# Photon-Ghost coupling?

1. Jun 23, 2007

### astros

Hello; By using the gauge d.A+A.A=0 to find the quantum Lagrangian density, I found a term of Ghost-Photon coupling.does it have a physical significance? And why one does not find it in the gauge d.A=0? Thank you.

2. Jun 23, 2007

### dextercioby

Are you sure the resuling term in the Lagr. density doesn't reduce itself to a 4-div or it's not s-exact term (perhaps modulo d) ?

3. Jun 23, 2007

### olgranpappy

No; It has a mathematical importance in that you will of course need to consider diagrams with ghosts in order to get the correct answer when you compute matrix elements. But the ghosts themselves are non-physical for the usual reasons, thus their "coupling" to any real particles is non-physical.

4. Jun 23, 2007

### olgranpappy

P.S. It should be obvious that your choice of gauge can have no physical significance and so too for a "coupling" which disappears upon a change of gauge.

5. Jun 23, 2007

### olgranpappy

Sorry to keep replying to this thread, but after further consideration it is not even clear to me that the choice d.A+A.A=0 is appropriate at all.

One might first be a little nervous about such a choice because it is non-linear. Then, when attempting to get rid of superflouous gauge degrees of freedom with the Fadeev-Popov proceedure I find that I cannot seperate out these degrees of freedom. How exactly did you proceed in applying this gauge fixing proceedure, astro?

6. Jun 24, 2007

### astros

Hello,
I proceed as usual, I found L=c_bar(d.d+2.A.d)c (excuse for latex!!) wich is not independant and hence can not be absorbed in the coefficient of the generating fonctional! one have then diagrams 2photons-ghost, I think that it's essential to have this ghost coupling to renormalize these diagrams!!!

7. Jun 24, 2007

### olgranpappy

what's "c"? a ghost? Why did you have to introduce ghosts at all? what does the rest of the lagrangian look like?

8. Jun 28, 2007

### astros

Hello, thank you for your interest:
I have introduced ghosts to eliminate the Fadeev-Popov determinant, by using an integral over fermionic numbers (the ghosts, c), it is the same method as for the gauge d.A=0 except that in this case one finds free ghosts.

9. Jun 28, 2007

### olgranpappy

Hi again, I mentioned d.A+A.A=0 gauge to a friend and he suggested that perhaps you would be interested in something called "background field gauge" (see, e.g. Peskin and Schroder) which uses a gauge which looks like d.A+A.A=0, i.e. looks like D.A=0 but the covarient derivative is evaluated at the "background field."

10. Jun 28, 2007

### kharranger

There should be nothing wrong with using a dumb choice of gauge where the ghosts do not decouple. The ghosts have no physical significance whether or not they decouple. They don't appear in initial or final states. In calculating gauge invariant quantities, you will have to include diagrams with the ghosts and work harder, but the answers should be the same as in the usual lorentz gauge.