 Solution

An ordinary bosonic Gaussian functional integral produces an inverse determinant. An integral over independent Grassmann fields $c,\bar{c}$ instead produces a determinant, allowing a local action to represent the Faddeev-Popov determinant. With one convenient antighost sign convention,Multiplying the operator by the field-independent factor $-i$ changes only the normalization of its determinant. After integration by parts, an equivalent density isThus the Faddeev-Popov ghost fields have a massless kinetic term and a photon-ghost-antighost interaction proportional to $\lambda$. They are auxiliary anticommuting scalar fields, not physical external particles. Closed ghost loops carry the minus sign of Grassmann variables. Replacing $\bar{c}$ by $-\bar{c}$ reverses both displayed signs without changing any observable; both the kinetic and interaction conventions must be changed together.