma_{a|x}\}$ to be reproducible by any local hidden state model $\sigma_{a|x}=\sum_\lambda p(\lambda)\,p(a|x,\lambda)\,\rho_\lambda$ ; it is a directional, one-sided device-independent resource lying strictly between entanglement and Bell nonlocality.

Where this lives today

Steering has moved from a footnote in the EPR debate to a working tool in the quantum-network toolbox. Its flagship application is one-sided device-independent quantum key distribution: because a steering inequality bounds an eavesdropper's information while trusting only the receiver's detectors, it delivers security under a realistic asymmetry — a well-characterized server talking to a client whose hardware it cannot vouch for — and tolerates substantially more noise than fully device-independent QKD. Experimentally, EPR steering has been demonstrated across more than a kilometre of optical fibre, with many measurement settings closing the detection-efficiency loophole one-sidedly, and with continuous-variable light where the Reid criterion on conditional quadrature variances, $\Delta_{\inf}^2(x_B)\,\Delta_{\inf}^2(p_B)<1/4$ , is the working witness. The 2012 realization of genuine one-way steering — a Gaussian state steerable from Alice to Bob but provably not back — turned the directionality from theorem into laboratory fact. On the theory side, steering is now a full resource theory: robustness and steering weight are computed by off-the-shelf semidefinite solvers, and the equivalence between steering and measurement incompatibility has become a two-way street for certifying quantum devices. The open frontiers — the exact POVM steering threshold, steering in networks with independent sources, and self-testing from steering — keep it a live subject.

The physical picture

Hold the archetypal Bell pair $|\Phi^+\rangle=\tfrac1{\sqrt2}(|00\rangle+|11\rangle)$ in mind. Bob's reduced state is the maximally mixed $\rho_B=\mathbb 1/2$ — the dead centre of the Bloch sphere, carrying no bias in any direction. Yet the instant Alice measures and announces her outcome, Bob's qubit collapses to a pure state on the surface. If Alice measured $Z$ , Bob is steered to a pole, $|0\rangle$ or $|1\rangle$ ; if she measured $X$ , he is steered to the equator, $|+\rangle$ or $|-\rangle$ . Same fixed $\rho_B$ , two utterly different pure-state decompositions — and Alice picks which one after the pair has separated. The first figure draws this carving.

Figure

Alice's measurement choice carves Bob's fixed reduced state ρ_B = 𝟙/2 into two incompatible pure-state ensembles — Z sends Bob to the poles 0 and 1 while X sends him to the equator points +x and −x — yet Bob trusts his own lab and Alice is a black box that only emits a label a

Now ask the operational question that turns a curiosity into a resource. Could a sceptic explain Alice's carvings without any spooky action — by insisting that Bob's qubit was already in some definite hidden state $\rho_\lambda$ drawn from a fixed catalogue, and that Alice's box merely reports a hint about which one? That is a local-hidden-state model. When such a model exists, the state is unsteerable and no quantum magic is needed; when no catalogue can reproduce all of Alice's carvings at once, the state is steerable. This is a strictly one-sided test: Bob writes down real density operators because he trusts his lab, while Alice's box is never trusted at all. The second figure places steering in its family — the Werner state makes the whole hierarchy quantitative on a single axis.

Figure

The three resources on the Werner-state visibility axis — separable below V = 1/3, entangled but unsteerable on 1/3 < V ≤ 1/2, steerable but Bell-local on 1/2 < V ≤ 1/√2 ≈ 0.707, and CHSH-violating above 0.707

The trust asymmetry is the conceptual heart, and it explains why steering is directional. Entanglement is a property of the density operator, symmetric under swapping the labels A and B. Bell nonlocality treats both boxes as black, again symmetric. But steering forces the steering party to defeat every LHS model living on the other party's Hilbert space — and that task depends on the steered party's reduced state and on the incompatibility of the steering party's effective measurements. For an asymmetric state those differ between the two directions, so the LHS problem can be solvable one way and unsolvable the other. That is one-way steering, a phenomenon with no analogue in either neighbour.

The core equations

Everything starts with what Bob actually holds. When Alice applies a POVM element $M_{a|x}$ and reports her outcome $a$ , Bob's conditional state is obtained by tracing Alice out — but left unnormalized, so that its trace records how often that outcome occurs. The full list over all $a$ and $x$ is the assemblage, and it is the complete data of a steering experiment.

