Saturday, November 25, 2017

Competent formal theorists know they can't rely on the existence of the Lagrangian

An Ising model example of non-Lagrangian CFT methods

Japanese string theorist Judži Tačikawa (100% racially clean Czech nationalist Mr Tomio Okamura got over 10% in Czech elections so, as Luboshi Nakamotl, I plan to take over Japan with my Japanese nationalist party as a revenge) gave a talk somewhere at IPMU where he mentioned quantum field theories that don't have a Lagrangian – see the last page – while many courses use obsolete textbooks that pretend otherwise, namely that the Lagrangians are enough. This has attracted some interest of the TRF readers. The interest is fun but I am disappointed that something that I consider basic conceptual lore of modern physics is still so utterly unknown in the broader community of people who are interested in physics.

Lagrange. I admit it was my nickname during the introduction summer camp preparing and baptizing the soon-to-be freshmen at Faculty of Mathematics and Physics at the Charles University. The camps take place in Albeř, a village in Czech Canada on the Austrian border (not to be confused with Czech Switzerland on the Saxon i.e. East German border LOL).

Five days ago, I discussed some general and perhaps surprising insights about symmetry in Nature that theoretical physicists made in recent decades. In quantum gravity, exact symmetries have to be gauge symmetries while gauge symmetries may be emergent and their identity generally depends on the point in the configuration space. Everyone who hypes his theory of quantum gravity with global symmetries is probably an incompetent amateur. Everyone who assumes one particular gauge group for his theory of everything, including quantum gravity, is an incompetent amateur, and so on.

It's similar with the Lagrangian. Theoretical physicists who are up to their job know that Lagrangians are extremely useful but they simply cannot assume that the theories constructed from Lagrangians exhaust the list of all interesting, relevant, promising, or possible quantum theories of Nature. And even more obviously, they know that the methods and descriptions based on the Lagrangian aren't the only ones they have to master and use – simply because they're using different methods than the Lagrangian ones all the time.

So everyone who assumes that every quantum theory – it doesn't have to be one with gravity – must start with a Lagrangian is obviously not an expert who understands the lessons from recent decades in physics.

OK, what is the Lagrangian \(L\)? It's the density of the action \(S\) in time. In other words, it's the integrand in the integral over time that defines the action:\[

S = \int dt\,L

\] In field theory, we integrate not only over time but over the whole spacetime. Equivalently, \(L\) may be written as an integral over (e.g. three-dimensional) space. So we end up with\[

S = \int dt\,dx\,dy\,dz\,{\mathcal L}.

\] Here, the integrand \({\mathcal L}\) is known as the Lagrangian density although many of us often call it just the "Lagrangian", too.

This link between the action and the Lagrangian is inseparable. For this reason, any approach to a problem based on the action is exactly the same thing as the approach based on the Lagrangian. The two concepts only differ by having or omitting an integral over time (or spacetime)! Physics students tend to use the term "Lagrangian" for these discussions because "action" is reserved for discussions about Hollywood action movies. ;-)

Fine. What is the action or, equivalently, what is the Lagrangian? The action \(S\) is something that is extremized when the dynamical equations of motion are satisfied. It means that all the laws of physics that have the form of differential equations for positions \(x(t)\) or fields \(\phi(x,y,z,t)\) may be written simply as\[

\delta S = 0.

\] The action is extremized which means that its variation is zero. Using the differential calculus and a formula for \(S\) as a function of degrees of freedom such as \(x(t)\) or \(\phi(x,y,z,t)\), we can derive the differential, "Euler-Lagrange" equations for the degrees of freedom that are obeyed in Nature classically. Many fundamental classical theories may be written using this "principle of least action", some can't. When you add things like friction or other irreversible terms, it becomes harder or impossible to use the action. But fundamentally, there's no friction in the laws of physics and the action often works great.

In mechanics, the Lagrangian – the integrand in the action – is \(L=T-U\), the difference between the kinetic and potential energy, at least in the simplest examples. This \(L\) is related to the Hamiltonian \(H\) in another description (using the Poisson brackets) by the Legendre transform. Note that \(H=T+U\) in the simplest examples. I don't want to extend this basic discussion of actions too far. You're supposed to know at least something about the actions, otherwise stop reading because this isn't supposed to be a text helping tutors with troubled schoolkids at a high school.

