# 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

1. 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

1. Is there at least some way to "formally" justify transformation to $$T,\theta$$ integrals from $$\int De\,D\chi$$?
• Maybe 19.7 from amazon.com/Integrals-Quantum-Mechanics-Statistics-Financial/dp/… will be helpful Jan 6, 2020 at 12:34
• See also references (31-39) from arxiv.org/abs/hep-th/0101036 Jan 6, 2020 at 12:59
• Most of the references are already familiar to me, yet they don't address the questions I'm interested in. I'll check out the other ones, thanks. Jan 6, 2020 at 13:58
• For sign conventions, are you following a reference? Jan 12, 2020 at 17:21
• @Qmechanic no, I wasn't really following a reference; I wanted to find a way to derive it myself from the very beginning without relying on supersymmetry, as Polyakov does Jan 12, 2020 at 19:35

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).

• Thanks a lot for the answer! I am somewhat familiar with Faddeev-Popov procedure. The approach I know is inserting "Faddeev-Popov unity" in the integral (as is, an integral over diff. group with gauge-fixing delta function and determinant), than one proceeds with change of variables in functional integral and so the gauge is fixed. Do you perhaps know of any references in which this approach is followed for this particular problem and the correct result is obtained? Because this one kinda seems more intuitive to me than what Polyakov discusses, but I wasn't able to make it work here. Jan 7, 2020 at 14:49
• Yes - the references are Appendices B and C of the paper referenced above. In this case both of the determinants that appear are relatively trivial, involving the determinant of the first or second derivative operator on periodic or antiperiodic functions.
– lux
Jan 16, 2020 at 3:53
• What further details are you looking for in an answer to this question?
– lux
Jan 18, 2020 at 16:45
• I am looking to the most detailed answer possible; there is a lot of points I still don't feel like any of the references elaborated enough on. It would be too much text to fit in the comment, so here's a Google Drive link to a file with some of my questions: drive.google.com/open?id=1p5h9RvS0B2XNjm7kwr0wpPokm5wX_cez . I do not want to waste another people's time on my misunderstandings, but it would be really helpful if you or someone else could comment on them. Jan 19, 2020 at 8:53
• OK I'll take a look sometime soon :-)
– lux
Jan 20, 2020 at 0:42

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.