# Some basic QFT

1. Sep 27, 2010

### Hurkyl

Staff Emeritus
I think I've cleared up a fundamental misunderstanding I've had for a while, and want to get confirmation I'm right. Is the following statement true?

In QFT, a quantum field is not part of the state of a quantum system, and is not an aspect of reality that needs to be measured to be known. Instead, the field is known a priori, and instead is (a field of) observables that one may measure about the quantum system.​

I had been confused on this point since in classical mechanics, fields were always part of the physical system. Then, when I see quantum field being defined as operators -- especially with emphasis on the resemblance to wave-functions -- I was confused because the formalism only provides for measurements to be done on a state space, not on an operator algebra!

But now I realize "the value of the field at a point" is being treated as an observable (like position in OQM) rather than as an aspect of the quantum state, and things now start to make more sense.

I just want to make sure this is the right kind of sense before I start molding my understanding to it.

2. Sep 27, 2010

### xepma

Yes, this is an aspect of QFT which confused me for a long time as well, even though the statement is fairly trivial:

Quantum fields are operators and have nothing to do with the state of the system itself.

They are the equivalent of the position and momentum operators in ordinary quantum mechanics. They are not the equivalent of the wavefunction.

One way to think of a field operator is as a probe: the combination $$\hat{\psi}^\dag\hat{\psi}(x)$$ "probes" the position x for a particle. In condensed matter physics this combination of the field operator is nothing but the density number operator.

I'm quite suprised how so many text books glance over this conceptual difficulty.

3. Sep 27, 2010

### dextercioby

I would say that in the <normal> QFT describing the Standard Model, we don't really care about the state of the system. We assume it to fit with our scattering theory standard in which particles are <free> at + and - infinity on the time axis.

The so-called quantum fields are just mathematical tools giving us after very long calculations the right scattering cross sections.

4. Sep 27, 2010

### peteratcam

What continues to confuse me is whether the field operators in QFT are supposed to be ladder operators are not.

It so happens that acting on a groundstate of a harmonic oscillator, $$x$$ and $$a^\dagger$$ do pretty much the same thing, but this is a groundstate property, the operators are quite different. I don't think enough distinction is made in intro QFT because they deal with vacuum states so often.

5. Sep 27, 2010

### dextercioby

The <ladder> operators are at the core of everything related to quantum mechanics, field theory included. They come from Lie algebra representation theory which is intimately related to symmetry. Since you can build quantum field theory from the symmetries of Minkowski flat spacetime, 'voila', you can't really escape the use of 'ladder' operators.

6. Sep 27, 2010

### kexue

But in standard QFT we have the Heisenberg picture, where the operator fields are objected to time-evolution which is governed by the Hamiltonian. The states, or more corectly the multi-particle states are frozen in time and are only acted on by the operator fields.

7. Sep 27, 2010

### peteratcam

Fine, but I mean there is an ambiguity in presentation:
Sometimes
$$[\phi(r),\phi(r')]=0; \quad [\phi^\dagger(r),\phi^\dagger(r')]=0;\quad[\phi(r),\phi^\dagger(r')]=\delta(r-r')$$
but sometimes
$$[\phi(r),\phi(r')]=0; \quad [\pi(r),\pi(r')]=0;\quad[\phi(r),\pi(r')]=i\hbar\delta(r-r')$$
(ie, hermitian-operator valued fields)

My point is that I get super-confused when the notational convention for field ladder operators borrows the same letter as used for the canonical coordinates. The confusion gets worse because acting on a free particle vacuum, the canonical operators create particles! Unless I'm being thick here...

8. Sep 27, 2010

### dx

The quantum 'field operators' $\Psi [/tex] and [itex] \Psi^{\dagger}$ are the position space versions of the creation and annihilation operators for the momentum modes. Thus they have a representation as

$$\Psi(x) = \int a^{+}_k e^{-ikx} dk$$

$$\Psi^{\dagger}(x) = \int a^{-}_k e^{ikx} dk$$

Just like $a^{+}_k$ creates a quantum in the k mode, $\Psi(x)$ creates a particle at x:

$$\Psi(x) |\Omega \rangle = \int e^{-ikx} a^{+}_k |\Omega \rangle dk = \int e^{-ikx}|k\rangle dk = |x \rangle$$

The field $\Psi$ is not itself exactly measurable, since its real and imaginary parts dont commute, but when there are a large number of quanta, it behaves like a classical wave. The number density of quanta in space can be shown to be $\Psi ^{\dagger} \Psi$, just like the corresponding expression for the creation and annihilation operators gives the occupation number of the momentum mode.

Last edited: Sep 27, 2010
9. Sep 28, 2010

### strangerep

For the single-particle case...
My understanding is that a "point" is a pair (q,p) in a 6D phase space. (Both the q and p are
3-vectors). The state of a classical system is given by a particular point in this phase space.
A classical observable is a particular function "f" on phase space. A measurement of the
observable "f" for a system which is in state (q,p) yields the number f(p,q).

I.e., it's the _function_ f as a whole which is the observable, not its value at any particular
point.

