Path integral with zero energy modes

Question

Consider the field integral for the partition function of a free non-relativistic electron in a condensed matter setting, i.e.

$$ Z = ∫D\bar\psi D\psi \exp\left(-\sum_{k,ω} \bar\psi_{k,ω} (-iω + \frac{k^2}{2m} - \mu) \psi_{k,ω}\right) $$

where the action is written in Fourier representation and $\mu$ denotes the chemical potential. Now, this integral is well known to be the determinant

$$ Z = \det\left(β\left(-iω+\frac{k^2}{2m} - \mu\right)\right) $$

which is equal to the product of all eigenvalues of the quadratic form in brackets.

But here is my problem:

How to calculate a quadratic path integral if the quadratic form has some eigenvalues that are equal to zero?

If the chemical potential $\mu$ is positive, then all all momenta with $\frac{k^2}{2m} = \mu$ (the Fermi surface) will represent an eigenvalue equal to zero and would force the determinant to become zero.

Of course, I could just drop the problematic eigenvalues from the determinant and call it a day, but unfortunately, I would like to understand quantum anomalies a bit better and

zero energy eigenmodes are important for understanding the axial quantum anomaly

for example of the $1+1D$ Schwinger model. Fujikawa's book on quantum anomalies argues that the axial anomaly comes from an asymmetry of zero modes of the Dirac operator, but I am very confused because a zero mode would make the determinant of the Dirac operator and hence the path integral vanish. How to make sense of this?

Answer 1

Quillen generalized the definition of the determinant of an oparator to a form applicable to operators with zero modes, between finite or infinite dimensional Hilbert spaces:

$D: \mathrm{H_1} \rightarrow \mathrm{H_2}$

According to this generalization, the determinant is not a C-number but an element of a one dimensional vector space :

$\mathrm{Det}(D) = (\wedge^{top}( \mathrm{H_1}/\mathrm{ker}(D)))^{\dagger} \wedge^{top}\mathrm{img}(D))$

Where $\wedge^{top}$ denotes the top wedge product. This basically means that we do not include the zero modes in the eigenvalue product. For example consider a three dimensional matrix $A$ without zero modes, then its determinant according to Quillen is:

$\mathrm{Det}(A) = e_1^{\dagger} \wedge e_2^{\dagger} \wedge e_3^{\dagger} \wedge A e_1 \wedge A e_2 \wedge A e_3 = \mathrm{det}(A) e_1^{\dagger} \wedge e_2^{\dagger} \wedge e_3^{\dagger} \wedge e_1 \wedge e_2 \wedge e_3 $

Where $\mathrm{det}$ is the conventional matrix determinant. Notice that the Quillen determinant in this case is just the conventional determinant multiplied by the one dimensional unit vector $e_1^{\dagger} \wedge e_2^{\dagger} \wedge e_3^{\dagger} \wedge e_1 \wedge e_2 \wedge e_3 $.

Now, it is not difficult verify that the determinant of a diagonal matrix $ A = \mathrm{diag} [ \lambda_1, \lambda_2, , 0]$ with zero eigenvalues will be just the product of its nonvanishing eigenvalues times the unit vector composed from the top wedge product the einvectors with nonvanishing eigenvalues:

$\mathrm{Det}(A) =\lambda_1 \lambda_2,e_1^{\dagger} \wedge e_2^{\dagger} \wedge e_1 \wedge e_2 \equiv det^{'}(A) e_1^{\dagger} \wedge e_2^{\dagger} \wedge e_1 \wedge e_2$

Where $ det^{'}(A)$ is the determinant on the subspace excluding the zero modes. Please notice that now $e_3$ disappeared from the top wedge product.

Relation to anomalies:

The scalar value $\lambda_1 \lambda_2$ of the Quillen determinant is basis dependent, because if one applies a unitary transformation:

$ A \rightarrow U^{\dagger} A U$

Only the full top wedge product $e_1^{\dagger} \wedge e_2^{\dagger} \wedge e_3^{\dagger} \wedge e_1 \wedge e_2 \wedge e_3 $ is invariant but not the partial one: $e_1^{\dagger} \wedge e_2^{\dagger} \wedge e_1 \wedge e_2 $ .Thus the scalar value of the determinant changes.

Thus in this case:

$\mathrm{Det}(U^{\dagger} A U) =c(A, U) \mathrm{det^{'}}(A)e_1^{\dagger} \wedge e_2^{\dagger} \wedge e_1 \wedge e_2 $

Where $c(A, U) $ is a scalar depending on $A$ and $U$. Consequently, the Quillen determinant is not invariant under unitary transformations.

Applying two consecutive unitary transformations one observes that the additional scalar must satisfy the relation:

$ c(A, UV) = c(V^{\dagger} A V, U) c(A, V)$

This relation is called the one cocycle condition.

This phenomenon occurs when $\mathrm{D}$ is a Dirac operator in the background of a gauge field. Due to the fact that there exist zero modes, a unitary transformation on the spinors and the gauge fields gives rise to a scalar multiple to the determinant stemming from the anomaly. Basically, there is one type of function of a gauge field and a unitary operator which satisfies the one cocycle condition (up to a constant multiple).

Please see the following lecture notes and the following article by M. Blau for further reading.