In Minkowski spacetime, the commutator of the Klein-Gordon field operator with itself at different spacetime points evaluates to the advanced minus retarded Green's function of the classical theory,

\begin{align} [\phi (x),\phi (y)]=\langle 0|[\phi (x),\phi (y)]|0\rangle =G_A(x-y)-G_R(x-y), \end{align} which vanishes for spacelike separations. [I use the convention that the K-G Green's functions are defined by $(\partial^2+m^2)G(x-y)=-i\delta^{(4)}(x-y)$.]

Because of the $SO(4)$-isometries of Euclidean spacetime, there is no invariant notion of time direction and, indeed, all separations are spacelike. For this reason, I would naively expect the Euclidean K-G Green's function (which vanishes at infinity) is unique--i.e., there are no "advanced" or "retarded" Euclidean Green's functions--and the field commutator should, then, vanish for $all$ Euclidean spacetime separations.

More explicitly, the Euclidean 2-point function goes as

\begin{align} \langle 0|\phi (x)\phi(y) |0\rangle=\int \frac{d^4p}{(2\pi)^4}\frac{e^{ip\cdot(x-y)}}{p^2+m^2}, \end{align} so the Euclidean commutator is

\begin{align} [\phi (x),\phi(y)]=\int \frac{d^4p}{(2\pi)^4}\left\{\frac{e^{ip\cdot(x-y)}}{p^2+m^2}-\frac{e^{-ip\cdot(x-y)}}{p^2+m^2}\right\}=0, \end{align} where the last equality follows from the fact that the momentum "volume" measure is invariant under $p\to -p$ in Euclidean spacetime.

Is this conclusion and reasoning correct?

  • 2
    $\begingroup$ I'm not sure what kind of answer you're expecting for this, but the reasoning appears to be correct to me if it matters. $\endgroup$
    – Darkseid
    Commented Oct 4, 2017 at 22:43
  • 1
    $\begingroup$ @Darkseid Sorry, if my intentions were ambiguous in some way. I have seen statements about the commutator in Minkowski explicitly in many sources--and corresponding discussions about causality, but I cannot find a reference that gives an analogous treatment for the Euclidean theory. I was genuinely uncertain and trying to reason out what it should be. $\endgroup$
    – user143410
    Commented Oct 4, 2017 at 22:48

2 Answers 2


I doubt that your reasoning goes through without caveats. It's true that when you analytically continue the Minkowski spacetime two-point function to imaginary time, you get a function (the two-point Schwinger function $S(x,y)$) which is symmetric in the two Euclidean spacetime points and is covariant with respect to Euclidean isometries. So if you define $$ [\phi (x),\phi(y)]:=S(x,y)-S(y,x) $$ for $x$, $y$ being Euclidean spacetime points, then you have $[\phi (x),\phi(y)]=0$.

However, it is not clear to me that the left-hand side of the definition above can be interpreted as the commutator of two well-defined operators (or operator-valued distributions). For example, let $\phi$ be a free scalar field, defined initially in Minkowski spacetime by its mode expansion $$\phi(t,\mathbf{x})=\int\frac{d^{3}k}{(2\pi)^{3}}\frac{1}{\sqrt{2\omega(\mathbf{k})}}\left(a(\mathbf{k})e^{-i\omega t+i\mathbf{k}\cdot\mathbf{x}}+a^{\dagger}(\mathbf{k})e^{i\omega t-i\mathbf{k}\cdot\mathbf{x}}\right).$$

If we formally continue this to $t=-i\tau$, we see that the modes of positive (negative) frequency diverge badly for $\tau<0$ ($\tau>0$). (Note: This divergent behavior of the modes for imaginary times also puts your manipulations of the integral representation of the commutator into question.)

One can still make sense of expressions like $\langle 0|\phi (-i\tau_2,\mathbf{x}_2)\phi(-i\tau_1,\mathbf{y}_1) |0\rangle$ as long as $\tau_2>\tau_1$ and the momentum integrals are understood to be taken only at the end. Using this formalism, one can equate this expression to a path integral in Euclidean spacetime with two "operator insertions": $$ \langle 0|\phi (-i\tau_2,\mathbf{x}_2)\phi(-i\tau_1,\mathbf{x}_1) |0\rangle = \frac{\int \mathcal D \phi \exp{\left(-S_E[\phi]\right)\phi(x_2)\phi(x_1)}}{\int \mathcal D \phi \exp{\left(-S_E[\phi]\right)}} $$

The path integral automatically takes care of the ordering in imaginary time and we can therefore use it to extend the meaning of the left-hand side to arbitrary $\tau$-ordering, which gives us the Schwinger two-point function for this field expressed as a path integral. Still, this does not mean that we have made sense of the field operator by itself for non-zero imaginary times.