\sigma_{a|x} \;=\; \mathrm{Tr}_A\big[(M_{a|x}\otimes \mathbb 1)\,\rho_{AB}\big],\qquad p(a|x)=\mathrm{Tr}\,\sigma_{a|x} \qquad (1)

where: $\rho_{AB}$ — the shared bipartite state; $M_{a|x}\ge0$ — Alice's POVM element for outcome $a$ at setting $x$ , with $\sum_a M_{a|x}=\mathbb 1$ ; $\sigma_{a|x}$ — Bob's subnormalized conditional state; $p(a|x)=\mathrm{Tr}\,\sigma_{a|x}$ — the probability Alice reports $a$ . Reading: the normalized state Bob would reconstruct is $\rho_{a|x}^B=\sigma_{a|x}/p(a|x)$ , but keeping the weight attached is what lets the assemblage encode both which states appear and how likely they are. The sceptic's alternative — the story with no steering — is the local-hidden-state model. A single classical variable $\lambda$ is drawn; it fixes a pre-existing state $\rho_\lambda$ that Bob genuinely holds, and it feeds Alice's box a response function $p(a|x,\lambda)$ that fabricates her statistics without ever touching Bob's system.

\sigma_{a|x} \;=\; \sum_{\lambda} p(\lambda)\, p(a|x,\lambda)\, \rho_\lambda \qquad \forall\,a,x \qquad (2)

where: $\lambda$ — classical hidden variable with distribution $p(\lambda)$ ; $p(a|x,\lambda)$ — Alice's stochastic response function, the "black box"; $\rho_\lambda\ge0,\ \mathrm{Tr}\,\rho_\lambda=1$ — the pre-assigned hidden state on Bob's side. Reading: a state is unsteerable from A to B exactly when a model of the form (2) reproduces the whole assemblage; steering is the certified non-existence of any such $\{p(\lambda),\rho_\lambda,p(a|x,\lambda)\}$ . Two constraints define which collections $\{\sigma_{a|x}\}$ are physically legal in the first place, and the second of them is precisely no-signalling. It is the reason steering can never carry a message: summing an assemblage over Alice's outcomes must return the same operator whatever setting she chose.

\sigma_{a|x}\ge 0,\qquad \sum_a \sigma_{a|x} \;=\; \sum_a \sigma_{a|x'} \;=\; \rho_B \quad \forall\,x,x' \qquad (3)

where: $\sigma_{a|x}\ge0$ — every conditional operator is positive; $\rho_B=\sum_a\sigma_{a|x}$ — Bob's reduced state, independent of Alice's setting $x$ . Reading: Alice controls only the decomposition of the fixed $\rho_B$ ; she cannot move $\rho_B$ itself, so no protocol built on steering transmits information. Any LHS model automatically satisfies (3), because $\sum_a p(a|x,\lambda)=1$ collapses the $x$ -dependence. To turn steering into an experiment we need a witness computed from measured numbers. The simplest is the two-setting linear steering inequality, which uses just two of Bob's trusted, mutually unbiased Pauli observables. Its local-hidden-state bound is $\sqrt2$ ; quantum mechanics on a Bell pair reaches $2$ .

S_2 \;=\; \langle A_1 B_1\rangle + \langle A_2 B_2\rangle \;\le\; \sqrt2 \qquad\text{(LHS)},\qquad S_2^{\,\mathrm{QM}}=2 \qquad (4)

where: $A_x=\pm1$ — Alice's black-box outcome for setting $x$ ; $B_1,B_2$ — Bob's trusted orthogonal Pauli observables (e.g. $Z$ and $X$ ), measured on the conditional states; $\sqrt2$ — the LHS ceiling. Reading: a measured $S_2\in(\sqrt2,2]$ certifies steering with Alice's device left entirely uncharacterized. (The four-term CHSH combination $\langle A_1B_1\rangle+\langle A_1B_2\rangle+\langle A_2B_1\rangle-\langle A_2B_2\rangle$ is not a steering witness at $\sqrt2$ : its LHS value already reaches $2$ , so only its Tsirelson-level violation above $2$ signals anything.) Two settings are weak; adding more of Bob's directions tightens the test. The $n$ -setting linear steering inequalities of Cavalcanti, Jones, Wiseman and Reid average $n$ correlators against a shrinking bound $C_n$ , and only the continuum of all directions recovers the true projective threshold.

S_n \;=\; \frac1n\sum_{k=1}^{n}\langle A_k B_k\rangle \;\le\; C_n,\qquad C_2=\tfrac1{\sqrt2},\ \ C_3=\tfrac1{\sqrt3},\ \ C_\infty=\tfrac12 \qquad (5)

where: $B_k$ — $n$ trusted Pauli directions on Bob's side; $C_n$ — the LHS bound, decreasing from $1/\sqrt2\approx0.707$ at $n=2$ toward $1/2$ as $n\to\infty$ . Reading: for the Werner state $S_n=V$ , so $n=2$ settings detect only $V>0.707$ , $n=3$ settings reach $V>0.577$ , and the full sphere pins the exact projective threshold $V>1/2$ . For a quantitative resource measure rather than a yes/no witness, mix the assemblage with noise until steering just disappears; the least noise needed is the steering robustness, a convex program that is zero exactly for unsteerable assemblages.

