Categories: Physics, Quantum mechanics.

# Equation-of-motion theory

In many-body quantum theory, equation-of-motion theory is a method to calculate the time evolution of a system’s properties using Green’s functions.

Starting from the definition of the retarded single-particle Green’s function $G_{\nu \nu'}^R(t, t')$, we simply take the $t$-derivative (we could do the same with the advanced function $G_{\nu \nu'}^A$):

\begin{aligned} i \hbar \pdv{G^R_{\nu \nu'}(t, t')}{t} &= \pdv{\Theta(t \!-\! t')}{t} \Expval{\comm{\hat{c}_\nu(t)}{\hat{c}_{\nu'}^\dagger(t')}_{\mp}} + \Theta(t \!-\! t') \pdv{}{t}\Expval{\comm{\hat{c}_\nu(t)}{\hat{c}_{\nu'}^\dagger(t')}_{\mp}} \\ &= \delta(t \!-\! t') \Expval{\comm{\hat{c}_\nu(t)}{\hat{c}_{\nu'}^\dagger(t')}_{\mp}} + \Theta(t \!-\! t') \Expval{\Comm{\dv{\hat{c}_\nu(t)}{t}}{\hat{c}_{\nu'}^\dagger(t)}_{\mp}} \end{aligned}

Where we have used that the derivative of a Heaviside step function $\Theta$ is a Dirac delta function $\delta$. Also, from the second quantization, $\expval{\comm{\hat{c}_\nu(t)}{\hat{c}_{\nu'}^\dagger(t')}_{\mp}}$ for $t = t'$ is zero when $\nu \neq \nu'$.

Since we are in the Heisenberg picture, we know the equation of motion of $\hat{c}_\nu(t)$:

\begin{aligned} \dv{\hat{c}_\nu(t)}{t} = \frac{i}{\hbar} \comm{\hat{H}_0(t)}{\hat{c}_\nu(t)} + \frac{i}{\hbar} \comm{\hat{H}_\mathrm{int}(t)}{\hat{c}_\nu(t)} \end{aligned}

Where the single-particle part of the Hamiltonian $\hat{H}_0$ and the interaction part $\hat{H}_\mathrm{int}$ are assumed to be time-independent in the Schrödinger picture. We thus get:

\begin{aligned} i \hbar \pdv{G^R_{\nu \nu'}}{t} &= \delta_{\nu \nu'} \delta(t \!-\! t')+ \frac{i}{\hbar} \Theta(t \!-\! t') \Expval{\Comm{\comm{\hat{H}_0}{\hat{c}_\nu} + \comm{\hat{H}_\mathrm{int}}{\hat{c}_\nu}}{\hat{c}_{\nu'}^\dagger}_{\mp}} \end{aligned}

The most general form of $\hat{H}_0$, for any basis, is as follows, where $u_{\nu' \nu''}$ are constants:

\begin{aligned} \hat{H}_0 = \sum_{\nu' \nu''} u_{\nu' \nu''} \hat{c}_{\nu'}^\dagger \hat{c}_{\nu''} \quad \implies \quad \comm{\hat{H}_0}{\hat{c}_\nu} = - \sum_{\nu''} u_{\nu \nu''} \hat{c}_{\nu''} \end{aligned}

Using the commutator identity for $\comm{A B}{C}$, we decompose it like so:

\begin{aligned} \comm{\hat{H}_0}{\hat{c}_\nu} &= \sum_{\nu' \nu''} u_{\nu \nu''} \comm{\hat{c}_{\nu'}^\dagger \hat{c}_{\nu''}}{\hat{c}_\nu} = \sum_{\nu' \nu''} u_{\nu' \nu''} \Big( \hat{c}_{\nu'}^\dagger \comm{\hat{c}_{\nu''}}{\hat{c}_\nu} + \comm{\hat{c}_{\nu'}^\dagger}{\hat{c}_\nu} \hat{c}_{\nu''} \Big) \end{aligned}

Bosons have well-known commutation relations, so the result follows directly:

\begin{aligned} \comm{\hat{H}_0}{\hat{b}_\nu} &= \sum_{\nu' \nu''} u_{\nu' \nu''} \Big( \hat{b}_{\nu'}^\dagger \comm{\hat{b}_{\nu''}}{\hat{b}_\nu} + \comm{\hat{b}_{\nu'}^\dagger}{\hat{b}_\nu} \hat{b}_{\nu''} \Big) = - \sum_{\nu''} u_{\nu \nu''} \hat{b}_{\nu''} \end{aligned}

Fermions only have anticommutation relations, so a bit more work is necessary:

\begin{aligned} \comm{\hat{H}_0}{\hat{f}_{\!\nu}} &= \sum_{\nu' \nu''} u_{\nu' \nu''} \Big( \hat{f}_{\!\nu'}^\dagger \comm{\hat{f}_{\!\nu''}}{\hat{f}_{\!\nu}} + \comm{\hat{f}_{\!\nu'}^\dagger}{\hat{f}_{\!\nu}} \hat{f}_{\!\nu''} \Big) \\ &= \sum_{\nu' \nu''} u_{\nu' \nu''} \Big( \hat{f}_{\!\nu'}^\dagger \acomm{\hat{f}_{\!\nu''}}{\hat{f}_{\!\nu}} - 2 \hat{f}_{\!\nu'}^\dagger \hat{f}_{\!\nu} \hat{f}_{\!\nu''} + \acomm{\hat{f}_{\!\nu'}^\dagger}{\hat{f}_{\!\nu}} \hat{f}_{\!\nu''} - 2 \hat{f}_{\!\nu} \hat{f}_{\!\nu'}^\dagger \hat{f}_{\!\nu''} \Big) \\ &= \sum_{\nu' \nu''} u_{\nu' \nu''} \Big( \delta_{\nu \nu'} \hat{f}_{\!\nu''} - 2 \acomm{\hat{f}_{\!\nu'}^\dagger}{\hat{f}_{\!\nu}} \hat{f}_{\!\nu''} \Big) = - \sum_{\nu''} u_{\nu \nu''} \hat{f}_{\!\nu''} \end{aligned}

