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+---
+title: "Equation-of-motion theory"
+date: 2021-11-08
+categories:
+- Physics
+- Quantum mechanics
+layout: "concept"
+---
+
+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](/know/concept/greens-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](/know/concept/heaviside-step-function/) $\Theta$
+is a [Dirac delta function](/know/concept/dirac-delta-function/) $\delta$.
+Also, from the [second quantization](/know/concept/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](/know/concept/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:
+
+$$\begin{aligned}
+    \hat{H}_0
+    = \sum_{\nu' \nu''} u_{\nu' \nu''} \hat{c}_{\nu'}^\dagger \hat{c}_{\nu''}
+    \quad \implies \quad
+    \comm{\hat{H}_0}{\hat{c}_\nu}
+    = - \sum_{\nu''} u_{\nu \nu''} \hat{c}_{\nu''}
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-commH0"/>
+<label for="proof-commH0">Proof</label>
+<div class="hidden">
+<label for="proof-commH0">Proof.</label>
+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:
+
+$$\begin{aligned}
+    \comm{\hat{H}_0}{\hat{f}_{\!\nu}}
+    &= \sum_{\nu' \nu''} u_{\nu' \nu''} \Big( \hat{f}_{\!\nu'}^\dagger \comm{\hat{f}_{\!\nu''}}{\hat{f}_{\!\nu}}
+    + \comm{\hat{f}_{\!\nu'}^\dagger}{\hat{f}_{\!\nu}} \hat{f}_{\!\nu''} \Big)
+    \\
+    &= \sum_{\nu' \nu''} u_{\nu' \nu''} \Big( \hat{f}_{\!\nu'}^\dagger \acomm{\hat{f}_{\!\nu''}}{\hat{f}_{\!\nu}}
+    - 2 \hat{f}_{\!\nu'}^\dagger \hat{f}_{\!\nu} \hat{f}_{\!\nu''}
+    + \acomm{\hat{f}_{\!\nu'}^\dagger}{\hat{f}_{\!\nu}} \hat{f}_{\!\nu''}
+    - 2 \hat{f}_{\!\nu} \hat{f}_{\!\nu'}^\dagger \hat{f}_{\!\nu''} \Big)
+    \\
+    &= \sum_{\nu' \nu''} u_{\nu' \nu''} \Big( \delta_{\nu \nu'} \hat{f}_{\!\nu''}
+    - 2 \acomm{\hat{f}_{\!\nu'}^\dagger}{\hat{f}_{\!\nu}} \hat{f}_{\!\nu''} \Big)
+    = - \sum_{\nu''} u_{\nu \nu''} \hat{f}_{\!\nu''}
+\end{aligned}$$
+</div>
+</div>
+
+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](/know/concept/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.