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

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


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

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

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

  • $\begingroup$ $\int \frac{d^3 p}{(2\pi)^3 2p_0}e^{-ip(x-y)}$ is just the propagator of your theory. $\endgroup$ Commented Mar 24, 2021 at 10:47
  • $\begingroup$ 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 $J$. $\endgroup$
    – bodokaiser
    Commented Mar 24, 2021 at 10:49
  • $\begingroup$ For spacelike separated events we have $D(x-y)\sim e^{-mr}$ but this suggests that eq. (7) is not strictly zero? $\endgroup$
    – bodokaiser
    Commented Mar 24, 2021 at 10:57

1 Answer 1


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.

  • $\begingroup$ Can you give examples for non-local operators/observables? The Hamilton operator does not annihilate the vacuum but is non-local same with momentum and number operator. $\endgroup$
    – bodokaiser
    Commented Mar 24, 2021 at 12:57
  • $\begingroup$ @bodokaiser The number operator in a free field theory does annihilate the vacuum. (I restricted this statement to free field theory because I'm not sure what "number operator" would mean more generally.) So do the momentum operators and Hamiltonian, if we use the standard convention for the overall constant terms in those operators, because the vacuum is invariant under Poincaré transformations. Loosely speaking, the vacuum is an "eigenvector" of the momentum operators and Hamiltonian with eigenvalue zero. $\endgroup$ Commented Mar 24, 2021 at 13:04
  • $\begingroup$ thanks for the example! Can you recommend some literature (or review papers) on AQFT? I read a bit in Haag's and Bogulubov's book but they quickly get very mathematical. I am most interested in the physical aspects of AQFT. $\endgroup$
    – bodokaiser
    Commented Mar 24, 2021 at 13:10
  • 1
    $\begingroup$ @bodokaiser Review papers include arxiv.org/abs/math-ph/0411072 and arxiv.org/abs/math-ph/0602036. Araki's book Mathematical Theory of Quantum Fields (1999) and Horuzhy's book Introduction to Algebraic Quantum Field Theory (1990) were both helpful to me. I wouldn't say that any of these papers/books are less mathematical than Haag's book, but at least they give some different perspectives on the subject. $\endgroup$ Commented Mar 24, 2021 at 13:33
  • $\begingroup$ @bodokaiser If I remember right, Araki's book is more basic and pedagogical than Horuzhy's book, despite their titles. $\endgroup$ Commented Mar 24, 2021 at 14:02

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