# Quantum corrections to force laws

1. Aug 26, 2009

### RedX

For the free photon-field:

$$Z[J]=exp(\frac{i}{2} \int d^4x \int d^4y J^\mu(x)\Delta_{\mu \nu}} (x-y) J^\nu(y))$$

where $$\Delta_{\mu \nu}}=g_{\mu \nu}/(k^2-i\epsilon)$$ is the free-photon propagator. This leads to Coloumb's law for the electrostatic energy between two charges at $$x_1$$ and $$x_2$$ when you plug in $$J(x)=e_1\delta(\vec{x}-\vec{x_1})+e_2\delta(\vec{x}-\vec{x_2})$$.

What happens if you use the full-photon propagator $$\Delta_{\mu \nu}}=g_{\mu \nu}/(k^2-\Pi(k^2)-i\epsilon)$$ instead? Since $$\Pi(k^2)$$ is of the order of the fine-structure constant when the fine-structure constant is small, wouldn't you get Coloumb's law plus a small correction?

Or is this cheating, and you need to include the full interaction Lagrangian in Z[J]:

$$Z[J]=exp(i \int d^4t \mathcal L_I(\frac{1}{i}\frac{\delta}{\delta J(t})) exp(\frac{i}{2} \int d^4x \int d^4y J^\mu(x)\Delta_{\mu \nu}} (x-y) J^\nu(y))$$

to get the correction to Coloumb's law? Or is this all bad anyways because in the end, we will treat the surviving source functions, J(x), the electrons, as classical electrons, stationary point particles with delta functions? The correct fully quantum-mechanical way would be to just amputate the sources and plug in the external momenta instead of plugging in classical sources?

Because it seems to me that the only difference between non-Abelian gauge theories and Abelian ones are self-interactions of the boson field. If we can treat the self-interaction of the gluons as corrections to the propagator, then why can't we calculate perturbatively the gluon force?

2. Aug 26, 2009

### Ben Niehoff

I thought the reason was that the strong coupling constant is NOT small? Isn't it on the order 1?

Another part of the issue is that in non-Abelian gauge theory, the charge carried by the fermions is not gauge-invariant. So you can't just put in sources where you have "one red quark here, one green quark there". The color charges on each quark are in fact a function of time (and space, if they were to move).

3. Aug 28, 2009

### RedX

I meant at high energies, when the strong coupling constant gets smaller.

Are you sure the fermion charges can change? I thought only the gluon charges could change? Wait, doesn't Noether's theorem say something that it's all conserved? I'm confused now.

4. Aug 28, 2009

### Ben Niehoff

To find the conserved currents, vary the Lagrangian by a global, infinitesimal gauge transformation. Or do it the following, easier way. Define the fermion current as

$$j_a^\mu = -g \bar \psi \gamma^\mu T_a \psi$$

Note that the current is in the adjoint representation. Anyway, vary the Lagrangian in the standard way to obtain the equations of motion, and you should get

$$j_a^\nu = D_\mu F_a^{\mu\nu}$$

where D is the gauge covariant derivative. Taking another gauge covariant derivative on both sides, you should obtain

$$D_\nu j_a^\nu = D_\nu D_\mu F_a^{\mu\nu} = 0$$

So, this current is gauge covariant, but NOT conserved, because the gauge covariant derivative has an extra term. Writing out the gauge covariant derivative in full,

$$j_a^\nu = \partial_\mu F_a^{\mu\nu} + gf^{abc} A^b_\mu F_c^{\mu\nu}$$

$$j_a^\nu - gf^{abc} A^b_\mu F_c^{\mu\nu} = \partial_\mu F_a^{\mu\nu}$$

Now take derivatives on both sides to get

$$\partial_\nu (j_a^\nu - gf^{abc} A^b_\mu F_c^{\mu\nu}) = \partial_\nu \partial_\mu F_a^{\mu\nu} = 0$$

and so the conserved current is

$$j_a^\nu - gf^{abc} A^b_\mu F_c^{\mu\nu}$$

which is NOT gauge invariant. This makes sense, because the gluon field carries charge, too, so to have total conservation you must take it into account.

In particular, though, you can't assign arbitrary color charges to things, except as initial conditions (allowing them to change afterward). Any object with a free color index is not gauge-invariant, and therefore not strictly physical. The only physical quantities are the Wilson loop integrals.