14 The OCB Causal Game

Chapter 13 leaned on two numbers without earning them: the classical bound $P_{\mathrm{win}} \leq 3/4$, and the quantum switch’s score $P_{\mathrm{win}} = (2+\sqrt{2})/4 \approx 0.854$. They are the numerical content of the whole causal type hierarchy — the gap between them is why Switch(d) deserves to be its own type. This chapter derives both.

They come from a game. Oreshkov, Costa & Brukner introduced it in the same 2012 paper that gave us process matrices, and it plays the role for causal structure that the CHSH game plays for entanglement: a task with a sharp classical ceiling that a quantum resource provably beats.

14.1 The Rules

Two laboratories, $A$ and $B$. In each round:

A referee generates two independent uniform bits, $x$ and $y$. Bit $x$ is delivered to lab $A$, bit $y$ to lab $B$. Neither lab sees the other’s bit.
Each lab performs one local quantum operation — an instrument chosen in advance as a function of its own input bit — and reports a single output bit: $a$ from $A$, $b$ from $B$.
The round is won if \[ a \oplus b \;=\; x \wedge y . \]

So for three of the four input pairs ($x\wedge y = 0$) the labs must produce equal bits; for $x = y = 1$ they must produce different bits. The score is the win probability averaged over the four equally likely inputs: \[ P_{\mathrm{win}} \;=\; \tfrac{1}{4}\sum_{x,y \in \{0,1\}} P\big(a \oplus b = x\wedge y \,\big|\, x, y\big). \]

Here is the twist that makes it a causal game rather than a nonlocality game. The two labs are not simply spacelike-separated with a shared state. They are embedded in a process — some background arrangement of wires, channels, and systems that routes quantum information between them. The labs agree in advance on their instruments and on that process. What they are not given is a definite causal order between $A$ and $B$: the process is allowed to be any valid process matrix.

The question is exactly the question of Chapter 13, now made operational: does having a definite causal order (or a classical mixture of orders) limit what the labs can score?

14.2 Why Definite Order Caps the Score at 3/4

Fix a definite causal order, say $A \prec B$: information may flow from $A$’s lab to $B$’s, never the reverse. (The $B \prec A$ case is identical by symmetry, and Chapter 13 established that a classical mixture of the two is a convex combination, so whatever bound holds for each order holds for the mixture.)

The intuition is a see-saw. In the order $A \prec B$, lab $B$ is downstream: the system it operates on has already passed through $A$, so $B$’s statistics can be correlated with what $A$ did. Lab $A$ is upstream: nothing it touches has been near $B$, so $A$’s output cannot reflect $B$’s input $y$ in any way. Whatever coordination the win condition demands, at most one of the two labs can be adapting to the other; the other is working blind.

Make this precise. Under $A \prec B$, lab $B$ cannot signal to lab $A$, so $A$’s output marginal is independent of $y$: \[ P(a \mid x, y) \;=\; P(a \mid x). \] Now look at the win condition column by column:

$(x,y)$	$x \wedge y$	required relation
$(0,0)$	$0$	$a = b$
$(0,1)$	$0$	$a = b$
$(1,0)$	$0$	$a = b$
$(1,1)$	$1$	$a \neq b$

The three “$a = b$” rows want the labs correlated the same way; the last row wants them anti-correlated. Consider the two rows with $x = 1$: $(1,0)$ needs $a = b$ and $(1,1)$ needs $a \neq b$, but $A$’s bit $a$ has the same distribution in both rows (it depends only on $x = 1$, not on $y$). So the burden of switching from “agree” to “disagree” falls entirely on $B$ — and $B$ must do it knowing $y$ but not $x$. When $B$ sees $y = 1$ it does not know whether it is in row $(0,1)$ (agree) or row $(1,1)$ (disagree). Any fixed response to $y=1$ therefore gets one of those two rows wrong.

Carrying the bookkeeping through all four rows — a short calculation, or OCB’s causal inequality applied to the cone characterisation from Chapter 13 — gives \[ \boxed{P_{\mathrm{win}}^{\text{causally separable}} \;\leq\; \tfrac{3}{4}.} \] The bound is tight: with order $A \prec B$, let $A$ output $a = 0$ always and let $B$ output $b = 0$ when $y = 0$ and a fair coin when $y = 1$. Rows $(0,0)$, $(1,0)$ win outright; rows $(0,1)$, $(1,1)$ win with probability $1/2$ each. Average: $\tfrac14(1 + 1 + \tfrac12 + \tfrac12) = \tfrac34$.

