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A first-class constraint is typically associated with a gauge redundancy. In order to account for this in the path integral, we simply integrate over only gauge-inequivalent configurations. This is usually done by the Faddeev-Popov trick, $$\int_{\mathcal{A}/H} \mathcal{D} A = \int_{\mathcal{A}} \mathcal{D} A \, \delta[F(A)] \, \det \frac{\partial F(A^h)}{\partial h} \,\, e^{iS[A]}$$ where an integral over the space $\mathcal{A}/H$ of gauge-inequivalent configurations can be rewritten over the space of all gauge configurations (which is much simpler to evaluate), at the expense of new terms appearing that depend on the gauge-fixing function $F$.

I'm wondering how we would account for second-class constraints in the path integral. Naively I imagine we could just ignore them completely, because second-class constraints only need to hold on-shell, i.e. for solutions to the classical equations of motion. Thus we could just integrate over configurations that don't obey the constraints in the quantum path integral.

Is this the right way to think about second-class constraints, or is something more complicated needed? If so, why is this never mentioned in standard QFT textbooks?

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The formal Hamiltonian phase space path integral for a theory with second-class constraints $\theta_{\alpha}$ is given by

$$ Z~=~\int \!{\cal D}\frac{q}{\sqrt{\hbar}} {\cal D}\frac{p}{\sqrt{\hbar}} ~e^{\frac{i}{\hbar}S_H} ~\underbrace{{\rm Pf}(\{\theta_{\alpha},\theta_{\beta}\} )}_{\text{Pfaffian}}\prod_{\gamma}\delta(\theta_{\gamma}). $$

The Pfaffian is needed in order to make the path integral $Z$ invariant under reparametrizations $\theta_{\alpha}\longrightarrow \theta^{\prime}_{\alpha}=\Lambda_{\alpha}{}^{\beta} \theta_{\beta}$ of the second-class constraints.

References:

  1. P. Senjanovic, Path integral quantization of field theories with second-class constraints, Ann. Phys. 100 (1976) 227; eqs. (66)-(67).
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    $\begingroup$ Can you explain why the usual QFT textbooks don't seem to mention this? I thought there were second-class constraints in Yang-Mills. $\endgroup$
    – knzhou
    Mar 21, 2018 at 18:36
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    $\begingroup$ There are no second-class constraints in Yang-Mills theory. $\endgroup$
    – Qmechanic
    Mar 21, 2018 at 19:02

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