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This is a basic qft question. I am looking for the condition on a free scalar $\phi$ of mass $m$ in Euclidean space such that it satisfies the Klein-Gordon equation.

The Euclidean space Klein-Gordon operator is $(-\nabla^2+m^2)\phi(x)=0$. In momentum space, this becomes $(p^2+m^2)\tilde{\phi}(p)=0$, which seems to imply that $p^2=-m^2$ is the condition I'm looking for. However, this result seems nonsensical to me as a Euclidean vector can't have negative norm.

Is there some minus sign subtlety I'm messing up? Or is there a deeper reason to think that Euclidean space scalars can't satisfy the classical equation of motion?

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    $\begingroup$ That $p^2$ is the Wick rotated 4-momentum. The energy term turns purely imaginary, and using that, the vector would have the negative norm. $\endgroup$
    – naturallyInconsistent
    Commented Jul 2 at 5:39
  • $\begingroup$ If the energy term is purely imaginary, then when we calculate the "norm" wouldn't we multiply it by its conjugate, getting rid of the minus sign? It's not as if complex vectors can have negative norm generally speaking. $\endgroup$
    – Sam
    Commented Jul 2 at 5:44
  • $\begingroup$ You are simply getting lost in the woods. The real physics that we really want to care about is the Minkowski version; the Euclidean version is mathematical fudgery just to make it work, and happens to be physically realised in statistical thermodynamics version of QFT. And it also happens to be better to mathematically prove results with. Sadly, with fudgery, you often do have to throw away some sense-making stuff. This is one of those things. $\endgroup$
    – naturallyInconsistent
    Commented Jul 2 at 5:54

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Yes, OP is right: In Euclidean (E) signature, the mass-shell condition is $-E_E^2=E^2_M={\bf p}^2+m^2$, and we're solving an elliptic boundary value problem, which is different from the hyperbolic boundary value problem in Minkowski (M) signature.

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  • $\begingroup$ How interesting...Thank you for laying it out for me so plainly. I've never seen it described like this. But how can we reconcile this fact with $p^2=-m^2$, assuming the vector $p=(\mathbf{p}, E_E)=(\mathbf{p}, iE_M)$? $\endgroup$
    – Sam
    Commented Jul 2 at 5:46
  • $\begingroup$ From a pedagogical perspective, it is easier to explain the Wick rotation using the $(-,+,+,+)$ Minkowski signature convention, cf. e.g. this Phys.SE post. $\endgroup$
    – Qmechanic ♦
    Commented Jul 2 at 7:00

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