Substituting this into $G_{\nu \nu'}^R$’s equation of motion, we recognize another Green’s function $G_{\nu'' \nu'}^R$:

\begin{aligned} i \hbar \pdv{G^R_{\nu \nu'}}{t} &= \delta_{\nu \nu'} \delta(t \!-\! t') + \frac{i}{\hbar} \Theta(t \!-\! t') \bigg( \Expval{\comm{\comm{\hat{H}_\mathrm{int}}{\hat{c}_\nu}}{\hat{c}_{\nu'}^\dagger}_{\mp}} - \sum_{\nu''} u_{\nu \nu''} \Expval{\comm{\hat{c}_{\nu''}}{\hat{c}_{\nu'}^\dagger}_{\mp}} \bigg) \\ &= \delta_{\nu \nu'} \delta(t \!-\! t') + \frac{i}{\hbar} \Theta(t \!-\! t') \Expval{\comm{\comm{\hat{H}_\mathrm{int}}{\hat{c}_\nu}}{\hat{c}_{\nu'}^\dagger}_{\mp}} + \sum_{\nu''} u_{\nu \nu''} G_{\nu''\nu'}^R(t, t') \end{aligned}

Rearranging this as follows yields the main result of equation-of-motion theory:

\begin{aligned} \boxed{ \sum_{\nu''} \Big( i \hbar \delta_{\nu \nu''} \pdv{}{t} - u_{\nu \nu''} \Big) G^R_{\nu'' \nu'}(t, t') = \delta_{\nu \nu'} \delta(t \!-\! t') + D_{\nu \nu'}^R(t, t') } \end{aligned}

Where $D_{\nu \nu'}^R$ represents a correction due to interactions $\hat{H}_\mathrm{int}$, and also has the form of a retarded Green’s function, but with $\hat{c}_{\nu}$ replaced by $\comm{-\hat{H}_\mathrm{int}}{\hat{c}_\nu}$:

\begin{aligned} \boxed{ D^R_{\nu'' \nu'}(t, t') \equiv - \frac{i}{\hbar} \Theta(t \!-\! t') \Expval{\comm{\comm{-\hat{H}_\mathrm{int}(t)}{\hat{c}_\nu(t)}}{\hat{c}_{\nu'}^\dagger(t')}_{\mp}} } \end{aligned}

Unfortunately, calculating $D_{\nu \nu'}^R$ might still not be doable due to $\hat{H}_\mathrm{int}$. The key idea of equation-of-motion theory is to either approximate $D_{\nu \nu'}^R$ now, or to differentiate it again $i \hbar \idv{D_{\nu \nu'}^R}{t}$, and try again for the resulting corrections, until a solvable equation is found. There is no guarantee that that will ever happen; if not, one of the corrections needs to be approximated.

For non-interacting particles $\hat{H}_\mathrm{int} = 0$, so clearly $D_{\nu \nu'}^R$ trivially vanishes then. Let us assume that $\hat{H}_0$ is also time-independent, such that $G_{\nu'' \nu'}^R$ only depends on the difference $t - t'$:

\begin{aligned} \sum_{\nu''} \Big( i \hbar \delta_{\nu \nu''} \pdv{}{t} - u_{\nu \nu''} \Big) G^R_{\nu'' \nu'}(t - t') = \delta_{\nu \nu'} \delta(t - t') \end{aligned}

We take the Fourier transform $(t \!-\! t') \to (\omega + i \eta)$, where $\eta \to 0^+$ ensures convergence:

\begin{aligned} \sum_{\nu''} \Big( \hbar \delta_{\nu \nu''} (\omega + i \eta) - u_{\nu \nu''} \Big) G^R_{\nu'' \nu'}(\omega) = \delta_{\nu \nu'} \end{aligned}

If we assume a diagonal basis $u_{\nu \nu''} = \varepsilon_\nu \delta_{\nu \nu''}$, this reduces to the following:

\begin{aligned} \delta_{\nu \nu'} &= \sum_{\nu''} \Big( \hbar \delta_{\nu \nu''} (\omega + i \eta) - \varepsilon_\nu \delta_{\nu \nu''} \Big) G^R_{\nu'' \nu'}(\omega) \\ &= \Big( \hbar (\omega + i \eta) - \varepsilon_\nu \Big) G^R_{\nu \nu'}(\omega) \end{aligned}

For a non-interacting, time-independent Hamiltonian, we therefore arrive at:

\begin{aligned} \boxed{ G^R_{\nu \nu'}(\omega) = \frac{\delta_{\nu \nu'}}{\hbar (\omega + i \eta) - \varepsilon_\nu} } \end{aligned}

## References

1. H. Bruus, K. Flensberg, Many-body quantum theory in condensed matter physics, 2016, Oxford.