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Consider scattering some particles in a state collectively denoted by $i$ to a final state denote by $f$. The scattering amplitude, S-matrix is then defined by: $S_{fi}\equiv \langle f|e^{-iHt}|i\rangle$. We then separate the S-matrix into the identity and another part as $S_{fi}=\delta_{fi}+iT_{fi}$. The statement of unitarity is that $S^\dagger S=1$ which implies that $2{\rm Im}T=T^\dagger T$ which leads to the optical theorem and all that.

In field theory, what we calculate is the amplitude where we stick just $T$ between two states. That is, we only usually calculate the amplitudes where something interesting is happening.

In the study of effective field theories, I often see statements about the violation of unitarity which confuse me. For example, if we took a simple scalar field theory with a derivative interaction $\mathcal L=\frac{1}{2}(\partial\phi)^2+\lambda(\partial\phi)^4/\Lambda^4$ then we could calculate $2\to 2$ scattering and we'd find something like $\mathcal{M}_{2\to 2}\sim \lambda k^4/\Lambda^4$.

I've read and heard people say that for $k\gg \Lambda$, this leads to a violation of unitarity. I assume this means a violation of $2{\rm Im}T=T^\dagger T$. Why is this the case? Certainly the perturbative expansion breaks down in this regime but why is this connected to unitarity?

If the above example does violate unitarity, then what's the difference between the above and the case of a normal, non-derivative $\lambda\phi^4$ interaction and with $\lambda\gg 1$ the case above? The key thing seems to be that $\mathcal{M}$ gets really big in the derivative example, but this would also occur in $\lambda\phi^4$ and I would doubt that this latter theory has any violations of unitarity.

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We could say, that in the $2$ cases , we cannot make a perturbative treatment, because (for instance) the probability amplitude could have a modulus greater than one. And this would violate unitarity. – Trimok Dec 19 '13 at 10:45
So, I see that $|\mathcal{M}_{2\to 2}|^2=|T_{fi}|^2$ is greater than one, but I don't necessarily see that $2{\rm Im}T=T^\dagger T$ is violated. Is it obvious that this is true? – user26866 Dec 19 '13 at 17:25
As signaled by Luboš Motl, with the equation $T^\dagger T=i(T-T^\dagger)$, is "morally" equivalent to $|z|^2=2 Im (z)$. Now, writing $z= a+ib$, this gives : $a^2+b^2=2b$, so $b>0$. The equation could be written $(a-b)^2+2b(a-1)=0$, so $a<1$. The equation could also be written $a^2=b(2-b)$, so $b<2$. Finally, we have $|z|=\sqrt{a^2+b^2} \leq \sqrt 5$. So if $T$ "has" "values" "greater than" $\sqrt 5$, there is a contradiction. – Trimok Dec 20 '13 at 10:06

1 Answer 1

As Trimok said, the probability of scattering of some nicely focused packets will still go like the cross section and like $|{\mathcal M}|^2$. For bosons, there are no energy-dependent extra factors, so $|{\mathcal M}|^2$ itself has to be smaller than a number of order one for the probability to stay smaller than one.

This is related to $T^\dagger T=i(T-T^\dagger)$ because this equation implies inequalities for the scattering amplitudes, too. The mathematical essence of this statement is that for a complex variable, $|z|^2={\rm Im}(z)$ only has solutions is $|z|$ is smaller than a particular number of order one (which you may calculate). For too large values of $|z|$, like a million, it's clear that $|z|^2$ is vastly greater than ${\rm Im}(z)$ and the equation can't be satisfied. The inequality that $|z|$ has to be smaller than something is therefore morally equivalent to the statement that the probability never exceeds one and both propositions may be derived from the unitarity.

That's why "two to two" bosons' scattering amplitude that grows with the bosons' energy implies too fast a growth of the cross section, too fast a growth of the probability for some packets, and that violates the unitarity conditions because of the previous paragraph. Theories with derivative interactions break down at energies not too different from the Higgs mass.

But even if you deal with superficially renormalizable interactions, some cancellations are needed to preserve the unitarity. In fact, the classical electroweak Lagrangian may be derived from these conditions of "tree unitarity" combined with the basic beta-decay experimental data, see

Your argument that the growth doesn't matter because it's just like a large $\lambda\gg 1$ doesn't work because $\lambda \gg 1$ is impossible for the same reasons. A dimensionless coupling larger than one implies that the loop corrections are actually more important than the "leading", tree-level contribution. It turns out that the theory with a quartic interaction doesn't admit any consistent definition for $\lambda\gt \lambda_0 \sim O(1)$. That's why even for a low value of $\lambda$ which grows with energy due to the renormalization running, we ultimately encounter the "Landau pole" where the value of $\lambda$ grows too strong (and quickly infinite) and the theory becomes ill-defined. (The situation is different for gauge theories which may be consistent for a very large $g$, due to S-dualities etc.).

So a too large value of the scattering amplitude is always a problem.

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Thank you for the answer. So, for the $\lambda \phi^4$ case it appears to me that your statements about the theory being poorly defined are just due to issues in perturbation theory and hence they're more practical problems than fundamental ones, no? Do you think that even in principle it's impossible to solve $\lambda \phi^4$ for $\lambda\gg 1$? I could imagine that one day people would come up with a method for solving it outside of perturbation theory. – user26866 Dec 19 '13 at 17:20
Similarly, if you took some theory which is asymptotically free, say like the Gross-Neveu model, then you could imagine that below some energy scale $E$ the couplings at hand become larger than one. Would you then say that such a theory violates unitarity below energy $E$? – user26866 Dec 19 '13 at 17:23
Dear user, as I said, the theory at $\lambda\gg 1$ doesn't exist at all. There is no consistent theory at this coupling. This absence isn't an artifact of perturbation theory, it is an exact fact. It's like if you are asking about 10-meter tall people and suspect that we don't see them just because we don't have strong enough glasses for them. No, you should trust your eyes (perturbation theory), the 10-meter-tall people really don't exist. – Luboš Motl Dec 20 '13 at 15:01
Your wording is completely wrong - it seems that you don't want to understand the elementary words that I am saying - I believe - totally clearly. It is not that we can't "solve" the theory at $\lambda\gg 1$. The quantum theory doesn't exist at all so there is nothing to solve. It's like prime integers smaller than $-5$. It's not a matter of technology how to find them. Whatever you do, you won't find them. On the contrary, your illusion that the theory should exist for any lambda is a wrong artifact of your perturbative or classical reasoning! – Luboš Motl Dec 20 '13 at 15:04
Ok, and what about the Gross-Neveu example? – user26866 Dec 20 '13 at 15:26

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