14.2.1 Seeing the bound in QRL

Chapter 13 met the causally ordered processes as ProcessMatrix values. Their game scores are computed directly by causal_inequality_value():

from qrl.causal import identity_process, definite_order_process

identity_process(n_parties=2).causal_inequality_value()   # 0.5   — no signalling at all
definite_order_process('AB').causal_inequality_value()    # 0.625 — one-way channel A → B

The no-signalling process cannot even use the “downstream” adaptation — $B$ learns nothing about $A$ — and scores a flat $1/2$, no better than answering at random. The one-way channel $A \prec B$ does better, $0.625$, by letting $B$ adapt to $A$; but it is still short of $3/4$ because these specific processes are not using the optimal instruments. What matters is that every causally separable process, with every choice of instruments, stays at or below $3/4$. That ceiling is the whole point.

14.3 The Switch Strategy

Now embed the two labs in a quantum switch instead of a definite wiring. Recall its isometry (Chapter 13), for unitary operations $U_A, U_B$ on the target and a control qubit $C$: \[ V \;=\; (U_B U_A) \otimes |0\rangle\langle 0|_C \;+\; (U_A U_B) \otimes |1\rangle\langle 1|_C . \]

Set the control to $|+\rangle$. The process is now a coherent superposition of “$A$ then $B$” and “$B$ then $A$”. Neither lab is downstream; neither is upstream; the amplitudes for the two orders are both present and can interfere.

The labs’ strategy is the one already coded into QRL’s score operator:

On input $0$, measure the target in the $Z$ basis and report the outcome.
On input $1$, measure in the $X$ basis and report the outcome.

Why this beats $3/4$: the two orders differ by whether $A$’s operation or $B$’s comes last, and $Z$ and $X$ measurements do not commute, so the order genuinely changes the target the other lab sees. With the control in $|+\rangle$, the win amplitude picks up a contribution from each order, and — exactly as in the CHSH derivation — the two contributions add constructively along a direction $45°$ between the axes. The interference term carries a factor $\cos(\pi/4) = 1/\sqrt2$, and the score becomes \[ P_{\mathrm{win}} \;=\; \frac{1}{2} + \frac{1}{2\sqrt2} \;=\; \frac{2 + \sqrt2}{4} \;\approx\; 0.854 . \]

The same $\sqrt2$ that sets Tsirelson’s bound for CHSH sets the causal advantage here. It is not a coincidence: both are the operator norm of a sum of two anticommuting $\pm1$ observables.

14.4 Computing It in QRL

Chapter 13 asserted the number; the surface language computes it. Three lines:

ocb.qrl

let A : UniProc(2) = cptp([[1, 0], [0, 1]]) in   -- identity unitary
let B : UniProc(2) = cptp([[1, 0], [0, 1]]) in   -- identity unitary
switch(A, B, |+>)

$ qrl exec ocb.qrl
ocb.qrl  Process(2x2, P_win=0.8536, robustness=0.4142)

The program type-checks as Switch(2) — both operations are UniProc(2), so T-Switch applies — and the interpreter takes the E-Switch-Coherent reduction, builds the switch process matrix, and scores it. Note the operations are identity unitaries: as Chapter 13 stressed, the causal advantage is a property of the process, not of what the labs compute.

The same computation through the Python API, which exposes the intermediate objects:

from qrl.causal import QuantumSwitch, cptp_from_unitary
import numpy as np

sw = QuantumSwitch(
    channel_A=cptp_from_unitary(np.eye(2)),
    channel_B=cptp_from_unitary(np.eye(2)),
)

sw.causal_inequality_value()   # 0.8535533905932737  = (2 + √2)/4

14.5 The Score Operator

causal_inequality_value() is not a Monte Carlo simulation of the game — it evaluates it in closed form. The game’s entire payoff structure is packed into one Hermitian operator $S$, the score operator, built from the Choi–Jamiołkowski representations of the optimal instruments:

\[ S \;=\; \sum_{\substack{x,y,a,b \\ a \oplus b \,=\, x \wedge y}} J_A^{\,x,a} \otimes J_B^{\,y,b}, \]

where $J^{\,x,a}$ is the Choi operator of “measure in the $x$-basis, obtain outcome $a$”. Summing over exactly the winning $(x,y,a,b)$ tuples means that for any process matrix $W$, \[ P_{\mathrm{win}}(W) \;=\; \tfrac{1}{4}\,\mathrm{Tr}[\,W \, S\,]. \]

