Why do the path integrals of different inequivalent canonical quantisations (due to the choice of different Cauchy foliations) agree?

Question

Suppose we have a field theory on Minkowski spacetime. This theory can be quantised in several inequivalent ways:

We can foliate the spacetime into Cauchy hypersurfaces, and write the action in co ordinates of the foliation: $$S=\int dt \int d^3x L(\phi, \partial _{\mu}\phi, t).$$ Then we can switch to the Hamiltonian formulation and impose the CCR. In the Heisenberg picture, this gives us a spacetime dependent quantum field.

Now, the resulting quantum field and the spacetime commutation relations seem to be dependent on the initially chosen foliation of spacetime. But it can be shown (e.g. in this post) that the quantisation is independent of the foliation if the foliation satisfies a certain condition: the normal vector $n^{\mu}$ to the foliation should be future directed and have unit length

So the quantisations that do not satisfy the above condition presumably lead to inequivalent quantum fields, as in the resulting spacetime commutation relations of the inequivalent quantum theories, $f(x,y)=[\phi (x), \phi (y)]$, do not agree.

However, all these inequivalent quantum theories have been derived from the same action. So, on a formal level, their path integral formulations agree.

Where does the inequivalence of these quantisations show up in the path integral formalism?

Answer 1

For the path integral approach to quantization, I'll start the same way you did with the classical theory, and then work on the quantization part, and explain any bumps along the way.

For the path integral, we know the starting point is the partition function $Z[J]$, which we turn into a quantum theory by first starting which state is being projected onto the ends of the time contour (i.e. what vacuum or density matrix the functional integral is preparing). Second, we have to say along which contour in complex time the fields are continued such that time-ordering in the canonical manner is reproduced. (These choices are in the $\int [\mathcal{D}\Phi_a$] differential.)

In the path integral, it reproduces the canonical vacuum only after we attached a projector onto the vacuum. Namely, we can insert \begin{equation} |0\rangle\langle 0| = \lim_{\epsilon\rightarrow 0^+}e^{-\epsilon H}\:\:\text{or}\:\: e^{-\beta H/2} \end{equation} at the ends of the contour. This is performed by either the $i\epsilon$-prescription/Wick-rotation which sends the contour a distance into the lower (or upper) half plane at $t = \pm\infty$. Or, it is implemented via boundary conditions in Euclidean time in which the Euclidean half-plane configures the Minkowski vacuum, while a circular Euclidean segment of length $\beta$ prepares a thermal state.

Choosing a different foliation implies choosing a different notation of positive frequency, and hence a different projector in the path integral. The action $S$ remain unchanged, but the inequivalence shows up in the boundary conditions and in the Green's function.

As an example, the Minkowski vacuum is replicated by taking the Euclidean path integral on the half plane $t_E\in (-\infty,0]$. But, if we change the Euclidean path integral to be on a wedge with imaginary Rindler time modulo $2\pi/a$, the periodic identification is what allows thermal correlators.

Thus, the choice of state is hidden in the path integral within the choice of contour and boundary condition that prepares the vacuum. When the foliation fails the unit-future directed condition, the required projector is no longer one that produces the Minkowski vacuum, and the functional integral generates a different set of Green's functions showing the inequivalence of the quantizations.