--- title: "Lehmann representation" date: 2021-11-03 categories: - Physics - Quantum mechanics layout: "concept" --- In many-body quantum theory, the **Lehmann representation** is an alternative way to write the [Green's functions](/know/concept/greens-functions/), obtained by expanding in the many-particle eigenstates under the assumption of a time-independent Hamiltonian $\hat{H}$. First, we write out the greater Green's function $G_{\nu \nu'}^>(t, t')$, and then expand its expected value $\Expval{}$ (at thermodynamic equilibrium) into a sum of many-particle basis states $\Ket{n}$: $$\begin{aligned} G_{\nu \nu'}^>(t, t') = - \frac{i}{\hbar} \Expval{\hat{c}_\nu(t) \hat{c}_{\nu'}^\dagger(t')} &= - \frac{i}{\hbar Z} \sum_{n} \Matrixel{n}{\hat{c}_\nu(t) \hat{c}_{\nu'}^\dagger(t') e^{-\beta \hat{H}}}{n} \end{aligned}$$ Where $\beta = 1 / (k_B T)$, and $Z$ is the grand partition function (see [grand canonical ensemble](/know/concept/grand-canonical-ensemble/)); the operator $e^{\beta \hat{H}}$ gives the weight of each term at equilibrium. Since $\Ket{n}$ is an eigenstate of $\hat{H}$ with energy $E_n$, this gives us a factor of $e^{\beta E_n}$. Furthermore, we are in the [Heisenberg picture](/know/concept/heisenberg-picture/), so we write out the time-dependence of $\hat{c}_\nu$ and $\hat{c}_{\nu'}^\dagger$: $$\begin{aligned} G_{\nu \nu'}^>(t, t') &= - \frac{i}{\hbar Z} \sum_{n} e^{-\beta E_n} \Matrixel{n}{e^{i \hat{H} t / \hbar} \hat{c}_\nu e^{- i \hat{H} t / \hbar} e^{i \hat{H} t' / \hbar} \hat{c}_{\nu'}^\dagger e^{- i \hat{H} t' / \hbar}}{n} \\ &= - \frac{i}{\hbar Z} \sum_{n} e^{-\beta E_n} \Matrixel{n}{e^{i \hat{H} (t - t') / \hbar} \hat{c}_\nu e^{- i \hat{H} (t - t') / \hbar} \hat{c}_{\nu'}^\dagger}{n} \end{aligned}$$ Where we used that the trace $\Tr\!(x) = \sum_{n} \matrixel{n}{x}{n}$ is invariant under cyclic permutations of $x$. The $\Ket{n}$ form a basis of eigenstates of $\hat{H}$, so we insert an identity operator $\sum_{n'} \Ket{n'} \Bra{n'}$: $$\begin{aligned} G_{\nu \nu'}^>(t - t') &= - \frac{i}{\hbar Z} \sum_{n n'} e^{- \beta E_n} \Matrixel{n}{e^{i \hat{H} (t - t') / \hbar} \hat{c}_\nu e^{- i \hat{H} (t - t') / \hbar}}{n'} \Matrixel{n'}{\hat{c}_{\nu'}^\dagger}{n} \\ &= - \frac{i}{\hbar Z} \sum_{n n'} e^{-\beta E_n} \matrixel{n}{\hat{c}_\nu}{n'} \matrixel{n'}{\hat{c}_{\nu'}^\dagger}{n} e^{i (E_n - E_{n'}) (t - t') / \hbar} \end{aligned}$$ Note that $G_{\nu \nu'}^>$ now only depends on the time difference $t - t'$, because $\hat{H}$ is time-independent. Next, we take the [Fourier transform](/know/concept/fourier-transform/) $t \to \omega$ (with $t' = 0$): $$\begin{aligned} G_{\nu \nu'}^>(\omega) &= - \frac{i}{\hbar Z} \sum_{n n'} e^{-\beta E_n} \matrixel{n}{\hat{c}_\nu}{n'} \matrixel{n'}{\hat{c}_{\nu'}^\dagger}{n} \int_{-\infty}^\infty e^{i (E_n - E_{n'}) t / \hbar} \: e^{i \omega