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?

  • 4
    $\begingroup$ 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. $\endgroup$ – knzhou Feb 23 '17 at 5:30
  • 1
    $\begingroup$ Comment to the post (v3): Note that according to standard conventions there is no minus in eq. (1). $\endgroup$ – Qmechanic Feb 23 '17 at 11:39

Recall that in the derivation of $$ \langle0|0\rangle = \int\mathrm d\phi\ \exp\left[\ i\int\mathrm dx\ (L_0+L_\mathrm{int})\right], $$ the integral over $\mathrm dt$ is actually tilted in the complex plane, $$ \int\mathrm dt\equiv\lim_{T\to(1-i\epsilon)\infty}\int_{-T}^{+T}\mathrm dt $$

This is done to pick up the right boundary conditions (i.e., the vacuum state). See for example Peskin and Schroeder's book, page 284, the discussion below (9.17).

Therefore, in the real-time Minkowski formulation, a quadratic term with the wrong sign does also lead to a divergent integral. The divergence is much more transparent in the Euclidean formulation (which is in fact one of the main advantages of such a formulation: convergence properties are usually much more clear).

In this sense, the theory with negative $\lambda$ is intrinsically ill-defined, irrespective of whether you use the Minkowski or the Euclidean path integral. The ultimate reason is that, as knzhou mentions in the comments, a potential $V(\phi)=\frac12m^2\phi^2-\lambda\phi^4$ is unbounded from below, and hence $|0\rangle$ doesn't even exist to begin with. The theory is unstable (even though this is mostly invisible to perturbation theory: the Feynman rules are oblivious to the sign of $\lambda$).

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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