This is the generalised Born rule of Chapter 13 with $S$ in place of a product of local operations: the game score is linear in the process. That linearity is what turns the $3/4$ bound into a witness. Define \[ \Omega \;=\; \tfrac{3}{4}\,I \;-\; S . \] Then for any valid two-party qubit process, which Chapter 13’s normalisation condition fixes at $\mathrm{Tr}[W] = 4$, \[ \mathrm{Tr}[\Omega\, W] \;=\; \tfrac{3}{4}\,\mathrm{Tr}[W] \;-\; \mathrm{Tr}[S\,W] \;=\; 3 \;-\; 4\,P_{\mathrm{win}}(W). \] The classical ceiling $P_{\mathrm{win}} \leq 3/4$ is exactly $\mathrm{Tr}[\Omega\, W] \geq 0$. This is the operator Chapter 13 called causal_nonseparability_witness(), and its non-negativity on causally separable processes is the OCB causal inequality in its sharpest form.

from qrl.causal import causal_nonseparability_witness, identity_process, definite_order_process

Omega = causal_nonseparability_witness()

identity_process(n_parties=2).witness_value(Omega)    # 1.0   = 3 − 4(0.5)
definite_order_process('AB').witness_value(Omega)     # 0.5   = 3 − 4(0.625)

Both non-negative — as they must be for causally separable processes. For the switch, the relation $\mathrm{Tr}[\Omega\, W] = 3 - 4P_{\mathrm{win}}$ gives \[ \mathrm{Tr}[\Omega\, W_{\mathrm{sw}}] \;=\; 3 - 4\cdot\frac{2+\sqrt2}{4} \;=\; 1 - \sqrt2 \;\approx\; -0.414 \;<\; 0. \] Negative: the certificate of Chapter 13, now with its provenance. (The switch’s process matrix is expressed over target $\otimes$ control rather than $A_I \otimes A_O \otimes B_I \otimes B_O$, so witness_value cannot be applied to it directly — but the linear relation is basis-independent, and causal_inequality_value() returns the score that feeds it.)

14.6 Robustness: How Far Outside the Bound

A negative witness value says the switch is causally nonseparable. Robustness says by how much. It is the smallest fraction of maximally-mixed noise that, blended in, drags the process back onto the classical side: \[ r^*(W) \;=\; \min\Big\{\, r \geq 0 \;:\; \frac{W + r\,W_{\mathrm{id}}}{1 + r}\ \text{is causally separable} \,\Big\}. \]

Because the score is linear and $W_{\mathrm{id}}$ scores exactly $1/2$, the blended process scores \[ P_{\mathrm{win}}\!\left(\frac{W + r\,W_{\mathrm{id}}}{1+r}\right) \;=\; \frac{P_{\mathrm{win}}(W) + r/2}{1 + r}, \] and setting this equal to the threshold $3/4$ and solving for $r$ gives \[ r^*(W) \;=\; 4\left(P_{\mathrm{win}}(W) - \tfrac{3}{4}\right) \] whenever $P_{\mathrm{win}}(W) > 3/4$, and $0$ otherwise. For the switch: \[ r^* \;=\; 4\left(\frac{2+\sqrt2}{4} - \frac{3}{4}\right) \;=\; \sqrt2 - 1 \;\approx\; 0.414 . \]

W_sw = sw.process_matrix()
W_sw.causal_nonseparability_robustness()   # 0.4142135623730949  = √2 − 1

The switch tolerates roughly 41% admixed noise before its causal nonseparability is gone. And now the coincidence Chapter 13 flagged is explained: the witness value is $3 - 4P_{\mathrm{win}} = -(r^*)$ exactly, whenever the bound is exceeded. The two numbers $1 - \sqrt2$ and $\sqrt2 - 1$ are negatives of each other because they are the same linear function of $P_{\mathrm{win}}$, read with opposite sign.

14.7 So What?

A gate-based language can compute every number in this chapter. Give it the switch isometry $V$, let it build the Choi operator, contract it with $S$, and it will print $0.8536$. The number is not the point.

