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Vacuum stability in quantum field theory

+ 7 like - 0 dislike
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What exactly do people mean when they talk about the scale dependence of the effective potential ($V$)? I explain the motivation for my question (and hence my confusion) below. Please correct me as appropriate.

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If one defines the effective potential as the non-derivative part ($p^2 \rightarrow 0$) of the effective action, then something like the Callan-Symanzik equation will imply that
$$\frac{d}{d \log \mu} V = 0 \implies \left[ \frac{\partial}{\partial \log \mu} + \beta_i \frac{\partial}{\partial \lambda_i} + \gamma \frac{\partial}{\partial \log \phi} \right] V = 0$$ where we have removed the vacuum energy $V(\phi=0)$ from $V$, for this to hold.

If the value of the effective potential at any field value (and zero momentum) is scale independent, then shouldn't the vacuum be stable at all scales, if we know it's stable at some scale?

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All that I've said above holds so long as perturbation theory has been used correctly.

I've also seen some people "RG-improve" the effective potential by resumming the leading logs across all loop order. But then, the moment you do a partial resummation across loop orders, is there any reason for the thing you calculate to be scale independent? It's also not clear what the physical interpretation of such a scale-dependent quantity should be -- so why should one take the appearance of another vacuum seriously -- after all, you don't see any such thing when you do your calculations at some "low" scale -- and physical observables (at zero momentum) better not be RG-scale dependent?

asked Sep 5, 2014 in Theoretical Physics by Siva (710 points) [ revision history ]
edited Sep 17, 2014 by Siva

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