10
$\begingroup$

I have heard assertions that in QFT, the Higgs mechanism and Yukawa Couplings are calculated "at tree level" making the effects only classical in nature, not quantum. What does this mean, and is it accurate?

$\endgroup$

3 Answers 3

15
$\begingroup$

The question has two parts: (1) What does "tree level" mean, and why are tree-level effects said to be classical in nature? (2) Is the Higgs mechanism classical in nature?

Tree level $\to$ classical

Solving linear differential equations is easy. Solving nonlinear differential equations is hard, usually too hard. If the nonlinearities are small enough, then we can start with a solution of the linear version and add successively-refined corrections to account for the nonlinear part. That technique is called perturbation theory. The word "theory" here doesn't mean any hypothesis about how nature works. It's just a technique for doing calculations.

In field theory, both classical and quantum, perturbation theory can lead to cumbersome mathematical expressions. We use a graphical notation called Feynman diagrams to summarize those expressions. In quantum field theory, Feynman diagrams can have loops. In classical field theory, they can't. A diagram without any loops is called a tree. Even if we're using quantum field theory, we often use an approximation that only includes tree diagrams. Since those diagrams can also be generated by a classical field theory, we can call this approximation "tree level" or "classical."

Two caveats:

  • A classical approximation can also include nonperturbative effects — effects that are not visible in perturbation theory at all, no matter how far we iterate it. Example: instantons.

  • The kind of "classical" field theory that we would need to use to reproduce the Standard Model's tree diagrams still includes some ingredients — like fermions — that we would not ordinarily include in classical field theory. It's called classical (in the tree-diagram sense) because all of its observables commute with each other, even if it involves some otherwise distinctly quantum ingredients.

Is the Higgs mechanism a classical effect?

I'll answer a different question first, because it's easier: Is the Higgs mechanism a tree level effect?

The answer to that question is no. The Higgs mechanism rearranges the vacuum state compared to what the vacuum state would be if interactions between the Higgs field and gauge field were absent. Determining the vacuum state is the archetypical nonperturbative problem: it is extremely hard, usually too hard, even when the nonlinearities are very small. Perturbation theory does not work for determing the vacuum state. Sometimes, we can work around that obstacle using things like experience, intuition, and lots and lots of consistency-checks. That's how we "know" that the Higgs field in the Standard Model leads to the Higgs mechanism.

Once we know that the Higgs mechanism occurs, then we can use perturbation theory to study many other processes. In this case, we're using perturbation theory to add successive refinements to a modified ansatz that already accounts for the Higgs mechanism. We can use this modified perturbation theory to do systematic calculations for many different scattering processes, and the tree-level approximation is often good enough. But this only works because we already know about the Higgs mechanism, and we know that only thanks to nonperturbative insights.

Now, back to the original question: Is the Higgs mechanism a classical effect? Well, we can use intuition from classical field theory to anticipate the Higgs mechanism, which is exactly how most textbooks introduce the subject. That intuition is really just an analogy, not a true derivation, but maybe that's enough of a reason to justify saying that the Higgs mechanism is a "classical" thing. I would say it isn't, but I would say that without much conviction, because I rely on the classical intuition/analogy just as much as everybody else does.

$\endgroup$
4
  • 4
    $\begingroup$ Suppose you have an Abelian Higgs theory with a negative mass term and you have a large parameter multiplying your action (like large N or $\hbar^{-1}$). You can do an expansion in that parameter and look at the lowest order which is classical. Since you would see spontaneous symmetry breaking at lowest order, it's classical. $\endgroup$
    – octonion
    Mar 29, 2021 at 5:03
  • $\begingroup$ @octonion That's a good argument. Thinking of the Abelian Higgs model on a lattice, where the "bare" parameters are finite, having a Higgs potential with a double-well shape doesn't necessarily produce the Higgs mechanism unless the wells are sufficiently deep. That, and the fact that the Higgs phase can be analytically connected to the confinement phase, made me reluctant to say it's "classical." But there are counter-arguments to both of those things, and your large-$N$ argument is a very good one. $\endgroup$ Mar 29, 2021 at 12:50
  • $\begingroup$ My point is mostly that this is about saddle points. The larger that $N$ or $\hbar^{-1}$ gets the better a simple saddle point approximation gets. The naive thing to do in treating spontaneous symmetry breaking is to consider the minimum of the potential with a constant field, and that is exactly what a saddle point is. With your consideration on a lattice you are not taking the classical limit since you are summing up all sorts of configurations. If you did lattice theory with $\beta\rightarrow \infty$ that would be more what I'm talking about. $\endgroup$
    – octonion
    Mar 29, 2021 at 19:51
  • $\begingroup$ $\beta$ being an inverse temperature multiplying the action, thinking of the lattice theory more from a statistical field theory point of view. $\endgroup$
    – octonion
    Mar 29, 2021 at 19:53
10
$\begingroup$

I'll give a brief answer that is somewhat at odds with Chiral Anomaly's answer. When we do quantum field theory we expand about some 'saddle point' of the action, i.e. some place where the variation of the action vanishes and thus we can think of it as classical. When you are playing around with just the Lagrangian in quantum field theory, you are looking at these classical saddle points.

Often there is a saddle point where all of the fields vanish which is the ordinary thing we do in perturbation theory. But in the Higgs mechanism there is a non-trivial saddle where some fields get non-zero vacuum expectation values. This can't be seen (easily) by doing perturbation theory about the wrong saddle where all the fields vanish so in that sense it is 'nonperturbative' and that is what Chiral Anomaly's answer focuses on. But in another sense we are finding lowest order of the Higgs mechanism by just looking at the Lagrangian and as I am arguing in my first paragraph, that is as classical as it gets.

$\endgroup$
2
  • 1
    $\begingroup$ I don’t know if I agree that “leading-order” automatically implies “classical”. IIRC, even tree-level scattering amplitudes can contain nonzero powers of $\hbar$. Can’t the classical limit be completely trivial, and the leading nontrivial order already be semiclassical? $\endgroup$
    – tparker
    Mar 29, 2021 at 5:21
  • 1
    $\begingroup$ If I consider a classical field theory based on some action that involves the Higgs mechanism, would I still see something like a Higgs mechanism? Of course, the textbook treatment just looks at what the minimum of the potential is and makes an appropriate gauge transformation. $\endgroup$
    – octonion
    Mar 29, 2021 at 5:29
-2
$\begingroup$

What does "at tree level" mean? Look at this question. It means that you look only at diagrams that contain no "bubbles" attributable (in colloquial language) to "bubbles"", which are attributable {in colloquial language) to pairs of whatever particles-antiparticles. These particles-antiparticles are somehow connected more to quantum effects than the reactions you are looking at. They do happen though, but with a look probability, so they can be ignored. So one says that the process is classical (just as leaving out these bubbles in electron-electron scattering renders it "classical, meaning that, for example, the magnetic moment of the electron is exactly two, and not quantum corrected).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.