# Quantum fields are the basic constituents?

1. Dec 21, 2007

### pellman

Occasionally I come across statements to the effect that quantum field theory has replaced particles as the basic consituent of matter with continuous fields. Is this right?

From what I understand, a quantum field is an operator field. Which is to say--I would think--that by itself it is meaningless without something to operate on. And what do these operators operate on? Quantum states--essentially the same kind of quantum states that we encounter in quantum mechanics, right? So aren't we still talking particles?

Sometimes I read someone saying that particles are really just excitations of a field. Of an "operator field"?

Isn't it the case that the field really consists of raising and lowering operators that govern pair production/annihilation? So that, with the field "turned on", a quantum state consisting of an electron and a proton, for example, is no longer orthogonal to a state consisting of zero electron/positrons + 2 photons. That is, there is a non-zero probability of observing 2 photons instead of the electron+positron that you started with. But we're still just talking regular-old particle quantum states, right? or wrong?

Hoping someone will clarify this for me.

Todd

2. Dec 21, 2007

### jostpuur

This is a confusing topic. To me, saying that a field is operator valued, seems to be analogous to saying that the in QM the position of a particle has become operator valued. It could be right, but not very clear way of describing the QM.

Besides the operators, there can be explicit representations for the state. For particles we have wave functions, and for fields we have wave functionals.

Unfortunately I don't know this stuff very well myself yet, but to me the mere knowledge of the existence of the wave functionals clarified things a lot.

3. Dec 21, 2007

### blechman

Wow, that sounds a little melodramatic to me! QFT is nothing more or less than the mathematical framework that describes "particle" interactions, where by "particle" I mean photons, electrons, quarks, etc... This sounds like someone in 1800 saying that "Newton's laws have replaced levers and pullys!" Sounds a little ridiculous, doesn't it?!

I think you're hinting on what is called "second quantization." A classical field is like the usual vector potential of electrodynamics. We now promote that to an operator, so that it actually creates and destroys "quanta" of that field, aka, "photons":

$$A_\mu|0\rangle\propto|\epsilon_\mu,\vec{k},1\rangle$$

where $|0\rangle$ is the "vacuum state" and the state on the RHS is the state with one photon of momentum $\vec{k}$ and polarization $\epsilon_\mu$. These states form a sort of generalized Hilbert space called a "Fock Space." That's all there is to it!

Well, that's all there is to it, until you start actually trying to calculate stuff. Then it starts to get hairy. But that wasn't your question.

I'm not entirely sure I understand what you mean. I think you have the right idea, more or less. But I'm not sure what you mean when you say the field is "turned on" - if there are photons, then the field is turned on! I think what you meant to say is when INTERACTIONS among the fields are turned on. Then what you're saying is true even in ordinary QM - perturbed states are not the same as the unperturbed ("free") states, and therefore what you think of as "free" photons/electrons/etc are no longer orthogonal. This is true in general perturbation theory.

I hope that helps!

4. Dec 21, 2007

### meopemuk

Hi pellman,

I agree completely that talking about "particles as excitations of quantum fields" is very confusing. One can reasonably take the position that quantum field theory is not much different from quantum mechanics. In QFT the state of the system is described as a unit vector in a Hilbert (Fock) space. This state gives the probability amplitude for finding 0,1,2,..,n,.. particles in the system. The biggest difference with respect to QM is that the number of particles is not fixed in QFT. They can be created and annihilated. In the Fock space one can define operators of observables (position, momentum, spin, etc) of individual particles as well as total observables of the system.

In this picture quantum fields don't have any preferred status or physical meaning. They are just convenient operator functions which are (sometimes) useful for writing expressions for certain operators of observables, e.g., the Hamiltonian. In fact, entire QFT can be formulated without invoking quantum fields at all. Relevant operators can be expressed as functions of particle creation and annihilation operators, which also are convenient formal mathematical objects without direct physical interpretation.

Eugene.

