The commutator of two free scalar fields at two spacetime points $x$ and $y$ given by $$[\phi(x),\phi(y)]=\int\frac{d^3p}{(2\pi)^32E_{\textbf{p}}}[e^{-ip\cdot (x-y)}-e^{ip\cdot (x-y)}]$$ is Lorentz-invariant. We can therefore compute this quantity in any inertial frame we like.

If it is zero in one frame it remains zero in every other frame. If the interval is spacelike i.e., $(x-y)^2<0$, one can perform a Lorentz transformation to go to a frame when the events can be made simultaneous i.e., $x^0=y^0$. Therefore, \begin{equation}[\phi(x),\phi(y)]=\int\frac{d^3\textbf{p}}{(2\pi)^32E_\textbf{p}}\sin(\textbf{p}\cdot (\textbf{x}-\textbf{y}))=0\end{equation} since the integrand is an odd function of $\textbf{p}$.

Now, this should be true in all other inertial frames connected by Lorentz transformation. Since a Lorentz transformation cannot keep a spacelike interval spacelike, we have shown that this vanishes identically for space-like separation $(x-y)^2<0$ i.e. outside the light cone.

Will the commutator $[\phi(x),\phi(y)]$ be zero if the ``mass-squared parameter" i.e. $m^2$ of the term $\frac{1}{2}m^2\phi^2$ of the Lagrangian is negative? If not, which part of this proof fails and why?

I expect this to violate causality i.e., the above commutator not to remain zero even for $(x-y)^20$ because a particle of negative mass corresponds to a propagation velocity faster than the speed of light in vacuum.

  • 1
    $\begingroup$ You should not expect tachyons to violate causality, see physics.stackexchange.com/q/166095/50583. $\endgroup$
    – ACuriousMind
    Dec 26, 2017 at 12:02
  • $\begingroup$ @ACuriousMind quest.ph.utexas.edu/sudarshan_tachyons.html#tech This article by E. C. G Sudarshan points out that for negative mass one ends up with tachyons for which $v^2>c^2$. Is it not a violation of relativity (and therefore, causality)? $\endgroup$
    – SRS
    Dec 28, 2017 at 8:52
  • $\begingroup$ No. Please read the answers to the question I linked. $\endgroup$
    – ACuriousMind
    Dec 28, 2017 at 10:44

2 Answers 2


Yes, it would be zero independently from the value of $m^2$ and also for values of $m^2$ depending on the point of the spacetime. However for negative $m^2$ you cannot quantize according to the usual standard way by assuming the existence of a Poincare' invariant vacuum state and building up the theory in the Fock space relying upon this state.

The procedure to be exploited is more complicated and is exactly the same as in curved (globally hyperbolic) spacetime where Poincare' symmetry does not exist.

First, you should assume the existence of an abstract unital $^*$-algebra of observables generated by the field operators. This algebra exists (up to isomorphisms) just in view of the CCR relations, which are consistent independently from the value of $m^2$, and the structure arises from the existence of a certain symplectic form in the space of classical solutions. The only requirement is that the field equations are linear and of hyperbolic type. Everything remains true for KG equation with every real value of $m^2$ and also for such term depending on the point in the spacetime.

Finally, you have to fix a state in algebraic sense, a positive normalized functional on the said algebra. The GNS construction permits us to eventually represent the theory in a Fock space, where the state is the vacuum vector.

If $m^2>0$, the algebraic state can be uniquely fixed in order to produce the standard representation in Minkowski spacetime. This choice is not possible for negative values.

From the physical side, however, the physical meaning of a reference vacuum which is not Poincare' invariant in the presence of geometric Poincare' invariance (algebrically implemented) is not obvious and would represent a spontaneously breaking of symmetry.

Possible reference: https://arxiv.org/abs/1412.5945


A free theory wit $m^2<0$ is not stable. It only makes sense if there is an interaction term, say $\lambda \phi^4$. Dues to spontaneous breaking of the symmetry, and for small perturbations around the new vacuum, one recovers a positive mass term.

If the symmetry is continuous, one also gets massless modes (Goldstone modes).


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