What happens with the Lagrangian and Hamiltonian in quantum mechanics? The Hamiltonian becomes an operator. Does the similar Lagrangian become an operator? The answer is No. Instead, the approach that uses the action in quantum mechanics is the Feynman sum over histories – or path integral. We directly calculate the complex transition probability amplitudes using the formula\[

{\mathcal A} = \int {\mathcal D}^n\phi_i(x,y,z,t)\,\exp[iS(\phi_j)/\hbar]

\] where some extra factors – the so-called "insertions" – may be added to the integrand if we calculate some expectation values or other things. This is an integral over the classical histories – the variables \(\phi_j\) we integrate over are classical degrees of freedom – but the result is interpreted quantum mechanically, as a probability amplitude. The integrand is a hugely oscillating complex number on the unit circle in the complex plane, because of that \(\exp(iS/\hbar)\) which is the exponential of a huge imaginary number. This huge number only matters "modulo \(2\pi\)" and this remainder of the division by \(2\pi\) is a "random number", one that is uniformly distributed over the circle.

Because these phases are so random, they mostly average out once we integrate (i.e. sum many of them). The only places in the infinite-dimensional configuration space parameterized by \(\phi_i(x,y,z,t)\) where they don't average out in the classical limit are places where \(\delta S=0\). Near such places, the point on the unit circle is almost constant, so lots of nearby histories support each other and constructive interference makes them important. And that's why quantum mechanics in Feynman's path integral description reduces to the principle of least action: the histories close to the history with the least action cooperate and constructively interfere, instead of cancelling each other – which is why they give a big collective contribution to the transition probabilities! That's why a quantum theory in the classical limit predicts that "the probability is close to one" that a particle will move along a trajectory that is rather close to the solution of classical equations.

So you should see that the fates of the Lagrangian and the Hamiltonian in quantum mechanics are very different. The former remains a classical function of classical variables while the other is an operator. That's why e.g. Cesaruliana's comment that one simply relates the Hamiltonian and Lagrangian by the Legendre transform is utterly naive. This transform is a concept in classical physics. In quantum mechanics, it is only useful as long as the quantum mechanical theory is "directly determined" by a classical theory, i.e. by its classical limit. That's often not the case.

As in so many cases, you can see that lots of the people don't really understand quantum mechanics. They are thinking classically and they assume that everything they know remains basically the same in quantum mechanics. But it doesn't. The quantum revolution has changed many things – qualitative features as well as quantitative predictions – dramatically. The classical theories are only good enough proxies near the classical limit but many phenomena predicted by quantum mechanics take place very far from any such limit.

Great. So there may be some general Hamiltonian operator which is a function of some fundamental non-commuting degrees of freedom. But because their commutators may be very large and in no way "tiny", they're not similar to any classical theory where the Hamiltonian is a function of commuting variables. So one can't identify this quantum mechanical theory with any classical theory which is why he can't use the Legendre transform. That means that he won't be able to find any Lagrangian description. That may obviously happen. It's common sense. An intelligent reader with an undergraduate degree in physics – or someone rightfully considering himself to be equivalent to that – must understand it. This paragraph is meant to be the full proof that the intelligent reader should be able to verify and confirm.

It makes no sense to sell the "theories without a Lagrangian" as some questionable heresy or miracle. It's totally obvious why you can't rely on the existence of the Lagrangian in quantum mechanics – because the Lagrangian and action are classical concepts and they're only usable in quantum mechanical theories that have some important classical theory to build upon.

How does it work in reality? Do we have theories without Lagrangians etc.? In quantum field theory, we may write the Lagrangian for the Standard Model etc. (see the mug above) so we may get rather far. But there are lots of analogous quantum field theories that simply don't allow you a similar straightforward treatment. If they're not weakly coupled in any limit, e.g. if it's impossible to set any \(g\ll 1\) for the coupling \(g\), your Lagrangian strategy may be in trouble. So strongly coupled and emergent theories may fail to have a good Lagrangian. This may include lots of situations in condensed matter physics.

But one theme that has been increasingly obvious since the 1970s is that fundamental theoretical particle physics and condensed matter physics are close relatives and some "surprising and important phenomena" that are discovered in one of them are extremely likely to be relevant in the other, too. We know that the Mexican hat potential is important to analyze phase transitions in condensed matter physics and thermodynamics; and it's important for the Higgs mechanism in particle physics. The renormalization group is also important in both, and so on and so on.

And indeed, genuine theoretical particle physicists simply know that all these general lessons that could be considered typical for condensed matter physics are true in fundamental particle physics and quantum field theory, too. We often see lots of quasiparticles. Whether a particle is a "real fundamental particle" or a "quasiparticle" may often be a subtle question. The Lagrangian description may often be absent.

