At $T=0$ in the non-interacting case the $1$-particle Green function for an electron in the excited state $\lambda$ (empty band) is of the form \begin{eqnarray} G^{(0)}(\lambda,t-t') = -i \theta(t-t') e^{-i\epsilon_\lambda(t-t')} \end{eqnarray} where $H_0 C_\lambda^\dagger|0\rangle = \epsilon_\lambda C_\lambda^\dagger|0\rangle$. This expression can be Fourier transformed into \begin{eqnarray} G^{(0)}(\lambda,E) &=& -i \int_0^\infty e^{i(E-\epsilon_\lambda + i\delta )t } dt \\ &=& \frac{1}{E-\epsilon_\lambda + i\delta} \ . \end{eqnarray}

The question I have is related to the meaning of $E$ for a non-interacting particle. Shouldn't a non-interacting particle just have the energy $E=\epsilon_\lambda$?

In the interacting case things appear more clear. The excited state $\lambda$ has a limited lifetime and hence there is an uncertainty for the energy $E$, hence the actual energy of the particle in a state is given by a distribution (spectral function). But for the non-interacting case the excited state has infinite lifetime, hence $E=\epsilon_\lambda$.


1 Answer 1


$E$ is not the energy per se of the particle, it is a Fourier parameter. What gives you the possible energy accessible to the particle is the spectral function $$\rho(E)=-{\rm Im}(G(E))\propto\delta(E-\epsilon_\lambda),$$ which is peaked at the accessible energy of the free particle $\epsilon_\lambda$.

  • $\begingroup$ Then is it correct to say that the Fourier expansion in 'energy space' is not an expansion in terms of the possible energies the particle may have. Instead it is an expansion in terms of a Fourier parameter with the dimension energy? It is interesting that $\theta(t)$ is the only reason why $G(\lambda,E) \not = \delta(E-\epsilon_\lambda)$ for the freely moving particle. $\endgroup$
    – DrCommando
    Nov 23, 2014 at 15:31
  • $\begingroup$ Yes, that's it. It's the fourier transform of a causal propagator, so there is more information than just the energy of the free particle. $\endgroup$
    – Adam
    Nov 23, 2014 at 15:47
  • $\begingroup$ Then the 'energy conservation' for knots in Feynman diagrams is purely related to the time reversal symmetry? Is there any other restriction that requires that frequency i.e. energy is conserved at knots in Feynman diagrams? $\endgroup$
    – DrCommando
    Nov 23, 2014 at 16:45
  • 1
    $\begingroup$ Energy conservation is related to time translation invariance of the interaction. If you have a specific question on that, I suggest that you open a new thread. $\endgroup$
    – Adam
    Nov 23, 2014 at 22:23

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