I am doing an exercise on solid state physics. Using Hartree-Fock formalism, and assuming a non-interacting electron system with the following grand canonical hamiltonean: $$\mathcal{H}= \sum_{k,\sigma} (\epsilon_k - \mu) c_{k \sigma}^\dagger c_{k \sigma}$$ with eigenstates defined as: $E_n = \cal{E} -\mu N_n$

I calculated the spectral density of this system, defined as: $$\mathcal{S}_\sigma (\omega, k)= \frac{1}{\Omega} \sum_b e^{- \beta K_b} \cdot \sum_a |\langle a | c_{k \sigma}| b\rangle|^2 \delta(E_b - E_a - \hbar w)$$ and by manupulating the terms, I eventually arrived at the following expression: $$\mathcal{S}_\sigma (\omega, k)= \frac{1}{\Omega} f(\epsilon_k) \delta(\epsilon_k - \mu- \hbar w)$$ where $f(\epsilon_k)$ represents the fermi function: $$f(\epsilon_k)= \frac{1}{1+ e^{\beta (\epsilon_k - \mu)}}$$

I am having some trouble with the next steps in the exercise. The next challenge is to analyze the meaning of these results at the limit $T \to 0$. Now, in this limit I know that:

  • $\mu = E_F$, the fermi energy of the system
  • $f(\epsilon_k) = \theta(\epsilon_k - \mu)$, where $\theta$ represents the heaviside step function (meaning all states are occupied up to the fermi level)

Using this, I know that as $T\to 0$, the spectral density of the system is only non-zero as $\epsilon_k \le \mu$, at which the function becomes simply:

$$\mathcal{S}_\sigma (\omega, k)= \frac{1}{\Omega} \delta(\epsilon_k - \mu- \hbar w)$$ But I don't understand what this means for the system and the electron occupation.


1 Answer 1


At T = 0 the system is in its ground state where all energy levels up to the Fermi level have full occupation and all of the energy levels above the Fermi level are completely unoccupied.

The spectral density essentially represents how easy it is for your system to exchange energy with the environment. At T= 0, because all lower energy states are filled, the system can only take energy from the environment (absorb), it cannot give energy (emit). The only excitations you have are electrons near the Fermi level absorbing some quanta of energy, and that is represented in your expression.

  • $\begingroup$ yes, but that we know from the Fermi function. I'm asking about the meaning of the result for the spectral density (which just becomes a delta function) $\endgroup$
    – Rye
    Jun 7, 2022 at 18:21
  • 1
    $\begingroup$ See edit above. Let me know if that helps. $\endgroup$
    – pmal
    Jun 7, 2022 at 18:59

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