Issues with the Operator to State map using Path Integral Suppose your QFT has a Hilbert space $\mathcal{H}$, and let $\text{End}(\mathcal{H})$ be the set of operators on $\mathcal{H}$. It is often stated that in QFT there is a map $$\mathcal{F}: \text{End}(\mathcal{H}) \rightarrow \mathcal{H}$$ which is straightforward to define in operator language: For any operator $\phi$, define $$\mathcal{F}(\phi) = \phi |0\rangle$$ where $|0\rangle$ is some canonical vacuum state.
However, what's confusing me is the description of this map using the path integral, which goes as follows: If your QFT is defined on a manifold $M$, with non-trivial boundary $\partial M$, then doing a path integral requires some boundary conditions, i.e one has to specify field configurations  on $\partial M$, $x|_{\partial M} = x_0$. It is then claimed that this space of boundary field configurations forms a Hilbert space $V$, and the path integral assigns to each operator a linear functional on $V$, a member of $V^*$ that is we have a map $$\text{PI}: \text{End}(\mathcal{H}) \rightarrow V^*.$$ This is defined as follows: Let $x_0 \in V$, and $\phi \in \text{End}(\mathcal{H})$ The linear functional is defined as $$\text{PI}(\phi) (x_0) = \int_{x(\partial M) = x_0} \mathcal{D}x \,\, e^{-S[x]} \, \varphi.$$ Here inside the path integral, $\varphi$ is the classical field corresponding to the operator $\phi$. From this it's clear that $\text{PI}$ assigns a number to a boundary field configuration, however there are several things that are unclear:


*

*What  is $V$ precisely? The "space of boundary field configurations" is a bit vague, because if your target space is $X$, then this space consists of all maps from $\partial M$ to $X$, which it isn't even clear is a vector space (how do we add two such maps?).

*Why does $V$ (or $V^*$) coincide with the QFT Hilbert space $\mathcal{H}$? For example, assume we have the quantum mechanical system of the harmonic oscillator with $M = [0,\infty)$. Then $\partial M = \{0\}$, and the space of boundary field configurations is simply $\mathbb{R}$, whereas the quantum mechanical Hilbert space of this system is $L^2(\mathbb{R})$. 

*Why is $\text{PI}(\phi)$ linear in boundary field configurations? This isn't clear from the definition at all.
I have looked at a couple of sources on this and all I can find is a one or two paragraph description of this map, which neglects all the issues I have mentioned above. Can anyone please answer my queries, or suggest a source which discusses them in some detail?
 A: *

*It is the space of field configurations, if you conceive of a field as a (possibly vector-valued) $p$-form on the boundary, which, as a section of an exterior algebra naturally carries a vector space structure. Think about it - the classical fields are always $\mathbb{R}^n$ or $\mathbb{C}^n$ valued (with subtleties because they may not only transform under spacetime transformations, but also under internal (gauge) symmetries), and hence you can simply add them pointwise. (Pointwise addition still works if they're the quantum operator-valued versions)

*Note that the space of field configurations is not the space of states - the state-field correspondence is, expect for special cases like 2D CFT, not a bijection. Instead, the space of states is given by the wavefunctionals - (nice, i.e. "square-integrable" or, stricter, tempered) functionals of the fields at fixed time. For your example, recall that QM can be viewed as $(0+1)$-dimensional QFT where the only spacetime coordinate is time and the fields are themselves the position coordinates $x(t)$, so the cobordant manifold we are looking it is simply an interval whose length is time, and whose two endpoints are the points $t_0,t_1$ the initial and final states are associated to. So, at such a point the fields are mere numbers, i.e. $\mathbb{R}$, and the wavefunctionals are $L^2(\mathbb{R})$ (or the Schwartz subspace of it if you want those that play nice with Fourier transform and don't have many domain issues for operators).

*It is probably best to observe that the linearity in the boundary holds in QM, where the path integral can actually be derived, and then postulate that the continuum limit process that gives the QFT path integral preserves that property. So, recall that, in QM, we get the path integral as
$$ \langle q_F \rvert \mathrm{e}^{\mathrm{i}Ht} \lvert q_I \rangle = \int_{q_I}^{q_F} \mathrm{e}^{\mathrm{i}S[q,\dot{q}]}\mathcal{D}q$$
and, therefore, we have that the path integral of $\langle q_F \rvert \mathrm{e}^{\mathrm{i}Ht} (\lvert q_{I,1} \rangle + \lvert q_{I,2} \rangle)$ will simply be the sum of the two path integrals with lower boundary $q_{I,1}$,$q_{I,2}$. Note that, in QM, the total boundary is $\{t_I\} \times \{t_F\}$, and you are feeding $q_F,q_I$ jointly as $x_0$ into the path integral in your formulation. In this case, the $\mathcal{H}$ in your OP would not really be "the Hilbert space of the theory", but $\mathcal{H}_\text{in}\otimes\mathcal{H}_\text{out}$ on which "fields" at $t_0$ and $t_1$ can act.
Altogether, I guess my answer says that you have to be more careful how you want $\mathcal{H}$ or $V$ or such to be defined. Usually, QFTs come with a notion of "in" and "out" states, separating $\partial M$ into two parts, making $M$ a cobordism, and giving us the notion of the propagator from one boundary to the other. When they don't, it is still often useful to first consider cobordisms and see if we can't glue the non-cobordant manifolds together out of the cobordisms (at least in 2D, this works very well, often meaning you only have to solve the theory on the three-holed sphere + find the gluing rules).
In general, you embed the space of boundary field configurations $V$ into the Hilbert space by the operator-state map $\mathcal{F}$, and then define the path integral on the image of that, whence it will then extend to the wavefunctionals on the fields that form the entire space of states (as being the integral kernel of the propagator, formally).