\mathrm{SR}(\{\sigma_{a|x}\}) \;=\; \min\Big\{\,t\ge0 \;:\; \tfrac{\sigma_{a|x}+t\,\tau_{a|x}}{1+t}\ \text{is LHS for some assemblage } \{\tau_{a|x}\}\Big\} \qquad (6)

where: $\mathrm{SR}$ — steering robustness; $\tau_{a|x}$ — an arbitrary noise assemblage; $t$ — mixing weight. Reading: $\mathrm{SR}>0\iff$ steerable, and $\mathrm{SR}$ is computed exactly by a semidefinite program whose dual is an optimal steering inequality — this SDP tractability is what makes steering sharper to certify than entanglement or nonlocality in general. The payoff is cryptographic. Because a steering witness bounds an eavesdropper's knowledge while trusting only Bob's detectors, it yields a positive key rate in one-sided device-independent QKD, sitting between full device independence (no trust, lowest rates) and standard QKD (full trust).

r_{\mathrm{key}} \;\ge\; 1 - h(Q_x) - I_{AE}^{\,\mathrm{steer}} \qquad\text{(1sDI-QKD)} \qquad (7)

where: $r_\mathrm{key}$ — asymptotic secret-key rate; $Q_x$ — quantum bit-error rate in basis $x$ ; $h(\cdot)$ — binary Shannon entropy; $I_{AE}^{\,\mathrm{steer}}$ — Eve's information, upper-bounded by the certified steering. Reading: a demonstrated steering inequality violation is enough to lower-bound $r_\mathrm{key}>0$ even if Alice's box was manufactured by the adversary.

Derivation AEPR and Schrödinger steering of a Bell pair

The goal is to reproduce, in modern assemblage language, the exact argument that started the subject: that measuring Alice's qubit in two different bases prepares two incompatible pure-state ensembles on Bob, that no single fixed catalogue of pre-existing states can reproduce both at once, and that despite this remote shaping Bob's reduced state never moves — so nothing is signalled. We take the maximally entangled state $|\Phi^+\rangle=\tfrac1{\sqrt2}(|00\rangle+|11\rangle)$ and steer it two ways.

Step 1. Steer in the $Z$ basis. Project Alice onto $|a\rangle$ , $a\in\{0,1\}$ : since $\langle a|_A|\Phi^+\rangle=\tfrac1{\sqrt2}|a\rangle_B$ , Bob's assemblage elements are $\sigma_{0|Z}=\tfrac12|0\rangle\langle0|$ and $\sigma_{1|Z}=\tfrac12|1\rangle\langle1|$ , each with weight $\tfrac12$ . Why: the perfect $Z\!-\!Z$ correlation of $|\Phi^+\rangle$ means Alice's $Z$ outcome pins Bob to the matching pole — Bob's conditional states are the $Z$ eigenstates.

Step 2. Steer in the $X$ basis. Writing $|\Phi^+\rangle=\tfrac1{\sqrt2}(|{+}{+}\rangle+|{-}{-}\rangle)$ gives $\langle\pm|_A|\Phi^+\rangle=\tfrac1{\sqrt2}|\pm\rangle_B$ , so now $\sigma_{+|X}=\tfrac12|+\rangle\langle+|$ and $\sigma_{-|X}=\tfrac12|-\rangle\langle-|$ . The reduced state is unchanged, $\sum_a\sigma_{a|X}=\tfrac12(|+\rangle\langle+|+|-\rangle\langle-|)=\tfrac{\mathbb 1}2=\rho_B$ , but the pure-state decomposition is completely different. Why: $|\Phi^+\rangle$ is basis-invariant up to conjugation, so it correlates $X$ with $X$ just as it correlates $Z$ with $Z$ — Alice's choice selects which decomposition of the same $\rho_B$ is revealed.

Step 3. Show the two ensembles have no common hidden catalogue. A single LHS model would demand one fixed ensemble $\{p(\lambda),\rho_\lambda\}$ from which both the $Z$ -conditional states $\{|0\rangle,|1\rangle\}$ and the $X$ -conditional states $\{|+\rangle,|-\rangle\}$ arise as post-selections. But $|+\rangle=\tfrac1{\sqrt2}(|0\rangle+|1\rangle)$ is a nontrivial superposition of the $Z$ eigenstates, not a coarse-graining of them; the four steered states are extreme points of two mutually unbiased bases with no shared refinement. Why: pre-existing states cannot be simultaneously sharp in two complementary bases, so no catalogue Bob "already held" can wear both faces on demand.

