2
$\begingroup$

I am confused by fermionic path integral used in Peskin and Schroeder. Equation (9.69) gives $$\Big(\prod_n\int d\bar{\theta}_nd\theta_n\Big)e^{-\bar{\theta}_iM_{ij}\theta_j}=\det M\tag{9.69}$$ But above equation (9.72), it seems $$\int \mathscr{D}\bar{\psi} \mathscr{D}\psi e^{i\bar{\psi}M\psi}=\det M\tag{9.72}$$ How are these consistent? By the definition of measure?

$\endgroup$
1
  • 2
    $\begingroup$ So your second equation seems to be referring to the statement in the paragraph before $(9.72)$ that says: "The denominator of this expression, according to (9.69), is $\det\left(i \partial_\mu\gamma^\mu-m\right)$." I think you can add an "up to a constant" there without impeding the result in $(9.72)$ to happen, as that additional $(-i)$ would produce, in both the numerator and the denominator, $\det(-i)$. Leaving aside mathematical formalities that are to be expected in functional integration, you can drop it. Is not worse than dropping $\det\left(i \partial_\mu\gamma^\mu-m\right)$ anyway :) $\endgroup$
    – secavara
    Commented Jan 6, 2021 at 13:18

1 Answer 1

4
$\begingroup$

If you mean that the factor of $i$ is inconsistent, note that $$ \det (i M) = i^d \det M $$ where $d = \dim M$ is the number of degrees of freedom that we integrate over. Hence, $i^d$ is a constant.

With path integrals, $d = \infty$ and it is not clear how to make sense of $i^d$. This problem can be traced back to the definition of the path integral.

In physics, we usually treat path integrals as abstract symbols that conveniently encode the properties that we expect the actual mathematically well-defined quantities (correlation functions) to have. The definitions of correlation functions naively don't depend on a normalization constant: $$ \left< \Omega \right> = \frac{\int D \phi e^{i S[\phi]} \Omega[\phi]}{\int D \phi e^{i S[\phi]}} = \frac{N \int D \phi e^{i S[\phi]} \Omega[\phi]}{N \int D \phi e^{i S[\phi]}}. $$

Hence, an arbitrary constant factor can be absorbed into the definition of the path integral measure.

Ofcourse, one has to always keep in mind that the value of the path integral is defined up to an arbitrary normalization constant. For example, saying that $\det M = 42$ in your example wouldn't mean anything physically relevant – I can always make it $\det M = 43$ by absorbing $N = 43/42$ in the path integral measure.

What is physically relevant is the dependence of $\det M(\varepsilon)$ on $\varepsilon$, where $\varepsilon$ is some parameter that is meaningful for your problem (e.g. in QFT path integrals this is usually the "source field"). One can compare e.g. $$ \frac{\det M(\varepsilon_1)}{\det M(\varepsilon_2)} = \frac{N \det M(\varepsilon_1)}{N \det M(\varepsilon_2)} $$ which is independent of the choice of $N$.

$\endgroup$

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.