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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?

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The $\omega$ are discrete Matsubara frequencies. In the case of fermions these can never be zero. For bosons, zero modes can indeed appear. This is related to Bose condensation. You can have anomalies in non-relativistic systems with a Fermi surface, but that's a more complicated story, see for example arxive:1203.2697. –  Thomas Dec 26 '12 at 1:00
    
Ok, so you're saying that the $ω_n$ are always odd multiples of $2π/T$ and $ω=0$ is never among them. But then, I don't understand how zero modes of the Dirac operator can influence the path integral and be responsible for the axial quantum anomaly. (Maybe I should spell out the latter in more detail.) –  Greg Graviton Dec 26 '12 at 10:50
    
You would first have to include chiral fermions and a coupling to gauge fields. –  Thomas Dec 27 '12 at 0:18

1 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.

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Thanks for your answer! If I understand that correctly, the unitary transformation $U$ is assumed to preserve $\ker D$? Otherwise, the transformed wedge product would not be a scalar multiple of the old one, but a multiple of, say $e_1^\dagger\wedge e_3^\dagger \wedge e_1 \wedge e_3$. (I'm thinking of a permutation $Ue_1 = e_1, Ue_2 = e_3, Ue_3 = e_2$) –  Greg Graviton Dec 27 '12 at 10:15
    
Thanks for the references as well. Unfortunately, I am unable to access the article by Blau from my university. :( –  Greg Graviton Dec 27 '12 at 10:16
    
@Greg, 1) The unitary transformation U does not need to preserve $\mathrm{ker}(D)$, sorry for not emphasizing that one must project on the top form after performing the action, because the top form subspace is one dimensional. I'll try to add in a few days an explicit computation of the scalar multiple and the cocycle condition. 2) I'll be happy to help if you need me to send a copy of the article. –  David Bar Moshe Dec 28 '12 at 5:10
    
1) Ah, I see. I would be very happy if you could elaborate on the projection. (For instance, I would have guessed that the top form is orthogonal to the forms with fewer unit vectors?) 2) That would be great! My email address can be found on my user profile. I would be grateful if you could find the time to send me a copy of the article. –  Greg Graviton Dec 28 '12 at 16:55
    
@Greg I can't see your E-mail, instead I placed the copy in a file exchange server, fileconvoy.com/… where it will be abailable in the next 7 days –  David Bar Moshe Dec 30 '12 at 4:34

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