Step 4. Confirm the SDP is infeasible and the state is steerable. Feasibility (2) with deterministic responses would require positive operators $\{\sigma_\lambda\}$ solving $\sigma_{a|x}=\sum_\lambda D(a|x,\lambda)\sigma_\lambda$ for both settings simultaneously; because the target conditional states are rank-one projectors onto complementary bases, no such positive $\{\sigma_\lambda\}$ exists, and the linear program returns infeasible. Hence $|\Phi^+\rangle$ is steerable, exactly as Schrödinger concluded. Why: infeasibility of the LHS SDP is the rigorous, computable stand-in for Schrödinger's verbal claim that Alice "controls" Bob's catalogue. *Verified: reduced state $\rho_B=\mathbb 1/2$ is $x$ -independent (no-signalling holds), while the $Z$ - and $X$ -conditional ensembles are complementary and share no LHS catalogue — reproducing Schrödinger's 1935 conclusion.*

Go deeper · Derivation B · 2 minThe local-hidden-state model for the Werner state at V = 1/2

Derivation BThe local-hidden-state model for the Werner state at V = 1/2

The goal is to build Wiseman, Jones and Doherty's explicit local-hidden-state model for the two-qubit Werner state and prove it reproduces every projective-measurement assemblage precisely up to visibility $V=\tfrac12$ , thereby pinning the exact steering threshold and demonstrating — through the gap with the entanglement threshold $V=\tfrac13$ — that some entangled states are provably unsteerable. The engine is a single geometric identity: the average of a uniform Bloch vector over a hemisphere equals half the pole direction.

Step 1. Compute the target assemblage. Tracing Alice's projector $\tfrac12(\mathbb 1\pm\hat a\cdot\vec\sigma)$ against $\rho_V=V|\Phi^+\rangle\langle\Phi^+|+(1-V)\tfrac{\mathbb 1}4$ gives $\sigma_{\pm|\hat a}=\tfrac14(\mathbb 1\pm V\,\hat a\cdot\vec\sigma)$ . The isotropic $\mathbb 1/4$ comes from the noise floor; the $V\,\hat a\cdot\vec\sigma$ term is the steering signal, shrinking linearly with visibility. Why: this is the exact object any LHS model must reproduce for all directions $\hat a$ simultaneously.

Step 2. Propose the hidden ensemble. Let $\lambda=\hat\lambda$ be a pure qubit Bloch direction drawn uniformly over the sphere, $p(\lambda)\,\mathrm d\lambda=\mathrm d\hat\lambda/4\pi$ , with hidden state $\rho_{\hat\lambda}=\tfrac12(\mathbb 1+\hat\lambda\cdot\vec\sigma)$ . This is a fixed catalogue of pure states, chosen once and for all, with no reference to what Alice will later measure. Why: an unsteerable model must present Bob with genuine pre-existing states, so we commit to the most symmetric such ensemble available.

Step 3. Fix Alice's black-box response. Let her box output the sign of the overlap between her measurement direction and the hidden vector, $D(\pm|\hat a,\hat\lambda)=\Theta(\pm\,\hat a\cdot\hat\lambda)$ — a deterministic hemisphere rule that needs no access to Bob. Why: the response function is the only freedom left, and a hemisphere rule is the natural way to fake a $\pm1$ correlation with the hidden direction.

Step 4. Recover the assemblage and read off the threshold. Integrating the hidden states over the " $+$ " hemisphere,

\int_{\hat a\cdot\hat\lambda>0}\frac{\mathrm d\hat\lambda}{4\pi}\,\tfrac12(\mathbb 1+\hat\lambda\cdot\vec\sigma)=\tfrac14\big(\mathbb 1+\tfrac12\,\hat a\cdot\vec\sigma\big),

because the hemispherical average of $\hat\lambda$ is $\langle\hat\lambda\rangle_+=\tfrac12\hat a$ (the perpendicular components cancel; the parallel mean is $\int_0^{\pi/2}\cos\theta\sin\theta\,\mathrm d\theta=\tfrac12$ ). This matches $\sigma_{+|\hat a}=\tfrac14(\mathbb 1+V\,\hat a\cdot\vec\sigma)$ exactly at $V=\tfrac12$ ; for $V<\tfrac12$ rescale the hidden states inward. Why: the model reproduces every projective assemblage for $V\le\tfrac12$ , so $\rho_V$ is unsteerable there, and no model survives above it — the threshold is sharp.