Last edited: Dec 21, 2007
5. Dec 21, 2007

### Peter Morgan

Partly. The interpretation of quantum field theory is less developed than that of nonrelativistic quantum mechanics. Firstly, the "particles" in quantum theory are not anything like classical particles, because of wave/particle duality, so there is a way in which there were already no particles to replace. Quantum field theory perhaps moves things slightly more towards waves, but there is still a vague idea of particles.

In a free quantum field -- which has no interactions -- there is an observable "number operator" which is usually said to be how many particles there are in a state, but there is no way to individuate the particles (Bose-Einstein and Fermi-Dirac statistics are already used in quantum mechanics, of course, but they are essentially required by the algebraic structure of quantum field theory). There has been great effort put into constructing acceptable particle position operators in quantum field theory, without success, which makes the idea of particles more numinous in quantum field theory than in nonrelativistic quantum mechanics.

Quantum fields are not at all like a continuous field. Properly speaking, a quantum field operator at a point is not well-defined; spoken of loosely, the expected value in the vacuum state of a quantum field operator at a point is zero, but the standard deviation of a quantum field operator at a point is infinity -- that's essentially why it's not well-defined at a point. That means that if we measured the field over and over again, almost every time we would get the value $$+\infty$$ or $$-\infty$$. Speaking even more loosely, at adjacent points the observed field would often be opposite sign infinities --- that's not what we call continuous!

A quantum field is properly thought of as an "operator-valued distribution". Suppose we average all those $$\pm\infty$$ values, then we might get a finite number, if there are about the same number of +'s and -'s. To get a mathematically well-defined object, we use a smooth function $$f(x)$$, which is called a "test function", to construct a "smeared" average, $$\hat\phi_f=\int f(x)\hat\phi(x) d^4x$$. In the vacuum state, for a free field, $$\hat\phi_f$$ generates a Gaussian distribution, with the variance a bilinear functional $$(f,f)$$ of $$f(x)$$. The inner product $$(f,g)$$, which for the free Klein-Gordon field is given by
$$(f,g)=\int \tilde f^*(k) \tilde g(k) 2\pi\delta^4(k^\mu k_\mu-m^2)\theta(k_0)d^4k,$$
in terms of fourier transforms of $$f(x)$$ and $$g(x)$$, determines most of the structure of the free quantum field. As we take different test functions, so we obtain different measurement results; the test functions can be thought of as descriptions of what measurement we make of the field.

I hope this might be approximately comprehensible, although getting myself to think intuitively in these terms, and being able to describe them as clearly as this (really, I think it looks clear) took me a few years of determined effort, but for interacting fields, almost all of this fairly elementary mathematical structure essentially disappears. Renormalization and all the mathematical formalism surrounding renormalization obscures almost all the structure. There are still large questions as to whether renormalization can possibly be well-defined, which I will summarize by saying (though there's part of me that's screaming NO!) that infinite numbers of virtual particles are created and annihilated every time a pair of (not-well-defined) particles interact. Sometimes people talk of particles being surrounded by a sea of virtual particles, but IMO this is a hopeless language game. This cries out for a mathematical formalism to be created that, like the elementary construction of the previous paragraph, does all the mathematical work without ever invoking infinity in any indecent ways. No-one has come close to a nice mathematical formalism so far, so no-one knows what idea of a particle there will be when someone does.

An observable $$\hat\phi_f$$ makes a connection with experiment only when we also specify the physical state.

Creation and annihilation operators allow us to construct different states. In the vacuum state $$\left|0\right>$$, the expected value of an operator $$\hat A$$ is $$\left<0\right|\hat A\left|0\right>$$; in the state $$a^\dagger_f\left|0\right>$$, the expected value of $$\hat A$$ is $$\left<0\right|a_f\hat A a^\dagger_f\left|0\right>$$. $$a_f$$ is a smeared creation operator, for a test function $$f(x)$$; the quantum field itself is $$\hat\phi_f=a_f+a^\dagger_{f^*}$$. People often use improper creation and annihilation operators $$a(k)$$, which are restricted to a specific wave number $$k_\mu$$, but these have to be used with care, because they are not strictly well-defined. The different states we can construct using creation operators can be used to model different experimental preparation procedures. Note that creation and annihilation operators are not observables; I haven't been able to think of a nice way to talk about the effects of creation operators and annihilation operators, but I think of creation operators introducing more complicated correlations between measurements, while annihilation operators reduce the complexity.

