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In my QFT course, we are looking at writing the most general lorentz invariant, renormalizable lagrangians with hermitian interactions.

However I have never seen the interaction in the title mentioned, where $\phi$ is a complex scalar field. To me it seems this would fulfill all the criteria: in 4 dimensions, it has mass dimension 3, so the theory would be super renormalizable. Is there a reason this is not allowed?

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    $\begingroup$ If $\phi^\dagger \phi (\phi + \phi^\dagger)$ is allowed, you are forfeiting particle conservation (global $U(1)$ symmetry). If so, why don't you consider $\phi + \phi^\dagger$ as well? If you go over to the dark side, go Darth Vader. $\endgroup$
    – MadMax
    Commented Feb 10, 2020 at 17:52
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    $\begingroup$ @MadMax $\phi+\phi^{\dagger}$ can be eliminated via the mass term by a shift $\phi\to \phi+\phi_0$. If you wanna go full Darth Vader you can add $(\phi^{\dagger})^2+\phi^2$ $\endgroup$
    – user245141
    Commented Feb 10, 2020 at 18:09
  • $\begingroup$ @yu-v, right, and $(\phi^{\dagger})^3+\phi^3$ to boot. $\endgroup$
    – MadMax
    Commented Feb 10, 2020 at 18:34

2 Answers 2

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it is odd under $\phi \to -\phi$ and therefore the energy will be unbounded from below, which is a requirement

Edit following discussion in comments: We demand that our theories will have an energy spectrum bounded from below. Taking the proposed term to be the only term in the potential energy will violate this, as it is odd under $\phi\to -\phi$. If we combine this with the knowledge that $\langle E \rangle \geq E_{\rm{gs}}$ for any state, where $E_{\rm{gs}}$ is the ground state energy, we see that if this is the highest-power term in our Lagrangian will lead to a configuration that will allow $\langle E \rangle \to -\infty$ (we will just push the VEV of $\phi$ to $\infty$ or $-\infty$).

However, if we have higher-power term with a finite positive coefficient, let's say $(\phi^{\dagger}\phi)^2$, then the energy can be bounded again from below.

One cal also consider such a term in the context of perturbation theory and calculate its effects and contributions, while keeping in mind that the underlying theory could be problematic without a necessary term that will prevent these divergences.

From symmetries point of views, as it was commented, this term violates the $U(1)$ symmetry for a complex field, which is something we would like to keep. But this in itself doesn't make such a term invalid. If you don't mind violating this symmetry, one can add other renormalizable terms (from dimension counting perspective) such as $\phi^2+\rm{h.c.}$.

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  • $\begingroup$ unsure this follows--you can have phi3 in 6D $\endgroup$
    – user21299
    Commented Feb 10, 2020 at 17:13
  • $\begingroup$ 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? $\endgroup$
    – don't train ai on me
    Commented Feb 10, 2020 at 17:13
  • $\begingroup$ Given @alexarvanitakis 's example, it would also be nice to have a source for this statement. $\endgroup$
    – don't train ai on me
    Commented Feb 10, 2020 at 17:17
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    $\begingroup$ 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. $\endgroup$
    – Seth Whitsitt
    Commented Feb 10, 2020 at 17:20
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    $\begingroup$ 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? $\endgroup$
    – don't train ai on me
    Commented Feb 10, 2020 at 17:27
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It seems like it's allowed to me. But this term precludes a straightforward coupling to a $U(1)$ gauge field, and we usually want to do that: under $\phi\to \exp(i\alpha)\phi$, this term is not invariant.

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    $\begingroup$ OP’s question doesn’t mention a gauge field. $\endgroup$
    – Prof. Legolasov
    Commented Feb 10, 2020 at 17:37
  • $\begingroup$ Yes, but QFT courses usually do $\endgroup$
    – user21299
    Commented Feb 10, 2020 at 17:38
  • $\begingroup$ sorry but isn’t that irrelevant (no pun intended)? This is a Q-A website. The question was about the possibility of having such an interaction term, not about coupling it to a gauge field $\endgroup$
    – Prof. Legolasov
    Commented Feb 10, 2020 at 17:39
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    $\begingroup$ I did say it's allowed, right? $\endgroup$
    – user21299
    Commented Feb 10, 2020 at 17:44

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