In Dysons Book "Quantum Field Theory", In the Section on Schwinger's action principle, he introduces Feynman's path integral method to obtain transition amplitudes:
$$ \langle \phi_2, t_2 | \phi_1, t_1 \rangle = \Sigma_H N e^{\frac{i}{\hbar}I_H} $$ H is a classic path of $\phi(t)$ beginning at $t_1$ with value $\phi_1$ and ending at $t_2$ with value $\phi_2$. The sum goes over all classical paths that satisfy these requirements, and $N$ is a normalization factor.
Here, $\langle \phi_2, t_2 |$ is supposed to be an eigenstate of the field operator $\hat{\phi}(t_2)$ with eigenvalue $\phi_2$ at a space like plane in the 4 dimensional spacetime for simplicity I choose the plane with $x^0 = t_2$, and thus wrote $t_2$ for simplicity.
In the same manner $| \phi_1, t_1 \rangle$ is an eigenstate at time $t_1$ (By "eigenstate at $t_1$", I want to express that it's an eigenstate of the operator at $t_1$, so the operator is $\hat{\phi}(t_1)$). Note that because the operator can change in time, $\langle \phi_1, t_1 | \phi_1, t_2 \rangle$ isn't necessarily 1.
I assume he is working in the Heisenberg picture here, because if he weren't, there should be a time evolution operator in between the two states, which apparently isn't there.
Afterwards, Dyson introduces matrix elements as an additional assumption: $$ \langle \phi_2, t_2 | \hat{o}(t)|\phi_1, t_1 \rangle = \Sigma_H N O[\phi_H(t)] e^{\frac{i}{\hbar}I_H} $$ Where $O[\phi_H(t)]$ is the classical observable, evaluated at time $t$ at field-value $\phi_H(t)$.
My Question: Can I derive this formula from the first formula given, that deals solely with the transition amplitudes of the states?
I tried to do so, but I always fail at some point: I assume completenes of the eigenvectors ${ |\phi_i \rangle }$ expand the $\mathbb{1}$ operator and put it in
$$ \begin{array}{ccl} \langle \phi_2, t_2 | \hat{o}(t)|\phi_1, t_1 \rangle & = & \langle \phi_2, t_2 | \mathbb{1} \hat{o}(t) \mathbb{1} |\phi_1, t_1 \rangle \\ & = & \Sigma_{i,j} \langle \phi_2, t_2 |\phi_i, t \rangle \langle \phi_i, t | \hat{o}(t) |\phi_j, t \rangle \langle \phi_j, t |\phi_1, t_1 \rangle \\ & = & \Sigma_{i,j} \langle \phi_i, t | \hat{o}(t) |\phi_j, t \rangle \ \Sigma_{H_1} N_1 e^{\frac{i}{\hbar}I_{H_1}} \Sigma_{H_2} N_2 e^{\frac{i}{\hbar}I_{H_2}} \end{array} $$ Here $H_1$ is a path that begins at time $t_1$ with value $\phi_1$, and ends at time $t$ with value $\phi_j$, same goes for $H_2$. If ${ |\phi_i \rangle }$ would be a set of eigenvectors of Operator $\hat{O}$, I'd be done:
$$ \Sigma_{i} \langle \phi_i, t | \hat{o}(t) |\phi_i, t \rangle \ \Sigma_{H_1} N_1 e^{\frac{i}{\hbar}(I_{H_1}+I_{H_2})} \Sigma_{H_2} N_2 =\Sigma_H N O[\phi_H(t)] e^{\frac{i}{\hbar}I_H} $$ Where I used that $I_{H_1} + I_{H_2} = I_H$, if $H_1$ ends at $\phi_i$ and $H_2$ begins at $\phi_i$, and I assumed $\langle \phi_i, t | \hat{o}(t) |\phi_i, t \rangle = O[\phi_H(t)]$.
I however fail to derive this equation to hold if the states are no eigenvectors of $\hat{O}$, and thus the question.