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Wizard
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- TL;DR Summary
- In the derivation of the free EM path integral, it seems to be take for granted that variations in the choice of gauge are functionally orthogonal to variations that are connected by a gauge transformation. I do not see why.
Hi there, I've been stuck on this issue for two days. I'm hoping someone knowledgeable can explain.
I'm working through the construction of the quantum path integral for the free electrodynamic theory. I've been following a text by Fujikawa ("Path Integrals and Quantum Anomalies") and also reading the introduction to this paper https://arxiv.org/pdf/1407.7256.pdf.
Let me explain the context first, and I'll get to the part I don't understand.
The approach being taken in Fujikawa is to fix the gauge of the vector potential ##A_\mu## by choosing the Coulomb gauge (## \partial_k A_k = 0 ##) and taking ##A_\mu(t, \vec{\infty})=0##. This gives the action which is valid for any field satisfying the Coulomb gauge:
$$ S = \int \mathrm d^4 x \Big[ \frac 1 2 \partial_\mu A_k \partial^\mu A_k - \frac 1 2 \partial_k A_0 \partial^k A_0 \Big] $$
So, now the quantum path integral is given by
$$ \mathcal Z = \int \mathcal D \bar{A}^c \exp\left\lbrace \frac i \hbar \int \mathrm d^4 x \Big[ \frac 1 2 \partial_\mu A_k \partial^\mu A_k - \frac 1 2 \partial_k A_0 \partial^k A_0 \Big] \right\rbrace $$
where ## \int \mathcal D \bar{A}^c ## refers to an integration over unique field configurations in the Coulomb gauge (multiple Coulomb-gauge fields can represent the same physical system via ## A_\mu \mapsto A_\mu + \partial_\mu w## with ## \partial_k \partial_k w = 0 ##). The measure ascribed to the differential ## \mathcal D \bar{A}^c ## is inherited from the ambient space of generic field configurations ## \mathcal D A_\mu ## which does not respect the duplicity of field configurations.
Now, we would like to write the integral in terms of ## \int \mathcal D A ##, which produces a functional determinant and dirac-delta distribution which must multiply the integrand. We later introduce some new integration variables so that we can absorb the functional determinant and the dirac-delta distribution into the exponential, and get an effective Lagrangian in the path integral, now written in normal form, which allows us to touch base with the operator formalism.
All good so far.
Here's the problem: the approach for writing ## \int \mathcal D A ## in terms of ## \int \mathcal D \bar{A}^c ##. The approach (https://arxiv.org/pdf/1407.7256.pdf equation 3) is to write
$$ A_\mu = \bar{A}_\mu + \partial_\mu w $$
or in Fujikawa's case
$$ A_k = \bar{A}^c_k + \partial_k w $$
where ## \bar{A}^c ## satisfies the Coulomb gauge. Then we say
$$ \int \mathcal D A \exp \left\lbrace \dots \right\rbrace = \int \mathcal D \bar{A}^c \int \mathcal D w \exp \left\lbrace \dots \right\rbrace \, .$$
But hold up. The equation above is only correct if the effect of variations in ##w## on the value of ##A## are orthogonal to the variations of ##\bar{A}^c##:
$$||\delta(A)||^2 = \sum_{k=1}^3 \int \mathrm d^4 x \Big[ ||\delta(\bar{A}^c_k)(x)||^2 + ||\delta(\partial_k w)(x)||^2 \Big] \, .$$
This is what Fujikawa claims in Path Integrals and Quantum Anomalies, eq. (3.28).
I cannot see why this should be true. In fact, if ## \partial_k \delta(\partial_k w) = 0 ## then it seems the two variations are certainly not orthogonal, as the gauge variation maintains the Coulomb gauge and could be replicated by a variation in ## \bar{A}^c ##. (reading this again, I realize that ## \bar{A}^c ## is restricted to fields not connected by gauge transformations by definition... I am really just confused in general here).
