It has been claimed (e.g. here) and apparently already been established, that the interval $x - y$ being (called) "spacelike" implies that $\bigl[\hat O (x),\, \hat O' (y)\bigr]=0$ for any two (not necessarily distinct) operators $\hat O$ and $\hat O'$ corresponding to physical observables evaluated at $x$ or at $y$, respectively:

$$\text{spacelike}( \, x - y \, ) \quad\implies\quad \Bigl( \forall \hat O \, \forall \hat O': \Bigl[\hat O (x),\, \hat O' (y)\Bigr] = 0 \Bigr). $$

Is the converse correct, too, that the vanishing commutators imply (or are sufficient for) the interval $x - y$ to be (called) "spacelike":

$$\Bigl(\forall \hat O \, \forall \hat O': \Big[\hat O (x),\, \hat O' (y)\Bigr] = 0 \Bigr) \quad\implies\quad \text{spacelike}( \, x - y \, ) \, ?$$

  • $\begingroup$ Suppose $x$, $y$ be timelike separated. From assumption that all fields at $x$ commute with all fields at $y$, it'll follow that all fields at $\Lambda x$ will commute with all fields at $\Lambda y$ (where $\Lambda$ is any Poincare transformation). If you assume covariance under scaling $x\rightarrow ax$ too then (from above assumption) you can prove a stronger result that fields at any two timelike separated points will commute. So at least for a nontrivial scale covariant theory all fields at two timelike separated points can't commute. $\endgroup$ – user10001 Sep 18 '12 at 0:00
  • $\begingroup$ @dushya: In trying to follow your argument I wonder whether there may be a mistake/typo especially in the concluding sentence. Surely, if all (pairs of) operators commute for all (pairs of) arguments then my question would be pointless. But do they? $\endgroup$ – user12262 Sep 18 '12 at 15:17
  • $\begingroup$ @use12262 I meant if (for two given timelike separated points $x$, and $y$) all fields at $x$ commute with all fields at $y$, and if your theory has scale covariance (besides Poincare covariance) then (it can be shown that) fields corresponding to any two timelike separated points will commute. $\endgroup$ – user10001 Sep 19 '12 at 0:21
  • $\begingroup$ @dushya: Does Arnold Neumaier's answer (now also) express what you're getting at? (I suppose so; and if so, please consider my comment there.) $\endgroup$ – user12262 Sep 19 '12 at 19:02

As far as I know, all quantum theories have non-compatible observables (i.e., their corresponding commutators do not vanish as long as the space-time points at which they are evaluated are causally related), therefore if at two given points every pair of observables commute, then these two point are separated by a space-like interval. This should actually be the intrinsic way in which one defines the light-cone in a quantum field theory (quantum theories which do not have local observables are perhaps subtle in this issue and I ignore how —if possible— one can define the light-cone in an intrinsic manner in these theories).

So yes, it seems the implication works in both ways.

  • $\begingroup$ @drake: Thanks, I'm glad about your answer, and that you appreciate the question for a definition in terms of intrinsic notions. I'll have to think about whether and how to further question your answer which had been accepted there: physics.stackexchange.com/question/36178 (especially regarding the possibility of "refactive index"-values differing from unity). $\endgroup$ – user12262 Sep 17 '12 at 23:17
  • $\begingroup$ @user12262 You're welcome. Refractive index may increase phase velocity, not propagation of information velocity. See physics.stackexchange.com/questions/34214/lambda-frac2hp/… (there my answer is totally correct regardless it does not have any up-vote) $\endgroup$ – Diego Mazón Sep 18 '12 at 0:49
  • $\begingroup$ @drake: While I liked your answer, your comment gives me pause ... Do you agree that by your answer it is established that $\forall \hat O \, \forall \hat O' \Bigl( \Big[\hat O (x),\, \hat O' (y)\Bigr] = 0 \Bigr) \Longleftrightarrow \text{spacelike}( \, x - y \, )$ ? If so, I'll best express my reservations in a proper (follow-up) question, which however may take a moment to formulate. p.s. I find your answer at physics.stackexchange.com/a/34214/10552 appropriate there, but I find it difficult to relate it to this: physics.stackexchange.com/a/36178/10552 $\endgroup$ – user12262 Sep 18 '12 at 15:03
  • $\begingroup$ @user12262 1)Yes, I guess my answer implies that. 2) Those answers are not related. I linked the first in order for you to see how the refractive index may affect the different notion of wave's speed. $\endgroup$ – Diego Mazón Sep 18 '12 at 18:16
  • $\begingroup$ @drake: (1) Great! And this makes your answer acceptable to me. (Full disclosure: I didn't consider any other merits of this or other answers which were provided here. I just found the answer by Arnold Neumaier below ... which may have implications to your answer there: physics.stackexchange.com/a/36178/10552 ) $\endgroup$ – user12262 Sep 18 '12 at 20:57

No, any operators that can be measured simultaniously in quantum mechanics will commute, eg two different components of an electric field, these commute at non-spacelike separation, excluding electron-positron effects.


The meaning of the causal commutation rules $[O_1(x),O_2(y)] = 0$ if $x$ and $y$ are spacelike separated is that pairs of fields can be prepared independently at the same time, with time interpreted in the reference frame of an arbitrary observer.

The assertion in your converse question is not true in conformal theories in even dimension, where as a form of Huyghen's principle, (necessarily massless) fields commute at spacelike and timelike separation.

On the other hand, Poincare covariance implies that even for a single massive field, $[O(x),O(y)]$ is zero in a full neighborhood of two points only if the latter are spacelike separated.

(I had claimed in the first version of my answer that a scalar field and its gradient don't commute at spacelike separation, but this is false, as can be seen by differentiation of the commutation relation with respect to $x$ or $y$. A more realistic example is QED, where the vector potential $A(x)$ is unobservable, whereas its exterior derivatives define the observable electric field $E(x)$ and the observable magnetic field $B(x)$, and these can indeed be prepared independently at the same time.)

  • $\begingroup$ @Arnold Neumaier: Would you please work out your (counter-)example "scalar field and its gradient" in some detail? I do appreciate (Boy, do I ever!) that a derivative cannot be evaluated given only one value; but what if the limit approaching $y$ is taken, while the other operator refers to $x$? Also, you may find the question and provided answers there physics.stackexchange.com/a/35942/10552 instructive, which gave rise to the question considered here. $\endgroup$ – user12262 Sep 18 '12 at 21:31
  • $\begingroup$ We are talking about physical observables (which are usually at least bilinears in fields), not fundamental fields. $\endgroup$ – Diego Mazón Sep 19 '12 at 0:51
  • $\begingroup$ @user12262: My answer was wrong, and I corrected it. $\endgroup$ – Arnold Neumaier Sep 19 '12 at 10:37
  • $\begingroup$ @Arnold Neumaier (1): Thanks for the clarification. Now, my question didn't restrict the applicable operators as "in conformal theories" and/or "in even dimension" ... (However, this was just in keeping with this answer: physics.stackexchange.com/a/35942/10552 ) $\endgroup$ – user12262 Sep 19 '12 at 18:51
  • $\begingroup$ @Arnold Neumaier (2/2): Concerning "field preparation": Is it correct to conclude, that pairs of fields which were prepared (or plainly encountered) with some (any definite) dependence on each other were necessarily in (timelike or at least lightlight) order?, regardless of whether such dependence was "by group", "by front", "by signal", "electro-dynamic", "by mass", or by whatever specifically? $\endgroup$ – user12262 Sep 19 '12 at 18:52

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