# QED question to field experts (like Tom, etc)

1. May 15, 2003

### Alexander

Kinda continuing discussing of how classic electrodynamics follows from quantum one.

I was always wondering myself what is more fundamental of the "trinity" (=charge, field, virtual carrier)?

My feeling is that electric charge is the one which is "responsible" for virtual photons (bunch of which is basicly what we label as "electric field").

Or can "field" be postulated and then both electric charge and a virtual photon (field state) derived ?

Or can everything be derived from postulating a virtual photon (some call them DC photons, by the way) to be a fundamental entity?

Anybody willing to toss/kick opinions or "established views" around?

2. May 15, 2003

### Hurkyl

Staff Emeritus
Sounds like a good excuse to try and get the QFT thread active again!

3. May 15, 2003

### Tom Mattson

Staff Emeritus
I think you can dismiss the "field" straight away. That is the classical picture.

It sounds like you are asking if QED can be derived from classical electrodynamics. If so, then the answer is "no". The quantization of the EM field is done (noncovariantly) by replacing the coefficients of the modes of oscillation of the vector potential with creation and annihilation operators, like so:

Aclassical=&Sigma;ck,&alpha;eik.r+c.c.

edit: The sum is taken over photon momenta k and photon polarization &alpha;

Let ck,&alpha;-->ak,&alpha;
and c*k,&alpha;-->a+k,&alpha;

Now, the vector potential is an operator. As I am sure you know, the creation and annihilation operators satisfy [a,a+]=1. In the limit of large numbers of photons, we can ignore the 1 on the right hand side, and we recover classical electrodynamics for macroscopic fields. But we cannot take any limit of classical electrodynamics to generate QED.

That is how it's done.

4. May 15, 2003

### Tom Mattson

Staff Emeritus
I'm working on it. I'm trying to get a set of notes written on the Grassman calculus, from the documents I presented in the QFT thread.

5. May 15, 2003

### Alexander

Ok, so prime candidates into "fundamental entity" are then charge and corresponding to it virtual carrier. Can one be postulated and another derived, (or they are inseparable)? And if one is more fundamental than the other, is not it the charge which shall be treated as more fundamental than a photon? (If not, then how is charge derived from photon)?

Last edited by a moderator: May 15, 2003
6. May 15, 2003

### Alexander

I think, it is really interesting question to discuss - what is a consequence of what (if this is more "separable" than chicken-egg dilemma).

7. May 16, 2003

### Tom Mattson

Staff Emeritus
I think they are inseperable.

Let's look at how the charge and the photon field enter the theory.

Free Field Lagrangian
I need the Lagrangian to be a Lorentz scalar, so the only one I can construct from the terms at my disposal is:

L=i&psi;+&gamma;0&gamma;&mu;&part;&mu;&psi;-m&psi;+&gamma;0&psi;

Next comes the issue of gauge invariance. The above Lagrangian is obviously invariant under a global gauge transformation of the form:

&psi;(x)-->ei&alpha;&psi;(x)
&psi;+(x)&gamma;0-->e-i&alpha;&psi;+(x)&gamma;0

where &alpha; is a real constant. We run into a problem when considering local gauge transformations, however.

Local Gauge Transformations
Local gauge transformations of the form:

&psi;(x)-->ei&alpha;(x)&psi;(x)
&psi;+(x)&gamma;0-->e-i&alpha;(x)&psi;+(x)&gamma;0

do leave the last term of the Lagrangian invariant, but not the first, and the problem is due to the derivative.

Explicitly:

&part;&mu;&psi;(x)-->&part;&mu;ei&alpha;(x)&psi;(x)

We have to use the product rule for the right hand side.

&part;&mu;ei&alpha;(x)&psi;(x)=ei&alpha;(x)&part;&mu;&psi;(x)+iei&alpha;(x)&psi;(x)&part;&mu;&alpha;(x)

Thus, it is clearly seen that the derivative term is not invariant under local gauge transformations. We can rescue the Lagrangian by introducing a covariant derivative D&mu; as follows:

D&mu;=&part;&mu;-ieA&mu;(x)

such that, under local gauge transformations, A&mu;(x) transforms as follows:

A&mu;(x)-->A&mu;(x)+(1/e)&part;&mu;&alpha;(x)

Now, let's see how the derm D&mu;&psi;(x) transforms under a local gauge transformation.

D&mu;&psi;(x)=&part;&mu;&psi;(x)-ieA&mu;(x)&psi;(x)
D&mu;&psi;(x)-->&part;&mu;ei&alpha;(x)&psi;(x)-ie(A&mu;(x)+(1/e)&part;&mu;&alpha;(x))ei&alpha;(x)&psi;(x)

We already know how the derivative term transforms, so we can simplify the above to read as:

D&mu;&psi;(x)-->(ei&alpha;(x)&part;&mu;&psi;(x)+ie&alpha;(x)&psi;(x)&part;&mu;&alpha;(x))-ieA&mu;(x)-iei&alpha;(x)&psi;(x)&part;&mu;&alpha;(x)

And now we see the utility of the field A&mu;(x): Its transformation property under local gauge transformations is exactly what is required to eliminate the term we picked up when transforming &part;&mu;&psi;(x).

edit: Eliminating the terms in red, we have:

D&mu;&psi;(x)-->ei&alpha;(x)D&mu;&psi;(x)

which is exactly what we need for local gauge invariance.
--end edit.

It turns out that A&mu;(x) is the photon field, and e is the electric charge. We have to introduce the charge when defining the covariant derivative and specifying the local gauge transformation property of A&mu;(x), so I do not see how they can be separated.

edit: superscript bracket,typo

Last edited: May 16, 2003
8. May 16, 2003

### instanton

Obviously, there is no true answer to the original question, and the tentative answer largely depends on personal taste. Here is my personal taste.

Since we do physics let's start from some observation we make. We know quantum theory, and special relativity works really well. Also, we observe "particles". If you combine first two ingredient it is quite very difficult to cook up a theory other than local quantum field theory. Now, be aware. When I said field theory here it doesn't have the same nuance as in classical field. Here, particle picture comes in. Local quantum field theory is, at least in spacetime with well defined asymptotic property, is describable through collisions of swarm of particles. In asymptotic region each particles are described by single particle wave function. Wigner showed that if we have Poincare symmetry we can use it to characterize single particle states. So, at least in asymptotic regime we have well defined quantum state of multiparticle system - simply a tensor product of single particle states.

Now, with mild additional assumptions like clustering property, you can show the simplest time evolution operator for the state is a combination of creation and anihilation operator. So, we will have a Hamiltonian that is a combination of them. we can add additional ingredient to this by adding more symmetry. Since we believe quantum theory it is rather be unitary. Simplest possible one is U(1). Whenever we have a continuous symmetry like U(1) there is a corresponding conserved charge - in this case it is called electric charge.

It turns out we can go backward a lot easier by quantizing classical relativistic field with U(1) symmetry, which is typical description in most text books.

So, to answer the original question we need all of them. Field is not tied to charge directly. If you want to describe any interaction relativistically you need field. Now, the character of the interaction is described by what kind of charge the field carries - Noether current. Finally, particle is what we observe in suffieciently far regieon although it seems this concept is really secondary one.

Instanton