What I've described above is enough different from what you've said here (which is an accurate reflection of the frequent -- but not universal -- confusion in the literature, I would say) that I don't know that it will clarify your understanding. On the other hand, it will certainly twist your head, which stands a chance of leading you to an understanding.

If this looks interesting despite being horribly different, and you feel like a challenge, you can see my Phys. Lett. A paper "A succinct presentation of the quantized Klein–Gordon field, and a similar quantum presentation of the classical Klein–Gordon random field" (or the Arxiv version). I warn you that I'm grinding several axes there, however, which you can see me really trying to sharpen on this topic.

Despite all the above, whether there are particles, fields, or neither or both at the lowest conceptual level of a fundamental physical theory, ultimately a decent physical theory will have to be able to describe states in which there are macroscopically localized objects -- it seems a reasonable requirement that tables and chairs should be described adequately by a fundamental physical theory. It is one of the embarrassments of fundamental theory that there are no bound states in the standard model of particle physics.

Since I started this, there have been two posts. I would say that neither really touches why QFT has driven off all attempts at detailed interpretations for 80 years, but they are characteristic of ways in which Physicists work with QFT, particularly quantum optics, without having to worry much about Foundations of Quantum Theory. I could take all sorts of issue with them, but I recommend them to you unless you have lots of time and no thesis supervisor to worry about.

6. Dec 21, 2007

### meopemuk

The Newton-Wigner position operator seems to be just fine.

Eugene.

7. Dec 21, 2007

### Peter Morgan

Indeed. Perfectly fine. However, the fact that it is not Lorentz invariant means that where a particle is likely to be on the Newton-Wigner approach depends on what inertial coordinate system is used. Deemed problematic by some, when discussing whether there is, in principle, such a thing as a particle in a Lorentz covariant formalism that has a Lorentz invariant dynamics.

8. Dec 21, 2007

### meopemuk

I am not sure what you mean by saying that Newton-Wigner position operator is not Lorentz invariant. It is true that different observers would see different wave functions in their respective Newton-Wigner position representations (e.g., a particle that looks localized for one observer may not be localized for another observer). However, this is true for other operators and representations as well. For example, a particle having one momentum p wrt one observer may have a different momentum p' wrt another observer.

Eugene.

9. Dec 21, 2007

### jostpuur

To those of you who insist that QFT is nothing more than a strange formulation of theory of quantum mechanical particles, I must wonder how you are deriving the classical fields out of them, the classical EM field being a concrete example.

Isn't the value of a classical field interpreted as an expectation value of the quantum mechanical field, given by

$$\langle\phi\rangle = \int\mathcal{D}\phi\;\phi|\Psi[\phi]|^2,$$

similarly as the position of a classical point like particle is interpreted as an expectation value of the quantum mechanical particle, given by

$$\langle x\rangle = \int d^3x\; x|\Psi(x)|^2.$$

10. Dec 21, 2007

### meopemuk

Yes, you are right that the particle picture of QED does not reduce to the Maxwell's field theory in the classical limit. Actually, it reduces (at least in the 2nd perturbation order and in the 1/c^2 approximation) to a field-less theory with Darwin-Breit direct interaction between charges. The Darwin-Breit theory is known to describe electromagnetic interactions very accurately. It also doesn't have various paradoxes characteristic to the Maxwell's approach (such as the violation of the Newton's 3rd law in magnetic interactions or the infinite field energy associated with a point particle). So, it is conceivable that one can construct an alternative to the Maxwell's theory for classical electrodynamics. I am currently working on a paper trying to explore this idea.

Eugene.

Last edited: Dec 21, 2007
11. Dec 22, 2007

### kdv

very good questions. Unfortunately, there are no clear answers. Almost every particle/condensed matter theorist has a different answer ("opinion" would be a better word). Even more unfortunately, it seems to be one of those topics that people tend to feel quite strict about, i.e. they often shut you off if you don't readily agree with their point of view (and sometimes get upset when you point out to some steps in their explanations which are not completely logical).

