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pontus | 2 years ago
There are then two natural questions: 1. if the perturbation series diverges, why doesn't the universe explode? and 2. If the series diverges, why can we use it at all?
Let's first talk about the first part: why doesn't the universe explode? Well, it's because the perturbation series is not actually what is going on, the real answer is the solution to the full set of equations. It's just that we're using a perturbation expansion as a crutch. It's sort of like if the universe's function is 1/(1-x) but we constantly insist on using 1+x+x^2+... Clearly the first function is completely well behaved at x=2 but the second one is not. If we notice that our series explodes for x=2 we should not immediately assume that the universe also must explode, it's just that our representation of the true physics is not faithful. This is perhaps a bad example because the series in question is convergent for some x, just not for x=2. The perturbation expansions in question are more subtle since the never converge.
This then leads into the second question: if the series diverges, how can we even use it? Well the idea here is that it's not just any divergent series (like my silly example with 1/(1-x) above) but rather an asymptomatic series. This means that as long as you truncate the series at some point it is in fact reasonably close to the target function for a sufficiently small value of the parameter. It's just that the more terms you want to include, the sooner the approximation breaks in terms of the parameter. So, if you want to include 10 terms it might be a decent approximation until x ~0.1 but if you include 100 terms it might only be a good approximation until x~0.01. Now, within the overlapping range (x<0.01) it's better to have 100 terms than 10 terms, so it's not like including more terms is bad in all ways. But you see the issue: if you include 1000 terms you get a better approximation for your function for values x<0.001 than you had with 100 terms but now your approximation breaks much sooner. If you want to include all the terms your approximation breaks the moment you leave the point x=0.
Why do we think that QFT perturbation theories generally have zero radius of convergence? Well, look at QED, the quantum theory of E&M. If the theory had any nonzero radius of convergence, that also means that the theory would need to make sense for negative coupling constants. However, what would E&M look like for negative coupling? Well, we'd still have electron/positron virtual pair creation from the vacuum since the interactions of the theory are still the same. However this time around they wouldn't attract each other anymore but instead repel each other causing an instability in the vacuum of the theory. We would just constantly be producing these particle/anti-particle pairs and they'd form two separate clusters where all the electrons attract each other and all the positions attract each other but they pairwise repel. In other words, the vacuum would break. This suggests that QED with a negative coupling constants doesn't make sense. But this contradicts the fact that the radius of convergence of the perturbative expansion is nonzero.
That's not to say that all QFTs must have zero radius of convergence, but similar arguments can (I think) be made for the type of QFTs that we actually see in nature.
codethief|2 years ago