**Young physicists in Austin, be careful about some toxic junk in your city**

Three weeks ago, in the article titled

Responsibilityphysicist Jacques Distler of UT Austin mentioned a statement by Sasha Polyakov that he was "responsible" for quantum field theory. That comment was particularly relevant when Distler taught an undergraduate particle physics course and was frustrated by the following:

The textbooks (and I mean all of them) start off by “explaining” that relativistic quantum mechanics (e.g. replacing the Schrödinger equation with Klein-Gordon) make no sense (negative probabilities and all that …). And they then proceed to use it anyway (supplemented by some Feynman rules pulled out of thin air).Did the following text defend the legitimacy of Distler's frustration? Well, partly... but I would pick the answer No if I had to.

This drives me up the fúçkïñg wall. It is precisely wrong.

There is a perfectly consistent quantum mechanical theory of free particles. The problem arises when you want to introduce interactions.

What's going on? Indeed, textbooks and instructors often – and, according to some measures, always – say that quantum mechanics of one particle ceases to behave well once you switch to relativity – to theories covariant under the Lorentz transformations.

Are these statements right? Are they wrong? And are the correct statements one can make important? It depends what exact statements you have in mind.

What Distler discusses is the

*existence*of the Hilbert space – and Hamiltonian – for one particle, e.g. the Klein-Gordon particle. Does it exist? You bet. If you believe that a Hilbert space of particles exists in free quantum field theory, do the following: Write a basis vector of that Hilbert space as the basis of a Fock space, i.e. in terms of the basis vector that are\[

a^\dagger_{\vec k_1} \cdots a^\dagger_{\vec k_n} \ket 0

\] And simply pick those basis vectors that contain exactly one creation operator. This one-particle subspace of the Hilbert space will evolve to itself under the empty-spacetime evolution operators. In fact, if you write the basis in the momentum basis as I did, the Hamiltonian for one real quantum of the real Klein-Gordon equation will be simply\[

H = \sqrt{|\vec k|^2 + m^2}.

\] This is something you may derive from quantum field theory. The operator above is perfectly well-defined in the momentum space. The energy is non-negative, the norms of states are positive, everything works fine.

So has Distler shown that all the statements of the type "one particle isn't consistent in relativistic quantum mechanics" are wrong?

Nope, he hasn't. In particular, he was talking about the statement

...replacing the [non-relativistic, e.g. one-particle] Schrödinger equation with Klein-Gordon make[s] no sense...But this statement is right at the level of one-particle quantum mechanics because his equation for the evolution of the wave function is

*not*the Klein-Gordon equation. You know, the Klein-Gordon equation is\[

\left(\frac{\partial^2}{\partial t^2} - \frac{\partial^2}{\partial x^2} - \frac{\partial^2}{\partial y^2} - \frac{\partial^2}{\partial z^2} + m^2 \right) \Phi = 0.

\] That's a nice, local – perfectly differential equation. On the other hand, the replacement for the non-relativistic Schrödinger equation\[

i\hbar\frac{\partial}{\partial t} \psi = -\frac{\hbar^2}{2m} \Delta \psi + V(x) \psi

\] that he derived and that describes the evolution of one-particle states was\[

i\hbar\frac{\partial}{\partial t} \psi = c \sqrt{m^2c^2-\hbar^2\Delta} \psi + V(x) \psi

\] Because the square root has a neverending Taylor expansion, the function of the Laplace operator is a terribly non-local "integral operator" acting on the wave function \(\psi(x,y,z,t)\) in the position representation. So this equation for one particle, even though it follows from the Klein-Gordon quantum field theory, doesn't have the nice and local Klein-Gordon form. It isn't pretty and it isn't fundamental. If you wrote this equation in isolation, you

*should*be worried that the resulting theory

*isn't relativistic*because relativity implies locality and this equation allows the localized wave function packet to spread superluminally!

