Computing functional determinant for Dirac fermions

Question

In the path integral formulation for quantum field theory, one often encounters functional determinants of operators, for example for a free scalar field $\log \det (\partial^2+m^2)$. For this example, the expression can be expressed as an integral over the $\log$ of the operator's eigenvalue. For the fermionic case, the operator is of the form $(i\partial\!\!\!/-m)$. What does one do with the spinor indices? A sketch of the method followed by an explicit final expression would be appreciated.

Answer 1

For a positive operator $A$ on a $N$-dimensional vector space, we may easily compute the determinant as following: let $\lambda_n$ be the eigenvalues of $A$, where degenerate eigenvalues are handled by including them several times into the set of eigenvalues. Then

$$ \det A = \prod_n \lambda_n = e^F \ ,$$

where $F$ is defined as a sum

$$ F = \sum_n \log |\lambda_n| \ . \tag A $$

This is well-defined since all $|\lambda_n| = \lambda_n > 0 $.

Now generalize to a self-adjoint, but not necessarily positive operator $A$, which is assumed to have no zero eigenvectors. We can write the determinant as follows:

$$ \det A = \prod_n \lambda_n = \prod_n \left[ |\lambda_n| \exp\!\left(i \pi \frac{\text{sign}(\lambda_n) - 1}{1}\right) \right] = e^F e^{i (\Phi-\pi N/2)} \ , $$

here $F$ is defined as in equation (A), and $\Phi$ is

$$ \Phi = \frac{\pi}{2} \sum_n \text{sign}(\lambda_n) \ .$$

Now we can try to generalize this two infinite dimensions. First of all it is useful not to consider the determinant of $A$, but the determinant of $i A$ instead, since the additional factor of $i$ cancels the $N$-dependence, which has no hope to be finite in the limit $N \rightarrow \infty$:

$$ \det( i A) = e^{F + i \Phi} \ .$$

Now to the definition of $F$ in the limit $N \rightarrow \infty$, for the case at hand, Dirac fermions. Defining $F$ in terms of the the bosonic determinant is straight-forward (ignoring UV-divergences):

$$ F = \sum_n \log|\lambda_n| = \frac{1}{2} \sum_n \log \lambda_n^2 = \frac{m_d}{2} \log \det ( - \partial^2 + m^2) \ .$$

Here $m_d$ is the multiplicity of the eigenvalues; for a Dirac fermion in 4 space-time dimensions $m_d = 4$.

The term $\Phi$ is more subtle in general. It is known in the literature as the $\eta$-invariant; in the case of a free Dirac fermion on Minkowski spacetime, it is zero. This you can see intuitively as follows: for $N$ finite, it measures the asymmetry between positive energy and negative energy modes. Since the operator $ i \partial + m$ has a chiral symmetry, there is no asymmetry. One should be careful with this reasoning though because it might fail in case of interacting theories, or theories coupled to gauge fields.