Are constant terms in second-quantization relevant? I have a rather broad question and a specific problem. Let's take a orthonormal single-particle basis $\{ \vert i \rangle \}$, a simple single-particle Hamiltonian 
$$\tilde{H} = \sum_{i, j} h_{i j} \vert i \rangle \langle j \vert$$
and its second-quantized form
$$ H = \sum_{i, j} h_{ij} a^\dagger_i a_j ~.$$
Now I add a constant $C$, i.e. $H_C = \sum_{i, j} h_{ij} a^\dagger_i a_j + C ~.$
The broad question is, if this constant is relevant. Does it have any effect? How does one deal with a constant operator in Fock space? What is the single-particle version of $\tilde{H}_C$? I was unable to find literature about this. Such constants sometimes appear when one wants to define a Hamiltonian with a particular symmetry, e.g. particle-hole symmetry, and as far as I can tell, these constants are ignored or considered unimportant.
The specific problem, why I want to know about the consequences of these constants, is the following. (Don't worry about the details, just concentrate on the Hamiltonian.) The single-particle imaginary-time Green's function is defined as
$$ G_{k l}(\tau) = - \frac{1}{Z} \mathrm{Tr}(e^{-\beta H} \mathcal{T} a_k[\tau] a^\dagger_k) ~,$$
with $Z = \mathrm{Tr}\left( e^{-\beta H} \right)$, $\beta = 1/T$ and $a_k[\tau] = e^{\tau H} a_k e^{-\tau H}$. It's Fourier transformation is defined as
$$ G_{k l}(i \omega) = \int_0^\beta d\tau e^{i \omega \tau} G_{k l}(\tau) ~.$$
They fulfill the relations
$$\begin{align}
\partial_\tau G_{k l}(\tau) &= -\delta(\tau) \delta_{k l} - \sum_{m} h_{k m} G_{m l}(\tau) \\
\delta_{k l} &= \sum_{m} (i\omega \cdot \delta_{k m} - h_{k m}) G_{m l}(i\omega) ~. \tag{1} \end{align}$$
We see that $G(i\omega)$ -- understood as a matrix -- is inverse to $Q$ with $Q_{k l} := (i\omega \cdot \delta_{k l} - h_{k l})$. Another approach is given by the resolvent $\mathcal{G}(z) = (z - \tilde{H})^{-1}$. Then $\mathcal{G}_{k l}(i \omega) = \langle k \vert (i \omega - \tilde{H})^{-1} \vert l \rangle$ which apparently satisfies Eq. (1) and therefore $\mathcal{G}_{k l}(i \omega) = G_{k l}(i\omega)$. This is a useful relation and used at several occasions in the previous research from students and my research.
Now what happens if we use $H_C$ instead of $H$ in the definitions of $G(\tau)$ and $\mathcal{G}$? Naively I would expect the constant to drop out from $G(\tau)$ as it appears in the enumerator and denominator and can be factored out. But I can't see something similar happen for $\mathcal{G}$. Does the equality not hold anymore?
 A: The constant $C$ is not a part of the single-particle Hamiltonian. It is what is called the vacuum energy, and has no observable effect unless we look at gravitation.
To be specific, adding this constant to the Hamiltonian makes $e^{-\beta H}\,\rightarrow e^{-\beta C}e^{-\beta H}$ and $Z=\rm{Tr}(e^{-\beta H})\, \rightarrow\, e^{-\beta C} Z$. That is, the same factor $e^{-\beta C}$ is introduced to both the numerator and denominator in the expression for the imaginary-time Green's function, and they cancel out.
Update: OP considered a Hamiltonian of the form $H = \sum_{ij} h_{ij} |i\rangle\langle j|$, which translates into $H = \sum_{ij} h_{ij} a_{i}^{\dagger} a_{j}$ in the second quantization. If we call this a single-particle term in the Hamiltonian, it is inescapable that an $n$-particle term should be what consists of $n$ creation and $n$ annihilation operators in its second quantized form.
According to this definition, an $n$-particle term acts on the subspace of the Fock space corresponding to (# of particles)$\ge n$. In this regard, an overall constant in the second-quantized Hamiltonian is a zero-particle term, or the vacuum energy.
As a concrete example, let's consider the following second quantized Hamiltonian:
\begin{equation}
H = C + \sum_{ij} h_{ij} a_{i}^{\dagger} a_{j} + \frac{1}{2}\sum_{ijkl} V_{ijkl} a_{i}^{\dagger} a_{j}^{\dagger}a_{l} a_{k}.
\end{equation}
Then, its first-quantized form when there are $N$ particles should be
\begin{equation}
H = C + \sum_{p=1}^{N}\sum_{ij} h_{ij} |i\rangle_{p}\langle j|_{p} + \sum_{p=1}^{N}\sum_{q=p+1}^{N} \sum_{ijkl}V_{ijkl} |i,j\rangle_{p,q}\langle k,l|_{p,q},
\end{equation}
where $|i\rangle_{p}$ is a state in the single-particle Hilbert space of the $p$th particle, whereas $|i,j\rangle_{p,q}$ is a state in the two-particle Hilbert space of the $p$th and $q$th particles ($p<q$).