What the statements mean is that if you want to use some nice and local equation for a wave function for one particle – i.e. if you literally want to replace Schrödinger's equation by the similar Klein-Gordon equation – you won't find a way to construct (in terms of local functions of derivatives etc.) the probability current and density etc. that would have the desired positivity properties etc. And this statement is just true and important!

If you want to return to simple, fundamental, justifiable, beautiful equations, you can indeed use the Klein-Gordon, Dirac, Maxwell, and other equations. But you must appreciate that they're equations for (field) operators, not for wave functions.

This statement is important because it's not just a mathematical one. It's highly physical, too. In particular, if you consider any relativistic quantum mechanical theory of particles – quantum field theory or something grander, like string theory – it's unavoidable that when you confine particles to the distance shorter than the Compton wavelength \(\hbar / mc\) of that particle, you will unavoidably have enough energy so that particle-antiparticle pairs will start to be produced at nonzero probabilities. And in relativity, it's normal for a particular to move by a speed comparable to the speed of light, and then its wavelength is comparable to the Compton wavelength. You can't really trust the one-particle theory at distances comparable to its normal de Broglie wavelength! So the theory is wrong in some very strong sense.

The antiparticles (which are the same with the original particle in the real Klein-Gordon case, just to be sure) inevitably follow from relativity combined with quantum mechanics, and so does the pair production of particles and antiparticles. This physical statement has lots of nearly equivalent mathematical manifestations. For example, local observables in a relativistic quantum theory

*have to be*constructed out of quantum fields. So the 1-particle Hilbert space doesn't have any truly local observables: You can't construct the Klein-Gordon field \(\Phi(x,y,z,t)\) out of operators acting on the 1-particle Hilbert space because the latter operators never change the number of particles while \(\Phi(x,y,z,t)\) does (by one or minus one – it's a combination of creation and annihilation operators). In fact, you can't construct the bilinears in \(\Phi\) and/or its derivatives, either, because while those operators in QFT contain some terms that preserve the number of particles, they also contain equally important terms that change the number of particles by two (particle-antiparticle pair production or pair annihilation) and those are equally important for obtaining the right commutators and other things. The mixing of creation operators for particles and the annihilation operators for antiparticles is absolutely unavoidable if you want to define observables at points (or regions smaller than the Compton wavelength).

There's one more statement that Distler made and that is really wrong. Distler wrote that the problems only begin when you start to consider

*interactions*– and from the context, it's clear that he meant interactions involving several quanta of quantum fields, several particles in the quantum field theory sense. But that's not true.

Problems of "one-particle relativistic quantum mechanics" already appear if you consider the behavior of the single particle in

*external classical fields*. Just squeeze a Klein-Gordon particle – e.g. a Higgs boson – in between two metallic plates whose distance is sent to zero. Will it make sense? No, as I mentioned, the walls start to produce particle-antiparticle quanta in general. Time-dependent Hamiltonians lead to particle production, if you wish. Similarly, if you place these particles in any external classical field, the actual Klein-Gordon field may react in a way to create particle pairs.

So the truncation of the Hilbert space of a quantum field theory to the one-particle subspace is inconsistent not only if you consider interactions of particles in the usual Feynman diagrammatic sense – but even if you consider the behavior of the particle in external classical fields. Whatever you try to with the particle that goes beyond the stupid simple single free-particle Hamiltonian will force you to acknowledge that the truncated one-particle theory is no good.

We want to do something more with the theory than just write an unmotivated non-local Hamiltonian of the kind \(H\sim \sqrt{m^2+p^2}\) if I use \(\hbar=c=1\) units here. And as soon as we do anything else – justify this ugly and seemingly non-local (and therefore seemingly relativity-violating) Hamiltonian by an elegant theory, study particle interactions, study the behavior of one particle in external classical fields – we just need to switch to the full-blown quantum field theory, otherwise our musings will be inconsistent.

