# AQFT: Microscopic causality (local commutativity) of the free Klein-Gordon field

Let us consider the positive and negative frequency part of the free Klein-Gordon field operator

$$\hat\phi(x) = \hat\phi^-(x) + \hat\phi^+(x) \tag{1}$$

where

\begin{aligned} \hat\phi^-(x) = \hat\phi^+(x)^\dagger && \hat\phi^+(x) = \int\frac{d^3p}{(2\pi)^3\sqrt{2p_0}} e^{+ipx} \hat{a}^\dagger(\boldsymbol{p}) \end{aligned} \tag{2}

with $$\hat{a}^\dagger(\boldsymbol{p})$$ being the creation operator and

$$p_0 = \omega(\boldsymbol{p}) = \sqrt{\boldsymbol{p}^2+m^2} \tag{3} .$$

According to axiomatic quantum field theory, we have the smeared Klein-Gordon field operator

$$\hat\phi^+[f] = \int d^4x\ f(x)\hat\phi(x) = \int\frac{d^3p}{(2\pi)^3\sqrt{2p_0}} \left( \int d^4x\ f(x) e^{+ipx} \right) \hat{a}^\dagger(\boldsymbol{p}) = \int\frac{d^3p}{(2\pi)^3\sqrt{2p_0}} f(p) \hat{a}^\dagger(\boldsymbol{p}) \tag{4}$$

with $$f$$ being an element of the Schwartz space.

Let now $$f,g$$ be two smearing functions then the commutator

\begin{aligned} \left[\phi^-[g],\hat\phi^+[f]\right] = \langle 0\vert\hat\phi^-[g]\hat\phi^+[f]\vert0\rangle &= \int\frac{d^3p}{(2\pi)^3\sqrt{2p_0}} \int\frac{d^3q}{(2\pi)^3\sqrt{2q_0}} f(p)g(q)^* \langle0\vert\hat{a}(\boldsymbol{q})\hat{a}^\dagger(\boldsymbol{p})\vert0\rangle \\ &= \int\frac{d^3p}{(2\pi)^32p_0} f(p)g(q)^* \\ &= \int\frac{d^3p}{(2\pi)^32p_0} \left(\int d^4x\ f(x)e^{-ipx}\right) \left(\int d^4y\ g(y)e^{-ipy}\right)^* \\ &= \int d^4x\ f(x) \int d^4y\ g(y) \int\frac{d^3p}{(2\pi)^32p_0}e^{-ip(x-y)} \end{aligned} \tag{5}

quantifies the independence of an measurement of $$g$$ and $$f$$. For instance, one of the Wightman axioms requires that $$\left[\phi^-[g],\hat\phi^+[f]\right]=0$$ iff the support of $$f(x)$$ and $$g(x)$$ is spacelike separated, i.e., $$f$$ and $$g$$ couldn't have been in causal contact.

How can I show the previous statement from eq. (5)?

In particular, how do I rewrite eq. (5) into a term integrating over spacelike and timelike regions? The term integrating over the timelike region should be quickly decay to zero (?) and the term integrating over the spacelike region should be zero if the support of $$f(x),g(x)$$ is spacelike separated.

For a massless Klein-Gordon field, we have $$p_0=\omega(\boldsymbol{p})=\Vert\boldsymbol{p}\Vert$$ and the momentum integral of eq. (5) can be rewritten

\begin{aligned} \int\frac{d^3p}{(2\pi)^32p_0}e^{-ip(x-y)} &= \frac{2\pi}{(2\pi)^3} \int_0^\infty dp\ p^2\frac{1}{2p}e^{-ipt} \int_0^\pi d\theta \sin\theta e^{+ipr\cos\theta} \\ &= \frac{1}{2(2\pi)^2} \int_0^\infty dp\ pe^{-ipt} \int_{-1}^{+1} du\ e^{+ipru} \\ &= \frac{1}{2(2\pi)^2} \int_0^\infty dp\ pe^{-ipt} \frac{e^{+ipr}-e^{-ipr}}{ipr} \\ &= \frac{1}{(2\pi)^2} \int_0^\infty dp\ e^{-ipt}\sin(pr) \end{aligned} \tag{6}

where $$r=\Vert\boldsymbol{x}-\boldsymbol{y}\Vert$$ and $$t=x^0-y^0$$. Now, the interpretation of the last integral is ambiguous. If the integration domain would be over the real axis, we would find

$$\int_{-\infty}^{+\infty} dp\ e^{-ipt}\sin(pr) = \int_{-\infty}^{+\infty} dp\ \frac{e^{-ip(t-r)}-e^{-ip(t+r)}}{2i} = \frac{\delta(t-r)-\delta(t+r)}{2i}$$

which appears to represent the surface of the lightcone (lightlike separated) events. However, as I wrote, it is unclear what to do when we only integrate over $$\mathbb{R}^+$$ (there are very different answers on mathoverflow to this: from that the integral is not defined over that we need a $$1/2$$ prefactor to it is the same as the delta distribution as long as $$t\pm r>0$$).

Furthermore, I am not quite sure how to proceed from

$$\left[\phi^-[g],\hat\phi^+[f]\right] \propto \int d^4x\ f(x) \int d^4y\ g(y) \frac{\delta(t\mp r)}{r} \tag{7}$$

• $\int \frac{d^3 p}{(2\pi)^3 2p_0}e^{-ip(x-y)}$ is just the propagator of your theory. – Jeanbaptiste Roux Mar 24 at 10:47
• True! Then $\langle g\vert f\rangle=\int d^4x d^4y f(x) D(x-y) g(y)$ which reminds me of the the effective action $W=-\frac{1}{2}\int d^4x d^4y J(x)D(x-y)J(y)$ we would find from the path integral $Z[J]=Z[0]e^{iW}$ with source terms $J$. – bodokaiser Mar 24 at 10:49
• For spacelike separated events we have $D(x-y)\sim e^{-mr}$ but this suggests that eq. (7) is not strictly zero? – bodokaiser Mar 24 at 10:57

The commutator shown in the question is not zero, and it shouldn't be zero. Even if $$f$$ and $$g$$ both have compact support in spacetime, $$\phi^-(g)$$ and $$\phi^+(f)$$ are not local operators, so we do not expect them to commute when the supports of $$f$$ and $$g$$ are spacelike separated.
Intuitively, they can't be local operators, because the definitions of the positive- and negative-frequency parts involve an integral over all time. A contribution from the past can contribute anywhere in the interior of its future light-cone, and a contribution from the future can contribute anywhere in the interior of its past light-cone, so integrating over all time means including contributions from all of space. That's why $$\phi^-(g)$$ and $$\phi^+(f)$$ are not local operators.
We can also see this through a general theorem, the Reeh-Schlieder theorem, which implies that a local operator cannot annihilate the vacuum state (in a relativistic QFT). Since the positive-frequency part of an operator does annihilate the vacuum state, it cannot be a local operator. Conversely, the original field operator $$\phi(x)$$, which is a local operator by definition, does not annihilate the vacuum state.