So both classical and quantum formalisms are basically about duals: an observable algebra
and its dual. In the classical case, the observable algebra means a (commutative) algebra
of functions on phase space, and its dual is the set of points in phase space. In the quantum
case, an observable is not necessarily commutative, and its dual is the set of states in a
Hilbert space. So the only difference between classical and quantum is whether the
algebra of observables is commutative.

In the QFT case, we have a inf-dim algebra of observables (operators), called quantum fields.
To construct the dual of the algebra, we postulate a single state (vacuum) and demand
that it be annihilated by certain members of the operator algebra. Then the Hilbert space
(ie Fock space) is constructed by the action of the other operators on our arbitrarily
chosen vacuum state.

(So yes, I think your indented statement sounds right. :-)

PS: Oh, and one should of course always remember that quantum fields are not
bona-fide "operators", but rather "operator-valued distributions".... :-)

Last edited: Sep 28, 2010
10. Sep 28, 2010

### strangerep

One must look at such issues in the context of the entire algebra, not just certain members
of it.

E.g., the Heisenberg Lie algebra is spanned by (x,p,1) but also by $(a,a^\dagger,1)$.
This is just a change of basis of the algebra. (A Lie algebra is just a vector space equipped
with a particular noncommutative product.)

What matters is which member of the algebra we choose to annihilate the vacuum.
It proves convenient to choose "a" in many cases.

11. Jan 25, 2011

### A. Neumaier

Your first formula usually has the a in place of phi - this is the CCR for creation/annihilation operators, while the second formula is the CCR for canonical quantum fields. if you distinguish the two, the confusion will be gone.

12. Jan 25, 2011

### A. Neumaier

In the Heisenberg picture (which is associated with 4D field theory, treating space and time on the same footing), the situation is as follows, both in the classical and in the quantum case:

States are fixed, and operators carry the dynamics. A field is the 4D generalization of a path. Fields and paths are functions defined on R^4 resp. R^1, with values in the *-algebra of observables. States are positive linear functionals <.> on this algebra, assigning to each observable A its expectation <A>. Thus a field Phi produces in a given state for each point x in space-time an expectation value <Phi(x)>.

In classical physics, the algebra of observables is a commutative algebras of functions. Pure states are algebra homomorphisms into the complex numbers. In this case <Phi(x)> is the value of the field in the state <.> at the point x - in classical mechanics one silently identifies the values with the observable Phi(x) --- then Phi(x) becomes state-dependent; it even determines (if considered for all x) the state when Phi has enough degrees of freedom. In particular changing the state changes this function, though the observable hasn't changed. Hence this identification is a bit fishy, but usually goes unnoticed.

In quantum mechanics, the algebra of observables is a non-commutative algebras of linear operators defined on a dense subspace of a Hilbert space. Pure states are mappings of the form <A> = psi^*A psi for some state vector of norm 1. Again, <Phi(x)> is the value of the field in the state <.> at the point x, but one cannot identify these values with the observable Phi(x), since they live in very different spaces.

This explains why the above confusion is quite understandable.

This is far from true. In scattering theory, the input to the S-matrix is the Schroedinger state at t=-inf, and the output is the state at t=+inf. One only doesn't care about the finite-time dynamics.

And quantum fields don't only determine the S-matrix - they determine everything, including the finite time dynamics (usually recovered in QFTs in terms of the closed-time-path functional integral).

13. Jan 25, 2011

### K^2

It's not that simple, really. Fields are quantum objects. However, these fields have same properties as energy states in basic QM. That means the particular configuration of field in entire space is just one component of the state. A full state is a superposition of different field configurations. Working with them in this format is extremely inconvenient.

So instead, you second-quantize the fields, you call the state in second-quantized coordinates your quantum state, and the field can then be written as a field operator. If you act with field operator on the second-quantized state, you get the exactly the same field description as in paragraph above. You end up with superposition of field configurations.

But it is exactly the same description of exactly the same system. One just happens to be more convenient mathematically.

14. Jan 26, 2011

### balces

I understand it the following way, maybe this helps you:

In QFT a field operator is built out of a linear combination (or integral) of annihilation and creation operators. So you could regard it as field specific amplitude which tells you how probable it is that a photon or some other field quantum of some momentum and spin is created or annihilated. Or let’s phrase it differently, that an electromagnetic interaction will happen. At the end this interaction will only have non zero amplitude if some other quantum field comes around which couples with the first field at a specific point in time and space.
The quantum states itself, the number states in the Fock space, do not encode this interaction information. They just tell you how much and which quanta might be available for interaction.
In classical electrodynamics the field concept is not so much different: Maybe remember when for the first time the field of a charge was introduced in class: It normally is defined as the force a standard charge would experience at some point in space and time. So it kind of encodes the possibility and the strength of interaction some kind of charge will experience when it comes around.
This time however, I do not have states which the field is operating upon. The field itself contains the information of its strength at a certain point in time and space.
So the field operators in QFT and the classical field are kind of relatives, though they are quite different mathematical objects and treated very differently.