Path integral for particle with spin and Dirac propagator I am currently trying to compute a path integral for fermion particle using the action, provided in chapter 9 of Polyakov "Gauge fields and strings", and show that it yields Dirac propagator in the end. I personally find this fact really fascinating, which is why I decided to look into this topic, but the derivation itself troubles me.
I start with the following action:
$$S = -\frac{1}{2} \int \limits_0^1 dt\,\left[\frac{\dot{q}^2}{e} + m^2 e - i(\psi_\mu \dot{\psi}^\mu - \psi_5 \dot{\psi}_5) + \frac{i\chi}{e} \left(\psi_\mu \dot{q}^\mu + me \psi_5 \right) \right]$$
(some sign conventions are different from Polyakov, but this doesn't really matter; here $\psi$ and $\chi$ are Grassmann-valued) and I want to calculate
$$
\int \frac{Dq\,D\psi\,De\,D\chi}{VolDiff}  \exp (iS)
$$
with boundary conditions
$$
q(0)=0,q(1)=x^\mu,\psi(1) = \Psi
$$
This integral is supposed to give me a symbol for evolution operator "$e^{-iHt}$". Integrals over $q$ and $\psi$ are easy to compute (one can do it even by discretization, as there is a "canonical" measure for both cases); the result could be written as
$$
\int d^Dp\, e^{-i(px)} \int \frac{De\,D\chi}{VolDiff} \cdot \\ \cdot \exp \left(-i \frac{p^2-m^2}{2}\left[\int d\tau\, e(\tau) - \frac{i}{8}\int d\tau_1\,d\tau_2\, \chi(\tau_1) \chi(\tau_2) \text{sign }(\tau_2 - \tau_1)  \right] \right) \cdot \\ \cdot \exp \left(\frac{1}{2} \int d\tau\,\chi(\tau) (p_\mu \Psi^\mu + m\Psi_5) \right)
$$
The troublesome part are the $\chi$ and $e$ fields. 
They are in a nontrivial representation of reparametrization group: if one changes $\tau \to f(\tau)$, they change according to $e \to e(f(\tau)) (df/d\tau)^{-1}$ and similarly for $\chi$. That is a serious problem for defining the discretized measure for them that respects that; naive approach - to do something like$$\int D\chi \to \lim \limits_{N \to \infty} \prod \limits_{i=1}^N \int d\chi_i$$ - (obviously) doesn't work (to be precise, it gives an additional factor of $\lim \limits_{N \to \infty} (p^2-m^2)^N$ )
Another point is that diffeomorphisms are symmetry of the action; which is why we should divide the result by the volume of diff. group, so that path integral won't diverge. Polyakov suggests that one can replace functional integrals modulo VolDiff by two ordinary integrals
$$
\int \frac{De\,D\chi}{VolDiff} \to \int \limits_0^\infty dT \int d\theta
$$
while "fixing the gauge" and replacing fields with constant factors $e\to T,\chi \to \theta$. The problem is that I have no idea how to show that (at least somewhat) rigorously. For pure bosonic case, Polyakov has a pretty beautiful calculation of Jacobian for transformation $De \to Df\,dT$, where $f$ are diffeomorphisms, showing that it is equal to unity. Yet I don't understand if it is possible to generalize this approach here. Moreover, I have problem with "gauge fixing" two fields at the same time to be constant: they change simultaneously under reparametrization. There is also an additional "supersymmetry" which could maybe help to do this, but I wasn't able to.
Summarising, I have the following questions. If someone can answer any of them, I'd be super grateful


*

*Is it possible to construct reasonable discretized measure for $e$ and $\chi$ which respects necessary symmetries and makes evident the VolDiff factor? (that would be ideal for me, as discretized integral is at least somehow well-defined, unlike the "formal" approach Polyakov uses)


or


*Is there at least some way to "formally" justify transformation to $T,\theta$ integrals from $\int De\,D\chi$?

 A: Building on the answer from @ACuriousMind I want to point out that the procedure being followed is the Faddeev-Popov gauge fixing. 
The symmetry transformations can be solved to put $e(t) = T$, a constant and $\chi(t) = 0$ (on the loop) or $\chi(t) = \theta$, constant (on the open line). $T$ and $\theta$ represent physically distinct configurations of the worldline fields (that is, distinguish configurations not related by gauge transformations) and are known as modular parameters.
The integral, $\int \mathscr{D}e(t) \mathscr{D}\chi(t)$ is divergent because it vastly overcounts independent / distinct configurations related by the reparameterisation and SUSY symmetries; hence you divide out by the volume of these symmetries (also formally infinite). The Fadeev-Popov procedure deals with this by gauge fixing, where the integral becomes
$$\int \mathscr{D}e(t) \mathscr{D}\chi(t) \longrightarrow \int \mathscr{D}f \int \mathscr{D}g \int dT \int d\theta \, \mu(T, \theta)$$
where $\int \mathscr{D}f = \textrm{vol}(D)$, $\int \mathscr{D} g = \textrm{vol}(S)$ give the volumes of the diffeomorphism and SUSY transformation groups that cancel the volumes on the denominator. Here, and it something that @ACuriousMind left out, the measure $\mu(T, \theta)$ is the measure on the moduli, and comes from the Faddeev-Popov determinant factor arising in the gauge fixing.  
In this case, the Faddeev-Popov determinant for gauge fixing $e(t) = T$ is equal to $1$ on the open line and $1 / T$ on the loop. For fixing $\chi(t) = \theta$ on the line we get a determinant factor of $1$ and for $\chi(t) = 0$ on the loop we get the same factor. In other words we have
$$ \int \mathscr{D}e(t) \mathscr{D}\chi(t) \,\Omega[e(t), \chi(t)] \longrightarrow \textrm{vol}(D) \textrm{vol}(S) \times \begin{cases} \int dT \int d\theta \, \Omega[T, \theta] & \textrm{Line}\\ \int \frac{dT}{T}\,  \Omega[T, 0] & \textrm{Loop} \end{cases}$$
for any functional $\Omega$ of these fields. 
Good references include Appendices B and C of https://arxiv.org/abs/1410.3288 or section 1.5.1, 1.5.2 of https://arxiv.org/abs/1512.08694 or the notes at http://www-th.bo.infn.it/people/bastianelli/2-ch6-FT2-2018.pdf (second 2.1). 
A: The gravity gauge multiplet $(e,\chi)$ enjoys a local supersymmetry with infinitesimal transformations
$$ \delta e = -2\mathrm{i}\epsilon\chi \quad \delta \chi = \frac{\mathrm{d}\epsilon}{\mathrm{d}\tau}, \tag{1}$$
where $\epsilon$ is a fermionic parameter. Together with ordinary reparametrization symmetry by some rescaling $f$, this gives us two free gauge parameters to fix the two fields $e$ and $\chi$ to constant values. Since we use (and thereby eliminate from the result) the reparametrization symmetry here, we no longer need to divide out the "diffeomorphisms" corresponding to it when path integrating.
What would be left to show is that the coupled system of equations for $(f,\epsilon)$ resulting from integrating eqs. $(1)$ and setting $(e,\chi)$ constant actually has solutions so that this fixing is possible.
