---
title: "Equation-of-motion theory"
firstLetter: "E"
publishDate: 2021-11-08
categories:
- Physics
- Quantum mechanics

date: 2021-11-08T18:09:29+01:00
draft: false
markup: pandoc
---

# 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](/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 $i \hbar \dv*{D_{\nu \nu'}^R}{t}$,
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.