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Consider a non-interacting fermion system with Hamiltonian \begin{equation} H = \sum_{\nu}\epsilon_{\nu}c^{\dagger}_{\nu}c_{\nu}, \end{equation} 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 \begin{equation} G^{R}(\nu,t) = -i\theta(t)\left\langle c_{\nu}(t)c^{\dagger}_{\nu}(0) \right\rangle \end{equation} and its Fourier transformation as $G^{R}(\nu,\omega)$ to obtain the spectral function \begin{equation} A(\nu,\omega) = -2 \text{Im} G^{R}(\nu,\omega), \end{equation} then the non-interacting spectral function in finite temperature is \begin{equation} A(\nu,\omega) = 2\pi \delta(\omega - \epsilon_{\nu}). \end{equation} 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?

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The relevant temperature analysis usually appears in many-body texts under the headline of Lehmann representation in the chapter on finite-temperature Green's functions. Since the Hamiltonian is non-interacting, it is simply a sum of delta-functions corresponding to single-particle excitations, weighted with Boltzmann factors. In other words, temperature affects the weight of the resonances.

One way for these delta-functions to broaden into continuous lines is accounting for the interactions. These are not necessarily the Coulomb interaction between the particles, but also interactions with phonons, photons, etc., although these usually get less attention in generic many-body texts. Note that the contributions of these interactions will include the distribution functions, that is they will be temperature-dependent, which is why broadening via the interactions can also be viewed as broadening due to finite temperature. Sometimes one ignores the details and simply includes these interactions in an ad-hoc way, by introducing finite linewidth of the levels.

Another way for smearing the delta functions is by taking the thermodynamic limit, where the energy spectrum becomes continuous, and the delta-functions merge into a continuous spectrum.

Interpreting finite-temperature spectral function in terms of a number of states is complicated, especially in presence of interactions. Usually, such interpretations are made for illustrative/explicative purposes, but they are approximate.

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  • $\begingroup$ Thanks for the answering. My current understanding is that we can have the following four cases: (1) non-interacting + zero temperature -> no broadening, (2) interacting + zero temperature -> broadening, (3) non-interacting + finite temperature -> no broadening, (4) interacting + finite temperature -> broadening, but the broadening can be different from case (2). Is this understanding correct? $\endgroup$
    – ocf001497
    Nov 7, 2020 at 1:56

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