# What is a quantum field?

I'm just tasting a bit of QFT and want to get started. I got stuck right at the start: what is a quantum field, and how should I look at it? This question can be a follow up of the What is a field, really? question.

For me a field can be seen as a function. So saying "vector field" is the shorthand of saying a "function that maps from whatever domain to vectors of some dimensions".

What's the type of the value this function maps to when speaking of quantum fields?

In this another question. They says it's operator valued. As far as I can remember operators can be seen as "ket-bras" so the matrix product of an infinitely long column and row vector. So an infinitely large matrix. Or in other words a function that takes 2 parameters and gives back a scalar. Is that right?

So if I'm right a quantum field is a "scalar field field". So a function that maps space-time coordinates to functions that maps 2 scalars to a scalar.

In this yet another question. It's said it's a scalar valued in path integral formalism. I'm yet to understand that formalism so far. But for now I don't see how a scalar and a function valued field be equivalent...

On the other hand just out of curiosity, how many fields do we need to deal with in QFT? In classical electromagnetics we had two vector fields the electric and magnetic and the Maxwell equations that describes the time evolution. After some googling I can see there are boson fields, fermion fields, higgs field, whatever field... Are all of these quantum fields (so operator valued)? Do we have the field equations that describe all the relationship between these fields (just like the Maxwell-equations do in EM)?

• A quantum field is a ill-defined mathematical concept. It can be made rigorous, but it is not the way a physicist would like. So a rather good answer could be: a field is an operator-valued distribution that somehow describe the physical content of a relativistic quantum mechanical system. Dec 29, 2014 at 16:32
• @Phoenix87: You just defined a quantum field properly: it is a distribution of operator values over a pre-defined geometry. Mathematicians may not like this approach (it has very serious problems with regards to the usual concepts of calculus that physicists like to apply to this object), but that is what we do and we get the correct results (i.e. the predictions match the experiments). Dec 29, 2014 at 16:36
• @CuriousOne What I have in mind is something more similar to a Segal field, which acts on states of a C*-algebra. To make contact with the operator-valued distributions and test functions one can take some Fourier transforms that send the state to the test function and the Segal field to said distribution. The action of the Segal field on the state (or perhaps the action of the state on the Segal field), which is a well defined evaluation of a linear functional over an element of a C*-algebra, becomes the usual field smeared-out with a test function. Dec 29, 2014 at 16:46
• @Phoenix: To the mathematically uninitiated that sounds a lot like the procedure of defining distributions as linear operators over spaces of test functions in functional analysis? I would very much expect that kind of procedure to be necessary to formalize the problems physicists like to gloss over both in qft as well as in classic field theory. Dec 29, 2014 at 16:57
• @Phoenix87: I certainly hope that we can find a way to get rid of these ugly descriptions. Like a whole generation of physicists I tend to hang on to what stuck to my brain early, but I would like to see a clean, hopefully algebraic and geometric description of QFT that does not suffer the conceptual clutter of the early approaches. Dec 29, 2014 at 17:24