Consider the 4-dimensional Lagrangian density with two real scalar fields $\phi_1$ and $\phi_2$: $$\mathcal{L}=\frac{1}{2}\partial_{\mu}\phi_1 \partial^{\mu}\phi_1 + \frac{1}{2}\partial_{\mu}\phi_2 \partial^{\mu}\phi_2 + \frac{1}{2}m^2\phi_1^2 - \lambda_1\phi_1^4-\lambda_2 \phi_1^2\phi_2^2-\lambda_3\phi_2^4$$ A question asks: Taking $m,\lambda_1,\lambda_3$ to be positive, what constraint must $\lambda_2$ satisfy in order to have a reasonable theory?

The only thing I can think of is that we must have $\lambda_2 > -\lambda_1-\lambda_3$ because if we went to Euclidean signature, in order for the 'path integral' (generating functional) $$\int D\phi_1\phi_2 \; \exp\left(-\frac{1}{2}\partial_{\mu}\phi_1 \partial^{\mu}\phi_1 -\frac{1}{2}\partial_{\mu}\phi_2 \partial^{\mu}\phi_2 + \frac{1}{2}m^2\phi_1^2 - \lambda_1\phi_1^4-\lambda_2 \phi_1^2\phi_2^2-\lambda_3\phi_2^4 \right)$$ to 'converge' we need the exponential to become negative as we 'integrate the fields to infinity'. In this case, as we 'send $\phi_1$ and $\phi_2$ to infinity' we need $\lambda_2 > -\lambda_1-\lambda_3$ for it to make sense.

I feel like this is not the whole story though and that I am missing a further constraint on $\lambda_2$ (perhaps on the relative sizes of $\lambda_2$ and $\lambda_1$). Does someone see anything else?

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  • $\begingroup$ Look up 'positive definite quadratic form'. You only checked that it's positive when $\phi_1 = \phi_2$, you need to check every direction. $\endgroup$ – knzhou Jun 30 '16 at 22:58
  • $\begingroup$ Not sure I understand. In the question $\lambda_1$ and $\lambda_3$ are taken to be positive. So if either $\phi_1 \rightarrow \infty$ or $\phi_2 \rightarrow \infty$ is 'faster' then I am guaranteed convergence. The only direction I am left to check (as far as I can see) then is precisely $\phi_1 = \phi_2$. Am I wrong? $\endgroup$ – Rudyard Jun 30 '16 at 23:20
  • $\begingroup$ This isn't a QFT question. Consider $20x^2 + 10y^2 - 29xy$. It's positive for $x \to \infty$, $y\to \infty$, and when $x = y$. Does that automatically mean it's positive everywhere in the plane? $\endgroup$ – knzhou Jun 30 '16 at 23:30
  • $\begingroup$ But it doesn't have to be positive everywhere (unless I'm mistaken). It just has to be positive 'at infinity' for the integral to converge. Using your example: $$\int dxdy \; exp \left( -20x^2 -10y^2+29xy \right)$$ converges. I put it as a qft question because I suspected there were different reasons to have constraints on $\lambda_2$ that I couldn't see. (thank you for replying btw) $\endgroup$ – Rudyard Jun 30 '16 at 23:36
  • $\begingroup$ Did you try doing the integral? It doesn't converge. It's not generally positive at infinity. You're only "going to infinity" in 3 different directions, while there are a continuum of possibilities. $\endgroup$ – knzhou Jun 30 '16 at 23:44

I consider the potential (slightly modifying your notation for the sake of clarity):

$$V= m_1^2\,\phi_1^2+m_2^2\,\phi_1^2 + \tfrac{1}{2}\lambda_1\,\phi_1^4+\tfrac{1}{2}\lambda_2\, \phi_2^4+\lambda_{12}\,\phi_1^2\phi_2^2\,\,,$$

then the conditions for boundedness from below are: \begin{eqnarray} \lambda_{1} &\geq& 0 \\ \lambda_{2} &\geq& 0 \\ \lambda_{12} &\geq& - \sqrt{\lambda_1 \lambda_2} \end{eqnarray}

These are derived quickly using the co-positivity of the matrix of the quartic couplings criterion (c.f., e.g. 1205.3781 and refs therein).

Note that if $m_{1}$ or $m_2$ can be negative then other constraints on the quartics should be derived to keep the masses positive.

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