One extra comment. I mentioned that the non-local differential operator allows the wave packet to spread superluminally. How is it possible that such a thing results from a relativistic theory? Well, quantum field theory has no problem with that because when you do any doable measurement, the processes in which a particle spreads in the middle gets combined with processes involving antiparticles. When you calculate the "strength of influences spreading superluminally", some Green's functions – which are nonzero for spacelike separations – will combine to the "commutator correlation function" which vanishes at spacelike separation. So the inseparable presence of antiparticles will save the locality for you. The truncation to particles-only (without antiparticles) would indeed violate locality required by relativity as long as you could experimentally verify it (you need at least some interactions of that particle with something else for that).

While Jacques is right about the possibility to truncate the Hilbert space of quantum field theories to the one-particle subspaces, he's morally wrong about all these big statements – and some of his statements are literally wrong, too. At least morally, the lore that drives him up the wall is right and there are ways to formulate this lore so that it is both literally true and important, too.

So students in Austin are encouraged to actively ignore their grumpy instructor's tirades against the quantum field theory lore and even more encouraged to understand in what sense the lore is true.

As I explain in the comments, many quantum field theory textbooks have wonderful explanations – usually at the very beginning – of the wisdom that Jacques Distler seems to misunderstand, namely why quantum fields and the mixing of sectors with different numbers of particles is unavoidable for consistency of quantum mechanics with special relativity.

The 2008 textbook by my adviser Tom Banks starts the explanation on Page 3, in section "Why quantum field theory?" It says that the probability amplitude for a particle emission at spacetime point \(x\) and its absorption at point \(y\) is unavoidably nonzero for spacelike separations and because it would only be only nonzero for one of the two time orderings of \(x,y\), and the ordering of spacelike-separated event isn't Lorentz-invariant, the Lorentz invariance would be broken and one must actually demand that only amplitudes where both orders are summed over are allowed. In other words, as argued on page 5, the only known consistent ways to solve this clash with the Lorentz invariance is to postulate that every emission source must also be able to act as an absorption sink and vice versa. When both terms are combined, the sum is still nonzero in the spacelike region but has no brutal discontinuities when the ordering gets reversed.

Also, when the particle carries charges, the emission and absorption in the two related processes must involve particles of opposite charges and one predicts (and Dirac predicted) the existence of antiparticles that are needed for things to work.

Weinberg QFT Volume 1 explains the negative probabilities and energies of the relativistic equations naively used instead of the non-relativistic Schrödinger equation on pages 7, 12, 15... Read it for a while. It's OK but, in my opinion, much less deep than Tom's presentation.

Peskin's and Schroeder's textbook on quantum field theory discusses the non-vanishing of the amplitudes in the spacelike region on page 14 and pages 27-28 discuss that the actual influence of one measurement on another is measured by the commutator of two field operators. And that vanishes for spacelike separations – again, because two processes that are opposite to each other are subtracted.

Without the mixing of creation operators (for particles) and annihilation operators (for antiparticles), you just can't define any observables that would belong to a point or a region and that would behave relativistically (respected the independence of observables that are spacelike separated). Quantum fields are the only known way to avoid this conflict between quantum mechanics and relativity. They are unavoidably superpositions of positive- and negative-energy solutions, and therefore are expanded in sums of creation and annihilation operators. That's why all local discussions make it necessary to allow emission and absorption at the same time – and, consequently, the combination of quantum mechanics and relativity makes it necessary to consider the whole Fock space with a variable number of particles. The one-particle truncation is inconsistent with relativistic dynamics such as time-dependent interactions, emission, or absorption.

In the mathematical language, fields and their functions are necessary for any local observables in relativistic quantum mechanical theories. They always contain terms that change the number of particles – except for the trivial constant operator \(1\). In the physical language, relativity and quantum mechanics simultaneously imply that emission and absorption are linked, antiparticle exists, and scattering amplitudes for particles and antiparticles have to obey identities such as the crossing symmetry.

The teaching of a quantum field theory course could be a good opportunity for Jacques to learn this basic stuff that is often presented on pages such as 3,5,7,12,14... of introductory textbooks.

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