0
$\begingroup$

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?

| cite | improve this question | | | | |
$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

| cite | improve this answer | | | | |
$\endgroup$

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.