Quantizing the electromagnetic field (without ghosts or gauge fixing terms) using path integrals is not possible because the propagator is not well defined. Textbooks such as P&S or Ashok Das say that the reason behind this is the gauge redundancy of the system, in the path integral, an infinite number of fields are being summed over which are gauge equivalent to one another. However I don't understand what makes gauge invariance so special. Even in a complex KG field we have a phase symmetry which leaves the action invariant. But upon summing over an infinite number of equivalent field configurations the path integral and propagator is well defined. Same holds for symmetry under rotations and boosts. Another issue is that gauge symmetry makes canonically quantizing a system difficult because there exist states of negative norm and as a result (atleast in QED) we need to make the physical states transverse. Why don't these problems arise in other systems with symmetries?


Spacetime-dependent symmetries are a problem for path integrals because they break the approximation scheme that's used to define them. In the finite-dimensional case, we can approximate an oscillatory integral by a sum over the stationary points of the action: $$ \int_{\mathbb{R}^{n}} d^{n}x\, e^{\frac{i}{\hbar}S(x)} \simeq (2\pi\hbar)^{n/2}e^{i\pi/4}\sum_{x \in Crit(S)} \lvert\det H_{x}S\rvert^{-1/2} e^{\frac{i}{\hbar}S(x)} $$ Here $H_{x}S$ means the Hessian of $S$ at the point $x$. To generalize this formula to the path integral, we replace the sum over the stationary points with an integral over the space of classical solutions, and we interpret $\det H_{x}S$ as a functional determinant.

If a theory has spacetime-dependent symmetries then this expression diverges, because the determinant of the Hessian vanishes and so $\lvert\det H_{x}S\rvert^{-1/2}$ blows up. The determinant of an operator is nonzero if and only if the operator is invertible, and if a theory has spacetime-dependent symmetries then the Hessian won't be invertible. For example, in electromagnetism the Hessian is $(\eta^{\mu\nu}\partial^{2} - \partial^{\mu}\partial^{\nu})$, so it's invertible if and only if Maxwell's equation $$ (\eta^{\mu\nu}\partial^{2} - \partial^{\mu}\partial^{\nu})A_{\nu} = j^{\mu} $$ has a unique solution for any source term $j^{\mu}$. It doesn't: if $A_{\mu}$ solves this equation then so will any gauge transform $A_{\mu} + \partial_{\mu}\alpha$. Another way to put this is that if gauge transformations are symmetries then the Cauchy problem for the electromagnetic potential isn't well-posed: fixing initial data on any Cauchy slice can't determine the state of the field at later times because there will always be a gauge transformation that's the identity on your Cauchy slice and nontrivial to the future.

So if a theory has spacetime-dependent symmetries then it will have a singular Hessian, and this breaks the formula we use to perturbatively define path integrals. This is why your other examples---the phase symmetry of the complex KG field, rotations, translations, boosts, etc.---aren't a problem: they're not spacetime-dependent. For example, the Hessian of the complex massless KG field is $\partial^{2}$, and it's invertible because $$ \partial^{2}\phi = j $$ gives a well-posed Cauchy problem.

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  • $\begingroup$ So if I extend your argument to a complex scalar field coupled to a gauge field (such that it is invariant under local spacetime dependent gauge transformations), the Cauchy problem will be ill-defined? This is because the symmetry is again spacetime dependent. $\endgroup$ – Sounak Sinha Dec 10 '19 at 10:38
  • $\begingroup$ @SounakSinha that's right. The relationship between the Cauchy problem and the path integral is a little more complicated when you introduce interactions, but the Cauchy problem still won't be well-posed and the path integral will still diverge because of it. $\endgroup$ – John Dougherty Dec 10 '19 at 11:30

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