My question is, given two space-time points $x^{\mu}$ and $y^{\mu}$, if the events that occur at these points are simultaneous, i.e. $x^{0}=y^{0}$, are the two events necessarily space-like separated? The reason I ask is that I'm trying to understand the notion of equal-time commutation relations in QFT (in which the commutator is non-zero in the case where $\mathbf{x}=\mathbf{y}$).

For example, if one has a field $\phi$ and its conjugate momentum $\Pi_{\phi}(y)$, then the commutation relation between them is given by $$[\phi (t,\mathbf{x}),\Pi_{\phi}(t,\mathbf{y})]=i\delta^{(3)}(\mathbf{x}-\mathbf{y})$$ Now is the reason for this being equal to a $\delta$-function because of locality? i.e. given that the two fields are evaluated at the same time, then as locality demands that they can only "communicate" if they are separated by a time-like separation, they must necessarily be evaluated at the same spatial point, as if $\mathbf{x}\neq\mathbf{y}$ then there would be a space-like separation between the two fields (as $\Delta s^{2}=(x^{0}-y^{0})^{2}-(\mathbf{x}-\mathbf{y})^{2}=-(\mathbf{x}-\mathbf{y})^{2}<0$), and they would therefore commute (in order to obey locality)?

  • $\begingroup$ For your first question: Yes, they are space-like separated (you gave the reason yourself later on). Also, your commutation relation is wrong: It should be the commutator of a field with its conjugate. $\endgroup$ – Noiralef Jul 6 '15 at 17:11
  • $\begingroup$ @Noiralef Apologies, have corrected it now :-) $\endgroup$ – Will Jul 6 '15 at 17:22
  • $\begingroup$ Pay attention to the fact that you have to smear the operators with suitable test functions in order for them to make sense. The $\delta$-function must be smeared, you cannot interpret that as something which is non-zero only at equal points, it's wrong (I cleared the answer below). $\endgroup$ – gented Jul 6 '15 at 17:40

One way to define spacelike separation in special relativity is that any two events are spacelike separated if and only if there exists a reference frame in which the two events have the same time coordinate. So yes, if $x^0 = y^0$ the separation is spacelike.

Alternatively you can work from the definition where two events are spacelike separated if (and only if) the interval between them has the same sign as the spatial components of the metric. In other words, if your metric convention is $(-1,1,1,1)$, a spacelike interval has $\Delta s^2 > 0$, or if you use $(1,-1,-1,-1)$, it has $\Delta s^2 < 0$. If $x^0 = y^0$, then clearly the interval is determined only by the spatial components, and will necessarily have the same sign.

  • $\begingroup$ Ah ok, that's what I thought. So is this the reason why commutators between fields are only non-zero if they are evaluated at the same spacetime point? Also, is what I put about locality correct? $\endgroup$ – Will Jul 6 '15 at 17:20

The reason why the commutation relations between a field and its conjugate at equal times are of the form $$ \left[\phi(t,\textbf{x}),\pi(t,\textbf{y})\right]=i\hbar\,\delta^{(3)}(\textbf{x}-\textbf{y}) $$ is only to mirror and copy the canonical hamiltonian commutation relations $[q_i,p_j]=i\hbar\,\delta_{ij}$. No causality is involved, rather it is somehow the definition of quantum field commutation relations, which are employed basically copying the ones from classical point particles theory and extending them to infinitely many degrees of freedom.

However, notice that those in the commutators are not operators, rather they are operator valued distributions which make sense as operators only when smeared with suitable test functions. Only then we can talk about causality, because we have a concrete definition of spatially separated domains. As such, the correct definition for causality in quantum field theory is $$ \left[\phi(f),\varphi(g)\right]=0\qquad\mathcal{D}(f)\cap\mathcal{D}(g)=\textrm{space-like}. $$ If you calculate the above in some simple cases, for example for the free Klein-Gordon field (or Dirac) you will see that the right hand side will produce a result which always vanishes for space-like distances, due to properties of the Lorentz transformations (or equivalently, cancellations in the integrals). For non-free theories the issue is much more subtle and usually commutation relations must be imposed rather than calculated (unless you know some particular tricks).

In a nutshell: causality refers to the operators smeared with test functions; it does not give information on the commutation relations of the operators at equal times, which are nothing but a mere extension of the $q,p$ canonical commutation relations. And yes, if $x^0=y^0$, as you pointed out, the two events are space-like separated since you have at least one reference frame (the one you started with) where they happen at the same time.

  • $\begingroup$ Ah ok. In the notes I've read on QFT they do that first, i.e. postulate that the commutators of quantum fields obey a "continuum limit" of the canonical commutation relations of quantum mechanics, but they then say that this is a statement about micro causality, i.e. two fields can only interact if they are separated by a timelike interval?! $\endgroup$ – Will Jul 6 '15 at 17:57
  • $\begingroup$ Yes, true, but in order to make precise mathematical sense you have to smear the fields with test functions, since they are operator valued distributions; the $\delta$ on the right hand side should warn you that the expression doesn't make sense per sé, as it is. $\endgroup$ – gented Jul 6 '15 at 18:05
  • $\begingroup$ So does it only make full sense when one operates on a test quantum state with the commutator? (I understand that the delta function isn't a function per se, but the limit of a distribution). $\endgroup$ – Will Jul 6 '15 at 18:15
  • $\begingroup$ No, that is not the point. The point is that $\phi(x)$ is not an operator, rather $\phi(f)=\int dx\,f(x)\phi(x)$ is. Once you smear the operators with functions in suitable domain then you can "localise" them and do whatever you want. $\endgroup$ – gented Jul 6 '15 at 19:39
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    $\begingroup$ It's not really true to say that the commutation relations of the quantum field are "only" to copy the commutation relations of the single-particle theory. The more fundamental reason to use those equations is that they work, i.e., because the Standard Model matches experiment so amazingly well. $\endgroup$ – Harry Johnston Jul 6 '15 at 22:44

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