path: root/source/know/concept/equation-of-motion-theory/index.md

---
title: "Equation-of-motion theory"
sort_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:

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


{% include proof/start.html id="proof-commutator" -%}
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:

$$\begin{aligned}
    \comm{\hat{H}_0}{\hat{f}_{\!\nu}}
    &= \sum_{\nu' \nu''} u_{\nu' \nu''} \Big( \hat{f}_{\!\nu'}^\dagger \comm{\hat{f}_{\!\nu''}}{\hat{f}_{\!\nu}}
    + \comm{\hat{f}_{\!\nu'}^\dagger}{\hat{f}_{\!\nu}} \hat{f}_{\!\nu''} \Big)
    \\
    &= \sum_{\nu' \nu''} u_{\nu' \nu''} \Big( \hat{f}_{\!\nu'}^\dagger \acomm{\hat{f}_{\!\nu''}}{\hat{f}_{\!\nu}}
    - 2 \hat{f}_{\!\nu'}^\dagger \hat{f}_{\!\nu} \hat{f}_{\!\nu''}
    + \acomm{\hat{f}_{\!\nu'}^\dagger}{\hat{f}_{\!\nu}} \hat{f}_{\!\nu''}
    - 2 \hat{f}_{\!\nu} \hat{f}_{\!\nu'}^\dagger \hat{f}_{\!\nu''} \Big)
    \\
    &= \sum_{\nu' \nu''} u_{\nu' \nu''} \Big( \delta_{\nu \nu'} \hat{f}_{\!\nu''}
    - 2 \acomm{\hat{f}_{\!\nu'}^\dagger}{\hat{f}_{\!\nu}} \hat{f}_{\!\nu''} \Big)
    = - \sum_{\nu''} u_{\nu \nu''} \hat{f}_{\!\nu''}
\end{aligned}$$
{% include proof/end.html id="proof-commutator" %}


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.