String theory is full of such descriptions and theories that don't start with a classical Lagrangian. The (2,0) superconformal theory in 5+1 dimensions is a great example of a non-gravitational theory (well, it's a boundary CFT dual for some AdS background of [gravitational] M-theory, too). And M-theory in particular offers us other examples in which you can't rely on the Lagrangian. But I wanted something much more elementary. It doesn't have exactly the properties I need – but I think that the reader should be able to learn the same lessons from the simpler example, anyway.

Take the two-dimensional Ising model, the scaling limit of it which is a conformal field theory. First, is it some "super recent" theory created last summer? No, it's not. The Ising model began with Ising's thesis in 1924. It's been around for 93 years. Various extra insights were added decades later but most of the truly important ones – even about the conformal limit – have been known for more than 30 years. It's simply no cutting-edge science anymore.

OK, does this Ising model have a Lagrangian? For most people who study the model using the "Zamolodčikov" methods, it's far from obvious. In the "Zamolodčikov" methods, we're really interested in the operators, their products, expectation values, and what happens when one operator's location rotates around another one in the 2D Euclideanized plane where the two-dimensional conformal field theory lives.

And this Ising model CFT has three primary fields: the identity \(1\) of dimension zero – every sane CFT has it; the spin field whose chiral dimension is \(\Delta=1/16\); and the fermion field \(\psi\) whose dimension is \(\Delta=1/2\), or some composite fields like the energy density that are made of it. The details depend on "primary fields under what" we discuss.

Does this Ising model CFT have a Lagrangian description? Well, yes. You may write it as a theory of a Majorana (real or Hermitian) fermion \(\psi\) living in the 2D space. Or you may write it as the bosonic \(\phi^4\) model. The Lagrangians are "straightforward" given these general words. The kinetic term for the bosonic or the fermionic field; decorated by the appropriate potential in the bosonic case.

Now, this is curious. We don't have one Lagrangian description but two Lagrangian descriptions. And they're very different. One of them says that the CFT is a theory of a fermion, another one says that it's a theory of a boson. Those look different. Why they can't be a single answer? Is it a boson, or a fermion? Well, both are equally correct. The fermion may be written schematically as \(\psi=\exp(\phi)\) while \(\partial\phi=\psi\partial \bar\psi\) etc. These identifications need to be clarified and all the renormalization and ordering issues have to be discussed and refined.

Imagine that there are exactly two Lagrangian descriptions for this CFT – which isn't really the case. Why two? Clearly, the law that there is "one" has been violated. The number of Lagrangian descriptions seems to be a random integer. If it can be two instead of one, it can surely be zero, right? And indeed, in other cases, it's zero.

But even in this case where some Lagrangian descriptions exist, we may see how weak and insufficient the Lagrangian description is. I mentioned that a primary field of the Ising model is the spin field \(\sigma(z)\) whose dimension is \(\Delta=1/16\). This statement really means that the field \(\sigma\) has the units of "mass to the one-sixteenth-th" power or what's the right way to attach "th" from the two sources (fraction; and adjective to count). Well, if we have a non-chiral CFT that contains both left-movers and right-movers, the relevant primary field is the product \(\sigma\bar\sigma\) of the left-moving and right-moving spin fields. At any rate, this spin field \(\sigma\) is as "fundamental" as the field \(\psi\).

If you hear it for the first time, you're probably shocked by the claim that its dimension is \(\Delta=1/16\). Why is it one over sixteen? Well, it may be derived using some – non-Lagrangian – methods of two-dimensional field theory. For physics types, this is a very important portion of quantum field theory that conceptually goes beyond the "textbook material of perturbative QFT" and I recommend you e.g. Volume I of Polchinski's "String Theory" for an excellent treatment of 2D conformal field theory for physicists. "Less physically" oriented readers may pick "the big yellow book" by Di Francesco and other authors, Conformal Field Theory.

The dimension \(\Delta=1/16\) may incidentally be expressed as\[

2\Delta &= \zav{ \frac 12 + \frac 32 + \frac 52 + \dots }-(1+2+3+\dots) =\\
&= \frac{1}{24} + \frac{1}{12} = \frac{1}{8}.

\] This is not just a random identity involving random seemingly divergent sums; it's a way to actually calculate the number for the first time. Whining directed against the fundamental identities that were used above won't be tolerated in the comments, I have wasted too much time with these clueless critics.

A funny thing about \(\Delta = 1/16\) is that it is not even a multiple of \(1/2\). Because the dimension of \(\psi\) is \(1/2\) and the derivatives add \(1\), you can't construct the field \(\sigma\) of dimension \(1/16\) as a polynomial of the "basic Majorana fermion" and the derivatives at all! So even though the Ising model "is" formally some quantum field theory of a Majorana fermion which does follow from a Lagrangian, one of the three most important, primary operators in the theory isn't even a function of the "fundamental degrees of freedom" used in the Lagrangian (and their derivatives)!

