Relativistic causality does not require every two-point function to vanish outside the light cone. It requires local observables to commute there = [ϕ(x),ϕ(y)]=0for spacelike separation.​

The key distinction is between:$$\text{correlation}$$

and$$\text{causal influence}.$$

A two-point function measures correlation. A commutator measures whether one local operation can affect another.

1. The two-point function can be nonzero

For a scalar field,$$W(x-y)=\langle0|\phi(x)\phi(y)|0\rangle.$$

For spacelike separation,$$(x-y)^2<0,$$

this generally does not vanish. For a massive free scalar field,$$W(x-y) = \frac{m}{4\pi^2r}K_1(mr), \qquad r=\sqrt{-(x-y)^2}.$$

At large distance,$$W(x-y)\sim e^{-mr}.$$

Thus the correlation is small but nonzero.

This means that measurements of the field at $x$ and $y$ can produce statistically correlated results. It does not mean that a particle or message traveled from one point to the other.

The quantum vacuum is an entangled state, not a classical empty space.

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2. The commutator measures causal influence

The field commutator is$$[\phi(x),\phi(y)] = \phi(x)\phi(y)-\phi(y)\phi(x).$$

For a free scalar field,$$\langle0|[\phi(x),\phi(y)]|0\rangle = W(x-y)-W(y-x).$$

Although the two terms are individually nonzero at spacelike separation, they are equal:$$W(x-y)=W(y-x).$$

Therefore,$$\boxed{ [\phi(x),\phi(y)]=0 \quad\text{when}\quad (x-y)^2<0. }$$

The nonzero correlations cancel in the commutator.

This condition is called microcausality.

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3. Why the commutator is the relevant quantity

Suppose Alice is at $x$ and Bob is at $y$, with the two points spacelike separated.

Alice performs a local operation represented schematically by$$U_x=e^{i\lambda\phi(x)}.$$

Bob measures a local observable $O(y)$. After Alice’s operation, Bob’s expectation value would be$$\langle O(y)\rangle’ = \langle0|U_x^\dagger O(y)U_x|0\rangle.$$

If Alice’s operator commutes with Bob’s observable,$$[U_x,O(y)]=0,$$

then$$U_x^\dagger O(y)U_x=O(y),$$

and therefore$$\boxed{ \langle O(y)\rangle’ = \langle O(y)\rangle. }$$

Nothing Alice does at $x$ can change the statistics Bob observes at $y$. Therefore, Alice cannot send information to Bob faster than light.

That is the operational meaning of relativistic causality.

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4. Correlation does not mean communication

Imagine Alice and Bob receive two correlated quantum systems. Their results may be strongly related, but Alice cannot choose her result and thereby control what Bob sees.

Bob’s local results remain statistically unchanged. Only later, after Alice and Bob communicate normally, can they compare their records and discover the correlation.

The same distinction applies to the field vacuum:$$\boxed{ \text{spacelike correlation exists} }$$

but$$\boxed{ \text{spacelike control or signaling does not}. }$$

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5. Why spacelike operators must commute

If $x$ and $y$ are spacelike separated, different inertial observers can disagree about their time ordering:

• One observer can see $x$ occur before $y$.
• Another can see $y$ occur before $x$.
• A third can see them occur simultaneously.

If the two operations commute,$$\phi(x)\phi(y)=\phi(y)\phi(x),$$

their order makes no physical difference. Every inertial observer predicts the same final result.

If they did not commute, the outcome could depend on which event was considered first, even though special relativity provides no observer-independent ordering for spacelike events.

Thus microcausality ensures consistency with relativity.

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6. Why the commutator vanishes for the Klein–Gordon field

At equal times, canonical quantization gives$$[\phi(t,\mathbf x),\phi(t,\mathbf y)]=0.$$

Every spacelike-separated pair of events can be transformed into a frame where the events are simultaneous. Because the scalar-field commutator is Lorentz covariant, the equal-time result implies$$[\phi(x),\phi(y)]=0$$

for every spacelike separation.

Explicitly,$$[\phi(x),\phi(y)] = \int\frac{d^3p}{(2\pi)^3\,2E_{\mathbf p}} \left[ e^{-ip\cdot(x-y)} – e^{ip\cdot(x-y)} \right].$$

For spacelike separation, the two contributions are equal and cancel.

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7. Inside the light cone

For timelike separation,$$(x-y)^2>0,$$

the commutator generally does not vanish:$$[\phi(x),\phi(y)]\neq0.$$

A causal signal traveling at or below the speed of light can connect the two events. Therefore, an operation at $y$ can in principle influence a later measurement at $x$.

The commutator’s structure is consequently:$$[\phi(x),\phi(y)] = \begin{cases} 0, & \text{spacelike separation},\\[4pt] \text{generally nonzero}, & \text{timelike or lightlike separation}. \end{cases}$$

The essential distinction

$$\boxed{ \langle0|\phi(x)\phi(y)|0\rangle\neq0 }$$

means:

The field fluctuations at $x$ and $y$ are correlated.

But$$\boxed{ [\phi(x),\phi(y)]=0 }$$

means:

An operation at $x$ cannot causally affect a measurement at $y$.

So relativistic causality does not demand an uncorrelated vacuum. It demands that spacelike-separated local operations cannot influence one another.