# What exactly is a Quantum Field

1. Jan 22, 2005

### Gold Barz

Is it literally a "field" that is undergoing quantum mechanics?

2. Jan 22, 2005

### vanesch

Staff Emeritus
Yes, this is a way to see it. It is the quantum description of the classical system "field". If you consider a classical system, and you describe the configuration space with variables q, and you identify their canonical momenta p, you can apply the machinery of QM. Well, that's what can be done with a quantum field: you take a classical field, and consider the amplitude of each mode to be a configuration variable q. After quantification, you have a quantum field. If the partial differential equation of the field is linear, you have a so-called free field, and guess what ? The dynamics of each mode is independent of another, and is the same as a harmonic oscillator. Quantifying it gives you the harmonic oscillator states: E = hbar omega (n + 1/2), for that mode.
So each mode can be in different quantum states, with equal energy jumps given by the above formula. In the case of a field equation which reduces to scalar components which follow the Klein-Gordon equation (such as the KG equation itself, or the dirac equation, or ... in fact all of the equations we know of), then you can do some calculations, and you will find that there are good reasons to find out that (dE)^2 = m^2 c^4 + (dp)^2 c^2 where dE is a change between two successive quantum levels of a mode, and dp the corresponding change in 3-momentum.
And that's how you get out "particles". In fact, the different quantum states of a mode seem to correspond to integer numbers of jumps of energy and momentum which satisfy the above relationship, "as if we had particles of mass m". And if you think about it, apart from other quantum numbers which I didn't mention here, that's the essence of the mechanical definition of a "particle": the relationship between the energy and the momentum.

For the EM field, m = 0, and you get the famous relations:
E = h nu and E = p c for... the photon :-)

So a photon is nothing else but "the difference in energy and momentum between two successive quantum states of the EM field".

Interactions are described by two fields, and by a non-linear term coupling them in each of the two otherwise linear field equations, for instance the product of the two fields. The study of this fills in the other 800 pages of a book on QFT, but the essence is that these interactions make the quantum fields have their modes "communicate": one mode of field 1 can decrease one level and a mode of field 2 can increase one level. That's how interactions seem to be described by the exchange of particles.

Of course, in the above I did away with spin, with bosons vs. fermions etc... but the general idea is there.

cheers,
Patrick.

3. Jan 22, 2005

### Janitor

And then when you get to the last page of the thick book on quantum field theory, you might possibly see a sentence to the effect: "Oh, by the way, the theory that you have now laboriously mastered is thought to be merely the low-energy approximation to string theory."

4. Jan 23, 2005

### EL

Great summary vanesch!

This is the formulation of QFT in terms of so called non-commuting operators, followed by for instance Mandl&Shaw.
However I would be very greatful if someone could write or link me to a similar summary of the path integral formulation of QFT.
It seems like most papers I have found uses the path integral formulation as starting point, and since I'm only familiar with Mandl&Shaw it would be interesting to get sort of an overwiev.

5. Jan 23, 2005

### marlon

https://www.physicsforums.com/journal.php?s=&journalid=13790&action=view [Broken]

I have also written a text on this for college some years ago...I just posted it into my journal, you can check it out if you want

regards
marlon

Last edited by a moderator: May 1, 2017
6. Jan 23, 2005

### lalbatros

Seems to me that a field needs quantum theory when measurements can perturb it. For example, measuring tiny lumps of electromagnetic field (photons) will perturb the field (absorb the photon for example).

7. Jan 23, 2005

### vanesch

Staff Emeritus
The path integral formalism is indeed computationally much more interesting and more used now. However, the canonical quantization is complementary to it: for instance the evident appearance of particle-like properties are harder to see in the path integral formalism.
The path integral formalism is in fact an extension of Feynman's view of quantum mechanics: there's no explicit structure with Hilbert space and so on. Feynman considered just one kind of values, these are the "inproducts of 2 states" ; the equivalent in Dirac notation of <a|b>. Indeed, if you know these numbers for all possible states a and b, you know all there is to know in QM. The Hilbert space and company are just a tool to get those numbers out.

Feynman then wrote down one formula: If A represents a classical begin situation, and B represents a classical end situation, then:

<A|B> = The sum over all possible paths through configuration space to get from A to B of the number exp(i S / hbar)

Here, S is the classically computed action of the path. You know that to each path from situation A to situation B, corresponds a classical action. The principle of minimum action then says that the true classical path taken by the system will be the one that makes S minimal (or extremal). Well, Feynman doesn't require you to take THAT path, but you take ALL paths, give them a weight equal to exp( i S / hbar), and add them all together (with the right measure !). That will give you the quantum-mechanical amplitude <A|B>.

For particle QM, you can rather easily work out that the Hilbert space formulation and the Feynman path formulation are identical.

For fields, this starts to get a life of his own, and in fact, you can avoid a lot of mathematical difficulties in the canonical approach by going through the path integral formulation. The Feynman diagrams one gets out of both methods are the same.

