# What exactly are ghosts?

1. Jul 24, 2008

### captain

what exactly are ghosts? i am guessing that they have something to do with overcounting feynman diagrams or something but i am still not sure. also is a so-called "longitudinal" free photon considered to be a ghost?

2. Jul 24, 2008

### ismaili

Re: ghosts

Please allow me to try to explain this, I have just studied about this in my study group these two days, so anybody is welcome to correct or supplement me.
In the path integral formalism of QFT, when we encounter a gauge theory, because the redundant degrees of freedom (the physics are not affected after gauge transformation), we have problems to deal with the partition function, from which all physical quantities are calculated.
The way to cure this is by inserting an identity
$$1 = \int Dg\delta(F)\det\left(\frac{\delta F}{\delta g}\right)$$
into the path integral.
The delta function serves as the gauge fixing, i.e. to avoid of calculation of redundant degrees of freedom. To deal with the determinant, since we still want certain perturbation theory, we try to write the determinant as some exponential. The way to do this is by introducing two Grassmanian fields, called ghost and anti-ghost. After writing the determinant in the form of exponential, the ghost and anti-ghost contributes to the Lagrangian, we can see from the quadratic kinetic operator that they propagate according to bosonic fields, since they are Gassmanian fields actually, so they don't have right statistics, or they are not casual fields.
Hence, they can be thought of as some mathematical artificial tools to deal with quantisaiton of gauge theory. In QED, you can see that the ghosts and anti-ghosts can be integrated out, so that we don't need them. But they are extremely important in the nonabelian case.
The longitudinal-photon state is not a physical state. The contribution to physically measurable quantities of longitudinal and time-like photon states would cancel each other.