I am revisiting Srednicki's book Chapter 77 and struggling to understand how you define the change of variables in the fermionic field integral

Srednicki defines the Jacobian functional matrix for the fermionic fields (under a $U(1)$ axial transform with a spacetime-dependent parameter $\alpha(x)$) as $$J(x,y) =\delta^4(x-y)\exp [-i\alpha(x)\gamma_5].\tag{77.17}$$

He then says that by using the identity $$\log(\det J)={\operatorname{Tr}}(\log J)$$ you get (for the correction to the integral measure of fermionic fields) $$(\det J)^{-2}=\exp\left[2i\int d^4x\alpha(x)\operatorname{Tr}\delta^4(x-x)\gamma_5\right].\tag{77.19}$$ Pardon me if this is a simple question, but can someone help me fill in the steps that take you from the Jacobian (top equation) to the bottom?


1 Answer 1


It all boils down to formally manipulating matrices with "continuous indices". The price to pay for these formal manipulations will be the need for regularization in the end.

If you have a diagonal matrix $M = \operatorname{diag}(\lambda_1,\dots,\lambda_n)$ and you have a function $f(x)$ you can define $f(M)$ to be the diagonal matrix $f(M)=\operatorname{diag}(f(\lambda_1),\dots, f(\lambda_n))$. This straightforwardly generalizes to operators and should be familar from Quantum Mechanics. Indeed, if ${\cal O}$ is an operator with basis of eigenvectors $|\lambda\rangle$ with $${\cal O}|\lambda\rangle=\lambda |\lambda\rangle,\tag{1}$$

we define $f({\cal O})$ by its action on this basis $$f({\cal O})|\lambda\rangle = f(\lambda)|\lambda\rangle,\tag{2}$$

and extension by linearity. All that said, observe that $J(x,y)$ given in your equation (77.17) can be thought of as a matrix whose entries are indexed by two continuous labels instead of the familiar discrete indices labeling rows and columns. In particular, if you see it like that, you will see that $\delta(x-y)$ is the continuous version of the identity matrix and therefore $J(x,y)$ is a diagonal matrix.

This means that $\log J$ will be the continuous matrix with entries $[\log J](x,y)$ given by $$[\log J](x,y) = \delta^{(4)}(x-y)\log \exp[-i\alpha(x)\gamma_5]=-\delta^{(4)}(x-y) i\alpha(x)\gamma_5\tag{3}.$$

With this result known use your un-numbered equation to write $$\det J = \exp[\operatorname{Tr}\log J]\Longrightarrow (\det J)^{-2} = \exp[-2 \operatorname{Tr}\log J]\tag{4}.$$

To finish, how do you take the trace of a matrix $M_{ij}$? Well, you basically sum the diagonal entries, i.e., we have $\operatorname{Tr} M = \sum_{i=1}^n M_{ii}$. Since you know the diagonal matrix $\log J$ from (3) you can construct the continuous version of the trace by integrating over $x$.

In other words $$\operatorname{Tr}\log J = \int d^4x\ \operatorname{Tr}[\log J](x,x) = -i\int d^4x\ \alpha(x) \operatorname{Tr}\delta^{(4)}(x-x)\gamma_5\tag{5}.$$

Observe that a trace remains on the integral because of the (finite) matrix that is there. Plugging this into (4) you find (77.19)

$$(\det J)^{-2} = \exp\left[2i\int d^4x\ \alpha(x)\operatorname{Tr}\delta^{(4)}(x-x) \gamma_5\right]\tag{7}$$

  • 1
    $\begingroup$ Thank you much. That analogy is very helpful! $\endgroup$
    – Cory
    Commented Dec 3, 2022 at 11:57
  • $\begingroup$ You're welcome ! I'm glad it helped. $\endgroup$
    – Gold
    Commented Dec 3, 2022 at 13:20
  • $\begingroup$ Quick follow up if you're willing (and forgive another stupid question), why is the delta function allowed to come out of the log as in your equation #3 above? $\endgroup$
    – Cory
    Commented Dec 4, 2022 at 20:54
  • 1
    $\begingroup$ The matrix analogy might help again. A diagonal matrix has entries of the form $M_{ij} = \delta_{ij} \lambda_i$. When that happens $f(M)$ is again diagonal and the entries are of the form $f(M)_{ij} = \delta_{ij} f(\lambda_i)$. Equation (3) is the analog for the continuous case. The delta function just tells that $\log J$ is still diagonal and we apply the function, in this case $\log$, to the diagonal entries. BTW, the questions are not stupid. Takes some time to get used to these formal manipulations. Don't worry, you'll get the hang of it. $\endgroup$
    – Gold
    Commented Dec 4, 2022 at 21:41
  • $\begingroup$ Boy howdy it does take some time. All my professors were pretty cavalier about the mathematics of things (and, to be fair, I could have studied more of the formal mathematics). At any rate thank you again. $\endgroup$
    – Cory
    Commented Dec 5, 2022 at 17:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.