# How do we define the Heisenberg picture within functorial/path integral QFT?

In the functorial approach to QFT, each Cauchy surface $$\Sigma$$ has an associated Hilbert space $$\mathcal{H}_\Sigma$$, and each pair of Cauchy surfaces $$\Sigma,\Sigma'$$ has an associated unitary $$U_{\Sigma\to\Sigma'}:\mathcal{H}_\Sigma\to\mathcal{H}_{\Sigma'}$$. These unitaries compose functorially. In practice, the unitaries are calculated by the path integral, at least formally.

I'm struggling to understand how the usual Heisenberg field operators are defined in this framework. Presumably, we define a commuting set of field operators on some initial slice $$\Sigma$$. These form an operator algebra which I'll call $$\mathcal{A}_\Sigma$$. In the spirit of the Heisenberg picture, we could then define the operator algebra associated to a different slice $$\Sigma'$$ to be $$\mathcal{A}_{\Sigma'} = U_{\Sigma\to\Sigma'} \mathcal{A}_{\Sigma} U_{\Sigma\to\Sigma'}^{\dagger}$$. This is nice, but it's not yet the full Heisenberg picture: we want to associate field operators to points, not just sets of field operators to Cauchy slices. Also, we'd like all of our Heisenberg field operators to act on one fixed Hilbert space, not different Hilbert spaces corresponding to each slice.

So it seems that in passing from the functorial/path integral picture to the Heisenberg picture, one must choose some way to identify points, and Hilbert spaces, on different slices. Essentially, we need to make some arbitrary choice of coordinates. However I've been unable to get such a procedure to work, since the definition of the field operators seems to end up depending on the coordinate choice, which is clearly wrong.

So: How are the Heisenberg field operators defined in the functorial/path integral QFT framework?

• This may be relevant Arxiv:0806.1079 Commented Jun 9, 2022 at 17:09
• I'm sure it is relevant, and I'd come across it in researching this question, but it's just too mathematically advanced for me at the moment (the huge commuting diagrams are quite scary). In any case, that paper is almost surely overkill when it comes to answering my fairly simple question. I'd be happy with a totally non-rigorous, plain-English sketch of an answer. Commented Jun 9, 2022 at 17:32
• Imho, calling this a "fairly simple question" is overly optimistic about the state of axiomatic/constructive QFT Commented Jun 9, 2022 at 19:56

Perhaps it may help to be slightly more general, defining Hilbert spaces and operators separately from the dynamics.

So, let's suppose we have a state $$|\psi\rangle$$ defined on one Cauchy surface $$\Sigma_0$$ in a Hilbert space $$\mathcal{H}_0$$, and an operator $$\mathcal{O}$$ which acts on a different surface $$\Sigma_t$$ (a local operator on $$\Sigma_t$$, say), so it's defined on a different Hilbert space $$\mathcal{H}_t$$. At this point, in general we may have no canonical way to identify $$\mathcal{H}_0$$ and $$\mathcal{H}_t$$ (for example, we might have a QFT on some unknown curved spacetime, and we're just given $$\Sigma_0$$ and $$\Sigma_t$$ as some spatial manifolds with no information about the spacetime between them). There's no way yet to compute the expectation value of $$\mathcal{O}$$ in $$|\psi\rangle$$.

To do that, we need more information: the time evolution' operator $$U=U_{\Sigma_0\to \Sigma_t}$$, an isomorphism $$\mathcal{H}_0\to \mathcal{H}_t$$. (For example, someone tells us about a spacetime bounded by $$\Sigma_t$$ and $$\Sigma_0$$, and $$U$$ is constructed from the path integral on that spacetime.) Now we can build something that makes sense: $$\langle \psi|U^\dagger \mathcal{O} U|\psi\rangle.$$

Now the "picture" just refers to two ways of splitting up this calculation. Schrödinger means that we "evolve the state", so we define $$|\psi(t)\rangle = U|\psi\rangle \in \mathcal{H}_t$$ and then compute $$\langle \psi(t)|\mathcal{O}|\psi(t)\rangle$$ using our original representation of $$\mathcal{O}$$. Heisenberg means that we instead "evolve the operator", $$\mathcal{O}(t) = U^\dagger \mathcal{O} U : \mathcal{H}_0\to \mathcal{H}_0$$ and compute $$\langle\psi|\mathcal{O}(t)|\psi\rangle$$.

