Handling the $\nabla \phi$ term in the Hamiltonian in a path integral Let the scalar Hamiltonian be of the form $H = \int d^3x \left [\hat{\pi}^2 + (\nabla \hat{\phi})^2 + m^2\hat{\phi}^2 \right ]$.
We wish to evaluate the quantity $\langle \phi_f | \exp(-iHt) | \phi_i \rangle$ via path-integral.
The standard procedure is to divide the time into $N$ slices, of width $\epsilon$, and insert a complete set of eigenstates $I = \int |\pi\rangle \langle \pi| \frac{d\pi}{2\pi} \times \int |\phi \rangle \langle \phi | d\phi$ between each time slice.
One then makes use of the relation $ \hat{\phi}(x)|\phi \rangle = \phi(x)|\phi\rangle$, to convert the Schroedinger picture operator $\hat{\phi}(x)$ to the scalar function $\phi(x)$ in the exponent of $\langle \phi_f | \exp(-iHt) | \phi_i \rangle$.  One can do a similar thing for $\hat{\pi}(x)$.
So far so good.  Now, my question is that we can use the relation $ \hat{\phi}(x)|\phi \rangle = \phi(x)|\phi(x)\rangle$ to convert $\hat{\phi}(x)$ to $\phi(x)$. But what about $\nabla \hat{\phi}$ ?  The eigenket, $|\phi\rangle$, is not an eigenket of $\nabla \hat{\phi}$.
How does one reduce the operator $\nabla \hat{\phi}(x)$ to a scalar $\nabla\phi(x)$?
Are we using the following trick?
$ \hat{\phi}(x)|\phi \rangle = \phi(x)|\phi\rangle$ ....................(1)
$ \hat{\phi}(x + \delta)|\phi \rangle = \phi(x + \delta)|\phi\rangle$ ........ (2)
Subtracting (1) from (2) and dividing by $\delta$, and taking the limit $\delta \rightarrow 0$, we get:
$ \nabla \hat{\phi}(x)|\phi \rangle = \nabla\phi(x)|\phi\rangle$.
Is the above reasoning correct?
 A: Yes, your reasoning is correct. To spell it out a bit more, the gradient $\nabla$ is an operator on position space, but doesn't act on Hilbert space. The field $\hat{\phi}(x)$ is an operator-valued function of position space (technically an operator valued distribution).
So we can write as an operator equation
\begin{equation}
\nabla \hat{\phi}(x) = \lim_{\epsilon\rightarrow 0} \frac{\hat\phi(x+\epsilon) - \hat\phi(x)}{\epsilon}
\end{equation}
Then we can act on a field eigenstate $|\lambda\rangle$ (I've changed the name from $\phi(x)$ to $\lambda(x)$ just to avoid notational confusion, but the label of the eigenstate doesn't matter). This state satisfies
\begin{equation}
\hat\phi(x) | \lambda \rangle = \lambda(x) | \lambda \rangle
\end{equation}
Note that I've chosen notation where the state $|\lambda\rangle$ does not have an explicit $x$ dependence. This is because the state is not a function of position, directly -- it is just a ray in Hilbert space. In other words, we don't have a different state at every position, there is just one state describing the whole system.
Putting this together, we can evaluate
\begin{eqnarray}
\nabla \hat{\phi}(x) | \lambda\rangle &=& \left[\lim_{\epsilon\rightarrow 0} \frac{\hat\phi(x+\epsilon) - \hat\phi(x)}{\epsilon} \right] | \lambda\rangle \\
&=& \lim_{\epsilon\rightarrow 0} \frac{\hat\phi(x+\epsilon) |\lambda\rangle - \hat{\phi}(x)|\lambda\rangle}{\epsilon} \\
&=& \left[\lim_{\epsilon\rightarrow 0} \frac{\lambda(x+\epsilon)  - {\lambda}(x)}{\epsilon} \right] | \lambda \rangle \\
&=& \nabla \lambda(x) | \lambda\rangle
\end{eqnarray}
In other words, an eigenstate $|\lambda\rangle$ of the field operator $\hat\phi(x)$ is also an eigenstate of the gradient-of-field operator $\nabla\hat\phi(x)$. If you think about it, it's actually hard to imagine that this would not be the case. In a state where you know the value of the field everywhere, surely you also know the value of the gradient of the field everywhere.
