What is the connection between Hilbert Space and path integrals? Given a space of states $|\rangle$, $|x\rangle$, $|x,y\rangle$, with the creation operators such as $\hat{\phi}(x)|y,z\rangle=|x,y,z\rangle$ for creating a particle at position $x$ and so on.
How does this relate to path integrals?
e.g.
$$\Delta(x,0;y,t) = \int \phi(x,0)\phi(y,t) \exp(i S[\phi] ) D\phi$$
How do we get this using creation and annihilation operators?
I tried setting $|\rangle = \Psi_0[\phi]$ for some ground state but $a^+(x)\Psi_0[\phi] \neq \phi(x)\Psi_0[\phi]$ so then I got stuck. Also because then I'd end up with the ground state inside the path integral which wouldn't right.
Edit
Let me clarify a bit.
Say you have a wave functional $\Psi[\phi,t]$ which satisfies the second quantized Shrodinger equation:
$$ i\frac{d}{dt}\Psi[\phi,t] = \left(-\frac{\delta^2}{\delta\phi(x)^2}-\nabla \phi(x)^2 +m^2\phi(x)^2\right) \Psi[\phi,t]$$
and it has a ground state of the form:
$$ \Psi_0[\phi] = \exp\left( \int \phi(x)s(x-y)\phi(y) \right)$$
This is the vacuum state $|>$ 
Now I want to find the Feynman propagator. So I want something like:
$$\Delta(x,0;y,t) = \int \Psi[\phi_0]\Phi[\phi_t] \exp(i S[\phi] ) D\phi$$
for some particular wave functionals. But I want to find the one particle states to put in. If I set $\Psi[\phi_t] = \phi_t(x)\Psi_0[\phi_t]$ for example then I get the ground state inside the integral which is not what I want. Is it correct to expand the wave functional as:
$$\Psi[\phi,t] = \left(\psi_t + \int \psi_t(x)\phi(x)dx^3 + \int \psi_t(x,y)\phi(x)\phi(y) dx^3 dy^3+...\right)\Psi_0[\phi]$$
where the terms correspond to the states $|x,y,..;t>$. Or should I be using operators like $a^+(x)$ where the hamiltonian is:
$$H=\int\{a^+(x),a^-(x)\}dx^3$$
and
$$a^\pm(x) = \frac{\delta}{\delta \phi(x)} \pm \int s(x-y) \phi(y) dy^3$$
Whichever way I look at it I always end up with the ground state in the integral instead of just:
$$\Delta(x,0;y,t) = \int \phi(x,0)\phi(y,t) \exp(i S[\phi] ) D\phi$$
See here it is just $\phi(x)$ not $\phi(x)\Psi_0[\phi] =\hat{\phi}(x)|>$. What am I doing wrong? How can I get rid of the ground state from the integral?
Edit 2
The only thing I can think of is that the ground state is not the same as the no-particle state. If the no-particle state is $\Psi_{NP}[\phi]=const$ then I think this solves it. Is this true or is the no-particle state equal to the ground state? But in that case what do the operators $a^\pm(x)$ correspond to? Do they mean excitations in the field as opposed to creation of particles?
 A: I have looked at this. Which seems to give more of a clue.
The ground state can be written as:
$$<0|\phi> = \int\limits^{\eta_0=\phi}e^{i S[\eta]} D\eta $$
The transition function can be written as:
$$<\phi|U(t,t')|\psi> = \int\limits^{\eta_t=\phi}_{\eta_{t'}=\psi}e^{i S[\eta]} D\eta $$
So:
$$D(x-y) = <0|\phi_0(x)\phi_t(y)|0> = <0|\phi_t>\phi_t(x)<\phi_t|U(t,t')|\phi_{t'}>\phi(y)<\phi|0> $$
$$= \int \left(\int\limits^{\eta_t=\phi_t}e^{i S[\eta]} D\eta \phi_t(x) \int\limits^{\eta_t=\phi_t}_{\eta_{t'}=\phi_{t'}}e^{i S[\eta]}  D\eta \phi_{t'}(y)\int\limits_{\psi_t=\phi_t}e^{i S[\eta]} D\eta \right) D\phi D\psi $$ 
$$= \int \phi_t(x)\phi_{t'}(y) e^{iS[\phi]} D\phi$$
More or less
A: I believe the answer is that the ground state is not the same as the no particle state. Therefor the wave functional should be expanded as:
$$\Psi[\phi,t] = \psi_t + \int \psi_t(x)\phi(x)dx^3 + \int \psi_t(x,y)\phi(x)\phi(y) dx^3 dy^3+...$$
without anything to do with the ground state. Where $\psi(x,y)$ for example is the amplitude for detecting particles at both x and y.
I believe the $a^\pm(x)$ are (de-)excitations in the field at $x$ and not creation/annihilation operators for particles.
I think this is where the confusion lay.
