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Determinant of the Dirac operator and discrete eigenvalues

+ 2 like - 0 dislike
156 views

Consider the determinant of the euclidean form of the Dirac operator:
$$\text{det}(iD), \quad iD = i\gamma_{\mu}(\partial_{\mu}+A_{\mu})$$

It is an elliptic operator, so has a discrete spectrum on compact manifolds. The euclidean manifold  $R^{4}$ is not a compact manifold. However, people typically writes
$$
\text{det}(iD) = \prod_{i = 1}^{\infty}\lambda_{i},
$$

where $\lambda_{i}$ defines the $i$th eigenvalue:
$$
iD\psi_{i} = \lambda_{i}\psi_{i}
$$

Why can they do this?

asked Jan 12 in Theoretical Physics by NAME_XXX (1,010 points) [ revision history ]

1 Answer

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If it is done in the framework of QFT, it is usually quite a formal expressions which then must be regularized in some way. The same story as for other determinants, traces, etc.

answered Jan 12 by Andrey Feldman (600 points) [ no revision ]
Most voted comments show all comments

But does this regularization imposethe discrete spectrum? Ii rather seems that when computing the Dirac determinant, we regularize alreary discrete spectrum from above. The fact that the spectrum is discrete is unexplained.

@NAME_XXX Usually one replaces a non-compact space by a compact one, which gives the former in the limit of infinite radius (say, $\mathbb{R}$ may be replaced by $S^1$), evaluates the index on this manifold, and then tends the radius to infinity.

@AndreyFeldman : I understand, thank You. It seems that for the case of 4D euclidean theory we replace $R^{4}$ by, say, $S^{4}$, which is compact manifold. But why do we believe that this projection is correct? I.e., does there exist the explanation arguing that we can project the manifold $R^{4}$ of the gauge theory with fermions on the manifold $S^{4}$?

@NAME_XXX It may be a kind of philosophic question. How do we know that the Path Integral is correct? Basically, it is because it works, and in all the experimentally verified situations it works correctly with enormous precision. 

@NAME_XXX In general, compactification of $\mathbb{R}^4$ to get $S^4$ is subtle -- it may give masses to massless fields, etc. If you are looking for some universal recipe, I'm afraid that it doesn't exist.

Most recent comments show all comments

@AndreyFeldman : but is there some reason which allows us to do that? All of the theory quantities must be projected on $S^{4}$ coordinates, and corresponding action must be invariant. 

@ArnoldNeumaier : but is there some projection rule? I.e., if I want to construct the theory on $S^{4}$, I need to construct corresponding quantities defined on $S^{4}$.

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