# Is there a way to justify perturbation theory in QFT?

I can accept that some problems do require a perturbation theory approach to be solved. The development of perturbation theory is perfectly rigorous. So at first, even though many people say that no one knows how to tackle QFT in an exact fashion needing to resort to perturbation theory, I didn't see this as a problem.

It turns out that recently I've watched a QFT lecture on youtube where the lecturer told that in QFT the interaction Hamiltonian has infinite norm, and hence the Dyson series diverges.

Now wait a minute. This invalidates perturbation theory completely.

Even in free QFT some infinities appear as divergent deltas, but with some sloopy plausibility arguments they are thrown out. This is already non rigorous and a problem, however, it can even be accepted.

Now trying to use perturbation theory in a scenario where it is really proved it doesn't work and certainly diverges is a whole different story. From that point onwards one is not doing mathematics anymore, just manipulating symbols according to a different set of rules than those of mathematics, because if one goes on with the series diverging and the method failing, nothing that point forward has any concrete meaning.

One might say "well, it matches experiment so we better not care about that", but from a theoretical point of view I think this really demans attention. I mean it is already a miracle that thing matches experiment, but nonetheless the theory is nonsense! How can a theorist be satisfied with such a thing?

My point here is: is there any known way to deal with that, even if not mainstream? How can use a method in a situation it is proven to diverge? I mean, QFT has a serious problem, there must be some work on this, after all the theory is taken seriously nonetheless. How can this theory be taken serious with an immense issue such as that?

EDIT: I believe the post needed some more precise statements. First thing: I might have had the wrong understanding of the discussion about the convergence of the series and the implications. Second, I'm taking a QFT course and nothing has been discussed about justification for the perturbative method insofar. Third, what I know from perturbation theory from non-relativistic QM, is the following: we have a hamiltonian $H_0$ which we know the eigenstates and eigenvalues, and the full hamiltonian is

$$H =H_0+V,$$

we then write $V = \lambda W$ where this $W$ is small, and $\lambda$ is a parameter characterizing the problem. In that setting, we can write down a solution in terms of a power series in $\lambda$. The convergence of the series is tied to the fact that $W$ is small. Now, so long as we have a convergent power series in $\lambda$, we can interpret cutting the series on a certain power $n$ of $\lambda$ as one approximate solution for when $\lambda$ is such that $\lambda^{k}$, for $k> n$ can be neglected.

This is true, because since $\lambda^k$ can be neglected for $k > n$, the following terms of the series are so small that we can neglect them and approximate the solution. So the conclusion: the exact solution would be obtained by the full series, but since we don't know how to compute it, we are able to get approximate solutions depending on the magnitude of $\lambda$, considering that $W$ is small and the following terms of the series will be small.

What I've heard is: this $W$ in QFT usually has infinite norm $\|W\|_{\infty}=\infty$. It certainly isn't small, and these arguments as above would fail. The series wouldn't converge and furthermore, we can't just neglected the further terms saying they are negligible. This discussion isn't present in most QFT textbooks I've seem, at least in the chapters I've read. Most of them just present the series without discussing this issue, which as I said, I found out watching some QFT lectures.

• Two comments: (1) asymptotic series, which diverge, can be dealt with perfectly rigorously in mathematics (although we don't know how to handle the ones in QFT). (2) physicists have never cared about being perfectly rigorous. Apr 16, 2017 at 17:25
• Possible duplicates: physics.stackexchange.com/q/27665/2451 , physics.stackexchange.com/q/6530/2451 and links therein. Apr 16, 2017 at 17:58
• Well, my point of view is that if the theory gives such good results, there must be some way of justifying it, even if we don't know it and don't particularly care. Also see this nice paper. Apr 16, 2017 at 18:27
• I believe there are two ways of being non rigorous. The first is using a non rigorous approach because the rigorous one is tedious. Manipulating differentials as infinitesimals and using Dirac's formalism in QM I consider to fall under this and I agree there is nothing wrong with it. We are not doing anything wrong, but just using a shortcut to avoid a tedious approach. The other is picking something proven not to work and pretend it is fine to manipulate it freely. Applying a result out of the validity hypothesis is not just lack of rigor, is just plain wrong IMHO, that's why I'm worried.
– Gold
Apr 16, 2017 at 19:18
• @user1620696, humans are divided into two categories. Those who say that your question is important and those who shove one century's value of open questions in fundamental physics under the carpet by claiming that "this does not matter" or "this cannot be given an answer" or "this is just a tool". John Bell's ghost will haunt all in the second category :p Apr 18, 2017 at 8:54