t} \dd{t} \end{aligned}$$ Here, we recognize the integral as a [Dirac delta function](/know/concept/dirac-delta-function/) $\delta$, thereby introducing a factor of $2 \pi$, and arriving at the Lehmann representation of $G_{\nu \nu'}^>$: $$\begin{aligned} \boxed{ G_{\nu \nu'}^>(\omega) = - \frac{2 \pi i}{Z} \sum_{n n'} e^{-\beta E_n} \matrixel{n}{\hat{c}_\nu}{n'} \matrixel{n'}{\hat{c}_{\nu'}^\dagger}{n} \: \delta(E_n - E_{n'} + \hbar \omega) } \end{aligned}$$ We now go through the same process for the lesser Green's function $G_{\nu \nu'}^<(t, t')$: $$\begin{aligned} G_{\nu \nu'}^<(t - t') &= \mp \frac{i}{\hbar Z} \sum_{n} \matrixel{n}{\hat{c}_{\nu'}^\dagger(t') \hat{c}_\nu(t) e^{-\beta \hat{H}}}{n} \\ &= \mp \frac{i}{\hbar Z} e^{-\beta E_n} \sum_{n n'} \matrixel{n}{\hat{c}_{\nu'}^\dagger}{n'} \matrixel{n'}{\hat{c}_\nu}{n} e^{i (E_{n'} - E_n) (t - t') / \hbar} \end{aligned}$$ Where $-$ is for bosons, and $+$ for fermions. Fourier transforming yields the following: $$\begin{aligned} G_{\nu \nu'}^<(\omega) &= \mp \frac{2 \pi i}{\hbar Z} \sum_{n n'} e^{-\beta E_n} \matrixel{n}{\hat{c}_{\nu'}^\dagger}{n'} \matrixel{n'}{\hat{c}_\nu}{n} \: \delta(E_{n'} - E_n + \hbar \omega) \end{aligned}$$ We swap $n$ and $n'$, leading to the following Lehmann representation of $G_{\nu \nu'}^<$: $$\begin{aligned} \boxed{ G_{\nu \nu'}^<(\omega) = \mp \frac{2 \pi i}{Z} \sum_{n n'} e^{-\beta E_{n'}} \matrixel{n}{\hat{c}_\nu}{n'} \matrixel{n'}{\hat{c}_{\nu'}^\dagger}{n} \: \delta(E_n - E_{n'} + \hbar \omega) } \end{aligned}$$ Due to the delta function $\delta$, each term is only nonzero for $E_n' = E_n + \hbar \omega$, so we write: $$\begin{aligned} G_{\nu \nu'}^<(\omega) = \mp \frac{2 \pi i}{\hbar Z} \sum_{n n'} e^{-\beta (E_n + \hbar \omega)} \matrixel{n}{\hat{c}_\nu}{n'} \matrixel{n'}{\hat{c}_{\nu'}^\dagger}{n} \: \delta(E_n - E_{n'} + \hbar \omega) \end{aligned}$$ Therefore, we arrive at the following useful relation between $G_{\nu \nu'}^<$ and $G_{\nu \nu'}^>$: $$\begin{aligned} \boxed{ G_{\nu \nu'}^<(\omega) = \pm e^{-\beta \hbar \omega} G_{\nu \nu'}^>(\omega) } \end{aligned}$$ Moving on, let us do the same for the retarded Green's function $G_{\nu \nu'}^R(t, t')$, given by: $$\begin{aligned} G_{\nu \nu'}^R(t \!-\! t') &= \Theta(t \!-\! t') \Big( G_{\nu \nu'}^>(t - t') - G_{\nu \nu'}^<(t - t') \Big) \\ &= - \frac{i}{\hbar Z} \Theta(t \!-\! t') \sum_{n n'} \matrixel{n}{\hat{c}_\nu}{n'} \matrixel{n'}{\hat{c}_{\nu'}^\dagger}{n} \Big( e^{-\beta E_n} \mp e^{- \beta E_{n'}} \Big) e^{i (E_n - E_{n'}) (t - t') / \hbar} \end{aligned}$$ We take the Fourier transform, but to ensure convergence, we must introduce an infinitesimal positive $\eta \to 0^+$ to the exponent (and eventually take the limit): $$\begin{aligned} G_{\nu \nu'}^R(\omega) &= - \frac{i}{\hbar Z} \sum_{n n'} \Big( ... \Big) \int_{-\infty}^\infty \Theta(t) e^{i (E_n - E_{n'}) t / \hbar} e^{i (\omega + i \eta) t} \dd{t} \\ &= - \frac{i}{\hbar Z} \sum_{n n'} \Big( ... \Big) \int_0^\infty e^{i (E_n - E_{n'}) t / \hbar} e^{i (\omega + i \eta) t} \dd{t} \\ &= - \frac{i}{\hbar Z} \sum_{n n'} \Big( ... \Big) \bigg[ \frac{\hbar e^{i (\hbar \omega + E_n - E_{n'}) t / \hbar} e^{- \eta t}}{i (\hbar \omega + E_n - E_{n'}) - \hbar \eta} \bigg]_0^\infty \end{aligned}$$ Leading us to the following Lehmann representation of the retarded Green's function $G_{\nu \nu'}^R$: $$\begin{aligned} \boxed{ G_{\nu \nu'}^R(\omega) = \frac{1}{Z} \sum_{n n'} \frac{\matrixel{n}{\hat{c}_\nu}{n'} \matrixel{n'}{\hat{c}_{\nu'}^\dagger}{n}}{\hbar (\omega + i \eta) + E_n - E_{n'}} \Big( e^{-\beta E_n} \mp e^{- \beta E_{n'}} \Big) } \end{aligned}$$ Finally, we go through the same steps for the advanced Green's function $G_{\nu \nu'}^A(t, t')$: $$\begin{aligned} G_{\nu \nu'}^A(t \!-\! t') &= \Theta(t' \!-\! t) \Big( G_{\nu \nu'}^<(t - t') - G_{\nu \nu'}^>(t - t') \Big) \\ &= \frac{i}{\hbar Z} \Theta(t' \!-\! t) \sum_{n n'} \matrixel{n}{\hat{c}_\nu}{n'} \matrixel{n'}{\hat{c}_{\nu'}^\dagger}{n} \Big( e^{-\beta E_n} \mp e^{- \beta E_{n'}} \Big) e^{i (E_n - E_{n'}) (t - t') / \hbar} \end{aligned}$$ For the Fourier transform, we must again introduce $\eta \to 0^+$ (although note the sign): $$\begin{aligned} G_{\nu \nu'}^A(\omega) &= \frac{i}{\hbar Z} \sum_{n n'} \Big( ... \Big) \int_{-\infty}^\infty \Theta(-t) e^{i (E_n - E_{n'}) t / \hbar} e^{i (\omega - i \eta) t} \dd{t} \\ &= \frac{i}{\hbar Z} \sum_{n n'} \Big( ... \Big) \int_{-\infty}^0 e^{i (E_n - E_{n'}) t / \hbar} e^{i (\omega - i \eta) t} \dd{t} \\ &= \frac{i}{\hbar Z} \sum_{n n'} \Big( ... \Big) \bigg[ \frac{\hbar e^{i (\hbar \omega + E_n - E_{n'}) t / \hbar} e^{\eta t}}{i (\hbar \omega + E_n - E_{n'}) + \hbar \eta} \bigg]_{-\infty}^0 \end{aligned}$$ Therefore, the Lehmann representation of the advanced Green's function $G_{\nu \nu'}^A$ is as follows: $$\begin{aligned} \boxed{ G_{\nu \nu'}^A(\omega) = \frac{1}{Z} \sum_{n n'} \frac{\matrixel{n}{\hat{c}_\nu}{n'} \matrixel{n'}{\hat{c}_{\nu'}^\dagger}{n}}{\hbar (\omega - i \eta) + E_n - E_{n'}} \Big( e^{-\beta E_n} \mp e^{- \beta E_{n'}} \Big) } \end{aligned}$$ As a final note, let us take the complex conjugate of this expression: $$\begin{aligned} \big( G_{\nu \nu'}^A(\omega) \big)^* = \frac{1}{Z} \sum_{n n'} \frac{\matrixel{n}{\hat{c}_{\nu'}}{n'} \matrixel{n'}{\hat{c}_\nu^\dagger}{n}}{\hbar (\omega + i \eta) + E_n - E_{n'}} \Big( e^{-\beta E_n} \mp e^{- \beta E_{n'}} \Big) \end{aligned}$$ Note the subscripts $\nu$ and $\nu'$. Comparing this to $G_{\nu \nu'}^R$ gives us another useful relation: $$\begin{aligned} \boxed{ G^R_{\nu \nu'}(\omega) = \big( G^A_{\nu' \nu}(\omega) \big)^* } \end{aligned}$$ ## References 1. H. Bruus, K. Flensberg, *Many-body quantum theory in condensed matter physics*, 2016, Oxford.