Is Hamiltonian a differential operator in second quantization? Normally, a free particle Hamiltonian is written
$$ \hat{H} = - \frac{\hbar^2}{2m} \Delta $$
which is a differential operator because Laplacian $\Delta$ is.
On the other hand, in second quantization notation a Hamiltonian for free particle system is
$$ \hat{H} = \sum_j \varepsilon_j a^\dagger_j a_j $$
where $j$ are states with energies $\varepsilon_j$ and creation operator $a^\dagger_j$. It is not obvious if this is a differential operator.
Foundation: For example, the result known as Lippmann-Schwinger equation assumes in its derivation that the Hamiltonian is a differential operator so Schrödinger equation can be solved as a differential equation. Is it true in second quantization notation, as well?
 A: The expression 
$$
\hat{H}=\sum_j \varepsilon_j\,a_j^\dagger a_j
$$
is not the most general expression for free particles hamiltonian because it implies that you already found the eigenvalues $\varepsilon_j$ and diagonalized $\hat{H}$, i.e. already solved the Schrödinger equation.
Maybe you should look at the problem in a different basis. Let say $\{\vert j\rangle\}_j$ is a single particle state basis. Then you can chose to express $\hat{H}$ in another one particle basis, the position state basis $\{\vert x\rangle\}_x$ for instance.
The associated change of basis on the $a$ and $a^\dagger$ operators is perfomed with :
$$
a(x)=\sum_j \langle x\vert j\rangle\,a_j\,.
$$
Conversly, one can change back to the previous basis with :
$$
a_j=\int\mathrm{d}x\,\langle x\vert j\rangle\,a(x)
$$
Then you get the general expression of $\hat{H}$ in such basis :
$$
\hat{H}=\int\mathrm{d}x\,a^\dagger(x)\left[\frac{\hat{p}^2}{2m}\right]a(x)
$$
Such expression gives you back the fact that $\hat{H}$ is kind of a sum on one particle differential operators , provided that :
$$
\hat{p}=-\mathrm{i}\hbar\partial_x
$$ 
A: A 1-particle Hilbert space (neglecting spin for simplicity) is usually modelled as $L^2(\mathbb R^3)$ which is a function space, and the Hamiltonian is a differential operator. N-particle Hilbert spaces are usually constructed as tensor products of this 1-particle Hilbert space, but there exists an isomorphism such that you can again interpret them as function spaces. E.g. for 2 particles you have
$L^2(\mathbb R^3) \otimes L^2(\mathbb R^3) \simeq L^2(\mathbb R^3 \times \mathbb R^3)$.
So you can again interpret the Hamiltonian as a differential operator on these spaces.
The Fock space on which your second-quantized operators act is modelled as a direct sum of all N-particle Hilbert spaces. But unfortunately I don't know a simple way how this Fock space could be modelled as a function space. So my guess is: No, the Hamiltonian in second quantization is not a differential operator.
A: The second-quantization hamiltonian you wrote above is the hamiltonian for a collection of independent harmonic oscillators, where the quantities oscillating are now the amplitudes of the normal modes of the field being quantized. So, yes, the hamiltonian is a differential operator on the space complex functions of $N$ variables, one for each normal mode. Explicitly:
$$
\hat{H}f(\phi_1,\ldots,\phi_i,\ldots,\phi_N) = \sum_j^N \varepsilon_j a^{\dagger}_j a_j f(\phi_1,\ldots,\phi_i,\ldots,\phi_N) = \sum_j^N \varepsilon_j(-\frac{d^2}{d\phi_j^2} + \phi_j^2 - \frac{1}{2}) f(\phi_1,\ldots,\phi_i,\ldots,\phi_N)
$$
(in suitably chosen units) where $\phi_i$ is the amplitude of the $i^{th}$ normal mode.