The point is the type. When ocb.qrl type-checks as Switch(2), the compiler has certified — before any number is computed — that this process lies outside the causally separable set. P_win > 3/4 is not a measured result to be checked against a threshold; it is a static consequence of the term having the type it has. That is the guarantee Chapter 15 makes precise and proves: the Soundness Theorem, which says the type checker’s certificate is never wrong.

14.8 Exercises

Exercise 1 — The classical bound by hand. Enumerate all deterministic strategies for the order $A \prec B$ in which $a$ depends only on $x$ and $b$ depends on $y$ and on $a$. There are few enough to list. Show that none exceeds $P_{\mathrm{win}} = 3/4$, and identify one that achieves it. Then explain in one sentence why allowing shared randomness cannot help.

Exercise 2 — Linearity of the score. Using causal_inequality_value(), compute $P_{\mathrm{win}}$ for the mixtures $q\,W_{A\prec B} + (1-q)\,W_{\mathrm{id}}$ at $q = 0, 0.25, 0.5, 0.75, 1$ (build each with ProcessMatrix, as in Chapter 13). Confirm the results lie on the straight line $P_{\mathrm{win}}(q) = \tfrac12 + \tfrac18 q$. Which property of $\mathrm{Tr}[W S]$ forces this?

Exercise 3 — Robustness from the score. Derive $r^* = 4(P_{\mathrm{win}} - 3/4)$ from scratch: write down the score of the noise-blended process, set it to the threshold $3/4$, solve for $r$. Then verify numerically that W_sw.causal_nonseparability_robustness() agrees with $4(\,$W_sw.causal_inequality_value()$ - 0.75)$.

Exercise 4 — Why $Z$ and $X$ (harder). The switch strategy measures in the $Z$ basis on input $0$ and the $X$ basis on input $1$. Replace the $X$-basis choice with a measurement at angle $\theta$ from $Z$ in the $Z$–$X$ plane, and (using the API to rebuild the score operator, or by hand) find $P_{\mathrm{win}}(\theta)$. Show it is maximised at $\theta = \pi/2$ and that the maximum is $(2+\sqrt2)/4$. Where does the $\sqrt2$ enter?

Exercise 5 — A game the switch cannot win (open-ended). The switch beats the classical bound for this win condition, $a \oplus b = x \wedge y$. Construct a different two-input, two-output win condition for which the quantum switch scores no better than the best causally ordered process. What feature of $a \oplus b = x \wedge y$ — think about which rows demand correlation versus anti-correlation — is the switch exploiting, and how would you design a predicate that removes it?

14.9 Next Steps

You now have the gap: $3/4$ for anything with a definite causal order, $(2+\sqrt2)/4$ for the switch, and the robustness $\sqrt2 - 1$ measuring the distance between them.

Chapter 15: QuantumSwitch Soundness — the theorem that connects this gap to the type system. It states and proves $\vdash e : \mathsf{Switch}(d) \Rightarrow \llbracket e \rrbracket \in \mathrm{CNS}_{2,d}$: every well-typed switch term denotes a process on the far side of the $3/4$ bound. And it explains why McBeth, Sabry, Heunen, and Dave — four serious quantum programming languages — cannot even state it.

See also:

Chapter 13: The Causal Type Hierarchy — process matrices, causal separability, and the causal witness used here
Chapter 2: Entanglement — the CHSH game, whose Tsirelson bound is the same $\sqrt2$

References for this chapter:

Oreshkov, Costa & Brukner (2012), Quantum correlations with no causal order, Nature Communications 3, 1092 — the causal game and the causal inequality.
Araújo, Branciard, Costa, Feix, Giarmatzi & Brukner (2015), Witnessing causal nonseparability, New Journal of Physics 17, 102001 — the score operator, the witness, and the robustness measure.
Chiribella, D’Ariano, Perinotti & Valiron (2013), Quantum computations without definite causal structure, Physical Review A 88, 022318 — the quantum switch as a higher-order operation.

\((x,y)\)	\(x \wedge y\)	required relation
\((0,0)\)	\(0\)	\(a = b\)
\((0,1)\)	\(0\)	\(a = b\)
\((1,0)\)	\(0\)	\(a = b\)
\((1,1)\)	\(1\)	\(a \neq b\)