# Choice of folliation in path integral

Assume we have a scalar field theory for a field $$\phi$$. Can we think of the Hilbert space as being spanned by states of the form $$|\varphi\rangle$$ for configurations $$\varphi\in C^\infty(\mathbb{R}^3)$$, defined by $$\phi(0,\vec{x})|\varphi\rangle=\varphi(\vec{x})|\varphi\rangle$$? In that case, is it true that $$\langle\varphi_1|U(0,t)|\varphi_2\rangle=\int\limits_{\{\varphi\in C^\infty([0,t]\times\mathbb{R}^3)|\varphi(0,\vec{x})=\varphi_1(\vec{x})\text{ and }\varphi(t,\vec{x})=\varphi_2(\vec{x})\}}\mathcal{D}\varphi e^{iS(\varphi)}~?$$ If that is the case, usual canonical quantization requires the choice of a time foliation to find the Hamiltonian. Where is that choice on the left hand side (apart from the limits of integration)? More importantly, after Wick rotation, Euclidean space does not have a preferred time direction. What happens with such a foliation?

I just want to get a less schematic understanding of the physics behind path integrals.

• A few questions for clarification: By "limits of integration," I assume you mean the boundary conditions on $\phi$ at times $0$ and $t$ (corresponding to the initial and final states), correct? Then I assume you're asking how the left hand side involves the time foliation between these two times, correct? Are you already comfortable that the right-hand side doesn't depend on such a foliation, and you're asking about the left hand side because $U$ is normally expressed as the exponential of a Hamiltonian? Apr 17, 2020 at 23:28
• Yeah, all are correct. Does the left hand side not depend on the foliation you used to quantize your theory? Apr 18, 2020 at 18:36
• Yes you are right! Thank you! So, going back to the question, I thought that a foliation was necessary not only for the definition of the Hamiltonian (and thus the vacuum state) but also for the commutation relations between the fields. Moreover, different foliations lead to inequivalent quantizations and thus different physics. Of course, this different descriptions would correspond to the physics studied by different observers adapted to the foliation. Was this understanding wrong? Because I cannot see how this reflects on the computations using path integrals. Apr 20, 2020 at 11:57
• Ah, okay, now I see where you're coming from. Different foliations will lead to different-looking physics in the Schrödinger picture, because the Schrödinger picture associates the state-vector with a spacelike hypersurface. But it's the same theory to matter what foliation we use to construct it or study it. (This is more clear in the Heisenberg picture.) By "theory" I mean an association between observables (operators on the Hilbert space) and regions of spacetime, such as $\phi(x,t)$. That association, and the relationships between those observables, are independent of any foliation. Apr 20, 2020 at 12:33
• You're asking good questions. Addressing everything that you're thinking about would be too much for one post, I think, but I can write an answer outlining why $\langle 1|U(0,t)|2\rangle$ is independent of the foliation between the initial/final times, using only the operators-on-states formulation, if that would be helpful. Apr 20, 2020 at 12:38

Part of the question is why the quantity $$\langle\varphi_1|U(0,t)|\varphi_2\rangle$$ is independent of how the spacetime between times $$0$$ and $$t$$ is foliated by spacelike slices. It must be, because this quantity can be written as a path integral (as shown in the OP) which is more clearly independent of the foliation.

But suppose we didn't know that it could be written as a path integral. How can we see more directly that $$\langle\varphi_1|U(0,t)|\varphi_2\rangle$$ is independent of the foliation? In the usual foliation, the Hamiltonian $$H$$ is independent of time, and we have $$U(0,t)=\exp(-iHt)$$. However, $$H$$ depends on the foliation, and for a more general foliation, $$H$$ is not independent of the time coordinate. So why is $$U(0,t)$$ be independent of the foliation?

