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I'm given the integral $$ Z[w] = \frac{1}{ (2 \pi)^{n/2}} \int d^n x \prod_{i=1}^n d \overline{\theta}_i d \theta_i \exp \left( - \overline{\theta}_i \partial_j w_i (x) \theta_j - \frac{1}{2} w_i(x) w_i(x) \right). $$ Here $w_i(x)$ are functions of the $n$ real variables. Also, $\theta_i$ and $\overline{\theta}_i$ are $n$ independent Grassmann variables. Using the well known result that $$ \int \prod_{i=1}^n d \overline{\theta}_i d \theta_i \exp(- \overline{\theta}_i B_{ij} \theta_j + \overline{\eta}_i \theta_i + \overline{\theta}_i \eta_i ) = \det(B) \exp(\overline{n}_i (B^{-1})_{ij} \eta_j)$$ I calculated that $$ Z[w] = 1. $$

Now, for the particular example of $w_i(x) = x_i + \frac{1}{2} g_{ijk} x_j x_k$ with $g_{ijk} = g_{ikj} = g_{kij}$ (assuming $g_{ijk}$ are small), I wish to find the Feynman rules for the perturbative expansion of $Z[w]$.

I'm not really sure how to start. Plugging in the definitions, I get $$ Z[w] = \frac{1}{ (2 \pi)^{n/2}} \int d^n x \prod_{i=1}^n d \overline{\theta}_i d \theta_i \exp \left(- \overline{\theta}_i \theta_i - \overline{\theta}_i g_{ijm} x_m \theta_i - \frac{1}{2} x_i x_i - \frac{1}{2} g_{ijk} x_i x_j x_k - \frac{1}{8} g_{ijk} g_{ilm} x_j x_k x_l x_m \right). $$ I guess I can neglect the quadratic term in $g$ in the exponent, since the coefficients are small.

I don't understand what my free theory is (I.e. I wish to do a perturbative expansion around the free theory). Is my free theory defined by the exponential $\exp(- \overline{\theta}_i \partial_j w_i(x) \theta_j )$ and is the 'interaction' term $exp(- \frac{1}{2} w_i(x) w_i(x))$? This doesn't seem right, since the first term already contains a coefficient in $g$.

For an evaluation of a two point function, like $\langle \theta_1 \overline{\theta}_2 \rangle$, should I then write $$ \langle \theta_1 \overline{\theta}_2 \rangle = \frac{1}{ (2 \pi)^{n/2}} \int d^n x \prod_{i=1}^n d \overline{\theta}_i d \theta_i \exp \left( - \overline{\theta}_i \partial_j w_i (x) \theta_j \right) (\theta_1 \overline{\theta}_2 ) ( 1 - \frac{1}{2} w_i(x) w_i(x) + \frac{1}{4} (w_i(x) w_i(x)) (w_j(x) w_j(x)) + \ldots ) $$ and try derive the Feynman rules from this expansion? How does an interaction vertex look like then, in a diagram?

I hope someone can help me out with this.

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  • $\begingroup$ Having done this question a few months ago, it is pretty weird. I think the easiest way is to not think about Feynman rules at first: just find some correlation functions explicitly and stare at the results, then use those to reverse-engineer the rules. $\endgroup$ – knzhou Apr 9 '18 at 17:19
  • $\begingroup$ Is my expression for the two point function correct? $\endgroup$ – Kamil Apr 9 '18 at 17:24
  • $\begingroup$ I think you have mixed up two different things: The integration over the Grassmann fields, $\theta_i$, and the integration over the $n$-dimensional $x$. There are no interaction (quartic or higher-order) terms between Grassmann fields. It is exactly solvable, so no perturbation is needed. First perform the Grassmann Gaussian integral, and then perform the integration over $x$. $\endgroup$ – AlQuemist Apr 17 '18 at 12:04

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