In conclusion, we have a collection of many possible Hilbert spaces $$\mathcal{H}_\Sigma$$, along with isomorphisms $$U_{\Sigma\to \Sigma'}$$ between them. The "Heisenberg picture" means that we make one particular choice $$\mathcal{H}_0$$ from this collection, and define a "Hiesenberg picture operator" simply as an operator on $$\mathcal{H}_0$$ (typically for operators which act locally on some other surface).

For a more concrete example and a slightly different perspective, suppose we're interested in some QFT on some static spacetime $$\Sigma\times \mathbb{R}$$, so $$\Sigma_t =\{(x,t):x\in\Sigma\}$$. We first define a Hilbert space of states on $$\Sigma$$, perhaps by something like Osterwalder-Schader, along with a collection of local operators $$\mathcal{O}(x)$$. We consider a one-parameter family $$\{\mathcal{H}_t:t\in\mathbb{R}\}$$ of copies of this Hilbert space. The Heisenberg operator $$\mathcal{O}(x,t)$$ is defined as $$\mathcal{O}(x)$$ acting on $$\mathcal{H}_t$$. Next, we introduce a one-parameter group of unitaries $$U(t)$$ that define isomorphisms between different $$\mathcal{H}_t$$. Finally, we reduce $$\{\mathcal{H}_t:t\in\mathbb{R}\}$$ to a single Hilbert space $$\mathcal{H}$$ by quotienting by time translation', defining $$|\psi_1\rangle \simeq U(t_1-t_2)|\psi_2\rangle$$ for $$|\psi_1\rangle \in \mathcal{H}_{t_1}$$ and $$|\psi_2\rangle \in \mathcal{H}_{t_2}$$: a state in $$\mathcal{H}$$ is an equivalence class under this relation. The definition of the Heisenberg picture operator $$\mathcal{O}(x,t)$$ descends to $$\mathcal{H}$$ in the obvious way.

The whole point of the Cauchy surfaces is to give you the initial data for the field equations, because you're presenting them as a Cauchy Problem. So, you solve the field equations using the Cauchy data as the initial/boundary data. Afterwards, to find the brackets, you substitute in the solution and evaluate the brackets on the Cauchy data.

There is, however, a slight complication.

For free, non-self-interacting, fields, everything works in a very straightforwardly fashion and you get the textbook results.

But for interacting fields, which includes the case of self-interacting fields (meaning: non-Abelian gauge fields), the field equations are non-linear. That introduces two issues: (1) operator ordering ambiguity that needs to be resolved. Differential equations go naturally with the Weyl ordering prescription (e.g. if $$[r,dr/dt]$$ is a c-number then $$d(r^2)/dt = r(dr/dt) + (dr/dt)r$$). Weyl ordering has also been discussed here. Doing a spot-check, the brackets for the fields and their gradients are all c-numbers, for the cases of interest like with the Standard Model Lagrangian or subsets thereof (that's going to get updated and moved over to GitHub or somewhere else because scribd has pulled the rug out and gone semi-garden-wall mode), though they are distributional, so Weyl ordering seems to be the natural choice ... but that gets to (2) the fields quantize to distributional operators, so non-linear combinations involving such things as multiplying delta functions by delta functions can arise. That goes outside what classical (i.e. Schwartz) distribution theory can handle. So, you need a non-linear distribution theory, the go-to choice being Colombeau theory. But nobody's ever yet successfully formulated an interacting quantum field theory for 3+1 dimensional spacetime by this or any other means; certainly not to write down actual field equations, and their solutions, non-linear, distributional and all.

Generally, all the textbook treatments have flaked out and run away from it into the safe-zone of the interaction picture. Few treatments of any kind attempt to tackle the issue head on, and certainly not in a straightforward fashion. There's one here "Quantum Field Theory In The Interaction Picture" that makes a start of it, without mucking up the works with category theory or the like (there's a time and place for everything, this is not it for that), but it doesn't handle interacting fields. Section 1.2 in the treatment does the very thing you're asking for ... for free, non-self-interacting fields ... to produce the above-mentioned textbook results.