## Intuition (Lorentzian case)

Here's some intuition. The Hamiltonian can be expressed in terms of the stress-energy tensor $$T^{\mu\nu}(x)$$. The components of the stress-energy tensor generate "local translations" in spacetime, and mutually commuting components can be applied in any order we want. The mutually commuting condition is ensured by spacelike separation, so we can apply "local time translations" in spacelike-separated regions in any order we want. This corresponds to the freedom to change the foliation by spacelike hypersurfaces. The sequence of slices must be traversed sequentially so that we don't violate the mutually-commuting constraint, but we can deform the slices without changing the answer as long as they remain spacelike (and as long as the initial and final slices don't change, of course).

## Explicit formulation (Lorentzian case)

In case the intuition isn't convincing, here's a more explicit version. It's still not rigorous, but maybe a real mathematician could turn this loose outline into a legitimate proof.


Since $$H$$ depends on $$x^0$$, we can no longer express $$U(0,t)$$ as a simple exponential of $$H$$, but we can define it by these conditions: $$\frac{d}{dx^0}U(0,x^0)=-iH(x^0)U(0,x^0) \hskip2cm U(0,0) = 1. \tag{2}$$ This implies $$U(0,t) = U(0,x^0)U(x^0,\,x^0+dx^0)U(x^0+dx^0,\,t). \tag{3}$$ That's useful because we can consider a foliation-change that only affects the middle factor, and we can take the interval $$dx^0$$ to be arbitrarily short, because any foliation can be reached from any other through a sequence of such small deformations. For infinitesimal $$dx^0$$, we have \begin{align} U(x^0,\,x^0+dx^0) &\approx 1 - iH(x_0)dx^0 \\ &= 1 - i\,dx^0\int_\Sigma\ \sqrt{|\det{g}|}\,T^{0\mu}n_\mu. \tag{4} \end{align} The change in (4) induced by an infinitesimal change in $$\Sigma$$ is \begin{align} \delta U(x^0,\,x^0+dx^0) &\approx - i\,dx^0\oint_{\pl M}\ \sqrt{|\det{g}|}\,T^{0\mu}n_\mu \\ &\approx - i\,dx^0\int_M\ \pl_\mu\big( \sqrt{|\det{g}|}\,T^{0\mu}\big) \tag{5} \end{align} where $$\pl M$$ is the closed hypersurface formed by $$\Sigma$$ and its infinitesimally deformed version and where $$M$$ is the part of spacetime bounded by $$\pl M$$. Now use the identity $$\pl_\mu\big( \sqrt{|\det{g}|}\,T^{0\mu}\big) = \sqrt{|\det{g}|}\,\nabla_\mu\,T^{0\mu} \tag{6}$$ where $$\nabla$$ is the Levi-Civita connection associated with $$g$$. (This is equation (3.4.10) in Wald's General Relativity.) This is zero because of the local conservation law $$\nabla_\mu T^{\mu\nu}=0$$, which is just the coordinate transform of the usual conservation law $$\partial_\mu T^{\mu\nu}=0$$ from Minkowski spacetime. So (5) is zero.

Altogether, this indicates that $$U(0,t)$$ is independent of how the spacetime between $$0$$ and $$t$$ is foliated by spacelike hypersurfaces. (I said "indicates" because this wasn't a rigorous proof, just an outline.)

## Wick rotation

After Wick rotation, we no longer have a distinction between timelike and spacelike in the path integral. What happens to $$\langle\varphi_1|U(0,t)|\varphi_2\rangle$$ in this case?

Recall that in the operator formulation in the Lorentzian case, we have two distinct notions of ordering: operator ordering, and time ordering. In the path integral, these two orderings are equated: the path integral formulation of a correlation function automatically corresponds to a time-ordered correlation function in the operator formulation. An interesting exercise is to think about how the canonical commutation relations are encoded in the path-integral formulation (hint: use explicit infinitesimal time-differences to dictate the order of the operators in the operator products) and then to think about what happens to that in the Wick-rotated case. I won't try to address that here, partly because this post is already long and partly because I'd need to think through all the implications of Wick rotation more carefully myself, but I'm pretty sure this would be an enlightening exercise.