# Broadening of spectral function: interaction and temperature effect

Consider a non-interacting fermion system with Hamiltonian $$$$H = \sum_{\nu}\epsilon_{\nu}c^{\dagger}_{\nu}c_{\nu},$$$$ where $$\nu$$ is some single-particle quantum number. It can be shown that even in finite temperature, if we define the retarded Green's function as $$$$G^{R}(\nu,t) = -i\theta(t)\left\langle c_{\nu}(t)c^{\dagger}_{\nu}(0) \right\rangle$$$$ and its Fourier transformation as $$G^{R}(\nu,\omega)$$ to obtain the spectral function $$$$A(\nu,\omega) = -2 \text{Im} G^{R}(\nu,\omega),$$$$ then the non-interacting spectral function in finite temperature is $$$$A(\nu,\omega) = 2\pi \delta(\omega - \epsilon_{\nu}).$$$$ Clearly, from this expression, it can be deduced that for non-interacting system, the spectral function won't be broadened due to finite temperature effect. And it is usually said in the literature that it is the many-body interaction that broadens the spectral function, making it not a delta function. However, since we define the retarded Green's function through a thermal average, in a more general case, the explicit form of $$A(\nu,\omega)$$ will contain temperature dependence (which can be confirmed by looking at any many-body physics book that gives the explicit form of $$A(\nu,\omega)$$). So my questions are the following:

1. Does the spectral function $$A(\nu,\omega)$$ depend on temperature in general (say a system with many-body interaction)?

2. What is the key factor for the broadening of the spectral function. I already know that interaction is capable to do so, what about finite temperature?

3. If the spectral function depend on temperature, can we say that $$A(\nu,\omega)$$ tells us how many single particle states are available at quantum number $$\nu$$, energy $$\omega$$ and a specific temperature $$T$$, which we used to do the thermal average?