So the whole derivation of Fujikawa and in the linked paper both rely on this idea that the gauge degree of freedom ## w ## is functionally orthogonal to the degrees of freedom which are not equivalent under a variation of gauge. But I just can't see any reason why that should be the case. Any help appreciated. Maybe there is something else I'm not understanding properly.
I'm working through the construction of the quantum path integral for the free electrodynamic theory. I've been following a text by Fujikawa ("Path Integrals and Quantum Anomalies") and also reading the introduction to this paper https://arxiv.org/pdf/1407.7256.pdf.
Let me explain the context first, and I'll get to the part I don't understand.
The approach being taken in Fujikawa is to fix the gauge of the vector potential ##A_\mu## by choosing the Coulomb gauge (## \partial_k A_k = 0 ##) and taking ##A_\mu(t, \vec{\infty})=0##. This gives the action which is valid for any field satisfying the Coulomb gauge:
$$ S = \int \mathrm d^4 x \Big[ \frac 1 2 \partial_\mu A_k \partial^\mu A_k - \frac 1 2 \partial_k A_0 \partial^k A_0 \Big] $$
So, now the quantum path integral is given by
$$ \mathcal Z = \int \mathcal D \bar{A}^c \exp\left\lbrace \frac i \hbar \int \mathrm d^4 x \Big[ \frac 1 2 \partial_\mu A_k \partial^\mu A_k - \frac 1 2 \partial_k A_0 \partial^k A_0 \Big] \right\rbrace $$
where ## \int \mathcal D \bar{A}^c ## refers to an integration over unique field configurations in the Coulomb gauge (multiple Coulomb-gauge fields can represent the same physical system via ## A_\mu \mapsto A_\mu + \partial_\mu w## with ## \partial_k \partial_k w = 0 ##). The measure ascribed to the differential ## \mathcal D \bar{A}^c ## is inherited from the ambient space of generic field configurations ## \mathcal D A_\mu ## which does not respect the duplicity of field configurations.
Now, we would like to write the integral in terms of ## \int \mathcal D A ##, which produces a functional determinant and dirac-delta distribution which must multiply the integrand. We later introduce some new integration variables so that we can absorb the functional determinant and the dirac-delta distribution into the exponential, and get an effective Lagrangian in the path integral, now written in normal form, which allows us to touch base with the operator formalism.
All good so far.
Here's the problem: the approach for writing ## \int \mathcal D A ## in terms of ## \int \mathcal D \bar{A}^c ##. The approach (https://arxiv.org/pdf/1407.7256.pdf equation 3) is to write
$$ A_\mu = \bar{A}_\mu + \partial_\mu w $$
or in Fujikawa's case
$$ A_k = \bar{A}^c_k + \partial_k w $$
where ## \bar{A}^c ## satisfies the Coulomb gauge. Then we say
$$ \int \mathcal D A \exp \left\lbrace \dots \right\rbrace = \int \mathcal D \bar{A}^c \int \mathcal D w \exp \left\lbrace \dots \right\rbrace \, .$$
But hold up. The equation above is only correct if the effect of variations in ##w## on the value of ##A## are orthogonal to the variations of ##\bar{A}^c##:
$$||\delta(A)||^2 = \sum_{k=1}^3 \int \mathrm d^4 x \Big[ ||\delta(\bar{A}^c_k)(x)||^2 + ||\delta(\partial_k w)(x)||^2 \Big] \, .$$
This is what Fujikawa claims in Path Integrals and Quantum Anomalies, eq. (3.28).
I cannot see why this should be true. In fact, if ## \partial_k \delta(\partial_k w) = 0 ## then it seems the two variations are certainly not orthogonal, as the gauge variation maintains the Coulomb gauge and could be replicated by a variation in ## \bar{A}^c ##. (reading this again, I realize that ## \bar{A}^c ## is restricted to fields not connected by gauge transformations by definition... I am really just confused in general here).
So the whole derivation of Fujikawa and in the linked paper both rely on this idea that the gauge degree of freedom ## w ## is functionally orthogonal to the degrees of freedom which are not equivalent under a variation of gauge. But I just can't see any reason why that should be the case. Any help appreciated. Maybe there is something else I'm not understanding properly.
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