Step 5. Locate the threshold in the hierarchy. Peres–Horodecki gives entanglement for $V>\tfrac13$ and Tsirelson gives CHSH violation for $V>1/\sqrt2$ . Thus $\tfrac13<V\le\tfrac12$ is entangled-yet-unsteerable and $\tfrac12<V\le1/\sqrt2$ is steerable-yet-Bell-local, proving both inclusions of the hierarchy strict. Why: the explicit model turns the abstract claim "entanglement ⊋ steering" into a concrete interval of counterexample states. *Verified: the hemispherical identity $\langle\hat\lambda\rangle_{\hat a\cdot\hat\lambda>0}=\hat a/2$ (checked numerically to 4 digits) forces the match at exactly $V=1/2$ , reproducing Wiseman–Jones–Doherty 2007; for all POVMs the bound softens to $V\approx0.4904$ .*

Go deeper · Derivation C · 1 minThe local-hidden-state bound of the two-setting steering inequality

Derivation CThe local-hidden-state bound of the two-setting steering inequality

The goal is to derive from pure Bloch-sphere geometry why the two-setting linear functional $S_2=\langle A_1B_1\rangle+\langle A_2B_2\rangle$ can never exceed $\sqrt2$ under any local-hidden-state model, why a Bell pair pushes it to $2$ , and hence why any measured value above $\sqrt2$ certifies steering while trusting only Bob's two Pauli observables. This makes precise the sense in which a steering inequality is a hyperplane separating the LHS set from the quantum one.

Step 1. Factor the correlators through a hidden state. In an LHS model $\langle A_x B_x\rangle=\sum_\lambda p(\lambda)\,a_x(\lambda)\,(\hat b_x\cdot\vec r_\lambda)$ , where $a_x(\lambda)=\pm1$ is Alice's deterministic reply and $\hat b_x\cdot\vec r_\lambda=\mathrm{Tr}[B_x\rho_\lambda]$ is the expectation of Bob's trusted Pauli $B_x$ on the hidden state $\rho_\lambda$ with Bloch vector $\vec r_\lambda$ . Why: only Bob's operators are physical here — Alice's contribution is a bare $\pm1$ label, and all the geometry lives in Bob's Bloch vector.

Step 2. Bound the per- $\lambda$ contribution. Writing $u=\hat b_1\cdot\vec r_\lambda$ and $w=\hat b_2\cdot\vec r_\lambda$ with $B_1\perp B_2$ orthogonal Paulis and $|\vec r_\lambda|\le1$ , we have $u^2+w^2\le1$ , and $S_2^\lambda=a_1u+a_2w\le|u|+|w|$ . Why: Alice's best deterministic strategy is to align each sign with its correlator, giving $|u|+|w|$ as the per-state ceiling.

Step 3. Apply Cauchy–Schwarz. For two numbers, $|u|+|w|\le\sqrt2\,\sqrt{u^2+w^2}\le\sqrt2$ , saturated at $u=w=1/\sqrt2$ . Averaging over $\lambda$ preserves the bound because it is uniform in $\lambda$ , so $S_2\le\sqrt2$ for every LHS model. Why: the orthogonality of Bob's settings caps the joint projection of a unit Bloch vector at $\sqrt2$ , which is the whole content of the classical ceiling.

Step 4. Reach the quantum value. On $|\Phi^+\rangle$ take $A_1$ correlated with $B_1=Z$ and $A_2$ with $B_2=X$ : perfect correlation gives $\langle A_1B_1\rangle=\langle A_2B_2\rangle=1$ , so $S_2=2>\sqrt2$ . Any experiment returning $S_2\in(\sqrt2,2]$ therefore rules out all LHS models. Why: the Bell pair's simultaneous perfect correlation in two complementary bases is impossible to fake with pre-existing states, and the gap $2-\sqrt2$ is the certifiable margin. *Verified: a Monte-Carlo maximization over unit Bloch vectors and Alice's signs returns $S_2^{\max,\mathrm{LHS}}=1.414$ (= √2) for the two-term functional and $2.000$ for the four-term CHSH combination, confirming which functional is the genuine √2 witness.*

Limiting cases

For the two-qubit Werner state $\rho_V = V\,|\Phi^+\rangle\langle\Phi^+| + (1-V)\,\tfrac{\mathbb 1}{4}$ , the three resources separate cleanly along the single visibility axis $V\in[0,1]$ :

Case	Result	Meaning
Separable	$V \le 1/3$	below the PPT threshold the state is separable, hence trivially unsteerable and Bell-local — no quantum correlation to certify
Entangled, unsteerable	$1/3 < V \le 1/2$	entangled (Peres–Horodecki) yet the explicit Wiseman–Jones–Doherty hidden-state model reproduces every projective assemblage — steering is strictly stronger than entanglement
Steerable, Bell-local	$1/2 < V \le 1/\sqrt2$	no LHS model survives (steerable) but the CHSH value $2\sqrt2\,V \le 2$ (Bell-local) — nonlocality is strictly stronger than steering
Bell nonlocal	$V > 1/\sqrt2\approx0.707$	CHSH value $2\sqrt2\,V>2$ : all three resources present at once, the fully quantum regime
POVM steering	$V \le 0.4904$	allowing general POVMs on Alice's side lowers the steering threshold slightly below $1/2$ ; the canonical value $1/2$ is for projective measurements

Common trap“Steering lets Alice send Bob a message, or it is just entanglement by another name.”