And in some proper counting, the whole dimension \(\Delta=1/16\) is some kind of a "quantum correction". Classically, fields have dimensions that are integer or half-integer (the units – the kilogram to the integral or half-integral power – may be determined by dimensional analysis involving the Lagrangian, from the knowledge that the action has to be dimensionless and the derivatives have the dimension of mass). Quantum mechanically, the true dimensions of the corresponding operators may be different – they are corrected by quantum corrections which are basically proportional to powers of \(\hbar\). The Lagrangian-based dimensional analysis no longer works so simply because fields at nearby or coincident points lead to short-distance effects and ultraviolet (UV) divergences and those need to be renormalized away etc. The quantum corrections to dimensions are known as "anomalous dimensions" etc. And the Ising model is a simple example of a theory where the full dimension \(\Delta = 1/16\) – basically a value comparable to one that can't be made "very small" – is a quantum correction. The spin field may be schematically written as \(\sigma=\exp(\phi/2)\) in the bosonic description ("square root of the fermion \(\psi\)") and naively classically, the exponential of anything should be dimensionless! But quantum mechanically, \(\sigma\) isn't dimensionless because \(1/16\) isn't zero.

If you learn some basics of the CFT methods from Polchinski's textbook, for example, you will agree that the reasons are obvious. Even to determine the existence of the field \(\sigma\) and its dimension \(\Delta=1/16\), you need to use clever methods such as the "operator product expansions" (OPEs) or the "state-operator correspondence" based on the conformal mapping of a cylinder to the whole complex plane, and other things. These methods don't use any Lagrangian – and therefore don't depend on the existence of a Lagrangian – at all.

There is a lot to learn and I don't want this blog post to compete with Polchinski's or Di Francesco's book – I would have no chance and Polchinski wrote his book after a decade of perfectionist work, anyway, while I only wanted to spend less than an hour of recreational enough Saturday work with this text today. ;-) But my point is that the Japanese speaker is in no way the first person who suggests that quantum field theories may work without a Lagrangian. And the general proposition is in no way new or recent.

Instead, it's something else – a part of the basic lore or conceptual material that folks who are interested in theoretical physics know already as students. Folks just shouldn't get a "formal" (not LHC analysis-oriented etc.) theoretical physics PhD without being aware of the things above. I only learned the 2-dimensional CFT methods "reasonably technically" when I was a graduate student but of course I was exposed to the general lessons – bosonization and fermionization (the shocking equivalence between theories of bosons and fermions), the state-operator correspondence, and other "non-naive" phenomena in theoretical physics – as an undergrad. Similar methods and ideas enter lots of the derivations, even rather basic ones, and you simply cannot get almost anywhere if you remain naive and superficial and you imagine that all QFT research is just some standardized treatment of some Lagrangian.

It simply isn't. The Lagrangians are only good for theories which have a good and important classical limit – which typically means quantum field theories that are weakly coupled. And the Lagrangian methods are only good for questions that look like slight generalizations of questions that already existed in the classical theories. But many quantum mechanical theories and many questions in them simply aren't just some tiny upgrades of their classical counterparts. That's why lots of the non-Lagrangian methods – and even theories admitting no Lagrangian methods – have been a standard part of the daily work of good theorists for decades.

Just like the evolved status of symmetries in Nature, the evolved status of the Lagrangian is something that a genuine theorist simply must know, and those who don't know such things may be instantly seen to be scammers who are only pretending to be competent physicists. It's worse than that. These scammers – who know nothing about modern physics and their mindless "let me write a Lagrangian, analyze it, and hype it" shows that they're just mediocre sycophants who would like their instructor's [organ] as students – are often presented as lonely geniuses. They're no geniuses, they're mediocre naive laymen who are hopelessly confined in the average Joes' if not Janes' group think, and every TRF reader with IQ above 110 must have understood the proof.

And that's the memo.

I started with a Japanese physicist. Here's something about another Japanese physicist. Willie Soon – whose background is tied to Indochina – sent me a letter by Feynman to Kojči Mano who considered himself a nameless man and congratulated to Feynman. Feynman wrote him that Mano isn't nameless to his wife and child and to Feynman, either, and Mano should look for his own problems (see Mano's publications) that are solvable – even if they look trivial – and may make Mano feel successful. Also, Feynman apologized that he picked a problem for Mano instead of letting him find one by himself. Well, that's great but the excessive modesty is way too normal in Japan so Feynman could have failed to appreciate the personality differences between the Westerners and the Japanese. I think that the Japanese must feel happy in their own way when they behave in those excessively modest ways so often.

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