It is true that some people make the quite strong claim that the truly fundamental entities are the fields. I have heard that often. On the other hand, it's important to say that in the end, whatever is observed is always a particle.

So why placing more importance on the fields? I guess that a valid analogy here would be electromagnetic fields in classical physics. The E amd B fields are never observed directly, on eonly observes their effects on charged particles. But the mathematical description of what is going on only makes sense when everything is described in terms of those mysterious fields and very quickly the fields get a life of their own and physicists think of them as fundamental entities. It's not impossible that a more fundamental theory of physics would dispense entirely of electromagnetic fields! But the maths become much more transparent when expressed in terms of the fields, at least in the way the present theories of physics are expressed (for example when working out how electromagnetic phenomena differ when viewed from different frames).

I think this is what happened with quantum fields.

yes, we are still talking particles. Notice that perturbative calculations in QFT always involve calculating S matrix elements, that is probabilities of transitions between states of specific number fo free particles.
Yes, you are right. When interactions are present, states with definite number of particles are no longer eigenstates of the full Hamiltonian so there can be transitions between states with different particles (as long, of course, as certain selection rules are respected including conservation of energy and others depending on the interaction).

But you are correct that in the end, we are always observing particles.

12. Dec 23, 2007

### pellman

Thanks, guys.

Here is an example of the sort of thing I'm talking about.

What is Quantum Field Theory, and What Did We Think It Is? by Weinberg. Bottom of page 1: "For quite a
long time many physicists thought that the world consisted of both fields and particles: the electron is a particle, described by a relativistically invariant
version of the Schr¨odinger wave equation, and the electromagnetic
field is a field, even though it also behaves like particles."

and later on page 2 "In its mature form, the idea of quantum
field theory is that quantum fields are the basic ingredients of the universe,
and particles are just bundles of energy and momentum of the fields."

13. Dec 23, 2007

### George Jones

Staff Emeritus
From Wald's Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics:

"Quantum field theory is a quantum theory of fields, not particles. Although in appropriate circumstances a particle interpretation of the theory may be available, the notion of 'particles' plays no fundamental role either in the formulation or interpretation of the theory."

14. Dec 23, 2007

### pellman

Thank you for the informative post, Peter.
This is the most difficult part for me. What does the expected value of an operator mean? If a quantum field is an operator-valued field, then what does "measuring" it mean? Apparently, you mean some procedure where we associate the operator-value with a number (which turns out to be infinity). But what is that procedure?

Is what you are describing related to LSZ reduction? I am reading Srednicki's Quantum Field Theory. When deriving the LSZ reduction formula (pg 49), the first thing he does is smear over the creation operator with a sort of arbitrary function, similar to what you describe.

Yeah, it sounds like so many more Ptolemaic epicycles to me. I'm one of those who thinks that mathematical formalism you mention will have to be something radically different than current quantum theory.

Thank you. I am very interested and look forward to corresponding further.

Now you're really cracking me up. I didn't know this. The mighty standard model can't get two particles to stick together? Hilarious.

Last edited: Dec 23, 2007
15. Dec 23, 2007

### Peter Morgan

Now you're talking about the measurement theory of quantum field theory. Good luck, because IMO there isn't one worth the name. First, construct a model of the thing you are measuring that occupies all of Minkowski space. Second, construct a measurement apparatus that is not in that Minkowski space. Model that measurement apparatus by an operator, which is constructed using creation and annihilation operators that are related to different space-time regions as necessary. Sounds like the sort of thing God might do, but not Physicists. Of course there are all sorts of pragmatic justifications for this that I am perfectly and entirely happy with, LSZ reduction amongst them, but as a fundamental, true, never to be surpassed theory, IMO quantum field theory not so good. IMO.

