# Useage of the term field in QFT

1. May 31, 2012

### kith

useage of the term "field" in QFT

Wikipedia defines a field as "a physical quantity associated with each point of spacetime". So contrary to a particle, where physical quantities are associated with properties like position or momentum, the field itself is a physical quantity. (This definition restricts us to measureable fields like the em field but this is not important for my issue.)

In ordinary QM, we talk about states of particles and observables which correspond to physical quantities. So if fields are physical quantities, they need to correspond to observables in QFT, which they do. But the states of what are we then looking at? The states of fields also.

So QFT seems to use the term "field" in two distinct meanings:
1) Fields as physical systems which have no classical counterpart
2) Field operators which correspond to the classical fields

2. Jun 1, 2012

### Demystifier

Re: useage of the term "field" in QFT

The particle-number operator and the field operator are operators that
1) act in the same Hilbert space, but
2) do not commute

Consequently
1) a particle state and a field state are states in the same Hilbert space, but
2) a particle-number eigenstate is not a field eigenstate

3. Jun 1, 2012

### kith

Re: useage of the term "field" in QFT

How would you call the physical system which is described by this Hilbert space?

4. Jun 1, 2012

### Demystifier

Re: useage of the term "field" in QFT

That's a tricky question, because whatever name I choose, it can be misleading. But nevertheless, I would call it a system with an indefinite number of particles.

5. Jun 1, 2012

### kith

Re: useage of the term "field" in QFT

This sounds good. So using your terminology, it doesn't make sense to talk about "states of the field". Instead, "field states" are special states of our system (the eigenstates of the field observable).

Unfortunately, there doesn't seem to be a consensus regarding this. I recently read a paper from Glauber, where he talks about number states as "states of the field".

Last edited: Jun 1, 2012
6. Jun 1, 2012

### Demystifier

Re: useage of the term "field" in QFT

Well, physicists often use bad terminology, especially in quantum physics which lacks clear intuitive picture of the abstract concepts involved.

7. Jun 1, 2012

### kith

Re: useage of the term "field" in QFT

Yes. It is just sometimes hard to see if the problem is your understanding or the terminology of the others.

8. Jun 1, 2012

### Demystifier

Re: useage of the term "field" in QFT

You are welcome!

9. Jun 4, 2012

### A. Neumaier

Re: useage of the term "field" in QFT

Both classically and in quantum mechanics, an observable is something which gets a particular value in each state of the system. In classical mechanics, it is a deterministic value, in quantum mechanics, it is the expectation value.

Thus a classical point particle has the observables p and q, which depend on the state of the system, which changes with time. usually only the time dependence is written down explicitly, giving values p(t) and q(t). Similarly, a classical field F has the value F(x) which depends not only on x but also on the state of the system, and hence changes with time. Again this dependence on the state is suppressed and only the dependence on time is written, giving values for F(x,t).

In quantum mechanics, things are the same except that you need to replace definite values by expectation values:
$$\langle p(t)\rangle=\psi(t)^*p\psi(t)$$
$$\langle q(t)\rangle=\psi(t)^*q\psi(t)$$
$$\langle F(x,t)\rangle=\psi(t)^*F(x)\psi(t)$$

In the two cases, the (quantum or) classical physical system itself ''has'' or ''is described by'' a (mean) position and (mean) momentum, resp. a (mean) field.

In sloppy terminology, one can say in the second case that the system ''is'' the field, as there is no explicitly named carrier (like the particle in the first case), so (pars pro toto) the field stands for its carrier (the unnamed ''system''). But this is the case both in the classical and in the quantum mechanical situation.

10. Jun 5, 2012

### kith

Re: useage of the term "field" in QFT

But what about other observables? Why should the system "be" the electromagnetic field and not any observable of the system?

Thanks for pointing that out. I am not familiar with "states" of the classical em field. Is such a state simply specified by the field values and their derivatives at every space point at a given time? So the state space is something like an uncountable infinite dimensional phase space and the dynamics is "ordinary" Hamiltonian dynamics?

Last edited: Jun 5, 2012
11. Jun 5, 2012

### A. Neumaier

Re: useage of the term "field" in QFT

In QFT, the only observables are fields and functions of the fields. For example, particles are elementary excitations of the fields, hence particle properties are properties of the fields.
Of course in QED there are two fields so the system is ''the electromagnetic field + the electron/positron field''.
Unlike in classical mechanics, quantum mechanical states are not objects in phase space.

As everywhere in quantum mechanics, a pure state is a norm 1 vector of the underlying Hilbert space, and a mixed state a positive semidefinite Hermitian operator of trace 1 acting on this space. An example of a pure state is the vacuum state. In it all field expectations vanish, but correlation functions are nonzero. Other familiar examples of pure states are coherent states, the most classical-like states of the electromagnetic field, realized to fairly high accuracy in laser beams. Familiar examples of mixed states are the thermal states associated with a system in equilibrium (such as a photon gas, discussed in the context of black body radiation).

Unfortunately, the complete state space of QED is only poorly understood - it is not a Fock space (which, by Haag's theorem, only describes free theories), and we know very little about what it is instead. But qualitatively, you may consider a state to be a physically consistent assignment of (distribution-valued) expectation values to all field operators and products of field operators at arbitrary space-time arguments. These become true expectation values if appropriately smeared over space-time regions. The problem lies in specifying precisely the meaning of ''physically consistent''.