I would like to calculate $\langle\bar{\psi}\psi\rangle$ in free theory. I start from the following generating functional: $$Z[J]=\int\mathcal{D}[\bar{\psi},\,\psi]\exp\left(i\int d^dx\,[\bar{\psi}(i\gamma\partial-m)\psi+J\bar{\psi}\psi]\right)\tag{1}$$ and conclude that $$\langle\bar{\psi}\psi\rangle=\frac{1}{\mathcal{Z}}\left.\frac{\partial}{\partial J}Z[J]\right|_{J=0}=\frac{\partial}{\partial J}\left(\ln Z[J]\right).\tag{2}$$ Then, the path integral is gaussian, therefore $$Z[J]=\det(i\gamma\partial -m+J)\tag{3}$$ and $\ln\det=\mathrm{Tr}\ln$, where $\mathrm{Tr}$ is the trace over all indices. I don't understand how to deal with obtain functional determinant. If it has the form $\det(i\gamma\partial -m +J)/\det(i\gamma\partial-m)$, it will be more clear. May be I am wrong in my derivations?

In fact, it is still unclear. I can rewrite obtained result as $$\frac{1}{2}\mathrm{Tr}\ln((i\gamma\partial-m+J)(-i\gamma\partial-m+J))=\frac{1}{2}\mathrm{Tr}\ln(\partial^2+m^2+2mJ+J^2),$$ and then try to factor out $\partial^2+m^2$, expand $\ln(1+\text{smth})$ and calculate $\mathrm{Tr}$ in momentum space. But it seems wrong.

  • $\begingroup$ Quick hint : Jacaobi formula for derivative of a determinant : $\frac{\delta}{\delta J(x)}\ln\det[A[x,x']]=\textbf{Tr}[\sum_{x_{1}^{}}^{}A[x,x_{1}]_{}^{-1}\frac{\delta}{\delta J(x)}A[x_{1}^{},x']]$. $\endgroup$ – Sunyam Mar 21 '19 at 16:01
  • $\begingroup$ @Sunyam can You give some details? $\endgroup$ – Artem Alexandrov Mar 21 '19 at 16:05
  • $\begingroup$ @ArtemAlexandrov Functional generalization of matrix Jacobi formula. $\endgroup$ – Sunyam Mar 21 '19 at 16:10
  • $\begingroup$ A side note. If you prefer bi-local source (as in 2PI, two particle irreducible scheme), a source term like $J(x, y)\bar{\psi(x)}\psi(y)$ is a more general approach. $\endgroup$ – MadMax Mar 21 '19 at 21:13
  • $\begingroup$ @MadMax, thx! My interest was only $\langle\bar{\psi}(x)\psi(x)\rangle$, therefore I just use $J\bar{\psi}\psi$ $\endgroup$ – Artem Alexandrov Mar 21 '19 at 21:17

Use $$ {\rm det}(\gamma \partial+m+J)= {\rm det}(\gamma \partial+m){\rm det}(1+ J(\gamma\partial +m)^{-1}) $$ and the propagator $$S(x,x') =(\gamma\partial +m)_{x,x'}^{-1}=\int \frac {d^dp}{(2\pi)^d} e^{ip(x-x')}\frac{-i\gamma p +m}{p^2+m^2} $$

$$ \ln {\rm det}(1+JS)={\rm Tr}\ln(1+JS)= {\rm Tr}\left\{ \sum_{n=1}^\infty \frac {(-1)^{n+1}} n (JS)^n\right\}. $$ to get (from the first term in the sum) that $$ \langle\bar\psi(x)\psi(x)\rangle= {\rm tr} (S(x,x)) =\int \frac {d^dp}{(2\pi)^d} \frac{m}{p^2+m^2}. $$ I've used Euclidean signature everywhere as it makes no difference when you calculate time independent stuff.

PS: the quantity calculated is not a "current." That would be $\langle\bar\psi(x)\gamma^\mu\psi(x)\rangle$

  • $\begingroup$ is the expansion of $\ln$ correct? it seems that there should be $(-1)^{(n+1)}$ $\endgroup$ – Artem Alexandrov Mar 21 '19 at 14:09

First we have to correct the generating functional. External currents are linear in the fields, $$Z[\eta,\bar{\eta}] = \int \mathcal{D}[\psi]\mathcal{D}[\bar{\psi}]\exp\left(i\int d^4x\; \bar{\psi}(i\gamma^\mu{\partial_\mu}-m)\psi + \bar{\eta}\psi + \bar{\psi}\eta + i\varepsilon \bar{\psi}\psi \right),$$ (where the $i\varepsilon$ comes from the conditions at $t\rightarrow\pm\infty$ on the fields and is related to time ordering). One can then compute the two-point function in the free theory following the usual prescription (also including time ordering). $$\langle T\left(\psi(x)\bar{\psi}(y)\right)\rangle = \frac{1}{Z[0]}\frac{\delta^2 Z[\eta,\bar{\eta}]}{\delta\bar{\eta}(x)\delta\eta(y)}\bigg|_{\eta,\bar{\eta}=0}$$ You should arrive to the usual fermionic propagator while the functional determinant will be cancelled by the normalization (which is formally infinite).

Nevertheless, in other cases such as 1-loop effective field theory you do obtain functional determinants in your computations frequently and in general they are hard to compute. For this there are several methods (see e.g. Gelfand-Yaglom's theorem). Probably the most useful way in your case is to think of the determinant as the product of eigenvalues, some care must be taken when there are zero or negative eigenvalues which will complicate the integration, but this would be a whole other question. For now you should take home that functional determinants come from an extension of the common notion of determinants, but to make sense of it we might need to employ different techniques to compute/regularize the divergent results.

  • $\begingroup$ Why I cannot introduce $J\bar{\psi}\psi$? I am interested in equal-time average $\langle\bar{\psi}\psi\rangle$. $\endgroup$ – Artem Alexandrov Mar 21 '19 at 16:26
  • $\begingroup$ I just wanted to be a bit more general, it might work for the coincident limit I guess, but you should still be careful about the ordering and the fact that the $\psi$'s are Grassmanian variables. $\endgroup$ – ohneVal Mar 21 '19 at 17:53

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