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Why the following interaction, in QFT, $$\displaystyle{\cal L}_{\rm int} ~=~\frac{\lambda}{4!}\phi^4$$ where $\lambda$ is positive, represents a theory that is unstable (or unbounded from below as it is usually said in textbooks).

How can can one show that explicitly?

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See e.g. A. Zee, QFT in a Nutshell, p.174. –  Qmechanic Jul 31 '12 at 10:04
    
@Qmechanic I do not understand what Zee means exactly. He simply says one sign means repulsion, an opposite sign means attraction. Do we consider repulsion an unstable solution? why so? –  Revo Jul 31 '12 at 21:12
    
Please note that the coupling constant $\lambda$ in OP's question (v2) is opposite of what is used in Zee's book and drake's answer (v2). –  Qmechanic Jul 31 '12 at 21:30
    
Revo: No, it is attraction that corresponds to an unstable situation. –  Qmechanic Jul 31 '12 at 21:33
    
@Qmechanic OK, thanks. So my question is why the unstable solution is problematic (regardless of whether it means repulsion or attraction, since repulsion or attraction each is physical. What is wrong with either one? aren't both equally physical?) –  Revo Jul 31 '12 at 21:36

2 Answers 2

The reason attractive $\lambda \phi^4$ is unphysical is because a sufficient density of the $\phi$ particles has a self interaction which compensates for their mass-energy, so it is less energy to make a condensate of particles with a large density than to leave the vacuum alone.

This means that the vacuum will spontaneously decay by a monstrous explosion in a bubble into a state where the field is rolling off to plus or minus infinity. To see this, you can consider the energy of the classical field state

$$ \phi = C $$

which is

$$ a C^2 - \lambda C^4 $$

and is unbouded below. The problem with turning this rigorous (although it is completely persuasive) is that it is hard to write down wave-functions for quantum fields which are finite energy density. You usually use the path-integral to define this. While it is physically obvious that a wavefunction for the field $\phi$ which is peaked at a large constant value will have arbitrarily negative energy, constructing such a thing is a nightmare, because you need to control the short-distance correlations in the wavefunction to make sure that they don't have infinite energy in he ultraviolet, which is a pain.

But there is a simple way around this which is how everyone analyzes vacuum stability today, after Coleman. Use the path integral to show that there is an instanton which leads to vacuum decay. In this case, the Euclidean action is

$$ |\partial_t \phi|^2 + |\nabla \phi| + a \phi^2 - \lambda \phi^4 $$

You then note that this can be viewed as a classical system with a potential

$$ V(\phi) = -|\nabla\phi|^2 - a \phi^2 + \lambda \phi^4 $$

and the classical equations of motion for this thing has a closed zero energy solution where $\phi$ oscillates to a big value in a region, until it hits the $\lambda\phi^4$ wall, and comes back. The contribution of the instanton is to give a rate for nucleation out of the $\phi=0$ vacuum, in a way calculated in detail in Coleman's "Aspects of Symmetry". The essential point is that fluctuations around the zero energy solution have one negative eigenvalue, so a negative determinant, so that the square root of the determinant has an imaginary part, leading to slow oscillatory behavior in imaginary time, which is a decay rate in real time.

But I will use it in a much easier way to argue that the theory doesn't have a stable vacuum. Supposing the vacuum is stable, then the vacuum wavefunction is the probability of finding a given $\phi$ configuration in any constant time-slice in imaginary time, using the imaginary time action. This is a well-known path-integral relation.

But the imaginary time probability distribution for field values is of the form $e^{- S}$ where S is unbounded below! So the field $\phi$ has no ground state wavefunction which is normalized in the usual sense. I gave this answer, even though Drake's was sufficient, because you don't seem convinced.

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I think it is in general better to argue from increasing of entropy rather than from decreasing of energy since energy is conserved because of time translational invariance. I know this is a small thing but sometimes brings confusion. –  drake Aug 4 '12 at 6:04
    
@drake: I agree, perhaps that wasn't clear in the first parts of the answer. But the instanton conserves energy, dumping more and more heat into a bubble-wall moving out, so you are right. The argument I gave is from non-normalizability of the Euclidean probability distribution, which contradicts the well-definedness of the Euclidean theory, and this is doesn't involve dynamics, it just says "no vacuum here". –  Ron Maimon Aug 4 '12 at 6:30

$\cal H \sim \frac{\lambda}{4!} \phi ^4 $ (note that this term goes with the opposite sign in the Lagrangian). $\lambda$ has to be real because of unitarity and has to be positive because of vacuum-stability or, equivalently, since the Hamiltonian must be bounded from bellow. If $\lambda$ were negative, the larger the value of $\phi$, the more negative the Hamiltonian and therefore non-vacuum state (or ground state) could exist.

Now, at the quantum level, even if $\lambda$ is positive at the classical level the vacuum can be unstable or metastable. This can happen if the renormalization group drives the classical positive value to a negative one. To know if this is the case you have to know the complete theory. For instance, the measured value of the quartic self-coupling of the Higgs field leads to a non-stable vacuum at high enough energies. See this: Measured Higgs mass and vacuum stability

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In pure $\phi^4$ theory, the RG flow can't cross zero, at this point the theory becomes noninteracting, and the RG stops. In the standard model, there are other interactions to keep the Higgs coupling flowing. The wrong-sign theory, the unstable one, is asymptotically free (this was discovered by Symanzik around 1970 and motivated him to use this theory as a model for deep-inelastic scattering). –  Ron Maimon Jul 31 '12 at 7:12
    
@RonMaimon "Unstable" in what sense?unstable in the sense described by drake above? this is the point that is not clear to me. They say it is unstable, in whatever sense fine, but then Symanzik has used it to build a physical model as you said. Instability was not a problem then ?! In this paper, page L12, the author indicates that saying the theory is unstable is not justified cfif.ist.utl.pt/~kleefeld/public_html/hep_th_0506142_final.pdf –  Revo Jul 31 '12 at 21:05
    
@RonMaimon "The statements of Gross, Wilczek, and Politzer (which are reflected in the reasoning of [6, 7] and unfortunately shared by a great majority of contemporary scientists due to the way theoretical physics is presently taught in textbooks), which were interestingly written after the publication of Symanzik’s manuscript [23], made use of the here not applicable assumption of an underlying Hermitian quantum field theory and were obviously more guided by intuition rather than a rigorous proof." –  Revo Jul 31 '12 at 21:06
    
@Revo: The instability is real, but hard to demonstrate rigorously because of the difficulty with wavefunctions of quantum fields. I will provide a proof. Symanzik was doing perturbation theory, and was always aware that the model was unstable, but he knew something like this was responsible for deep-inelastic, so he studied it anyway. –  Ron Maimon Aug 4 '12 at 4:33

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