Timeline for Is $\phi^\dagger \phi (\phi + \phi^\dagger)$ a possible interaction for scalar fields in QFT?
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| Feb 11, 2020 at 16:59 | comment | added | user21299 | @Prof.Legolasov: Yes indeed one has to check stability carefully. I didn't mention that here because it seems tricky even perturbatively. Clearly there's issues at tree level if the OP's term is the only term present, but a $\phi^3$ interaction appears to generate $\phi^4$ terms at e.g. 1-loop which will have a stabilising effect. The only reference I can find that calculates that is one I can't read: osti.gov/etdeweb/biblio/6551029 | |
| Feb 11, 2020 at 14:58 | vote | accept | don't train ai on me | ||
| Feb 11, 2020 at 14:51 | comment | added | user245141 | @doublefelix done | |
| Feb 11, 2020 at 14:51 | history | edited | user245141 | CC BY-SA 4.0 |
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| Feb 11, 2020 at 14:26 | comment | added | don't train ai on me | @yu-v would you be willing to edit the answer to summarize the results from the discussion? I think it would make a nice contribution for anyone else who finds the question. Or else if you don't mind I might edit it myself. | |
| Feb 10, 2020 at 21:45 | review | Low quality answers | |||
| Feb 10, 2020 at 22:12 | |||||
| Feb 10, 2020 at 17:43 | comment | added | Prof. Legolasov | @alexarvanitakis boundedness from below of the Hamiltonian isn’t a consistency requirement for perturbative expansions, but it is for a constructive QFT model. It is usually taken to be an extra requirement that we want a perturbative expansion to satisfy in order to be a description of physics. | |
| Feb 10, 2020 at 17:35 | comment | added | user21299 | @Prof.Legolasov I can't name a corresponding constructive QFT model (but I don't really follow that literature anyway). To clarify my above comment: phi3 in 6d is power-counting renormalisable (unless I just power-counted wrong) and generally healthy perturbatively to my knowledge. As always one has to check stability etc. carefully. But you generally cannot trust blanket statements like "odd interactions break your theory": for example you can make any theory cubic by introducing enough auxiliary fields. | |
| Feb 10, 2020 at 17:34 | comment | added | don't train ai on me | Got it, so you are imagining $\phi(x)^3$ as a classical potential (as opposed to operator), and remarking that it is unbounded from below, though this is remedied if a $\phi^4$ term is added. Thank you :) | |
| Feb 10, 2020 at 17:32 | comment | added | Seth Whitsitt | @doublefelix I suppose my thinking is that if part of your potential rises to $+\infty$, then if the potential is purely odd potential it will be unbounded from below by taking $\phi \rightarrow -\phi$. The more general issue is just that your potential cannot be unbounded from below, and that's easy enough to check for your example (or any example with a scalar field potential). | |
| Feb 10, 2020 at 17:27 | comment | added | don't train ai on me | To be clear.. you mean that a term may be odd under $\phi \to -\phi$, but the entire potential may not be, or else the hamiltonian is unbounded from below. Did I interpret that correctly? | |
| Feb 10, 2020 at 17:20 | comment | added | Seth Whitsitt | The potential does not need to be even under $\phi \rightarrow -\phi$, but if cannot be unbounded from below, and the fact that this is odd under that transformation implies that it is unbounded. If you added a term $(\phi^{\dagger} \phi)^2$ to your interaction term, the theory is fine, and the potential is neither odd nor even. | |
| Feb 10, 2020 at 17:18 | comment | added | Prof. Legolasov | @alexavarnitakis on the level of perturbation theory only, or is there a constructive QFT model? | |
| Feb 10, 2020 at 17:18 | comment | added | user245141 | following @alexarvanitakis comment I would be more hesitant, but I would say that if the Lagrangian has terms which can get a negative expectation value then the ground-state energy would tend to minus infinity (the energy spectrum will be unbounded from below). I can always choose a configuration $\phi$ such that $\langle E \rangle \to -\infty$ which implies unbounded energy. Don't know how it works in 6D | |
| Feb 10, 2020 at 17:17 | comment | added | don't train ai on me | Given @alexarvanitakis 's example, it would also be nice to have a source for this statement. | |
| Feb 10, 2020 at 17:13 | comment | added | don't train ai on me | Interesting, so what is the general statement here? Any scalar field (real or not) should be invariant under $\phi \to -\phi$? Is it easy to give an intuition why oddness in this sense would cause the energy to be unbounded from below? | |
| Feb 10, 2020 at 17:13 | comment | added | user21299 | unsure this follows--you can have phi3 in 6D | |
| Feb 10, 2020 at 17:11 | history | answered | user245141 | CC BY-SA 4.0 |