# Sign of the coupling constant for a $\phi^4$ interaction

The vacuum-vacuum transition for a simple bosonic $$\phi^4$$ theory is typically written as $$\langle0|0\rangle = \int[D\phi]\ \exp\left[i\int (L_0+L_\mathrm{int}) d^4x \right], \tag{1}$$ Where $$L_0$$ is our typical free part of the Lagrangian and $$L_\mathrm{int}$$ can be written as $$+\lambda \phi^4$$. At a first glance the sign of $$\lambda$$ shouldn't matter (I wouldn't think) as it won't make our complex exponential ill behaved.

However if we Wick rotate to imaginary time our vacuum-vacuum transition, or the partition function $$Z$$, takes on the form: $$Z = \int[D\phi]\exp\left[-(S_{E0}+S_\mathrm{int}) \right], \tag{2}$$ where $$S_{E0}$$ is the free part of our Euclidean action and $$S_\mathrm{int}$$ is the interaction part. Now we see that if $$S_\mathrm{int}$$ < 0 then our partition function will blow up at large field values and will be ill behaved even though the corresponding object in real time has no immediate issues whatsoever.

Does our analytic continuation to imaginary time not work for this case of $$\lambda<0$$? I don't understand why our theory might work in real time but then describe nonsense in imaginary time. Can anyone explain to me what I am missing here?

• I'm not sure this issue has anything to do with the Wick rotation. If you give $\lambda$ the wrong sign, the Hamiltonian is unbounded below. That's a problem, Wick rotation or not. If you fix it (e.g. with a $\lambda \phi^6$ interaction) then the Wick rotated action is fine as well. Feb 23, 2017 at 5:30