Spectral function with negative value How does one understand a negative value in the spectral function
$$\chi=-\mathrm{Im(G)}$$ 
where $G$ is the Green function and $\chi$ is a spectral function?
 A: Actually, I guess details aren't necessary. Working pretty much directly from Xiao-Gang Wen's many-body textbook:
A general (zero-temperature) spectral function of a two-operator correllator can be defined as:
\begin{equation}
A_+(\omega)=\sum_m \delta(\omega-(\epsilon_m-\epsilon_0)) \langle 0|O_2 |\psi_m \rangle \langle \psi_m|O_1 |0 \rangle
\end{equation}
So for each energy eigenstate $m$ it is the product of two amplitudes, representing the probability amplitude to go from the vacuum to that state using the operator $O_1$ or $O_2$. If these operators are different, this can certainly take a negative or complex value that is just the product of two phases. However, we normally use the spectral function for a creation operator of a momentum eigenstate, $\psi_k^\dagger$, in which case this reduces to:
\begin{equation}
A_{k,+}(\omega)=\sum_m \delta(\omega-(\epsilon_m-\epsilon_0)) |\langle \psi_m| \psi_k^\dagger |0 \rangle|^2
\end{equation}
This expression is always positive or zero. If you integrate over $k$, it becomes the density of states.
A: The Green's function and the spectral function can be defined as
$$ G(\mathbf r,\mathbf r',\omega) = \sum_n \frac{f_n(\mathbf r)f_n^*(\mathbf r')}{\omega - \epsilon_n + i\eta}, $$
where $\eta \rightarrow 0^+$, and
$$ A(\mathbf r,\mathbf r',\omega) = \sum_n f_n(\mathbf r)f_n^*(\mathbf r') \delta(\omega - \epsilon_n). $$
In order to get from one to the other we make use of the identity
$$ \mathrm{Im} \frac{1}{x+i\eta} = -\frac{\eta}{x^2+\eta^2} = -\pi\delta(x). $$
In general, both $A$ and $G$ are complex but if $\mathbf r=\mathbf r'$ then we get
$$ A(\mathbf r,\mathbf r,\omega) = -\frac{1}{\pi} \mathrm{Im} G(\mathbf r,\mathbf r,\omega). $$
More generally, regarding $G(\mathbf r, \mathbf r', \omega)$ and $A(\mathbf r, \mathbf r', \omega)$ as the matrix elements of operators $\hat G(\omega)$ and $\hat A(\omega)$,
$$\hat A(\omega) = −\frac{1}{2\pi i} \left( \hat G(\omega) - \hat G^\dagger(\omega) \right)$$
and $\hat A(\omega)$ is a positive semi-definite operator.
If you forgot the minus sign, then the spectral function operator would have negative values, which would be unphysical.