Both readings are wrong. First, steering never signals: Bob's reduced state $\rho_B=\sum_a\sigma_{a|x}$ is fixed independently of Alice's setting $x$ , so with no classical announcement from Alice, Bob sees nothing change — she controls only which decomposition of a fixed $\rho_B$ is revealed once he learns her outcome, a freedom already guaranteed by the GHJW theorem. Second, steering is not a synonym for entanglement: it is strictly stronger (every steerable state is entangled, but Werner states with $1/3<V\le1/2$ are entangled and provably unsteerable) and strictly weaker than Bell nonlocality (Werner states with $1/2<V\le1/\sqrt2$ steer but do not violate CHSH). And unlike its symmetric neighbours, steering is directional — a state can steer A→B yet not B→A — because the resource is defined relative to which party is trusted.

Worked examples

1 · Placing a Werner state in the hierarchy (fully worked)

Take the Werner state at visibility $V=0.6$ . Is it entangled? Steerable? Bell-nonlocal? And how many measurement settings does Bob need to certify the steering with a linear inequality?

Step 1. Entanglement. The Peres–Horodecki threshold is $V>1/3\approx0.333$ . Since $0.6>0.333$ , the state is entangled. Concretely, the partial transpose of $\rho_{0.6}$ has minimum eigenvalue $\tfrac14(1-3V)=\tfrac14(1-1.8)=-0.2<0$ , so PPT fails and entanglement is confirmed.

Step 2. Steerability. The exact projective steering threshold is $V>1/2$ . Since $0.6>0.5$ , the state is steerable — no local-hidden-state model can reproduce all of its projective assemblages.

Step 3. Certifying it with a linear inequality. For the Werner state each aligned correlator is $\langle A_kB_k\rangle=V$ , so the $n$ -setting functional gives $S_n=V=0.6$ . Compare against $C_n$ : with two settings $C_2=1/\sqrt2\approx0.707$ , and $0.6<0.707$ — two settings fail to detect it. With three mutually unbiased settings $C_3=1/\sqrt3\approx0.577$ , and $0.6>0.577$ — three settings succeed. So Bob needs at least a three-measurement steering inequality; the two-setting test is simply too blunt for a state this close to the boundary.

Step 4. Bell nonlocality. The optimal CHSH value is $2\sqrt2\,V=2\sqrt2(0.6)=1.697$ . Since $1.697<2$ , the state is Bell-local — no CHSH violation is possible. Verdict: $V=0.6$ is entangled and steerable but Bell-local, living squarely in the gap $1/2<V<1/\sqrt2$ ; it is certified steerable by a 3-setting inequality but not a 2-setting one.

2 · Certifying steering from raw data

Bob measures his trusted Pauli observables $Z$ and $X$ on the states heralded by Alice's two black-box settings and estimates the correlators $\langle A_1B_1\rangle=0.90$ and $\langle A_2B_2\rangle=0.90$ . Is steering certified? What happens if depolarizing noise drops each correlator to $0.65$ ?

Step 1. Form the two-setting witness $S_2=\langle A_1B_1\rangle+\langle A_2B_2\rangle=0.90+0.90=\mathbf{1.80}$ .

Step 2. Compare to the local-hidden-state bound $\sqrt2\approx1.414$ . Since $1.80>1.414$ , steering is certified — and crucially, this conclusion trusts only Bob's calibration of $Z$ and $X$ ; Alice's box could be anything, even adversarial. The certified margin is $1.80-1.414=0.386$ .

Step 3. Now the noisy case: $S_2=0.65+0.65=\mathbf{1.30}$ . Since $1.30<1.414$ , this particular inequality does not certify steering. The state may still be steerable — a three-setting inequality (bound $2/\sqrt3\approx1.155$ on the summed form) or the full LHS-SDP could still catch it — but the two-setting test has run out of power. The lesson mirrors Example 1: a failed linear inequality is inconclusive, never a proof of unsteerability.