To go off world for a moment. A model is a model. It's not about measurement exactly, and it also doesn't reproduce the world exactly. It is a free creation of the human mind that we also find useful, somehow. The connection of a model that we have created to the world is mediated by our interests, skills, etc. The connection is pragmatic, not absolute, true, etc. IMO. I personally like beauty most as a criteria, then usefulness. Both are in the eye of the beholder, both are multifaceted. Not all Physicists think this way! If measurement operators make no sense to you, then the theory is no use to you unless you make a connection between it and the world in a different way. That I find measurement operators a helpful way to think about QFT doesn't mean you have to; indeed many Physicists seem to prefer to think in terms that I find impenetrable, S-matrices, Feynman diagrams, infinite seas of particles, non-visual algebraic structure, etc., etc. As I understand other Physicists' understandings, however, I believe that I'm not alone in taking measurement operators in the way I do.

As a way to get your head around Measurement Operators and States, try getting your head around the detailed mathematical definition of Random Variables and of Probability Measures, which have a similar relationship to experiment. Indeed, the latter can be given an algebraic presentation as a commuting set of Measurement Operators and States over them.

I would expect that Srednicki is being mathematically careful. The LSZ formula takes the time-ordered correlation functions (which are, properly speaking, distributions, hence the need for mathematical care, apply as directed) and produces S-matrix elements, which are more observable, in some sense, than the correlation functions.

For my current attempt -- no promises that you'll like it, but it is about to be published -- see my thread "Classical interacting random field models". It is somewhat different.

Of course my wording was not entirely proper. As I understand it, it would take a proof of color confinement for us to be able to be sure that we could in principle construct a stable localized particle. Actually constructing that model is another thing again (unless someone provides a constructive proof). Of course if we posit a central 1/r potential, we get the hydrogen atom, bound states etc., but that's a long way down the road from QCD. However QCD is close to a popular books read for me.

16. Dec 23, 2007

### Son Goku

Disclaimer: I will only discuss the free klien-gordon field for simplicity.

Remember that $$\phi(x)$$ is the analogue of the position operator in standard quantum mechanics. Now for non-relativistic QM $$\left<\psi\right|X\left|\psi\right>$$ is the expectation value of position. Which means if you measure the positions of particles all prepared in the state $$\left|\psi\right>$$ the values measured will tend to clusture around $$\left<\psi\right|X\left|\psi\right>$$. Now the standard deviation calculates the average difference between a measured values and the expectation values. Basically it measures how tightly the measured values for position cluster around $$\left<\psi\right|X\left|\psi\right>$$.

Similarly $$\left<0\right|\phi(x)\left|0\right>$$ should give you the expectation of the field value at the point x, when it is in the ground state. If we compute it, it is equal to 0. As we would expect, the field should be 0 in the ground state. However the standard deviation is undefined (infinite). This is because $$\phi(x)$$ isn't really a "proper" operator, you need to integrate it against some function before it becomes well defined. However this subtlety rarely matters.

Isn't it more a case of bound states being difficult to analyse. For instance QED allows the formation of positronium, it's just that the standard Dyson perturbation series approach is useless, since positronium it isn't really a scattering process.

17. Dec 23, 2007

### Peter Morgan

It depends what you mean by rarely. Indeed $$\left<0\right|\phi(x)^2\left|0\right>$$ is not well-defined. Nothing that is quadratic in the field at a point is. The free Hamiltonian is quadratic in the quantum field operator, so that it's only by subtracting infinity that we obtain a well-defined operator. Interacting Hamiltonians are cubic or higher degree in the field, not well-defined at all, hence regularization and renormalization (or some other mathematical refinement) is necessary if we take a Hamiltonian or Lagrangian approach to defining the structure of a quantum field. We're used enough to how we deal with the difficulties that it's not unreasonable to say that "this subtlety rarely matters", but perhaps something else should be added, such as "so long as the rules to be found in every text book on QFT are followed".

$$\left<0\right|\phi(x)\phi(y)\left|0\right>,x\ne y$$ is OK, so is anything that is multilinear in the field, at an arbitrary finite number of points, but
$$\left<0\right|\phi(x)\phi(y)\left|0\right>,x\ne y$$ increases without bound as $$x\rightarrow y$$.