There exist a large body of literature on the topic of divergent asymptotic series. This article gives an overview of the theory from a practical point of view. This article focuses on methods that can be applied to asymptotic series of which all the terms are known, or at least the coefficients of the late terms are known to some leading approximation. In QFT one typically has just a few terms of a perturbation expansion, but even in that case one can apply certain mathematical methods to resum the series.

So, a typical problem is then that given a perturbative expansion of some function $f(g)$ in powers of some coupling $g$ we want to know the behavior of $f(g)$ for large $g$, but for large $g$ the series starts to diverge really fast, and we only have a few terms. This is not as hopeless as it looks, let's consider the following example. The logarithm of the factorial function $\log(n!)$ has the following asymptotic expansion for large $n$:

$$\log(n!) = n\log(n) - n + \frac{1}{2}\log(2\pi n) + \frac{1}{12 n} - \frac{1}{360 n^3} + \mathcal{O(n^{-3})}$$

Suppose that these are all the terms that we know about. We want to extract the behavior of the factorial function near $n = 0$ using only the given terms. We obviously cannot set $n = 0$, the individual terms will already diverge. But what we can do is to extrapolate to $n = 0$ using only larger values for $n$ where the series does make sense. E.g. there is no problem with inserting even not so large values like $n = 1$ in the series, moreover you can put $n = 1+\epsilon$, expand in powers of $\epsilon$ and then set $\epsilon = -1$ to jump to $n = 0$. The validity of such methods then depend on assuming that the function you are dealing with is analytic in a neighborhood of $n = 0$, and the fact that you have to use a divergent series around infinity to get there doesn't make the approximations invalid.

A more sophisticated way to get to the behavior near $n = 0$ is to apply a conformal transform to the expansion parameter. If we put:

$$n = \frac{1-z}{p z}$$

and expand in powers of $z$, we get:

$$\begin{split} f(z) =& \frac{1}{2}\log\left(\frac{2\pi}{p}\right)+\frac{\log(p)}{p} +\frac{1-\log(p)}{p z}+\left(\frac{1}{p}-\frac{1}{2}\right)\log(z) -\frac{\log(z)}{p z} +\\&\left(\frac{p}{12}+\frac{1}{2 p}-\frac{1}{2}\right)z + \left(\frac{p}{12}+\frac{1}{6 p}-\frac{1}{4}\right)z^2 + \left(-\frac{p^3}{360}+\frac{p}{12}+\frac{1}{12 p}-\frac{1}{6}\right) z^3+\mathcal{O}(z^4) \end{split}$$

There is now no problem with inserting $z=1$ in the series which corresponds to $n = 0$. The result will depend on the auxiliary parameter $p$, the optimal choice of this parameter is to choose it in such a way that the series has the best convergence. A good choice is obtained by setting the coefficient of the last term equal to zero, and then evaluating the coefficient of the coefficient of the previous term to see which solution makes that coefficient the smallest. This yields good results because typically the error is of the order of the last omitted term. Another interpretation is that the answer doesn't depend on $p$, but now you do have a $p$ dependence due to only having a finite number of terms. If you then make the last terms as small as possible, you're going to get a result that for other values of $p$ would have had to come from higher order terms. It's also worthwhile to consider the earlier terms to see if a particular value yields a more robust series.

In this case you find that $p = 0.863778859012849315368798819337$ looks to be the best value to use, it would be the the second choice when considering the value of the coefficient of $z^2$, but the other solution that makes this coefficient the smallest yields a series with a coefficient that is slightly increasing before becoming zero for the $z^3$ term.