Note that "all possible paths in configuration space", when dealing with fields, is a mindboggling set of possibilities! You have to consider ALL different field configurations (which do NOT obey the classical equations) that smoothly connect the initial state A to the final state B. With each of these configurations corresponds an action of the field, and then you have to integrate over it. This is the famous path integral.
The only integral one can solve analytically that way is the gaussian integral, which corresponds to the free field case.

cheers,
Patrick.

8. Jan 23, 2005

### JesseM

As I understand it, there are many situations in quantum field theory where the only method that works to calculate anything is a perturbation series...in these cases, do physicists know how to write down some equations which, if solved, would give an exact solution without a need for a perturbation series, so the problem is just that they don't yet have a way to solve the equations exactly? Or is it actually the case that physicists don't even know how to write down the laws of nature in a nonperturbative form, in some instances?

Also, does quantum field theory involve any new observables beyond those used in nonrelativistic QM? Can you measure the strength of a field at a given point as in classical field theories?

9. Jan 23, 2005

### vanesch

Staff Emeritus
I'm not an expert on this, but I think that the situation is the following, more or less. The canonical approach, although naively ok, falls on its mathematical face. I think that the main problem there is Haag's theorem, which says that the only consistent QFT is a free field theory ; more knowledgeable people can probably explain you in more detail what the problem is. In fact, it is even a small miracle that the perturbative part comes out unharmed.
The Feynman path integral is in principle "ok" because the formulation includes the full interacting theory from the start (the full action, with all interactions). But the Feynman path integral ALSO falls on its mathematical face, because as far as I know, no mathematician has ever come up with a consistent theory of the measure of this big space of paths.
However, there is some more hope here, because if you slice up spacetime on a kind of grid, then that measure becomes definable. I'm absolutely no expert here, but I think that's the basis of things like lattice QCD.
Even the perturbative solution fails, in that it is at best an asymptotic series, which almost surely diverges.

So, you can say that in QFT, we know the equations "in principle" but they don't make mathematical sense (yet?) the way they are written down. We also know techniques to solve them, which also don't make mathematical sense. But out come numbers which compare very well with experiment

Well, position becomes a parameter just as time, and looses its "observable" character. The field values at a point are (highly singular) observables. But the philosophy of QFT is different from that in QM. It adapts to the mathematical difficulties and the experimental situation, and in fact, most of what you learn in QFT turns around one problem:
calculating the in product of an incoming state of free particles at t = -infinity with the outgoing state of free particles at t = +infinity. These in-products are sometimes called the S-matrix elements, and Feynman diagrams are a technique to calculate a series devellopment of these S-matrix elements.
You are much less "master of the situation" in QFT than in NR QM.

cheers,
Patrick.

10. Jan 23, 2005

### dextercioby

The 2 widely used methods of doing QFT are not really equivalent,in the sense of using them interchangebly to get the same result...The fall of the canonical formalism and the widely usage of path integral quantization was marked by the discovey in 1954 of the so-called Yang-Mills fields,that is gauge fields with nonabelian gauge algebra...And then came the electroweak interactions the QCD and the SM and people could not possibly use the canonical quantization,because they could no longer solve the classical field equations.Those equations were nonlinear and coupled...
However,the original field theory (QED) is still being taught worldwide in both formulations and i was taught both.I felt very comfortable doing derivatives & finding Green functions and generating functionals into different orders (of perturabtion theory) rather than applying the horrible Wick's theorem and working with (noncommuting) operators and all that horse**** in the canonical approach...
I don't care whether the mathematics (behind path integrals) is uncertian.I care it works...

Daniel.

11. Jan 23, 2005

### vanesch

Staff Emeritus
I have never seen it done, but I wonder if you cannot still use the canonical approach in non-abelian theories by "fixing the gauge" ? You know, like when working in the Coulomb gauge to quantify EM canonically. Ok, you loose all the explicit symmetry (gauge invariance and relativity), and consider the non-linear terms as self interactions (like in phi^4 theory). Calculations would probably be horrible that way...

cheers,
Patrick.

12. Jan 24, 2005

### dextercioby

I'm well aware of the existance of BRST canonical quantization techniques...Yet,since being taught QFT via path-integral,i was taught both BRST and Batalin & Vilkovisky quantization via path integral...

Daniel.

13. Jan 28, 2005

### EL

Thanks vanesch!
This confirmes I had understood it correctly.
Have found a quite nice book in "QFT in a nutshell" by Zee, which is like an overwiev. Someone who can suggest some more complete texts about the path integral approach to QFT?

14. Jan 28, 2005

### dextercioby

Chronologically:
Feynman & Hibbs (1965),Faddeev-Slavnov (1974),P.Ramon (1981 & 1989) and Bailin & Love (1986 & 1993).

There are others,too...But i found these ones to be the most relevant.

Daniel.

15. Jan 29, 2005

### EL

Thanks Daniel.