# Factors of k0

Need to be very careful where these come in.

For the Wronskian, one gets a factor of pz from the differentiation with respect to z. Therefore the wronskian itself should be multiplied by k0. For the residue, dW/dq gets a factor of 1/k0 from the differentiation with respect to q. Therefore the Wronskian derivative has no overall factor of k0.

Dimensionally, we can also work it out: Position space G(r,z,z') has dimensions of 1/m

Take a partial fourier transform: G(q,z,z') = \int d^2 r G(r, z,z') exp(iq.r) ---> G(q,z,z') has dimensions of m

We know that at a pole its of the form

G(q,z,z') = g0/(q-q0)

1/(q-q0) already has dimensions of m, thus the residue must be dimensionless.

Therefore, we must multiply the greens function by an overall factor of 1/k0 to get the correct dimension, but the residue should be left as is.