Physics of the portrait

Steering is the rare piece of quantum theory that carried a suggestive name for seventy years before anyone made it rigorous. The word, and the unease behind it, belong to Schrödinger — writing in the immediate aftermath of the EPR paper, in the same burst of work that gave us the cat.

Origins · 1 minErwin Schrödinger and how this was discovered

Erwin Schrödinger

1887–1961 · wave mechanics, the EPR debate, and the coining of “steering”

Erwin Schrödinger built wave mechanics in 1926 and spent the next decade as one of quantum theory's most penetrating critics. When Einstein, Podolsky and Rosen published their 1935 argument that quantum mechanics must be incomplete, Schrödinger answered with two papers in the Proceedings of the Cambridge Philosophical Society that named the two ideas at the heart of the debate. One introduced Verschränkung — entanglement — as "not one but rather the characteristic trait of quantum mechanics." The other introduced Steuerung, steering: his observation that by choosing her measurement, a distant experimenter can remotely "pilot" or "steer" the conditional state of a system she can no longer touch, disposing over its ensemble "at her mercy in spite of her having no access to it." Schrödinger found this so troubling that he doubted nature would permit it over large separations — modern experiments steer across kilometres of fibre and prove him wrong on that point, but right that he had isolated something new. His steering sat dormant until 2007, when Wiseman, Jones and Doherty recast it as a one-sided device-independent task with a semidefinite-programming backbone, and the word he coined became the name of a quantitative resource.

History

1935

— EPR argue quantum mechanics is incomplete; Schrödinger replies and names the effect Steuerung (“steering”)

1964

— Bell derives his inequality, sharpening nonlocality and, in hindsight, framing steering's stronger sibling

1989

— Werner constructs entangled states that admit a local hidden variable model

1993

— Hughston, Jozsa and Wootters prove the GHJW remote-ensemble theorem

2007

— Wiseman, Jones and Doherty give steering its operational, SDP-based definition and the entanglement–steering–nonlocality hierarchy

2009

— Saunders and co-workers demonstrate steering experimentally; CJWR publish the linear steering inequalities

2012

— Händchen and colleagues realize one-way steering with continuous-variable light

Going further

Texts. Nielsen & Chuang, Quantum Computation and Quantum Information, §2.4–2.5 (density operators, ensembles) and Ch. 12 (Bell/EPR) · Watrous, The Theory of Quantum Information, Ch. 6 (separability and measurements) · Wilde, Quantum Information Theory, Ch. 4 · Preskill, Lecture Notes on Quantum Computation, Ch. 4 (density operators and the GHJW theorem). Reviews: Cavalcanti & Skrzypczyk, Rep. Prog. Phys. 80, 024001 (2017); Uola, Costa, Nguyen & Gühne, Rev. Mod. Phys. 92, 015001 (2020).

Provenance. A verifier should reproduce: (1) the Werner thresholds $V>1/3$ (entanglement, Peres–Horodecki), $V>1/2$ (projective steering, Wiseman–Jones–Doherty 2007), $V>1/\sqrt2$ (CHSH via the value $2\sqrt2\,V$ , Tsirelson); (2) the hemispherical identity $\langle\hat\lambda\rangle_{\hat a\cdot\hat\lambda>0}=\hat a/2$ , hence the LHS-model match at exactly $V=1/2$ ; (3) the no-signalling sum $\sum_a\sigma_{a|x}=\rho_B$ from any LHS model; (4) the two-setting steering bound $S_2\le\sqrt2$ with quantum maximum $2$ , and the fact that the four-term CHSH combination instead has LHS value $2$ and so is not a steering witness at $\sqrt2$ ; (5) the CJWR $n$ -setting bounds $C_2=1/\sqrt2$ , $C_3=1/\sqrt3$ , $C_\infty=1/2$ . Original sources: Schrödinger, Proc. Camb. Phil. Soc. 31, 555 (1935); EPR, Phys. Rev. 47, 777 (1935); Wiseman, Jones & Doherty, Phys. Rev. Lett. 98, 140402 (2007); Cavalcanti, Jones, Wiseman & Reid, Phys. Rev. A 80, 032112 (2009). The POVM steering value $V\approx0.4904$ (Hirsch et al./Nguyen et al.) is quoted as approximate; the exact projective value is $1/2$ .

Prerequisites. Bell States and EPR Pairs · Separable versus Entangled States · Schmidt Decomposition · Bell's Theorem and Local Hidden Variables · CHSH Inequality and Its Quantum Violation. Leads to. Nonlocality and the No-Signalling Principle · Tsirelson's Bound · Entanglement Witnesses.

Ask the tutor: ask me to walk you through the Wiseman–Jones–Doherty hemisphere model line by line, or to show exactly why a three-setting inequality catches the $V=0.6$ Werner state that two settings miss.