The automatic Euclidean time-ordering done by the path integral prevents us from using it to calculate the general matrix elements of the field commutator for unequal Euclidean times. Maybe this is the reason why one usually does not find a calculation of this commutator for Euclidean spacetime in textbooks. (At least I found none in the texts I have available.)

For 2-dimensional conformal field theories one finds calculations of commutators $[T,\phi(y)]$ between the field $\phi$ and a conserved charge $T$ making use of the operator product expansion (OPE) in Euclidean spacetime ([1], [2]). Such commutators can be non-zero even if the operator $T$ is built from the field and its derivatives. Your conclusion that the commutator $[\phi(x), \phi(y)]$ is identically zero on Euclidean spacetime would also imply that all commutators of quantities built out of the fields and their derivatives vanish on Euclidean spacetime, wouldn't it?

Schwinger mentioned the formal possibility to construct fully (anti)commuting field operators for Euclidean spacetime at the end of [3]. I'm not sure what to make of that.

A related question about field operators and two-point functions in Euclidean spacetime has been discussed on mathoverflow: https://mathoverflow.net/q/237647

[1] K. Becker, M. Becker, J. H. Schwarz. String theory and M-theory: A modern introduction. Cambridge University Press, 2007. p. 64f.

[2] J. Polchinski. String Theory. Vol. 1 An Introduction to the Bosonic String. Cambridge University Press, 1998. p. 55.

[3] J. Schwinger, 1958. Four-dimensional euclidean formulation of quantum field theory. Proceedings, Conference: C58-06-30, p.134-140.


Yes, Euclidean commutators vanish.

This is pretty obvious in the path integral formalism: You can compute the matrix elements of the commutator from the correlation functions: $\langle \hat{\phi}(x)\hat{\phi}(y) ... \rangle = \int \phi(x)\phi(y) ... e^{-S(\phi)}d\phi$. The integral involves only commuting variables, so you can switch $x$ and $y$.

The connection between Euclidean correlation functions and Minkowski ones is slightly subtle. The Euclidean correlation functions are analytic functions of several variables $x$, $y$, and can be extended to the complexification of a Cartesian product of several copies of Euclidean space. The corresponding Cartesian products of Minkowski space also sit within this complexification. The Minkowski correlation functions are, in fact, boundary values of the analytic extension of the Euclidean correlation functions. But which time-ordering of Minkowski correlation functions you get depends on how you approach the boundary.

  • 6
    $\begingroup$ I'm not sure how the path integral formalism makes this "obvious" for the Euclidean case. We know they don't commute in Minkowski at timelike separations, but that's not obvious to me based on your current path integral explanation. $\endgroup$
    – user143410
    Commented Oct 5, 2017 at 2:07
  • $\begingroup$ Why is this not obvious? The Euclidean path integral involves only commuting quantities. (It's a statistical field theory.) So you can freely permute the observables. There's nothing more to it. $\endgroup$
    – user1504
    Commented Oct 5, 2017 at 18:19
  • 4
    $\begingroup$ The Minkowski path integral also only involves commuting quantities--the fields are just functionals inside the path integral--and yet the Minkowski fields DON'T commute at timelike separations inside correlation functions. Therefore, it is not at all obvious to me how writing down the path integral formula itself helps one see that the Euclidean fields DO commute at all separations. You are claiming what is to be shown. $\endgroup$
    – user143410
    Commented Oct 5, 2017 at 18:45
  • 2
    $\begingroup$ I see what you're objecting to. My answer is that Minkowski path integrals don't really exist in the same way that Euclidean ones do, and that the formal manipulation you're describing isn't valid. To define them you have to add some additional regularizing info (e.g., the $+i\epsilon$ prescription). This additional regularization information encodes the non-commutativity, and generally turns out to be equivalent to specifying which branch of the analytic extension of the Euclidean correlation function you're on. $\endgroup$
    – user1504
    Commented Oct 5, 2017 at 18:52
  • 1
    $\begingroup$ @user1504 I believe it would help greatly if you could include how exactly noncommutativity arises through this mechanism in your answer. $\endgroup$ Commented Oct 6, 2017 at 0:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.