So, if we then put $p = 0.863778859012849315368798819337$, we can put $z = 1$ to estimate $0!$, but we can do more. We can expand our function of $z$ around $z = 1$, by putting $z = 1-u$ and expanding in powers of $u$. If we then also expand $n$ in powers of $u$, we can invert the series to express $u$ in powers of $n$, so we then get to a series of $\log(n!)$ in powers of $n$. The result is:

$$\log(n!)= 0.0002197 - 0.577755 n + 0.823213 n^2 - 0.401347 n^3 +\cdots$$

The exact expansion with the coefficients to 8 significant figures is:

$$\log(n!) = -0.57721566 n + 0.82246703 n^2-0.40068563 n^3+\cdots$$

Clearly, a great deal of information can be extracted from divergent series, even if you just have a few terms. Mathematical rigor should be used to your advantage to extract as much information as possible, you should not be intimidated by mathematical rigor pointing to roadblocks.

I think you completely misinterpret the notion of divergence of a perturbation series. Suppose the result is known up to the second order, so one has something like $f\left(\epsilon\right)=1+a\epsilon+b\epsilon^{2}+\dots$ for the solution when $\epsilon\ll 1$. That's fine, and once you calculated the constants $a$ and $b$, you usually see that $b\ll a$, so you believe you got a nice perturbative expansion.

What you can prove with the perturbation expansion is that for some points, you will get a next-order term larger than the previous one, say the third one $c\gg b$. The full series will diverge, yet a solution with only $a$ and $b$ are good approximation of the solution.

This lacks rigor, but you have to know that there is no converging series in interesting problems of physics... all problems of interest are plagued with this problem. It was discovered by Poincaré at the beginning of the 20-th century when he tried to resolve the 3-body problem (say the motion of the moon cycling around the earth cycling around the sun).

Usually mathematicians would have written the solution up to the third order

$$f\left(\epsilon\right)=1+a\epsilon+b\epsilon^{2}+\mathcal{O}\left(\epsilon^{3}\right)$$

because they know that the next order is a small quantity. They have tools to evaluate all higher orders (usually an integral form of the complete solution when the problem is a differential equation for example) and to compare it with the first terms. Physicists write instead

$$f\left(\epsilon\right)=1+a\epsilon+b\epsilon^{2}+\cdots$$ because they usually have no idea when the perturbation theory will break up.

Mathematicians proved that $\mathcal{O}\left(\epsilon^{3}\right)$ (to the third order for our example here) is in fact large, and may even diverge for some physical applications. And so what ? Physicists believe that one simply has to forget these higher orders terms, since higher orders terms may well be controlled by more refined theory, taking into account extra interactions which were neglected in the diverging model. In any case, everyone is somehow correct, since no-one knows how the complete model of the entire universe looks like ...

So at the heart of perturbation theory lies the very profound method of physics: one takes a model, one calculates some outcomes, one confronts with experiments... again and again.

• where does renormalization enter in this argumetn? I thought it was invented in order to avoid the infinities. Is it a different story? see physics.umd.edu/courses/Phys851/Luty/notes/renorm.pdf Apr 18, 2017 at 8:48
• IMHO, this answer completely neglects the nature of asymptotic series, which are the core of the question. Also, the claims about "physicists being unable to do things rigorously" is completely unjustified. Apr 18, 2017 at 8:50
• @annav Indeed, I completely forget renormalisation, thanks for pointing this out. It's a shame indeed. Feel free to complete my answer. I had the feeling my answer gently introduces the concept of perturbation series. It seems the OP never opened a book on QFT so I just gave handwaving arguments. Apr 18, 2017 at 9:11
• @AccidentalFourierTransform Indeed, I didn't wanted to enter too much details, feel free to post your own answer to complete mine. I had the feeling my answer gently introduces the concept of perturbation series. It seems the OP never opened a book on QFT so I just gave handwaving arguments. Apr 18, 2017 at 9:12