How is steering different from entanglement and from Bell nonlocality?

All three are certificates of quantum correlation, and they differ by how much you trust the devices. Entanglement is a property of the density operator, certified by full tomography with both devices trusted; it is symmetric in the two parties. Bell nonlocality treats both devices as uncharacterized black boxes and asks whether the input–output statistics beat every local-hidden-variable model; it too is symmetric. Steering sits between them: one device (Bob's) is trusted and characterized, the other (Alice's) is a black box. This one-sided trust makes steering strictly stronger than entanglement and strictly weaker than Bell nonlocality, and — because the roles are asymmetric — it is the only one of the three that can hold in one direction but not the other.

If Alice “steers” Bob's state, why isn't that faster-than-light signalling?

Because Alice changes nothing that Bob can detect on his own. His reduced state is $\rho_B=\sum_a\sigma_{a|x}$ , and the no-signalling condition built into any valid assemblage guarantees this is the same operator for every setting $x$ Alice might choose. So if Bob simply measures his qubit without any word from Alice, his statistics are identical whatever she did. What Alice controls is only how a fixed $\rho_B$ decomposes into conditional states — information Bob can act on solely after receiving Alice's outcome through an ordinary classical channel, which travels no faster than light. Steering shapes correlations, not marginals.

What exactly is a local-hidden-state model, and how does it differ from a hidden-variable model?

A local-hidden-variable (LHV) model — the object Bell's theorem rules out — assigns classical response functions to both parties: $p(a,b|x,y)=\sum_\lambda p(\lambda)p(a|x,\lambda)p(b|y,\lambda)$ . A local-hidden-state (LHS) model is a hybrid: Alice still gets a classical response function $p(a|x,\lambda)$ , but Bob is handed a genuine quantum state $\rho_\lambda$ that he measures with real quantum mechanics. The steering question is whether Bob's assemblage can be written $\sigma_{a|x}=\sum_\lambda p(\lambda)p(a|x,\lambda)\rho_\lambda$ . Because Bob's side is quantum and only Alice's is classical, the LHS constraint is weaker than LHV but stronger than "just entangled" — which is exactly why steering nests between nonlocality and entanglement.

Why is deciding steerability easier than deciding entanglement?

For finitely many settings and outcomes, Alice's response functions can be taken deterministic, leaving only finitely many hidden variables $\lambda$ . The existence of an LHS model then becomes a finite-dimensional semidefinite feasibility program — find positive operators $\{\sigma_\lambda\}$ satisfying linear constraints — which is solved exactly in polynomial time, and whose dual solution is an optimal steering inequality. Deciding separability, by contrast, has no such general efficient characterization; it is NP-hard in the dimension. So even though steering is a subtler physical notion than entanglement, the one-sided structure makes it the more computationally tractable of the two.

What does “one-way steering” mean, and why can't entanglement do it?

One-way steering is a state that is steerable from Alice to Bob but provably not from Bob to Alice — the resource exists in one direction only. It is possible because steering is defined relative to which party is trusted: steering A→B means defeating every LHS model on Bob's Hilbert space, a task that depends on Bob's reduced state and on the incompatibility of Alice's effective measurements, and these generally differ from the reverse problem. Entanglement and Bell nonlocality cannot show this asymmetry because both are invariant under swapping the party labels. The first experimental one-way steering used a carefully engineered Gaussian state of light (Händchen et al., 2012), and qubit examples followed.

How many measurement settings do I actually need to detect steering?

It depends on how strong the steering is. Linear steering inequalities with $n$ settings get progressively more powerful — the Werner-state detection threshold drops from $V>0.707$ at two settings to $V>0.577$ at three and toward the true $V>0.5$ only as $n\to\infty$ — so a weakly steerable state near the boundary can slip past a two-setting test yet be caught by a three- or many-setting one. A failed linear inequality is therefore never a proof of unsteerability; only the full local-hidden-state semidefinite program (or an optimal inequality drawn from its dual) settles the question definitively. In practice, experiments trade more settings for a smaller certifiable margin and closure of the detection-efficiency loophole.

Is steering related to measurement incompatibility?

Yes, and the connection is exact for the standard scenario: a bipartite state is steerable by Alice only if her measurements are incompatible (not jointly measurable), and conversely any set of incompatible measurements can steer some state. Intuitively, if Alice's measurements were jointly measurable, a single "parent" measurement would produce all her outcomes at once, and its post-measurement states would furnish a ready-made LHS catalogue for Bob — no steering. This equivalence has become a practical bridge: certifying steering certifies incompatibility one-sidedly, and results about one transfer to the other. It is one reason steering has become central to device certification rather than a foundational curiosity.